97. Exercises - 97.6. Mathematical Treatment of Measurement Results
(b) This lumber can be used as vertical studs, which are typically placed 16.0 in. apart. What is that distance in centimeters?
86. The mercury content of a stream was believed to be above the minimum considered safe -1 part per billion (ppb) by weight. An analysis indicated that the concentration was 0.68 parts per billion. What quantity of mercury in grams was present in 15.0 L of the water, the density of which is $0.998 \mathrm{~g} / \mathrm{ml}$ ? $\left(1 \mathrm{ppb} \mathrm{Hg}=\frac{1 \mathrm{ng} \mathrm{Hg}}{1 \mathrm{~g} \text { water }}\right)$
87. Calculate the density of aluminum if $27.6 \mathrm{~cm}^{3}$ has a mass of 74.6 g .
88. Osmium is one of the densest elements known. What is its density if 2.72 g has a volume of $0.121 \mathrm{~cm}^{3}$ ?
89. Calculate these masses.
(a) What is the mass of $6.00 \mathrm{~cm}^{3}$ of mercury, density $=13.5939 \mathrm{~g} / \mathrm{cm}^{3}$ ?
(b) What is the mass of 25.0 mL octane, density $=0.702 \mathrm{~g} / \mathrm{cm}^{3}$ ?
90. Calculate these masses.
(a) What is the mass of $4.00 \mathrm{~cm}^{3}$ of sodium, density $=0.97 \mathrm{~g} / \mathrm{cm}^{3}$ ?
(b) What is the mass of 125 mL gaseous chlorine, density $=3.16 \mathrm{~g} / \mathrm{L}$ ?
91. Calculate these volumes.
(a) What is the volume of 25 g iodine, density $=4.93 \mathrm{~g} / \mathrm{cm}^{3}$ ?
(b) What is the volume of 3.28 g gaseous hydrogen, density $=0.089 \mathrm{~g} / \mathrm{L}$ ?


97. Exercises - 97.6. Mathematical Treatment of Measurement Results
(b) What is the volume of 3.28 g gaseous hydrogen, density $=0.089 \mathrm{~g} / \mathrm{L}$ ?
92. Calculate these volumes.
(a) What is the volume of 11.3 g graphite, density $=2.25 \mathrm{~g} / \mathrm{cm}^{3}$ ?
(b) What is the volume of 39.657 g bromine, density $=2.928 \mathrm{~g} / \mathrm{cm}^{3}$ ?
93. Convert the boiling temperature of gold, $2966^{\circ} \mathrm{C}$, into degrees Fahrenheit and kelvin.
94. Convert the temperature of scalding water, $54^{\circ} \mathrm{C}$, into degrees Fahrenheit and kelvin.
95. Convert the temperature of the coldest area in a freezer, $-10^{\circ} \mathrm{F}$, to degrees Celsius and kelvin.
96. Convert the temperature of dry ice, $-77^{\circ} \mathrm{C}$, into degrees Fahrenheit and kelvin.
97. Convert the boiling temperature of liquid ammonia, $-28.1^{\circ} \mathrm{F}$, into degrees Celsius and kelvin.
98. The label on a pressurized can of spray disinfectant warns against heating the can above $130^{\circ} \mathrm{F}$. What are the corresponding temperatures on the Celsius and kelvin temperature scales?
99. The weather in Europe was unusually warm during the summer of 1995. The TV news reported temperatures as high as $45^{\circ} \mathrm{C}$. What was the temperature on the Fahrenheit scale?


97. Exercises - 97.6. Mathematical Treatment of Measurement Results
60 1•Exercises

Figure 2.1 Analysis of molecules in an exhaled breath can provide valuable information, leading to early diagnosis of diseases or detection of environmental exposure to harmful substances. (credit: modification of work by Paul Flowers)


98. CHAPTER OUTLINE - 98.1. Early Ideas in Atomic Theory
2.2 Evolution of Atomic Theory
2.3 Atomic Structure and Symbolism
2.4 Chemical Formulas

INTRODUCTION Lung diseases and lung cancers are among the world's most devastating illnesses partly due to delayed detection and diagnosis. Most noninvasive screening procedures aren't reliable, and patients often resist more accurate methods due to discomfort with the procedures or with the potential danger that the procedures cause. But what if you could be accurately diagnosed through a simple breath test?

Early detection of biomarkers, substances that indicate an organism's disease or physiological state, could allow diagnosis and treatment before a condition becomes serious or irreversible. Recent studies have shown that your exhaled breath can contain molecules that may be biomarkers for recent exposure to environmental contaminants or for pathological conditions ranging from asthma to lung cancer. Scientists are working to develop biomarker "fingerprints" that could be used to diagnose a specific disease based on the amounts and identities of certain molecules in a patient's exhaled breath. In Sangeeta Bhatia's lab at MIT, a team used substances that react specifically inside diseased lung tissue; the products of the reactions will be present as biomarkers that can be identified through mass spectrometry (an analytical method discussed later in the chapter). A potential application would allow patients with early symptoms to inhale or ingest a "sensor" substance, and, minutes later, to breathe into a detector for diagnosis. Similar research by scientists such as Laura López-Sánchez has provided similar processes for lung cancer. An essential concept underlying this goal is that of a molecule's identity, which is determined by the numbers and types of atoms it contains, and how
they are bonded together. This chapter will describe some of the fundamental chemical principles related to the composition of matter, including those central to the concept of molecular identity.


99. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- State the postulates of Dalton's atomic theory
- Use postulates of Dalton's atomic theory to explain the laws of definite and multiple proportions

The earliest recorded discussion of the basic structure of matter comes from ancient Greek philosophers, the scientists of their day. In the fifth century BC, Leucippus and Democritus argued that all matter was composed of small, finite particles that they called atomos, a term derived from the Greek word for "indivisible." They thought of atoms as moving particles that differed in shape and size, and which could join together. Later, Aristotle and others came to the conclusion that matter consisted of various combinations of the four "elements"-fire, earth, air, and water-and could be infinitely divided. Interestingly, these philosophers thought about atoms and "elements" as philosophical concepts, but apparently never considered performing experiments to test their ideas.

The Aristotelian view of the composition of matter held sway for over two thousand years, until English schoolteacher John Dalton helped to revolutionize chemistry with his hypothesis that the behavior of matter could be explained using an atomic theory. First published in 1807, many of Dalton's hypotheses about the microscopic features of matter are still valid in modern atomic theory. Here are the postulates of Dalton's atomic theory.

1. Matter is composed of exceedingly small particles called atoms. An atom is the smallest unit of an element that can participate in a chemical change.
2. An element consists of only one type of atom, which has a mass that is characteristic of the element and is the same for all atoms of that element (Figure 2.2). A macroscopic sample of an element contains an incredibly large number of atoms, all of which have identical chemical properties.


99. LEARNING OBJECTIVES -
FIGURE 2.2 A pre-1982 copper penny (left) contains approximately $3 \times 10^{22}$ copper atoms (several dozen are represented as brown spheres at the right), each of which has the same chemical properties. (credit: modification of work by "slgckgc"/Flickr)
3. Atoms of one element differ in properties from atoms of all other elements.
4. A compound consists of atoms of two or more elements combined in a small, whole-number ratio. In a given compound, the numbers of atoms of each of its elements are always present in the same ratio (Figure 2.3).

FIGURE 2.3 Copper(II) oxide, a powdery, black compound, results from the combination of two types of atoms-copper (brown spheres) and oxygen (red spheres)-in a 1:1 ratio. (credit: modification of work by "Chemicalinterest"/Wikimedia Commons)
5. Atoms are neither created nor destroyed during a chemical change, but are instead rearranged to yield substances that are different from those present before the change (Figure 2.4).

The elements copper and oxygen

The compound copper(II) oxide

FIGURE 2.4 When the elements copper (a shiny, red-brown solid, shown here as brown spheres) and oxygen (a clear and colorless gas, shown here as red spheres) react, their atoms rearrange to form a compound containing copper and oxygen (a powdery, black solid). (credit copper: modification of work by http://images-ofelements.com/copper.php)

Dalton's atomic theory provides a microscopic explanation of the many macroscopic properties of matter that you've learned about. For example, if an element such as copper consists of only one kind of atom, then it cannot be broken down into simpler substances, that is, into substances composed of fewer types of atoms. And if atoms are neither created nor destroyed during a chemical change, then the total mass of matter present when matter changes from one type to another will remain constant (the law of conservation of matter).


101. Testing Dalton's Atomic Theory -
In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by these symbols violate any of the ideas of Dalton's atomic theory? If so, which one?


102. Solution -
The starting materials consist of two green spheres and two purple spheres. The products consist of only one green sphere and one purple sphere. This violates Dalton's postulate that atoms are neither created nor destroyed during a chemical change, but are merely redistributed. (In this case, atoms appear to have been destroyed.)


103. Check Your Learning -
In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by these symbols violate any of the ideas of Dalton's atomic theory? If so, which one?


104. Answer: -
The starting materials consist of four green spheres and two purple spheres. The products consist of four green spheres and two purple spheres. This does not violate any of Dalton's postulates: Atoms are neither created nor destroyed, but are redistributed in small, whole-number ratios.

Dalton knew of the experiments of French chemist Joseph Proust, who demonstrated that all samples of a pure compound contain the same elements in the same proportion by mass. This statement is known as the law of definite proportions or the law of constant composition. The suggestion that the numbers of atoms of the elements in a given compound always exist in the same ratio is consistent with these observations. For example, when different samples of isooctane (a component of gasoline and one of the standards used in the octane rating system) are analyzed, they are found to have a carbon-to-hydrogen mass ratio of 5.33:1, as shown in Table 2.1.

Constant Composition of Isooctane

| Sample | Carbon | Hydrogen |
| :---: | :---: | :---: | Mass Ratio.

TABLE 2.1

It is worth noting that although all samples of a particular compound have the same mass ratio, the converse is not true in general. That is, samples that have the same mass ratio are not necessarily the same substance. For example, there are many compounds other than isooctane that also have a carbon-to-hydrogen mass ratio of 5.33:1.00.

Dalton also used data from Proust, as well as results from his own experiments, to formulate another interesting law. The law of multiple proportions states that when two elements react to form more than one compound, a fixed mass of one element will react with masses of the other element in a ratio of small, whole
numbers. For example, copper and chlorine can form a green, crystalline solid with a mass ratio of 0.558 g chlorine to 1 g copