Number of half-lifes = ( 100 / 25 ) = 4 20g -----> 40g ----> 80g-----> 160g -----> 320g Original mass X = 320g Radium has a half-life of 1620 years. (i)What is half-life? The half-life period is the time taken for a radioactive nuclide to spontaneously decay/ disintegrate to half its original mass/ amount b)If one milligram of radium contains 2.68 x 10 18 atoms ,how many atoms disintegrate during 3240 years. Number of half-lifes = ( 3240 / 1620 ) = 2 1 mg ---1620---> 0.5mg ---1620----> 0.25mg If 1mg -> 2.68 x 1018 atoms Then 0.25 mg -> ( 0.25 x 2.68 x 1018 ) = 6.7 x 1017 Number of atoms remaining = 6.7 x 1017 Number of atoms disintegrated = (2.68 x 1018 - 6.7 x 1017 ) = 2.01 x 1018 The graph below shows the mass of a radioactive isotope plotted against time
17 Using the graph, determine the half – life of the isotope From graph 10 g to 5 g takes 8 days From graph 5 g to 2.5 g takes 16 – 8 = 8 days Calculate the mass of the isotope dacayed after 32 days Number of half lifes= 32/8 = 4 Original mass = 10g 10g—1st -->5g—2nd-->2.5—3rd –>1.25—4th -->0.625 g Mass remaining = 0.625 g Mass decayed after 32 days = 10g - 0.625 g = 9.375g A radioactive isotope X2 decays by emitting two alpha (a) particles and one beta (β) to form 214 83Bi (a)Write the nuclear equation for the radioactive decay 21286 X -> 214 83Bi + 242He + 0-1 e (b)What is the atomic number of X2? 86 (c) After 112 days, 1/16 of the mass of X2 remained.


The half-life period is the time taken for a radioactive nuclide to spontaneously decay/ disintegrate to half its original mass/ amount b)If one milligram of radium contains 2.68 x 10 18 atoms ,how many atoms disintegrate during 3240 years. Number of half-lifes = ( 3240 / 1620 ) = 2 1 mg ---1620---> 0.5mg ---1620----> 0.25mg If 1mg -> 2.68 x 1018 atoms Then 0.25 mg -> ( 0.25 x 2.68 x 1018 ) = 6.7 x 1017 Number of atoms remaining = 6.7 x 1017 Number of atoms disintegrated = (2.68 x 1018 - 6.7 x 1017 ) = 2.01 x 1018 The graph below shows the mass of a radioactive isotope plotted against time
17 Using the graph, determine the half – life of the isotope From graph 10 g to 5 g takes 8 days From graph 5 g to 2.5 g takes 16 – 8 = 8 days Calculate the mass of the isotope dacayed after 32 days Number of half lifes= 32/8 = 4 Original mass = 10g 10g—1st -->5g—2nd-->2.5—3rd –>1.25—4th -->0.625 g Mass remaining = 0.625 g Mass decayed after 32 days = 10g - 0.625 g = 9.375g A radioactive isotope X2 decays by emitting two alpha (a) particles and one beta (β) to form 214 83Bi (a)Write the nuclear equation for the radioactive decay 21286 X -> 214 83Bi + 242He + 0-1 e (b)What is the atomic number of X2? 86 (c) After 112 days, 1/16 of the mass of X2 remained. Determine the half life of X2
18 1—x-> 1 /2 –x-> 1 /4 –x-> 1 /8–x-> 1 /16 Number of t 1 /2 in 112 days = 4 t 1 /2 = 112 = 28 days 4 1.Study the nuclear reaction given below and answer the questions that follow.


Determine the half life of X2
18 1—x-> 1 /2 –x-> 1 /4 –x-> 1 /8–x-> 1 /16 Number of t 1 /2 in 112 days = 4 t 1 /2 = 112 = 28 days 4 1.Study the nuclear reaction given below and answer the questions that follow. 126 C --step 1-->127 N --step 2--> 1211Na (a)126 C and 146 C are isotopes. What does the term isotope mean? Atoms of the same element with different mass number /number of neutrons. (b)Write an equation for the nuclear reaction in step II 127 N -> 1211Na + 0 -1e (c)Give one use of 146 C Dating rocks/fossils: Study of metabolic pathways/mechanisms on plants/animals Study the graph of a radioactive decay series for isotope H below. (a) Name the type of radiation emitted when isotope (i) H changes to isotope J. Alpha-Mass number decrease by 4 from 214 to 210(y-axis) atomic number decrease by 2 from 83 to 81(x-axis) (ii) J changes to isotope K Beta-Mass number remains 210(y-axis) atomic number increase by 1 from 81 to 82(x-axis). 19 (b) Write an equation for the nuclear reaction that occur when isotope (i)J changes to isotope L 21081 J -> 21084L + 3 0 -1e (i)H changes to isotope M 21483 H -> 20682M + 3 0 -1e + 2 4 2He Identify a pair of isotope of an element in the decay series K and M Have same atomic number 82 but different mass number K-210 and M-206 a)A radioactive substance emits three different particles. Identify the particle: (i)with the highest mass. Alpha/ α (ii) almost equal to an electron Beta/ β 1.a)State two differences between chemical and nuclear reactions(2mks) (i) Nuclear reactions mainly involve protons and neutrons in the nucleus of an atom.Chemical reactions mainly involve outer electrons in the energy levels an atom. (ii) Nuclear reactions form a new element.


Identify the particle: (i)with the highest mass. Alpha/ α (ii) almost equal to an electron Beta/ β 1.a)State two differences between chemical and nuclear reactions(2mks) (i) Nuclear reactions mainly involve protons and neutrons in the nucleus of an atom.Chemical reactions mainly involve outer electrons in the energy levels an atom. (ii) Nuclear reactions form a new element. Chemical reactions do not form new elements (iii) Nuclear reactions mainly involve evolution/production of large quantity of heat/energy.Chemical reactions produce or absorb smaller quantity of heat/energy. (iv)Nuclear reactions are accompanied by a loss in mass /mass defect. Chemical reactions are not accompanied by a loss in mass. (v)Rate of decay/ disintegration of nuclide is independent of physical conditionsThe rate of a chemical reaction is dependent on physical conditions of temperature/pressure/purity/particle size/ surface area b)Below is a radioactive decay series starting from 21483 Bi and ending at 20682 Pb. Study it and answer the question that follows. 20 Identify the particles emitted in steps I and III (2mks) I - α-particle III - β-ray ii)Write the nuclear equation for the reaction which takes place in (a) step I 21483Bi -> 21081Bi + 4 2 He (b) step 1 to 3 21483Bi -> 21081Bi + 4 2 He + 2 0 -1 e (c) step 3 to 5 21082Pb -> 20682Pb + 4 2 He + 2 0 -1 e (c) step 1 to 5 21483Bi -> 20682Pb + 2 4 2 He + 3 0 -1 e The table below give the percentages of a radioactive isotope of Bismuth that remains after decaying at different times. Time (min) 0 6 12 22 38 62 100 Percentage of Bismuth 100 81 65 46 29 12 3 i)On the grid below , plot a graph of the percentage of Bismuth remaining(Vertical axis) against time. 21 ii)Using the graph, determine the: I. Half – life of the Bismuth isotope II.


Time (min) 0 6 12 22 38 62 100 Percentage of Bismuth 100 81 65 46 29 12 3 i)On the grid below , plot a graph of the percentage of Bismuth remaining(Vertical axis) against time. 21 ii)Using the graph, determine the: I. Half – life of the Bismuth isotope II. Original mass of the Bismuth isotope given that the mass that remained after 70 minutes was 0.16g (2mks) d) Give one use of radioactive isotopes in medicine (1mk) 14.a)Distinguish between nuclear fission and nuclear fusion. (2mks) Describe how solid wastes containing radioactive substances should be disposed of. (1mk) b)(i)Find the values of Z1 and Z2 in the nuclear equation below Z1 1 94 140 1 U + n -> Sr + Xe + 2 n 92 0 38 Z2 0 iii)What type of nuclear reaction is represented in b (i) above? A radioactive cobalt 6128Co undergoes decay by emitting a beta particle and forming Nickel atom, Write a balanced decay equation for the above change 1 mark If a sample of the cobalt has an activity of 1000 counts per minute, determine the time it would take for its activity to decrease to 62.50 if the half-life of the element is 30years 2 marks Define the term half-life. The diagram below shows the rays emitted by a radioactive sample
22 a) Identify the rays S,R and Q S- Beta ( β )particle/ray R- Alpha (α )particle/ray Q- Gamma (y )particle/ray b) State what would happen if an aluminium plate is placed in the path of ray R,S and Q: R-is blocked/stopped/do not pass through Q-is not blocked/pass through S-is blocked/stopped/do not pass through (c)The diagram bellow is the radioactive decay series of nuclide A which is 24194Pu.Use it to answer the questions that follow. The letters are not the actual symbols of the elements. (a)Which letter represent the : Explain.


The diagram below shows the rays emitted by a radioactive sample
22 a) Identify the rays S,R and Q S- Beta ( β )particle/ray R- Alpha (α )particle/ray Q- Gamma (y )particle/ray b) State what would happen if an aluminium plate is placed in the path of ray R,S and Q: R-is blocked/stopped/do not pass through Q-is not blocked/pass through S-is blocked/stopped/do not pass through (c)The diagram bellow is the radioactive decay series of nuclide A which is 24194Pu.Use it to answer the questions that follow.

The letters are not the actual symbols of the elements.

(a)Which letter represent the : Explain.

(i)shortest lived nuclide L-has the shortest half life (ii)longest lived nuclide
23 P-Is stable (iii) nuclide with highest n/p ratio L-has the shortest half life thus most unstable thus easily/quickly decay/disintegrate (iv) nuclide with lowest n/p ratio P-is stable thus do not decay/disintegrate (b)How long would it take for the following: (i)Nuclide A to change to B 10 years (half life of A) (ii) Nuclide D to change to H 27days +162000years+70000years+16days 232000 years and 43 days (iii) Nuclide A to change to P 27days +162000years+70000years+16days 232000 years and 43 days Study THE END


12.0.0 GAS LAWS (15 LESSONS) (a)Gas laws 1. Matter is made up of small particle in accordance to Kinetic Theory of matter: Naturally, there are basically three states of matter: Solid, Liquid and gas: (i)A solid is made up of particles which are very closely packed with a definite/fixed shape and fixed/definite volume /occupies definite space. It has a very high density. (ii) A liquid is made up of particles which have some degree of freedom. It thus has no definite/fixed shape. It takes the shape of the container it is put. A liquid has fixed/definite volume/occupies definite space. (iii)A gas is made up of particles free from each other. It thus has no definite /fixed shape. It takes the shape of the container it is put. It has no fixed/definite volume/occupies every space in a container. 2.Gases are affected by physical conditions. There are two physical conditions: (i)Temperature (ii)Pressure 3. The SI unit of temperature is Kelvin(K). Degrees Celsius/Centigrade(oC) are also used. The two units can be interconverted from the relationship: oC + 273= K K -273 = oC Practice examples 1. Convert the following into Kelvin. (i) O oC oC + 273 = K substituting : O oC + 273 = 273 K (ii) -273 oC oC + 273 = K substituting : -273oC + 273 = 0 K (iii) 25 oC oC + 273 = K substituting : 25 oC + 273 = 298 K (iv) 100 oC oC + 273 = K substituting : 100 oC + 273 = 373 K
2. Convert the following into degrees Celsius/Centigrade(oC).


Convert the following into Kelvin. (i) O oC oC + 273 = K substituting : O oC + 273 = 273 K (ii) -273 oC oC + 273 = K substituting : -273oC + 273 = 0 K (iii) 25 oC oC + 273 = K substituting : 25 oC + 273 = 298 K (iv) 100 oC oC + 273 = K substituting : 100 oC + 273 = 373 K
2. Convert the following into degrees Celsius/Centigrade(oC). (i) 10 K K -273 = oC substituting: 10 – 273 = -263 oC (ii) (i) 1 K K -273 = oC substituting: 1 – 273 = -272 oC (iii) 110 K K -273 = oC substituting: 110 – 273 = -163 oC (iv) -24 K K -273 = oC substituting: -24 – 273 = -297 oC The standard temperature is 273K = 0 oC. The room temperature is assumed to be 298K = 25oC 4. The SI unit of pressure is Pascal(Pa) / Newton per metre squared (Nm-2) . Millimeters’ of mercury(mmHg) ,centimeters of mercury(cmHg) and atmospheres are also commonly used. The units are not interconvertible but Pascals(Pa) are equal to Newton per metre squared(Nm-2). The standard pressure is the atmospheric pressure. Atmospheric pressure is equal to about: (i)101325 Pa (ii)101325 Nm-2 (iii)760 mmHg (iv)76 cmHg (v)one atmosphere. 5. Molecules of gases are always in continuous random motion at high speed. This motion is affected by the physical conditions of temperature and pressure. Physical conditions change the volume occupied by gases in a closed system. The effect of physical conditions of temperature and pressure was investigated and expressed in both Boyles and Charles laws. 6.


Physical conditions change the volume occupied by gases in a closed system. The effect of physical conditions of temperature and pressure was investigated and expressed in both Boyles and Charles laws. 6. Boyles law states that “the volume of a fixed mass of a gas is inversely proportional to the pressure at constant/fixed temperature ” Mathematically:
Volume α 1 (Fixed /constant Temperature) Pressure V α 1 (Fixed /constant T) ie PV = Constant(k) P From Boyles law , an increase in pressure of a gas cause a decrease in volume. i.e doubling the pressure cause the volume to be halved. Graphically therefore a plot of volume(V) against pressure (P) produces a curve. V P Graphically a plot of volume(V) against inverse/reciprocal of pressure (1/p) produces a straight line V 1/P For two gases then P1 V1 = P2 V2 P1 = Pressure of gas 1 V1 = Volume of gas 1 P2 = Pressure of gas 2 V2 = Volume of gas 2
Practice examples: 1. A fixed mass of gas at 102300Pa pressure has a volume of 25cm3.Calculate its volume if the pressure is doubled. Working P1 V1 = P2 V2 Substituting :102300 x 25 = (102300 x 2) x V2 V2 = 102300 x 25 = 12.5cm3 (102300 x 2) 2. Calculate the pressure which must be applied to a fixed mass of 100cm3 of Oxygen for its volume to triple at 100000Nm-2. P1 V1 = P2 V2 Substituting :100000 x 100 = P2 x (100 x 3) V2 = 100000 x 100 = 33333.3333 Nm-2 (100 x 3) 3.A 60cm3 weather ballon full of Hydrogen at atmospheric pressure of 101325Pa was released into the atmosphere. Will the ballon reach stratosphere where the pressure is 90000Pa?


Calculate the pressure which must be applied to a fixed mass of 100cm3 of Oxygen for its volume to triple at 100000Nm-2. P1 V1 = P2 V2 Substituting :100000 x 100 = P2 x (100 x 3) V2 = 100000 x 100 = 33333.3333 Nm-2 (100 x 3) 3.A 60cm3 weather ballon full of Hydrogen at atmospheric pressure of 101325Pa was released into the atmosphere. Will the ballon reach stratosphere where the pressure is 90000Pa? P1 V1 = P2 V2 Substituting :101325 x 60 = 90000 x V2 V2 = 101325 x 60 = 67.55 cm3 90000 The new volume at 67.55 cm3 exceed ballon capacity of 60.00 cm3.It will burst before reaching destination. 7.Charles law states that“the volume of a fixed mass of a gas is directly proportional to the absolute temperature at constant/fixed pressure ” Mathematically: Volume α Pressure (Fixed /constant pressure) V α T (Fixed /constant P) ie V = Constant(k) T From Charles law , an increase in temperature of a gas cause an increase in volume. i.e doubling the temperature cause the volume to be doubled. Gases expand/increase by 1/273 by volume on heating.Gases contact/decrease by 1/273 by volume on cooling at constant/fixed pressure. The volume of a gas continue decreasing with decrease in temperature until at -273oC /0 K the volume is zero. i.e. there is no gas. This temperature is called absolute zero. It is the lowest temperature at which a gas can exist. Graphically therefore a plot of volume(V) against Temperature(T) in: (i)oC produces a straight line that is extrapolated to the absolute zero of -273oC . V -273oC 0oC T(oC) (ii)Kelvin/K produces a straight line from absolute zero of O Kelvin V 0 T(Kelvin)
For two gases then V1 = V2 T1 T2 T1 = Temperature in Kelvin of gas 1 V1 = Volume of gas 1 T2 = Temperature in Kelvin of gas 2 V2 = Volume of gas 2 Practice examples: 1.


It is the lowest temperature at which a gas can exist. Graphically therefore a plot of volume(V) against Temperature(T) in: (i)oC produces a straight line that is extrapolated to the absolute zero of -273oC . V -273oC 0oC T(oC) (ii)Kelvin/K produces a straight line from absolute zero of O Kelvin V 0 T(Kelvin)
For two gases then V1 = V2 T1 T2 T1 = Temperature in Kelvin of gas 1 V1 = Volume of gas 1 T2 = Temperature in Kelvin of gas 2 V2 = Volume of gas 2 Practice examples: 1. 500cm3 of carbon(IV)oxide at 0oC was transfered into a cylinder at -4oC. If the capacity of the cylinder is 450 cm3,explain what happened. V1 = V2 substituting 500 = V2 T1 T2 (0 +273) (-4 +273) = 500 x (-4 x 273) = 492.674cm3 (0 + 273) The capacity of cylinder (500cm3) is less than new volume(492.674cm3). 7.326cm3(500-492.674cm3)of carbon(IV)oxide gas did not fit into the cylinder. 2. A mechanic was filling a deflated tyre with air in his closed garage using a hand pump. The capacity of the tyre was 40,000cm3 at room temperature. He rolled the tyre into the car outside. The temperature outside was 30oC.Explain what happens. V1 = V2 substituting 40000 = V2 T1 T2 (25 +273) (30 +273) = 40000 x (30 x 273) = 40671.1409cm3 (25 + 273) The capacity of a tyre (40000cm3) is less than new volume(40671.1409cm3). The tyre thus bursts. 3. A hydrogen gas balloon with 80cm3 was released from a research station at room temperature. If the temperature of the highest point it rose is -30oC , explain what happened.


3. A hydrogen gas balloon with 80cm3 was released from a research station at room temperature. If the temperature of the highest point it rose is -30oC , explain what happened. V1 = V2 substituting 80 = V2 T1 T2 (25 +273) (-30 +273) = 80 x (-30 x 273) = 65.2349cm3
(25 + 273) The capacity of balloon (80cm3) is more than new volume (65.2349cm3). The balloon thus remained intact. 8. The continuous random motion of gases differ from gas to the other.The movement of molecules (of a gas) from region of high concentration to a region of low concentration is called diffusion. The rate of diffusion of a gas depends on its density. i.e. The higher the rate of diffusion, the less dense the gas. The density of a gas depends on its molar mass/relative molecular mass. i.e. The higher the density the higher the molar mass/relative atomic mass and thus the lower the rate of diffusion. Examples 1.Carbon (IV)oxide(CO2) has a molar mass of 44g.Nitrogen(N2)has a molar mass of 28g. (N2)is thus lighter/less dense than Carbon (IV)oxide(CO2). N2 diffuses faster than CO2. 2.Ammonia(NH3) has a molar mass of 17g.Nitrogen(N2)has a molar mass of 28g. (N2)is thus about twice lighter/less dense than Ammonia(NH3). Ammonia(NH3) diffuses twice faster than N2. 3. Ammonia(NH3) has a molar mass of 17g.Hydrogen chloride gas has a molar mass of 36.5g.Both gases on contact react to form white fumes of ammonium chloride .When a glass/cotton wool dipped in ammonia and another glass/cotton wool dipped in hydrochloric acid are placed at opposite ends of a glass tube, both gases diffuse towards each other. A white disk appears near to glass/cotton wool dipped in hydrochloric acid. This is because hydrogen chloride is heavier/denser than Ammonia and thus its rate of diffusion is lower . The rate of diffusion of a gas is in accordance to Grahams law of diffusion.


A white disk appears near to glass/cotton wool dipped in hydrochloric acid. This is because hydrogen chloride is heavier/denser than Ammonia and thus its rate of diffusion is lower . The rate of diffusion of a gas is in accordance to Grahams law of diffusion. Grahams law states that: “the rate of diffusion of a gas is inversely proportional to the square root of its density, at the same/constant/fixed temperature and pressure” Mathematically R α 1 and since density is proportional to mass then R α 1 √ p √ m For two gases then: R1 = R2 where: R1 and R2 is the rate of diffusion of 1st and 2nd gas. √M2 √M1 M1 and M2 is the molar mass of 1st and 2nd gas. Since rate is inverse of time. i.e. the higher the rate the less the time: For two gases then: T1 = T2 where: T1 and T2 is the time taken for 1st and 2nd gas to diffuse. √M1 √M2 M1 and M2 is the molar mass of 1st and 2nd gas. Practice examples: 1. It takes 30 seconds for 100cm3 of carbon(IV)oxide to diffuse across a porous plate. How long will it take 150cm3 of nitrogen(IV)oxide to diffuse across the same plate under the same conditions of temperature and pressure.


Molar mass CO2 = 44g Molar mass HCl = 36.5g T CO2 = √ molar mass CO2 => 200 seconds = √ 44.0 T HCl √ molar mass HCl T HCl √ 36.5 T HCl = 200seconds x √ 36.5 = 182.1588 seconds √ 44.0 3. Oxygen gas takes 250 seconds to diffuse through a porous diaphragm. Calculate the molar mass of gas Z which takes 227 second to diffuse. Molar mass O2 = 32g Molar mass Z = x g T O2 = √ molar mass O2 => 250 seconds = √ 32.0 T Z √ molar mass Z 227seconds √ x √ x = 227seconds x √ 32 = 26.3828 grams 250 4. 25cm3 of carbon(II)oxide diffuses across a porous plate in 25seconds. How long will it take 75cm3 of Carbon(IV)oxide to diffuse across the same plate under the same conditions of temperature and pressure.


Molar mass O2 = 32g Molar mass Z = x g T O2 = √ molar mass O2 => 250 seconds = √ 32.0 T Z √ molar mass Z 227seconds √ x √ x = 227seconds x √ 32 = 26.3828 grams 250 4. 25cm3 of carbon(II)oxide diffuses across a porous plate in 25seconds. How long will it take 75cm3 of Carbon(IV)oxide to diffuse across the same plate under the same conditions of temperature and pressure. (C=12.0,0=16.0) Molar mass CO2 = 44.0 Molar mass CO = 28.0 Method 1 25cm3 CO takes 25seconds 75cm3 takes 75 x25 = 75seconds 25
T CO2 = √ molar mass CO2 => T CO2seconds = √ 44.0 T CO √ molar mass CO 75 √ 28.0 T CO2 =75seconds x √ 44.0 = 94.0175 seconds √ 28.0 Method 2 25cm3 CO2 takes 25seconds 1cm3 takes 25 x1 = 1.0cm3sec-1 25 R CO2 = √ molar mass CO => x cm3sec-1 = √ 28.0 R CO √ molar mass CO2 1.0cm3sec-1 √ 44.0 R CO2 = 1.0cm3sec-1 x √ 28.0 = 0.7977cm3sec-1 √ 44.0 0.7977cm3 takes 1 seconds 75cm3 takes 75cm3 = 94.0203seconds 0.7977cm3
13.0.0 THE MOLE-FORMULAE AND CHEMICAL EQUATIONS (40 LESSONS) Introduction to the mole, molar masses and Relative atomic masses 1. The mole is the SI unit of the amount of substance. 2. The number of particles e.g. atoms, ions, molecules, electrons, cows, cars are all measured in terms of moles. 3. The number of particles in one mole is called the Avogadros Constant. It is denoted “L”.


3. The number of particles in one mole is called the Avogadros Constant. It is denoted “L”. The Avogadros Constant contain 6.023 x10 23 particles. i.e. 1mole = 6.023 x10 23 particles = 6.023 x10 23 2 moles = 2 x 6.023 x10 23 particles = 1.205 x10 24 0.2 moles = 0.2 x 6.023 x10 23 particles = 1.205 x10 22 0.0065 moles = 0.0065 x 6.023 x10 23 particles = 3.914 x10 21 3. The mass of one mole of a substance is called molar mass. The molar mass of: (i)an element has mass equal to relative atomic mass /RAM(in grams)of the element e.g. Molar mass of carbon(C)= relative atomic mass = 12.0g 6.023 x10 23 particles of carbon = 1 mole =12.0 g Molar mass of sodium(Na) = relative atomic mass = 23.0g 6.023 x10 23 particles of sodium = 1 mole =23.0 g Molar mass of Iron (Fe) = relative atomic mass = 56.0g 6.023 x10 23 particles of iron = 1 mole =56.0 g (ii)a molecule has mass equal to relative molecular mass /RMM (in grams)of the molecule. Relative molecular mass is the sum of the relative atomic masses of the elements making the molecule. The number of atoms making a molecule is called atomicity. Most gaseous molecules are diatomic (e.g. O2, H2, N2, F2, Cl2, Br2, I2)noble gases are monoatomic(e.g. He, Ar, Ne, Xe),Ozone gas(O3) is triatomic e.g.


Most gaseous molecules are diatomic (e.g. O2, H2, N2, F2, Cl2, Br2, I2)noble gases are monoatomic(e.g. He, Ar, Ne, Xe),Ozone gas(O3) is triatomic e.g. Molar mass Oxygen molecule(O2) =relative molecular mass =(16.0x 2)g =32.0g 6.023 x10 23 particles of Oxygen molecule = 1 mole = 32.0 g
2 2 Molar mass chlorine molecule(Cl2) =relative molecular mass =(35.5x 2)g =71.0g 6.023 x10 23 particles of chlorine molecule = 1 mole = 71.0 g Molar mass Nitrogen molecule(N2) =relative molecular mass =(14.0x 2)g =28.0g 6.023 x10 23 particles of Nitrogen molecule = 1 mole = 28.0 g (ii)a compound has mass equal to relative formular mass /RFM (in grams)of the molecule. Relative formular mass is the sum of the relative atomic masses of the elements making the compound. e.g.


It is the simplest whole number ratios in which atoms of elements combine to form the compound. 2.It is mathematically the lowest common multiple (LCM) of the atoms of the elements in the compound 3.Practically the empirical formula of a compound can be determined as in the following examples. To determine the empirical formula of copper oxide (a)Method 1:From copper to copper(II)oxide Procedure. Weigh a clean dry covered crucible(M1).Put two spatula full of copper powder into the crucible. Weigh again (M2).Heat the crucible on a strong Bunsen flame for five minutes. Lift the lid, and swirl the crucible carefully using a pair of tong. Cover the crucible and continue heating for another five minutes. Remove the lid and stop heating. Allow the crucible to cool. When cool replace the lid and weigh the contents again (M3). Sample results Mass of crucible(M1) 15.6g Mass of crucible + copper before heating(M2) 18.4 Mass of crucible + copper after heating(M3) 19.1 Sample questions 1. Calculate the mass of copper powder used. Mass of crucible + copper before heating(M2) = 18.4 Less Mass of crucible(M1) = - 15.6g Mass of copper 2.8 g 2. Calculate the mass of Oxygen used to react with copper. Method I Mass of crucible + copper after heating(M3) = 19.1g
7 7 Mass of crucible + copper before heating(M2) = - 18.4g Mass of Oxygen = 0.7 g Method II Mass of crucible + copper after heating(M3) = 19.1g Mass of crucible = - 15.6g Mass of copper(II)Oxide = 3.5 g Mass of copper(II)Oxide = 3.5 g Mass of copper = - 2.8 g Mass of Oxygen = 0.7 g 3.


Mass of crucible + copper before heating(M2) = 18.4 Less Mass of crucible(M1) = - 15.6g Mass of copper 2.8 g 2. Calculate the mass of Oxygen used to react with copper. Method I Mass of crucible + copper after heating(M3) = 19.1g
7 7 Mass of crucible + copper before heating(M2) = - 18.4g Mass of Oxygen = 0.7 g Method II Mass of crucible + copper after heating(M3) = 19.1g Mass of crucible = - 15.6g Mass of copper(II)Oxide = 3.5 g Mass of copper(II)Oxide = 3.5 g Mass of copper = - 2.8 g Mass of Oxygen = 0.7 g 3. Calculate the number of moles of: (i) copper used (Cu = 63.5) number of moles of copper = mass used => 2.8 = 0.0441moles Molar mass 63.5 (ii) Oxygen used (O = 16.0) number of moles of oxygen = mass used => 0.7 = 0.0441moles Molar mass 16.0 4. Determine the mole ratio of the reactants Moles of copper = 0.0441moles = 1 => Mole ratio Cu: O = 1:1 Moles of oxygen 0.0441moles 1 5.What is the empirical, formula of copper oxide formed. CuO (copper(II)oxide 6. State and explain the observations made during the experiment. Observation Colour change from brown to black Explanation Copper powder is brown. On heating it reacts with oxygen from the air to form black copper(II)oxide 7. Explain why magnesium ribbon/shavings would be unsuitable in a similar experiment as the one above. Hot magnesium generates enough heat energy to react with both Oxygen and Nitrogen in the air forming a white solid mixture of Magnesuin oxide and magnesium nitride. This causes experimental mass errors. (b)Method 2:From copper(II)oxide to copper
8 8 Procedure. Weigh a clean dry porcelain boat (M1). Put two spatula full of copper(II)oxide powder into the crucible.


(b)Method 2:From copper(II)oxide to copper
8 8 Procedure. Weigh a clean dry porcelain boat (M1). Put two spatula full of copper(II)oxide powder into the crucible. Reweigh the porcelain boat (M2).Put the porcelain boat in a glass tube and set up the apparatus as below; jgthungu@gmail.com52HEATCopper(II)oxideHydrogen /Laboratory gas /Ammonia gas/Carbon(II)Oxidegas from a generatorExcess hydrogenburningDetermining empirical formula from copper(II)oxide to copper Pass slowly(to prevent copper(II)oxide from being blown away)a stream of either dry Hydrogen /ammonia/laboratory gas/ carbon(II)oxide gas for about two minutes from a suitable generator. When all the in the apparatus set up is driven out ,heat the copper(II)oxide strongly for about five minutes until there is no further change. Stop heating. Continue passing the gases until the glass tube is cool. Turn off the gas generator. Carefully remove the porcelain boat form the combustion tube. Reweigh (M3). Sample results Mass of boat(M1) 15.6g Mass of boat before heating(M2) 19.1 Mass of boat after heating(M3) 18.4 Sample questions
9 9 1. Calculate the mass of copper(II)oxide used. Mass of boat before heating(M2) = 19.1 Mass of empty boat(M1) = - 15.6g Mass of copper(II)Oxide 3.5 g 2. Calculate the mass of (i) Oxygen. Mass of boat before heating(M2) = 19.1 Mass of boat after heating (M3) = - 18.4g Mass of oxygen = 0.7 g (ii)Copper Mass of copper(II)Oxide = 3.5 g Mass of oxygen = 0.7 g Mass of oxygen = 2.8 g 3.


Mass of boat before heating(M2) = 19.1 Mass of empty boat(M1) = - 15.6g Mass of copper(II)Oxide 3.5 g 2. Calculate the mass of (i) Oxygen. Mass of boat before heating(M2) = 19.1 Mass of boat after heating (M3) = - 18.4g Mass of oxygen = 0.7 g (ii)Copper Mass of copper(II)Oxide = 3.5 g Mass of oxygen = 0.7 g Mass of oxygen = 2.8 g 3. Calculate the number of moles of: (i) Copper used (Cu = 63.5) number of moles of copper = mass used => 2.8 = 0.0441moles Molar mass 63.5 (ii) Oxygen used (O = 16.0) number of moles of oxygen = mass used => 0.7 = 0.0441moles Molar mass 16.0 4. Determine the mole ratio of the reactants Moles of copper = 0.0441moles = 1 => Mole ratio Cu: O = 1:1 Moles of oxygen 0.0441moles 1 5.What is the empirical, formula of copper oxide formed. CuO (copper(II)oxide 6. State and explain the observations made during the experiment. Observation Colour change from black to brown Explanation Copper(II)oxide powder is black. On heating it is reduced by a suitable reducing agent to brown copper metal. 7. Explain why magnesium oxide would be unsuitable in a similar experiment as the one above. Magnesium is high in the reactivity series. None of the above reducing agents is strong enough to reduce the oxide to the metal. 10 10 8.


Magnesium is high in the reactivity series. None of the above reducing agents is strong enough to reduce the oxide to the metal. 10 10 8. Write the equation for the reaction that would take place when the reducing agent is: (i) Hydrogen CuO(s) + H2(g) -> Cu(s) + H2O(l) (Black) (brown) (colourless liquid form on cooler parts ) (ii)Carbon(II)oxide CuO(s) + CO (g) -> Cu(s) + CO2(g) (Black) (brown) (colourless gas, form white ppt with lime water ) (iii)Ammonia 3CuO(s) + 2NH3(g) -> 3Cu(s) + N2 (g) + 3H2O(l) (Black) (brown) (colourless liquid form on cooler parts ) 9. Explain why the following is necessary during the above experiment; (i)A stream of dry hydrogen gas should be passed before heating copper (II) Oxide. Air combine with hydrogen in presence of heat causing an explosion (ii)A stream of dry hydrogen gas should be passed after heating copper (II) Oxide has been stopped. Hot metallic copper can be re-oxidized back to copper(II)oxide (iii) A stream of excess carbon (II)oxide gas should be ignited to burn Carbon (II)oxide is highly poisonous/toxic. On ignition it burns to form less toxic carbon (IV)oxide gas. 10. State two sources of error in this experiment. (i)All copper(II)oxide may not be reduced to copper. (ii)Some copper(II)oxide may be blown out the boat by the reducing agent. 4.Theoreticaly the empirical formula of a compound can be determined as in the following examples. (a)A oxide of copper contain 80% by mass of copper. Determine its empirical formula.


4.Theoreticaly the empirical formula of a compound can be determined as in the following examples. (a)A oxide of copper contain 80% by mass of copper. Determine its empirical formula. (Cu = 63.5, 16.0)
11 11 % of Oxygen = 100% - % of Copper => 100- 80 = 20% of Oxygen Element Copper Oxygen Symbol Cu O Moles present = % composition Molar mass 80 63.5 20 16 Divide by the smallest value 1.25 1.25 1.25 1.25 Mole ratios 1 1 Empirical formula is CuO (b)1.60g of an oxide of Magnesium contain 0.84g by mass of Magnesium. Determine its empirical formula(Mg = 24.0, 16.0) Mass of Oxygen = 1.60 – 0.84 => 0.56 g of Oxygen Element Magnesium Oxygen Symbol Mg O Moles present = % composition Molar mass 0.84 24 0.56 16 Divide by the smallest value 0.35 0.35 0.35 0.35 Mole ratios 1 1 Empirical formula is MgO (c)An oxide of Silicon contain 47% by mass of Silicon. What is its empirical formula(Si = 28.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2
12 12 (d)A compound contain 70% by mass of Iron and 30% Oxygen.


(Cu = 63.5, 16.0)
11 11 % of Oxygen = 100% - % of Copper => 100- 80 = 20% of Oxygen Element Copper Oxygen Symbol Cu O Moles present = % composition Molar mass 80 63.5 20 16 Divide by the smallest value 1.25 1.25 1.25 1.25 Mole ratios 1 1 Empirical formula is CuO (b)1.60g of an oxide of Magnesium contain 0.84g by mass of Magnesium. Determine its empirical formula(Mg = 24.0, 16.0) Mass of Oxygen = 1.60 – 0.84 => 0.56 g of Oxygen Element Magnesium Oxygen Symbol Mg O Moles present = % composition Molar mass 0.84 24 0.56 16 Divide by the smallest value 0.35 0.35 0.35 0.35 Mole ratios 1 1 Empirical formula is MgO (c)An oxide of Silicon contain 47% by mass of Silicon. What is its empirical formula(Si = 28.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2
12 12 (d)A compound contain 70% by mass of Iron and 30% Oxygen. What is its empirical formula(Fe = 56.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2 2.During heating of a hydrated copper (II)sulphate(VI) crystals, the following readings were obtained: Mass of evaporating dish =300.0g Mass of evaporating dish + hydrated salt = 305.0g Mass of evaporating dish + anhydrous salt = 303.2g Calculate the number of water of crystallization molecules in hydrated copper (II)sulphate(VI) (Cu =64.5, S = 32.0,O=16.0, H = 1.0) Working Mass of Hydrated salt = 305.0g -300.0g = 5.0g Mass of anhydrous salt = 303.2 g -300.0g = 3.2 g Mass of water in hydrated salt = 5.0g -3.2 g = 1.8g Molar mass of water(H2O) = 18.0g Molar mass of anhydrous copper (II)sulphate(VI) (CuSO4) = 160.5g Element/compound anhydrous copper (II) sulphate(VI) Oxygen Symbol Si O Moles present = composition by mass Molar mass 3,2 160.5 1.8 18 Divide by the smallest value 0.0199 0.0199 0.1 18 Mole ratios 1 5 The empirical formula of hydrated salt = CuSO4.5H2O
13 13 Hydrated salt has five/5 molecules of water of crystallizations 4.


Determine its empirical formula(Mg = 24.0, 16.0) Mass of Oxygen = 1.60 – 0.84 => 0.56 g of Oxygen Element Magnesium Oxygen Symbol Mg O Moles present = % composition Molar mass 0.84 24 0.56 16 Divide by the smallest value 0.35 0.35 0.35 0.35 Mole ratios 1 1 Empirical formula is MgO (c)An oxide of Silicon contain 47% by mass of Silicon. What is its empirical formula(Si = 28.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2
12 12 (d)A compound contain 70% by mass of Iron and 30% Oxygen. What is its empirical formula(Fe = 56.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2 2.During heating of a hydrated copper (II)sulphate(VI) crystals, the following readings were obtained: Mass of evaporating dish =300.0g Mass of evaporating dish + hydrated salt = 305.0g Mass of evaporating dish + anhydrous salt = 303.2g Calculate the number of water of crystallization molecules in hydrated copper (II)sulphate(VI) (Cu =64.5, S = 32.0,O=16.0, H = 1.0) Working Mass of Hydrated salt = 305.0g -300.0g = 5.0g Mass of anhydrous salt = 303.2 g -300.0g = 3.2 g Mass of water in hydrated salt = 5.0g -3.2 g = 1.8g Molar mass of water(H2O) = 18.0g Molar mass of anhydrous copper (II)sulphate(VI) (CuSO4) = 160.5g Element/compound anhydrous copper (II) sulphate(VI) Oxygen Symbol Si O Moles present = composition by mass Molar mass 3,2 160.5 1.8 18 Divide by the smallest value 0.0199 0.0199 0.1 18 Mole ratios 1 5 The empirical formula of hydrated salt = CuSO4.5H2O
13 13 Hydrated salt has five/5 molecules of water of crystallizations 4. The molecular formula is the actual number of each kind of atoms present in a molecule of a compound.


What is its empirical formula(Si = 28.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2
12 12 (d)A compound contain 70% by mass of Iron and 30% Oxygen. What is its empirical formula(Fe = 56.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2 2.During heating of a hydrated copper (II)sulphate(VI) crystals, the following readings were obtained: Mass of evaporating dish =300.0g Mass of evaporating dish + hydrated salt = 305.0g Mass of evaporating dish + anhydrous salt = 303.2g Calculate the number of water of crystallization molecules in hydrated copper (II)sulphate(VI) (Cu =64.5, S = 32.0,O=16.0, H = 1.0) Working Mass of Hydrated salt = 305.0g -300.0g = 5.0g Mass of anhydrous salt = 303.2 g -300.0g = 3.2 g Mass of water in hydrated salt = 5.0g -3.2 g = 1.8g Molar mass of water(H2O) = 18.0g Molar mass of anhydrous copper (II)sulphate(VI) (CuSO4) = 160.5g Element/compound anhydrous copper (II) sulphate(VI) Oxygen Symbol Si O Moles present = composition by mass Molar mass 3,2 160.5 1.8 18 Divide by the smallest value 0.0199 0.0199 0.1 18 Mole ratios 1 5 The empirical formula of hydrated salt = CuSO4.5H2O
13 13 Hydrated salt has five/5 molecules of water of crystallizations 4. The molecular formula is the actual number of each kind of atoms present in a molecule of a compound. The empirical formula of an ionic compound is the same as the chemical formula but for simple molecular structured compounds, the empirical formula may not be the same as the chemical formula.


What is its empirical formula(Fe = 56.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2 2.During heating of a hydrated copper (II)sulphate(VI) crystals, the following readings were obtained: Mass of evaporating dish =300.0g Mass of evaporating dish + hydrated salt = 305.0g Mass of evaporating dish + anhydrous salt = 303.2g Calculate the number of water of crystallization molecules in hydrated copper (II)sulphate(VI) (Cu =64.5, S = 32.0,O=16.0, H = 1.0) Working Mass of Hydrated salt = 305.0g -300.0g = 5.0g Mass of anhydrous salt = 303.2 g -300.0g = 3.2 g Mass of water in hydrated salt = 5.0g -3.2 g = 1.8g Molar mass of water(H2O) = 18.0g Molar mass of anhydrous copper (II)sulphate(VI) (CuSO4) = 160.5g Element/compound anhydrous copper (II) sulphate(VI) Oxygen Symbol Si O Moles present = composition by mass Molar mass 3,2 160.5 1.8 18 Divide by the smallest value 0.0199 0.0199 0.1 18 Mole ratios 1 5 The empirical formula of hydrated salt = CuSO4.5H2O
13 13 Hydrated salt has five/5 molecules of water of crystallizations 4. The molecular formula is the actual number of each kind of atoms present in a molecule of a compound. The empirical formula of an ionic compound is the same as the chemical formula but for simple molecular structured compounds, the empirical formula may not be the same as the chemical formula. The molecular formula is a multiple of empirical formula .It is determined from the relationship: (i) n = Relative formular mass Relative empirical formula where n is a whole number.


The molecular formula is the actual number of each kind of atoms present in a molecule of a compound. The empirical formula of an ionic compound is the same as the chemical formula but for simple molecular structured compounds, the empirical formula may not be the same as the chemical formula. The molecular formula is a multiple of empirical formula .It is determined from the relationship: (i) n = Relative formular mass Relative empirical formula where n is a whole number. (ii) Relative empirical formula x n = Relative formular mass where n is a whole number. Practice sample examples 1. A hydrocarbon was found to contain 92.3% carbon and the remaining Hydrogen. If the molecular mass of the compound is 78, determine the molecular formula(C=12.0, H =1.0) Mass of Hydrogen = 100 – 92.3 => 7.7% of Oxygen Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 92.3 12 7.7 1 Divide by the smallest value 7.7 7.7 7.7 7.7 Mole ratios 1 1 Empirical formula is CH The molecular formular is thus determined : n = Relative formular mass = 78 = 6
14 14 Relative empirical formula 13 The molecular formula is (C H ) x 6 = C6H6 2. A compound of carbon, hydrogen and oxygen contain 54.55% carbon, 9.09% and remaining 36.36% oxygen. If its relative molecular mass is 88, determine its molecular formula(C=12.0, H =1.0, O= 16.0) Element Carbon Hydrogen Oxygen Symbol C H O Moles present = % composition Molar mass 54.55 12 9.09 1 36.36 16 Divide by the smallest value 4.5458 2.2725 9.09 2.2725 2.2725 2.2725 Mole ratios 2 4 1 Empirical formula is C2H4O The molecular formula is thus determined : n = Relative formular mass = 88 = 2 Relative empirical formula 44 The molecular formula is (C2H4O ) x 2 = C4H8O2.


If the molecular mass of the compound is 78, determine the molecular formula(C=12.0, H =1.0) Mass of Hydrogen = 100 – 92.3 => 7.7% of Oxygen Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 92.3 12 7.7 1 Divide by the smallest value 7.7 7.7 7.7 7.7 Mole ratios 1 1 Empirical formula is CH The molecular formular is thus determined : n = Relative formular mass = 78 = 6
14 14 Relative empirical formula 13 The molecular formula is (C H ) x 6 = C6H6 2. A compound of carbon, hydrogen and oxygen contain 54.55% carbon, 9.09% and remaining 36.36% oxygen. If its relative molecular mass is 88, determine its molecular formula(C=12.0, H =1.0, O= 16.0) Element Carbon Hydrogen Oxygen Symbol C H O Moles present = % composition Molar mass 54.55 12 9.09 1 36.36 16 Divide by the smallest value 4.5458 2.2725 9.09 2.2725 2.2725 2.2725 Mole ratios 2 4 1 Empirical formula is C2H4O The molecular formula is thus determined : n = Relative formular mass = 88 = 2 Relative empirical formula 44 The molecular formula is (C2H4O ) x 2 = C4H8O2. 4.A hydrocarbon burns completely in excess air to form 5.28 g of carbon (IV) oxide and 2,16g of water. If the molecular mass of the hydrocarbon is 84, draw and name its molecular structure. Since a hydrocarbon is a compound containing Carbon and Hydrogen only.


4.A hydrocarbon burns completely in excess air to form 5.28 g of carbon (IV) oxide and 2,16g of water. If the molecular mass of the hydrocarbon is 84, draw and name its molecular structure. Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 => Molar mass of CO2 12 x 5.28 = 1.44g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 2.16 = 0.24g√ 18 Element Carbon Hydrogen
15 15 Symbol C H Moles present = mass Molar mass 1.44g 12 0.24g√ 1 Divide by the smallest value 0.12 0.12 0.24 0.12 Mole ratios 1 2√ Empirical formula is CH2√ The molecular formular is thus determined : n = Relative formular mass = 84 = 6√ Relative empirical formula 14 The molecular formula is (CH2 ) x 6 = C6H12. √ molecular name Hexene√/Hex-1-ene (or any position isomer of Hexene) Molecular structure H H H H H H H C C C C C C H√ H H H H 5. Compound A contain 5.2% by mass of Nitrogen .The other elements present are Carbon, hydrogen and Oxygen. On combustion of 0.085g of A in excess Oxygen,0.224g of carbon(IV)oxide and 0.0372g of water was formed.


√ molecular name Hexene√/Hex-1-ene (or any position isomer of Hexene) Molecular structure H H H H H H H C C C C C C H√ H H H H 5. Compound A contain 5.2% by mass of Nitrogen .The other elements present are Carbon, hydrogen and Oxygen. On combustion of 0.085g of A in excess Oxygen,0.224g of carbon(IV)oxide and 0.0372g of water was formed. Determine the empirical formula of A (N=14.0, O=16.0 , C=12.0 , H=1.0) Mass of N in A = 5.2% x 0.085 = 0.00442 g Mass of C in A = 12 x 0.224 = 0.0611g 44 Mass of H in A = 2 x 0.0372 = 0.0041g 18 Mass of O in A = 0.085g – 0.004442g = 0.0806g (Mass of C,H,O) => 0.0611g + 0.0041g = 0.0652g (Mass of C,H) 0.0806g (Mass of C,H,O)- 0.0652g (Mass of C,H) = 0.0154 g Element Nitrogen Carbon Hydrogen Oxygen Symbol N C H O Moles present = mass Molar mass 0.00442 g 14 0.0611g 12 0.0041g 1 0.0154 g 16 Divide by the smallest value 0.00032 0.00509 0.0041g 0.00096
16 16 0.00032 0.00032 0.00032 0.00032 Mole ratios 1 16 13 3 Empirical formula = C16H13NO3 (d)Molar gas volume The volume occupied by one mole of all gases at the same temperature and pressure is a constant.It is: (i) 24dm3/24litres/24000cm3 at room temperature(25oC/298K)and pressure(r.t.p). i.e.


22.4dm3 -> 6.0 x1023 2.24dm3 -> 2.24 x 6.0 x1023 22.4 =6.0 x1022 molecules = (CO2) = 3 x 6.0 x1022. = 1.8 x1023 atoms 2. 0.135 g of a gaseous hydrocarbon X on complete combustion produces 0.41g of carbon(IV)oxide and 0.209g of water.0.29g of X occupy 120cm3 at room temperature and 1 atmosphere pressure .Name X and draw its molecular structure.(C=12.0,O= 16.O,H=1.0,1 mole of gas occupies 24dm3 at r.t.p) Molar mass CO2= 44 gmole-1√ Molar mass H2O = 18 gmole-1√ Molar mass X = 0.29 x (24 x 1000)cm3 = 58 gmole-1√ 120cm3 Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 => Molar mass of CO2 12 x 0.41 = 0.1118g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 0.209 = 0.0232g√ 18 Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 0.g118 12 0.0232g√ 1 Divide by the smallest value 0.0093 0.0093 0.0232 0.0093√ Mole ratios 1 x2 2.5x2
18 18 2 5√ Empirical formula is C2H5√ The molecular formular is thus determined : n = Relative formular mass = 58 = 2√ Relative empirical formula 29 The molecular formula is (C2H5 ) x 2 = C4H10.√ Molecule name Butane Molecula structure H H H H H C C C C H√ H H H H (e)Gravimetric analysis Gravimetric analysis is the relationship between reacting masses and the volumes and /or masses of products.


= 1.8 x1023 atoms 2. 0.135 g of a gaseous hydrocarbon X on complete combustion produces 0.41g of carbon(IV)oxide and 0.209g of water.0.29g of X occupy 120cm3 at room temperature and 1 atmosphere pressure .Name X and draw its molecular structure.(C=12.0,O= 16.O,H=1.0,1 mole of gas occupies 24dm3 at r.t.p) Molar mass CO2= 44 gmole-1√ Molar mass H2O = 18 gmole-1√ Molar mass X = 0.29 x (24 x 1000)cm3 = 58 gmole-1√ 120cm3 Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 => Molar mass of CO2 12 x 0.41 = 0.1118g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 0.209 = 0.0232g√ 18 Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 0.g118 12 0.0232g√ 1 Divide by the smallest value 0.0093 0.0093 0.0232 0.0093√ Mole ratios 1 x2 2.5x2
18 18 2 5√ Empirical formula is C2H5√ The molecular formular is thus determined : n = Relative formular mass = 58 = 2√ Relative empirical formula 29 The molecular formula is (C2H5 ) x 2 = C4H10.√ Molecule name Butane Molecula structure H H H H H C C C C H√ H H H H (e)Gravimetric analysis Gravimetric analysis is the relationship between reacting masses and the volumes and /or masses of products. All reactants are in mole ratios to their products in accordance to their stoichiometric equation.


0.135 g of a gaseous hydrocarbon X on complete combustion produces 0.41g of carbon(IV)oxide and 0.209g of water.0.29g of X occupy 120cm3 at room temperature and 1 atmosphere pressure .Name X and draw its molecular structure.(C=12.0,O= 16.O,H=1.0,1 mole of gas occupies 24dm3 at r.t.p) Molar mass CO2= 44 gmole-1√ Molar mass H2O = 18 gmole-1√ Molar mass X = 0.29 x (24 x 1000)cm3 = 58 gmole-1√ 120cm3 Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 => Molar mass of CO2 12 x 0.41 = 0.1118g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 0.209 = 0.0232g√ 18 Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 0.g118 12 0.0232g√ 1 Divide by the smallest value 0.0093 0.0093 0.0232 0.0093√ Mole ratios 1 x2 2.5x2
18 18 2 5√ Empirical formula is C2H5√ The molecular formular is thus determined : n = Relative formular mass = 58 = 2√ Relative empirical formula 29 The molecular formula is (C2H5 ) x 2 = C4H10.√ Molecule name Butane Molecula structure H H H H H C C C C H√ H H H H (e)Gravimetric analysis Gravimetric analysis is the relationship between reacting masses and the volumes and /or masses of products. All reactants are in mole ratios to their products in accordance to their stoichiometric equation. Using the mole ration of reactants and products any volume and/or mass can be determined as in the examples: 1.


Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 => Molar mass of CO2 12 x 0.41 = 0.1118g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 0.209 = 0.0232g√ 18 Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 0.g118 12 0.0232g√ 1 Divide by the smallest value 0.0093 0.0093 0.0232 0.0093√ Mole ratios 1 x2 2.5x2
18 18 2 5√ Empirical formula is C2H5√ The molecular formular is thus determined : n = Relative formular mass = 58 = 2√ Relative empirical formula 29 The molecular formula is (C2H5 ) x 2 = C4H10.√ Molecule name Butane Molecula structure H H H H H C C C C H√ H H H H (e)Gravimetric analysis Gravimetric analysis is the relationship between reacting masses and the volumes and /or masses of products. All reactants are in mole ratios to their products in accordance to their stoichiometric equation. Using the mole ration of reactants and products any volume and/or mass can be determined as in the examples: 1. Calculate the volume of carbon(IV)oxide at r.t.p produced when 5.0 g of calcium carbonate is strongly heated.(Ca=40.0, C= 12.0,O = 16.0,1 mole of gas =22.4 at r.t.p) Chemical equation CaCO3(s) -> CaO(s) + CO2(g) Mole ratios 1: 1: 1 Molar Mass CaCO3 =100g Method 1 100g CaCO3(s) -> 24dm3 CO2(g) at r.t.p 5.0 g CaCO3(s) -> 5.0 g x 24dm3 = 1.2dm3/1200cm3 100g Method 2
19 19 Moles of 5.0 g CaCO3(s) = 5.0 g = 0.05 moles 100 g Mole ratio 1:1 Moles of CO2(g) = 0.05moles Volume of CO2(g) = 0.05 x 24000cm3 =1200cm3 /1.2dm3 2.


All reactants are in mole ratios to their products in accordance to their stoichiometric equation. Using the mole ration of reactants and products any volume and/or mass can be determined as in the examples: 1. Calculate the volume of carbon(IV)oxide at r.t.p produced when 5.0 g of calcium carbonate is strongly heated.(Ca=40.0, C= 12.0,O = 16.0,1 mole of gas =22.4 at r.t.p) Chemical equation CaCO3(s) -> CaO(s) + CO2(g) Mole ratios 1: 1: 1 Molar Mass CaCO3 =100g Method 1 100g CaCO3(s) -> 24dm3 CO2(g) at r.t.p 5.0 g CaCO3(s) -> 5.0 g x 24dm3 = 1.2dm3/1200cm3 100g Method 2
19 19 Moles of 5.0 g CaCO3(s) = 5.0 g = 0.05 moles 100 g Mole ratio 1:1 Moles of CO2(g) = 0.05moles Volume of CO2(g) = 0.05 x 24000cm3 =1200cm3 /1.2dm3 2. 1.0g of an alloy of aluminium and copper were reacted with excess hydrochloric acid.


Using the mole ration of reactants and products any volume and/or mass can be determined as in the examples: 1. Calculate the volume of carbon(IV)oxide at r.t.p produced when 5.0 g of calcium carbonate is strongly heated.(Ca=40.0, C= 12.0,O = 16.0,1 mole of gas =22.4 at r.t.p) Chemical equation CaCO3(s) -> CaO(s) + CO2(g) Mole ratios 1: 1: 1 Molar Mass CaCO3 =100g Method 1 100g CaCO3(s) -> 24dm3 CO2(g) at r.t.p 5.0 g CaCO3(s) -> 5.0 g x 24dm3 = 1.2dm3/1200cm3 100g Method 2
19 19 Moles of 5.0 g CaCO3(s) = 5.0 g = 0.05 moles 100 g Mole ratio 1:1 Moles of CO2(g) = 0.05moles Volume of CO2(g) = 0.05 x 24000cm3 =1200cm3 /1.2dm3 2. 1.0g of an alloy of aluminium and copper were reacted with excess hydrochloric acid. If 840cm3 of hydrogen at s.t.p was produced, calculate the % of copper in the alloy.(Al =27.0,one mole of a gas at s.t.p =22.4dm3 ) Chemical equation Copper does not react with hydrochloric acid 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g) Method 1 3H2(g) = 3 moles x (22.4 x 1000)cm3 => 2 x 27 g Al 840cm3 => 840cm3 x 2 x 27 = 0.675g of Aluminium 3 x 22.4 x 1000 Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g =0.325g of copper % copper = mass of copper x100% = 32.5% Mass of alloy Method 2 Mole ratio 2Al: 3H2 = 2:3 Moles of Hydrogen gas = volume of gas => 840cm3 = 0.0375moles Molar gas volume 22400cm3 Moles of Al = 2/3 moles of H2 => 2/3x 0.0375moles = 0.025moles Mass of Al = moles x molar mass =>0.025moles x 27 = 0.675g Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g = 0.325 g of copper % copper = mass of copper x100% = 32.5% Mass of alloy
20 20 (f)Gay Lussac’s law Gay Lussacs law states that “when gases combine/react they do so in simple volume ratios to each other and to their gaseous products at constant/same temperature and pressure” Gay Lussacs law thus only apply to gases Given the volume of one gas reactant, the other gaseous reactants can be deduced thus: Examples 1.


Calculate the volume of carbon(IV)oxide at r.t.p produced when 5.0 g of calcium carbonate is strongly heated.(Ca=40.0, C= 12.0,O = 16.0,1 mole of gas =22.4 at r.t.p) Chemical equation CaCO3(s) -> CaO(s) + CO2(g) Mole ratios 1: 1: 1 Molar Mass CaCO3 =100g Method 1 100g CaCO3(s) -> 24dm3 CO2(g) at r.t.p 5.0 g CaCO3(s) -> 5.0 g x 24dm3 = 1.2dm3/1200cm3 100g Method 2
19 19 Moles of 5.0 g CaCO3(s) = 5.0 g = 0.05 moles 100 g Mole ratio 1:1 Moles of CO2(g) = 0.05moles Volume of CO2(g) = 0.05 x 24000cm3 =1200cm3 /1.2dm3 2. 1.0g of an alloy of aluminium and copper were reacted with excess hydrochloric acid. If 840cm3 of hydrogen at s.t.p was produced, calculate the % of copper in the alloy.(Al =27.0,one mole of a gas at s.t.p =22.4dm3 ) Chemical equation Copper does not react with hydrochloric acid 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g) Method 1 3H2(g) = 3 moles x (22.4 x 1000)cm3 => 2 x 27 g Al 840cm3 => 840cm3 x 2 x 27 = 0.675g of Aluminium 3 x 22.4 x 1000 Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g =0.325g of copper % copper = mass of copper x100% = 32.5% Mass of alloy Method 2 Mole ratio 2Al: 3H2 = 2:3 Moles of Hydrogen gas = volume of gas => 840cm3 = 0.0375moles Molar gas volume 22400cm3 Moles of Al = 2/3 moles of H2 => 2/3x 0.0375moles = 0.025moles Mass of Al = moles x molar mass =>0.025moles x 27 = 0.675g Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g = 0.325 g of copper % copper = mass of copper x100% = 32.5% Mass of alloy
20 20 (f)Gay Lussac’s law Gay Lussacs law states that “when gases combine/react they do so in simple volume ratios to each other and to their gaseous products at constant/same temperature and pressure” Gay Lussacs law thus only apply to gases Given the volume of one gas reactant, the other gaseous reactants can be deduced thus: Examples 1. Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen.


1.0g of an alloy of aluminium and copper were reacted with excess hydrochloric acid. If 840cm3 of hydrogen at s.t.p was produced, calculate the % of copper in the alloy.(Al =27.0,one mole of a gas at s.t.p =22.4dm3 ) Chemical equation Copper does not react with hydrochloric acid 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g) Method 1 3H2(g) = 3 moles x (22.4 x 1000)cm3 => 2 x 27 g Al 840cm3 => 840cm3 x 2 x 27 = 0.675g of Aluminium 3 x 22.4 x 1000 Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g =0.325g of copper % copper = mass of copper x100% = 32.5% Mass of alloy Method 2 Mole ratio 2Al: 3H2 = 2:3 Moles of Hydrogen gas = volume of gas => 840cm3 = 0.0375moles Molar gas volume 22400cm3 Moles of Al = 2/3 moles of H2 => 2/3x 0.0375moles = 0.025moles Mass of Al = moles x molar mass =>0.025moles x 27 = 0.675g Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g = 0.325 g of copper % copper = mass of copper x100% = 32.5% Mass of alloy
20 20 (f)Gay Lussac’s law Gay Lussacs law states that “when gases combine/react they do so in simple volume ratios to each other and to their gaseous products at constant/same temperature and pressure” Gay Lussacs law thus only apply to gases Given the volume of one gas reactant, the other gaseous reactants can be deduced thus: Examples 1. Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen. Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l) Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 2.


If 840cm3 of hydrogen at s.t.p was produced, calculate the % of copper in the alloy.(Al =27.0,one mole of a gas at s.t.p =22.4dm3 ) Chemical equation Copper does not react with hydrochloric acid 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g) Method 1 3H2(g) = 3 moles x (22.4 x 1000)cm3 => 2 x 27 g Al 840cm3 => 840cm3 x 2 x 27 = 0.675g of Aluminium 3 x 22.4 x 1000 Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g =0.325g of copper % copper = mass of copper x100% = 32.5% Mass of alloy Method 2 Mole ratio 2Al: 3H2 = 2:3 Moles of Hydrogen gas = volume of gas => 840cm3 = 0.0375moles Molar gas volume 22400cm3 Moles of Al = 2/3 moles of H2 => 2/3x 0.0375moles = 0.025moles Mass of Al = moles x molar mass =>0.025moles x 27 = 0.675g Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g = 0.325 g of copper % copper = mass of copper x100% = 32.5% Mass of alloy
20 20 (f)Gay Lussac’s law Gay Lussacs law states that “when gases combine/react they do so in simple volume ratios to each other and to their gaseous products at constant/same temperature and pressure” Gay Lussacs law thus only apply to gases Given the volume of one gas reactant, the other gaseous reactants can be deduced thus: Examples 1. Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen. Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l) Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 2. Calculate the volume of air required to completely reacts with 50cm3 of Hydrogen.(assume Oxygen is 21% by volume of air) Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l)
21 21 Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 21% = 25cm3 100% = 100 x 25 = 21 3.If 5cm3 of a hydrocarbon CxHy burn in 15cm3 of Oxygen to form 10cm3 of Carbon(IV)oxide and 10cm3 of water vapour/steam, obtain the equation for the reaction and hence find the value of x and y in CxHy.


Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen. Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l) Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 2. Calculate the volume of air required to completely reacts with 50cm3 of Hydrogen.(assume Oxygen is 21% by volume of air) Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l)
21 21 Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 21% = 25cm3 100% = 100 x 25 = 21 3.If 5cm3 of a hydrocarbon CxHy burn in 15cm3 of Oxygen to form 10cm3 of Carbon(IV)oxide and 10cm3 of water vapour/steam, obtain the equation for the reaction and hence find the value of x and y in CxHy. Chemical equation: CxHy (g) + O2 (g) -> H2O(g) + CO2(g) Volumes 5cm3 : 15cm3 : 10cm3 : 10cm3 Volume ratios 5cm3 : 15cm3 : 10cm3 : 10cm3 (divide by lowest volume) 5 5 5 5 Reacting volume ratios 1volume 3 volume 2 volume 2 volume Balanced chemical equation: CxHy (g) + 3O2 (g) -> 2H2O(g) + 2CO2(g) If “4H” are in 2H2O(g) the y=4 If “2C” are in 2CO2 (g) the x=2 Thus(i) chemical formula of hydrocarbon = C2H4 (ii) chemical name of hydrocarbon = Ethene 4.100cm3 of nitrogen (II)oxide NO combine with 50cm3 of Oxygen to form 100cm3 of a single gaseous compound of nitrogen. All volumes measured at the same temperature and pressure. Obtain the equation for the reaction and name the gaseous product.


Chemical equation: CxHy (g) + O2 (g) -> H2O(g) + CO2(g) Volumes 5cm3 : 15cm3 : 10cm3 : 10cm3 Volume ratios 5cm3 : 15cm3 : 10cm3 : 10cm3 (divide by lowest volume) 5 5 5 5 Reacting volume ratios 1volume 3 volume 2 volume 2 volume Balanced chemical equation: CxHy (g) + 3O2 (g) -> 2H2O(g) + 2CO2(g) If “4H” are in 2H2O(g) the y=4 If “2C” are in 2CO2 (g) the x=2 Thus(i) chemical formula of hydrocarbon = C2H4 (ii) chemical name of hydrocarbon = Ethene 4.100cm3 of nitrogen (II)oxide NO combine with 50cm3 of Oxygen to form 100cm3 of a single gaseous compound of nitrogen. All volumes measured at the same temperature and pressure. Obtain the equation for the reaction and name the gaseous product. Chemical equation: NO (g) + O2 (g) -> NOx Volumes 100cm3 : 50cm3 : 100 Volume ratios 100cm3 : 50cm3 : 100cm3 (divide by lowest volume) 50 50 50 Reacting volume ratios 2volume 1 volume 2 volume Balanced chemical equation: 2 NO (g) + O2 (g) -> 2NO x(g) Thus(i) chemical formula of the nitrogen compound = 2 NO2 (ii) chemical name of compound = Nitrogen(IV)oxide 5.When 15cm3 of a gaseous hydrocarbon was burnt in 100cm3 of Oxygen ,the resulting gaseous mixture occupied70cm3 at room temperature and pressure. 22 22 When the gaseous mixture was passed through, potassium hydroxide its volume decreased to 25cm3.


Obtain the equation for the reaction and name the gaseous product. Chemical equation: NO (g) + O2 (g) -> NOx Volumes 100cm3 : 50cm3 : 100 Volume ratios 100cm3 : 50cm3 : 100cm3 (divide by lowest volume) 50 50 50 Reacting volume ratios 2volume 1 volume 2 volume Balanced chemical equation: 2 NO (g) + O2 (g) -> 2NO x(g) Thus(i) chemical formula of the nitrogen compound = 2 NO2 (ii) chemical name of compound = Nitrogen(IV)oxide 5.When 15cm3 of a gaseous hydrocarbon was burnt in 100cm3 of Oxygen ,the resulting gaseous mixture occupied70cm3 at room temperature and pressure. 22 22 When the gaseous mixture was passed through, potassium hydroxide its volume decreased to 25cm3. (a)What volume of Oxygen was used during the reaction.(1mk) Volume of Oxygen used =100-25 =75cm3√ (P was completely burnt) (b)Determine the molecular formula of the hydrocarbon(2mk) CxHy + O2 -> xCO2 + yH2O 15cm3 : 75cm3 15 15 1 : 3√ => 1 atom of C react with 6 (3x2)atoms of Oxygen Thus x = 1 and y = 2 => P has molecula formula CH4√ (g) Ionic equations An ionic equation is a chemical statement showing the movement of ions (cations and anions ) from reactants to products. Solids, gases and liquids do not ionize/dissociate into free ions. Only ionic compounds in aqueous/solution or molten state ionize/dissociate into free cations and anions (ions) An ionic equation is usually derived from a stoichiometric equation by using the following guidelines Guidelines for writing ionic equations 1.Write the balanced stoichiometric equation 2.Indicate the state symbols of the reactants and products 3.Split into cations and anions all the reactants and products that exist in aqueous state. 4.Cancel out any cation and anion that appear on both the product and reactant side. 5. Rewrite the chemical equation. It is an ionic equation.


5. Rewrite the chemical equation. It is an ionic equation. Practice (a)Precipitation of an insoluble salt
23 23 All insoluble salts are prepared in the laboratory from double decomposition /precipitation. This involves mixing two soluble salts to form one soluble and one insoluble salt 1. When silver nitrate(V) solution is added to sodium chloride solution,sodium nitrate(V) solution and a white precipitate of silver chloride are formed. Balanced stoichiometric equation AgNO3(aq) + NaCl(aq) -> AgCl (s) + NaNO3 (aq) Split reactants product existing in aqueous state as cation/anion Ag+(aq) + NO3- (aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq)+ NO3- (aq) Cancel out ions appearing on reactant and product side Ag+(aq) + NO3- (aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq)+ NO3- (aq) Rewrite the equation Ag+(aq) + Cl-(aq) -> AgCl(s) (ionic equation) 2. When barium nitrate(V) solution is added to copper(II)sulphate(VI) solution, copper(II) nitrate(V) solution and a white precipitate of barium sulphate(VI) are formed. Balanced stoichiometric equation Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq) Split reactants product existing in aqueous state as cation/anion Ba2+(aq) + 2NO3- (aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3- (aq)+ Cu2+(aq) Cancel out ions appearing on reactant and product side Ba2+(aq) + 2NO3- (aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3- (aq) + Cu2+(aq) Rewrite the equation Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ionic equation) 3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide.


Balanced stoichiometric equation AgNO3(aq) + NaCl(aq) -> AgCl (s) + NaNO3 (aq) Split reactants product existing in aqueous state as cation/anion Ag+(aq) + NO3- (aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq)+ NO3- (aq) Cancel out ions appearing on reactant and product side Ag+(aq) + NO3- (aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq)+ NO3- (aq) Rewrite the equation Ag+(aq) + Cl-(aq) -> AgCl(s) (ionic equation) 2. When barium nitrate(V) solution is added to copper(II)sulphate(VI) solution, copper(II) nitrate(V) solution and a white precipitate of barium sulphate(VI) are formed. Balanced stoichiometric equation Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq) Split reactants product existing in aqueous state as cation/anion Ba2+(aq) + 2NO3- (aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3- (aq)+ Cu2+(aq) Cancel out ions appearing on reactant and product side Ba2+(aq) + 2NO3- (aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3- (aq) + Cu2+(aq) Rewrite the equation Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ionic equation) 3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide. Balanced stoichiometric equation Pb(NO3)2(aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)
24 24 Split reactants product existing in aqueous state as cation/anion Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Cancel out ions appearing on reactant and product side Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Rewrite the equation Pb2+(aq) + 2I - (aq) -> PbI2 (s) (ionic equation) (b)Neutralization Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base.


When barium nitrate(V) solution is added to copper(II)sulphate(VI) solution, copper(II) nitrate(V) solution and a white precipitate of barium sulphate(VI) are formed. Balanced stoichiometric equation Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq) Split reactants product existing in aqueous state as cation/anion Ba2+(aq) + 2NO3- (aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3- (aq)+ Cu2+(aq) Cancel out ions appearing on reactant and product side Ba2+(aq) + 2NO3- (aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3- (aq) + Cu2+(aq) Rewrite the equation Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ionic equation) 3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide. Balanced stoichiometric equation Pb(NO3)2(aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)
24 24 Split reactants product existing in aqueous state as cation/anion Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Cancel out ions appearing on reactant and product side Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Rewrite the equation Pb2+(aq) + 2I - (aq) -> PbI2 (s) (ionic equation) (b)Neutralization Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base. (i)Reaction of alkalis with acids 1.Reaction of nitric(V)acid with potassium hydroxide Balanced stoichiometric equation HNO3(aq) + KOH (aq) -> H2O (l) + KNO3 (aq) Split reactants product existing in aqueous state as cation/anion H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Cancel out ions appearing on reactant and product side H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Rewrite the equation H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 2.Reaction of sulphuric(VI)acid with ammonia solution Balanced stoichiometric equation H2SO4(aq) + 2NH4OH (aq) -> H2O (l) + (NH4) 2SO4 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)
25 25 Rewrite the equation 2H+ (aq) + 2OH - (aq) -> 2H2O (l) H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 3.Reaction of hydrochloric acid with Zinc hydroxide Balanced stoichiometric equation 2HCl(aq) + Zn(OH)2 (s) -> 2H2O (l) + ZnCl 2 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Rewrite the equation 2H+(aq) + Zn(OH)2 (s) ->2H2O (l) + Zn 2+ (aq) (ionic equation) (h)Molar solutions A molar solution is one whose concentration is known.


Balanced stoichiometric equation Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq) Split reactants product existing in aqueous state as cation/anion Ba2+(aq) + 2NO3- (aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3- (aq)+ Cu2+(aq) Cancel out ions appearing on reactant and product side Ba2+(aq) + 2NO3- (aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3- (aq) + Cu2+(aq) Rewrite the equation Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ionic equation) 3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide. Balanced stoichiometric equation Pb(NO3)2(aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)
24 24 Split reactants product existing in aqueous state as cation/anion Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Cancel out ions appearing on reactant and product side Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Rewrite the equation Pb2+(aq) + 2I - (aq) -> PbI2 (s) (ionic equation) (b)Neutralization Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base. (i)Reaction of alkalis with acids 1.Reaction of nitric(V)acid with potassium hydroxide Balanced stoichiometric equation HNO3(aq) + KOH (aq) -> H2O (l) + KNO3 (aq) Split reactants product existing in aqueous state as cation/anion H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Cancel out ions appearing on reactant and product side H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Rewrite the equation H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 2.Reaction of sulphuric(VI)acid with ammonia solution Balanced stoichiometric equation H2SO4(aq) + 2NH4OH (aq) -> H2O (l) + (NH4) 2SO4 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)
25 25 Rewrite the equation 2H+ (aq) + 2OH - (aq) -> 2H2O (l) H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 3.Reaction of hydrochloric acid with Zinc hydroxide Balanced stoichiometric equation 2HCl(aq) + Zn(OH)2 (s) -> 2H2O (l) + ZnCl 2 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Rewrite the equation 2H+(aq) + Zn(OH)2 (s) ->2H2O (l) + Zn 2+ (aq) (ionic equation) (h)Molar solutions A molar solution is one whose concentration is known. The SI unit of concentration is Molarity denoted M.


Balanced stoichiometric equation Pb(NO3)2(aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)
24 24 Split reactants product existing in aqueous state as cation/anion Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Cancel out ions appearing on reactant and product side Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Rewrite the equation Pb2+(aq) + 2I - (aq) -> PbI2 (s) (ionic equation) (b)Neutralization Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base. (i)Reaction of alkalis with acids 1.Reaction of nitric(V)acid with potassium hydroxide Balanced stoichiometric equation HNO3(aq) + KOH (aq) -> H2O (l) + KNO3 (aq) Split reactants product existing in aqueous state as cation/anion H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Cancel out ions appearing on reactant and product side H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Rewrite the equation H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 2.Reaction of sulphuric(VI)acid with ammonia solution Balanced stoichiometric equation H2SO4(aq) + 2NH4OH (aq) -> H2O (l) + (NH4) 2SO4 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)
25 25 Rewrite the equation 2H+ (aq) + 2OH - (aq) -> 2H2O (l) H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 3.Reaction of hydrochloric acid with Zinc hydroxide Balanced stoichiometric equation 2HCl(aq) + Zn(OH)2 (s) -> 2H2O (l) + ZnCl 2 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Rewrite the equation 2H+(aq) + Zn(OH)2 (s) ->2H2O (l) + Zn 2+ (aq) (ionic equation) (h)Molar solutions A molar solution is one whose concentration is known. The SI unit of concentration is Molarity denoted M. Molarity may be defined as the number of moles of solute present in one cubic decimeter of solution.


(i)Reaction of alkalis with acids 1.Reaction of nitric(V)acid with potassium hydroxide Balanced stoichiometric equation HNO3(aq) + KOH (aq) -> H2O (l) + KNO3 (aq) Split reactants product existing in aqueous state as cation/anion H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Cancel out ions appearing on reactant and product side H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Rewrite the equation H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 2.Reaction of sulphuric(VI)acid with ammonia solution Balanced stoichiometric equation H2SO4(aq) + 2NH4OH (aq) -> H2O (l) + (NH4) 2SO4 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)
25 25 Rewrite the equation 2H+ (aq) + 2OH - (aq) -> 2H2O (l) H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 3.Reaction of hydrochloric acid with Zinc hydroxide Balanced stoichiometric equation 2HCl(aq) + Zn(OH)2 (s) -> 2H2O (l) + ZnCl 2 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Rewrite the equation 2H+(aq) + Zn(OH)2 (s) ->2H2O (l) + Zn 2+ (aq) (ionic equation) (h)Molar solutions A molar solution is one whose concentration is known. The SI unit of concentration is Molarity denoted M. Molarity may be defined as the number of moles of solute present in one cubic decimeter of solution. One cubic decimeter is equal to one litre and also equal to 1000cm3.


The SI unit of concentration is Molarity denoted M. Molarity may be defined as the number of moles of solute present in one cubic decimeter of solution. One cubic decimeter is equal to one litre and also equal to 1000cm3. The higher the molarity the higher the concentration and the higher/more solute has been dissolved in the solvent to make one cubic decimeter/ litre/1000cm3 solution. Examples 2M sodium hydroxide means 2 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter/ litre/1000cm3 uniform solution mixture of sodium hydroxide and water. 0.02M sodium hydroxide means 0.02 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter/ litre/1000cm3 uniform solution mixture of sodium hydroxide and water. “2M” is more concentrated than“0.02M”. Preparation of molar solution
26 26 Procedure Weigh accurately 4.0 g of sodium hydroxide pellets into a 250cm3 volumetric flask. Using a wash bottle add about 200cm3 of distilled water. Stopper the flask. Shake vigorously for three minutes. Remove the stopper for a second then continue to shake for about another two minutes until all the solid has dissolved. Add more water slowly upto exactly the 250 cm3 mark. Sample questions 1.Calculate the number of moles of sodium hydroxide pellets present in: (i) 4.0 g. Molar mass of NaOH = (23 + 16 + 1) = 40g Moles = Mass => 4.0 = 0.1 / 1.0 x 10 -1 moles Molar mass 40 (ii) 250 cm3 solution in the volumetric flask.


x 10-3moles 1000 1000 Mass of HNO3 =Moles x molar mass => 2.5 x 10-3 x 40 = 0.1 g 3. Calculate the volume required to dissolve : (a)(i) 0.25moles of sodium hydroxide solution to form a 0.8M solution Volume (in cm3) = moles x 1000 => 0.25 x 1000 = 312.5cm3 Molarity 0.8 (ii) 100cm3 was added to the sodium hydroxide solution above. Calculate the concentration of the solution. C1 x V1 = C2 x V2 where: C1 = molarity/concentration before diluting/adding water C2 = molarity/concentration after diluting/adding water V1 = volume before diluting/adding water V2 = volume after diluting/adding water => 0.8M x 312.5cm3 = C2 x (312.5 + 100) C2 = 0.8M x 312.5cm3 = 0.6061M
31 31 412.5 (b)(ii) 0.01M solution containing 0.01moles of sodium hydroxide solution . Volume (in cm3) = moles x 1000 => 0.01 x 1000 = 1000 cm3 Molarity 0.01 (ii) Determine the quantity of water which must be added to the sodium hydroxide solution above to form a 0.008M solution. C1 x V1 = C2 x V2 where: C1 = molarity/concentration before diluting/adding water C2 = molarity/concentration after diluting/adding water V1 = volume before diluting/adding water V2 = volume after diluting/adding water => 0.01M x 1000 cm3 = 0.008 x V2 V2 = 0.01M x 1000cm3 = 1250cm3 0.008 Volume added = 1250 - 1000 = 250cm3 (c)Volumetric analysis/Titration Volumetric analysis/Titration is the process of determining unknown concentration of one reactant from a known concentration and volume of another. Reactions take place in simple mole ratio of reactants and products.


Volume (in cm3) = moles x 1000 => 0.01 x 1000 = 1000 cm3 Molarity 0.01 (ii) Determine the quantity of water which must be added to the sodium hydroxide solution above to form a 0.008M solution. C1 x V1 = C2 x V2 where: C1 = molarity/concentration before diluting/adding water C2 = molarity/concentration after diluting/adding water V1 = volume before diluting/adding water V2 = volume after diluting/adding water => 0.01M x 1000 cm3 = 0.008 x V2 V2 = 0.01M x 1000cm3 = 1250cm3 0.008 Volume added = 1250 - 1000 = 250cm3 (c)Volumetric analysis/Titration Volumetric analysis/Titration is the process of determining unknown concentration of one reactant from a known concentration and volume of another. Reactions take place in simple mole ratio of reactants and products. Knowing the concentration/ volume of one reactant, the other can be determined from the relationship: M1V1 = M2V2 where: n1 n2 M1 = Molarity of 1st reactant M2 = Molarity of 2nd reactant V1 = Volume of 1st reactant V1 = Volume of 2nd reactant n1 = number of moles of 1st reactant from stoichiometric equation n2 = number of moles of 2nd reactant from stoichiometric equation Examples 1.Calculate the molarity of MCO3 if 5.0cm3 of MCO3 react with 25.0cm3 of 0.5M hydrochloric acid.(C=12.0 ,O =16.0) Stoichiometric equation: MCO3(s) + 2HCl(aq) -> MCl2(aq) + CO2(g) + H2O(l) Method 1
32 32 M1V1 = M2V2 -> M1 x 5.0cm3 = 0.5M x 25.0cm3 n1 n2 1 2 => M1 = 0.5 x 25.0 x1 = 1.25M / 1.25 moledm-3 5.0 x 2 Method 2 Moles of HCl used = molarity x volume 1000 => 0.5 x 25.0 = 0.0125 /1.25 x 10-2moles 1000 Mole ratio MCO3 : HCl = 1:2 Moles MCO3 = 0.0125 /1.25 x 10-2moles = 0.00625 / 6.25 x 10-3 moles 2 Molarity MCO3 = moles x 1000 => 0.00625 / 6.25 x 10-3 x 1000 Volume 5 = 1.25M / 1.25 moledm-3 2.


0.388g of a monobasic organic acid B required 46.5 cm3 of 0.095M sodium hydroxide for complete neutralization. Name and draw the structural formula of B Moles of NaOH used = molarity x volume 1000 => 0.095 x 46.5 = 0.0044175 /4.4175 x 10-3moles 1000 Mole ratio B : NaOH = 1:1
34 34 Moles B = 0.0044175 /4.4175 x 10-3moles Molar mass B = mass => 0.388 moles 0.0044175 /4.4175 x 10-3moles = 87.8324 gmole-1 X-COOH = 87.8324 where X is an alkyl group X =87.8324- 42 = 42.8324=43 By elimination: CH3 = 15 CH3CH2 = 29 CH3CH2 CH2 = 43 Molecula formula : CH3CH2 CH2COOH Molecule name : Butan-1-oic acid Molecular structure H H H O H C C C C O H H H H H 5. 10.5 g of an impure sample containing ammonium sulphate (VI) fertilizer was warmed with 250cm3 of o.8M sodium hydroxide solution.The excess of the alkali was neutralized by 85cm3 of 0.5M hydrochloric acid. Calculate the % of impurities in the ammonium sulphate (VI)fertilizer.


10.5 g of an impure sample containing ammonium sulphate (VI) fertilizer was warmed with 250cm3 of o.8M sodium hydroxide solution.The excess of the alkali was neutralized by 85cm3 of 0.5M hydrochloric acid. Calculate the % of impurities in the ammonium sulphate (VI)fertilizer. (N=14.0,S=32.0,O=16.0, H=1.0) Equation for neutralization NaOH(aq) + HCl(aq) -> NaOH(aq) + H2O(l) Mole ratio NaOH(aq):HCl(aq)= 1:1 Moles of HCl = Molarity x volume => 0.5 x 85 = 0.0425 moles 1000 1000 Excess moles of NaOH(aq)= 0.0425 moles Equation for reaction with ammonium salt 2NaOH(aq) + (NH4) 2SO4(aq) -> Na 2SO4(aq) + 2NH3 (g)+ 2H2O(l) Mole ratio NaOH(aq): (NH4) 2SO4(aq)= 2:1 Total moles of NaOH = Molarity x volume => 0.8 x 250 = 0.2 moles 1000 1000 Moles of NaOH that reacted with(NH4) 2SO4 = 0.2 - 0.0425 = 0.1575moles Moles (NH4) 2SO4 = ½ x 0.1575moles = 0. 07875moles Molar mass (NH4) 2SO4= 132 gmole-1
35 35 Mass of in impure sample = moles x molar mass =>0. 07875 x 132 = 10.395 g Mass of impurities = 10.5 -10.395 = 0.105 g % impurities = 0.105 x 100 = 1.0 % 10.5 Practically volumetric analysis involves titration. Titration generally involves filling a burette with known/unknown concentration of a solution then adding the solution to unknown/known concentration of another solution in a conical flask until there is complete reaction.


07875moles Molar mass (NH4) 2SO4= 132 gmole-1
35 35 Mass of in impure sample = moles x molar mass =>0. 07875 x 132 = 10.395 g Mass of impurities = 10.5 -10.395 = 0.105 g % impurities = 0.105 x 100 = 1.0 % 10.5 Practically volumetric analysis involves titration. Titration generally involves filling a burette with known/unknown concentration of a solution then adding the solution to unknown/known concentration of another solution in a conical flask until there is complete reaction. If the solutions used are both colourless, an indicator is added to the conical flask. When the reaction is over, a slight/little excess of burette contents change the colour of the indicator. This is called the end point. Set up of titration apparatus The titration process involve involves determination of titre. The titre is the volume of burette contents/reading before and after the end point. Burette contents/reading before titration is usually called the Initial burette reading. Burette contents/reading after titration is usually called the Final burette reading. The titre value is thus a sum of the Final less Initial burette readings. To reduce errors, titration process should be repeated at least once more. The results of titration are recorded in a titration table as below
36 36 Sample titration table Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution used(cm3) 20.0 20.0 20.0 As evidence of a titration actually done examining body requires the candidate to record their burette readings before and after the titration. For KCSE candidates burette readings must be recorded in a titration table in the format provided by the Kenya National Examination Council. As evidence of all titration actually done Kenya National Examination Council require the candidate to record their burette readings before and after the titration to complete the titration table in the format provided.


The results of titration are recorded in a titration table as below
36 36 Sample titration table Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution used(cm3) 20.0 20.0 20.0 As evidence of a titration actually done examining body requires the candidate to record their burette readings before and after the titration. For KCSE candidates burette readings must be recorded in a titration table in the format provided by the Kenya National Examination Council. As evidence of all titration actually done Kenya National Examination Council require the candidate to record their burette readings before and after the titration to complete the titration table in the format provided. Calculate the average volume of solution used 24.0 + 24.0 + 24.0 = 24.0 cm3 3 As evidence of understanding the degree of accuracy of burettes , all readings must be recorded to a decimal point. As evidence of accuracy in carrying the out the titration , candidates value should be within 0.2 of the school value . The school value is the teachers readings presented to the examining body/council based on the concentrations of the solutions s/he presented to her/his candidates. Bonus mark is awarded for averaged reading within 0.1 school value as Final answer. Calculations involved after the titration require candidates thorough practical and theoretical practice mastery on the: (i)relationship among the mole, molar mass, mole ratios, concentration, molarity. (ii) mathematical application of 1st principles. Very useful information which candidates forget appears usually in the beginning of the question paper as: “You are provided with…” All calculation must be to the 4th decimal point unless they divide fully to a lesser decimal point. Candidates are expected to use a non programmable scientific calculator. 37 37 (a)Sample Titration Practice 1 (Simple Titration) You are provided with: 0.1M sodium hydroxide solution A Hydrochloric acid solution B You are required to determine the concentration of solution B in moles per litre. Procedure Fill the burette with solution B. Pipette 25.0cm3 of solution A into a conical flask.


37 37 (a)Sample Titration Practice 1 (Simple Titration) You are provided with: 0.1M sodium hydroxide solution A Hydrochloric acid solution B You are required to determine the concentration of solution B in moles per litre. Procedure Fill the burette with solution B. Pipette 25.0cm3 of solution A into a conical flask. Titrate solution A with solution B using phenolphthalein indicator to complete the titration table 1 Sample results Titration table 1 Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution B used(cm3) 20.0 20.0 20.0 Sample worked questions 1. Calculate the average volume of solution B used Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3 3 3 2. How many moles of: (i)solution A were present in 25cm3 solution. Moles of solution A = Molarity x volume = 0.1 x 25 = 2.5 x 10-3 moles 1000 1000 (ii)solution B were present in the average volume. Chemical equation: NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l) Mole ratio 1:1 => Moles of A = Moles of B = 2.5 x 10-3 moles (iii) solution B in moles per litre.


How many moles of: (i)solution A were present in 25cm3 solution. Moles of solution A = Molarity x volume = 0.1 x 25 = 2.5 x 10-3 moles 1000 1000 (ii)solution B were present in the average volume. Chemical equation: NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l) Mole ratio 1:1 => Moles of A = Moles of B = 2.5 x 10-3 moles (iii) solution B in moles per litre. Moles of B per litre = moles x 1000 = 2.5 x 10-3 x 1000 = 0.1M Volume 20
38 38 (b)Sample Titration Practice 2 (Redox Titration) You are provided with: Acidified Potassium manganate(VII) solution A 0.1M of an iron (II)salt solution B 8.5g of ammonium iron(II)sulphate(VI) crystals(NH4)2 SO4FeSO4.xH2O solid C You are required to (i)standardize acidified potassium manganate(VII) (ii)determine the value of x in the formula (NH4)2 SO4FeSO4.xH2O. Procedure 1 Fill the burette with solution A. Pipette 25.0cm3 of solution B into a conical flask. Titrate solution A with solution B until a pink colour just appears. Record your results to complete table 1. Table 1:Sample results Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution A used(cm3) 20.0 20.0 20.0 Sample worked questions 1. Calculate the average volume of solution A used Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3 3 3 2. How many moles of: (i)solution B were present in 25cm3 solution.


Table 1:Sample results Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution A used(cm3) 20.0 20.0 20.0 Sample worked questions 1. Calculate the average volume of solution A used Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3 3 3 2. How many moles of: (i)solution B were present in 25cm3 solution. Moles of solution A = Molarity x volume = 0.1 x 25 = 2.5 x 10-3 moles 1000 1000 (ii)solution A were present in the average volume. Assume one mole of B react with five moles of B Mole ratio A : B = 1:5 => Moles of A = Moles of B = 2.5 x 10-3 moles = 5.0 x 10 -4 moles 5 5
39 39 (iii) solution B in moles per litre. Moles of B per litre = moles x 1000 = 2.5 x 10-3 x 1000 Volume 20 = 0.025 M /moles per litre /moles l-1 Procedure 2 Place all the solid C into the 250cm3 volumetric flask carefully. Add about 200cm3 of distilled water. Shake to dissolve. Make up to the 250cm3 of solution by adding more distilled water. Label this solution C. Pipette 25cm3 of solution C into a conical flask, Titrate solution C with solution A until a permanent pink colour just appears. Complete table 2. Table 2:Sample results Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution A used(cm3) 20.0 20.0 20.0 Sample worked questions 1.


Pipette 25cm3 of solution C into a conical flask, Titrate solution C with solution A until a permanent pink colour just appears. Complete table 2. Table 2:Sample results Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution A used(cm3) 20.0 20.0 20.0 Sample worked questions 1. Calculate the average volume of solution A used Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3 3 3 2. How many moles of: (i)solution A were present inin the average titre. Moles of solution A = Molarity x volume = 0.025 x 20 = 5.0 x 10-4 moles 1000 1000 (ii)solution C in 25cm3 solution given the equation for the reaction: MnO4- (aq) + 8H+(aq) + 5Fe2+ (aq) -> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) Mole ratio MnO4- (aq): 5Fe2+ (aq) = 1:5 => Moles of 5Fe2+ (aq) = Moles of MnO4- (aq) = 5.0 x 10-4 moles = 1.0 x 10 -4 moles 5 5 (iii) solution B in 250cm3. Moles of B per litre = moles x 250 = 1.0 x 10 -4 x 250 = 1.0 x 10 -3 moles Volume 25 3. Calculate the molar mass of solid C and hence the value of x in the chemical formula (NH4)2SO4FeSO4.xH2O.


Moles of solution A = Molarity x volume = 0.025 x 20 = 5.0 x 10-4 moles 1000 1000 (ii)solution C in 25cm3 solution given the equation for the reaction: MnO4- (aq) + 8H+(aq) + 5Fe2+ (aq) -> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) Mole ratio MnO4- (aq): 5Fe2+ (aq) = 1:5 => Moles of 5Fe2+ (aq) = Moles of MnO4- (aq) = 5.0 x 10-4 moles = 1.0 x 10 -4 moles 5 5 (iii) solution B in 250cm3. Moles of B per litre = moles x 250 = 1.0 x 10 -4 x 250 = 1.0 x 10 -3 moles Volume 25 3. Calculate the molar mass of solid C and hence the value of x in the chemical formula (NH4)2SO4FeSO4.xH2O. 40 40 (N=14.0, S=32.0, Fe=56.0, H=1.0 O=16.0) Molar mass = mass perlitre = 8.5 = 8500 g Moles per litre 1.0 x 10 -3 moles NH4)2SO4FeSO4.xH2O = 8500 284 + 18x =8500 8500 - 284 = 8216 = 18x = 454.4444 18 18 x = 454 (whole number) (c)Sample Titration Practice 3 (Back titration) You are provided with: (i)an impure calcium carbonate labeled M (ii)Hydrochloric acid labeled solution N (iii)solution L containing 20g per litre sodium hydroxide. You are required to determine the concentration of N in moles per litre and the % of calcium carbonate in mixture M. Procedure 1 Pipette 25.0cm3 of solution L into a conical flask. Add 2-3 drops of phenolphthalein indicator.


You are required to determine the concentration of N in moles per litre and the % of calcium carbonate in mixture M. Procedure 1 Pipette 25.0cm3 of solution L into a conical flask. Add 2-3 drops of phenolphthalein indicator. Titrate with dilute hydrochloric acid solution N and record your results in table 1(4mark) Sample Table 1 1 2 3 Final burette reading (cm3) 6.5 6.5 6.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 6.5 6.5 6.5 Sample questions (a) Calculate the average volume of solution N used 6.5 + 6.5 + 6.5 = 6.5 cm3 3 (b) How many moles of sodium hydroxide are contained in 25cm3of solution L Molar mass NaOH =40g Molarity of L = mass per litre => 20 = 0.5M
41 41 Molar mass NaOH 40 Moles NaOH in 25cm3 = molarity x volume => 0.5M x 25cm3 = 0.0125 moles 1000 1000 (c)Calculate: (i)the number of moles of hydrochloric acidthat react with sodium hydroxide in (b)above. Mole ratio NaOH : HCl from stoichiometric equation= 1:1 Moles HCl =Moles NaOH => 0.0125 moles (ii)the molarity of hydrochloric acid solution N. Molarity = moles x 1000 => 0.0125 moles x 1000 =1.9231M/moledm-3 6.5 6.5 Procedure 2 Place the 4.0 g of M provided into a conical flask and add 25.0cm3 of the dilute hydrochloric acid to it using a clean pipette.


Titrate with dilute hydrochloric acid solution N and record your results in table 1(4mark) Sample Table 1 1 2 3 Final burette reading (cm3) 6.5 6.5 6.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 6.5 6.5 6.5 Sample questions (a) Calculate the average volume of solution N used 6.5 + 6.5 + 6.5 = 6.5 cm3 3 (b) How many moles of sodium hydroxide are contained in 25cm3of solution L Molar mass NaOH =40g Molarity of L = mass per litre => 20 = 0.5M
41 41 Molar mass NaOH 40 Moles NaOH in 25cm3 = molarity x volume => 0.5M x 25cm3 = 0.0125 moles 1000 1000 (c)Calculate: (i)the number of moles of hydrochloric acidthat react with sodium hydroxide in (b)above. Mole ratio NaOH : HCl from stoichiometric equation= 1:1 Moles HCl =Moles NaOH => 0.0125 moles (ii)the molarity of hydrochloric acid solution N. Molarity = moles x 1000 => 0.0125 moles x 1000 =1.9231M/moledm-3 6.5 6.5 Procedure 2 Place the 4.0 g of M provided into a conical flask and add 25.0cm3 of the dilute hydrochloric acid to it using a clean pipette. Swirl the contents of the flask vigorously until effervescence stop.Using a 100ml measuring cylinder add 175cm3 distilled waterto make up the solution up to 200cm3.Label this solution K.Using a clean pipettetransfer 25.0cm3 of the solution into a clean conical flask and titrate with solution L from the burette using 2-3 drops of methyl orange indicator.Record your observations in table 2.


Mole ratio NaOH : HCl from stoichiometric equation= 1:1 Moles HCl =Moles NaOH => 0.0125 moles (ii)the molarity of hydrochloric acid solution N. Molarity = moles x 1000 => 0.0125 moles x 1000 =1.9231M/moledm-3 6.5 6.5 Procedure 2 Place the 4.0 g of M provided into a conical flask and add 25.0cm3 of the dilute hydrochloric acid to it using a clean pipette. Swirl the contents of the flask vigorously until effervescence stop.Using a 100ml measuring cylinder add 175cm3 distilled waterto make up the solution up to 200cm3.Label this solution K.Using a clean pipettetransfer 25.0cm3 of the solution into a clean conical flask and titrate with solution L from the burette using 2-3 drops of methyl orange indicator.Record your observations in table 2. Sample Table 2 1 2 3 Final burette reading (cm3) 24.5 24.5 24.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.5 24.5 24.5 Sample calculations (a)Calculate the average volume of solution L used(1mk) 24.5 + 24.5 + 24.5 = 24.5cm3 3 (b)How many moles of sodium hydroxide are present in the average volume of solution L used? Moles = molarity x average burette volume => 0.5 x 24.5 1000 1000 = 0.01225 /1.225 x 10-2 moles
42 42 (c) How many moles of hydrochloric acid are present in the original 200cm3 of solution K?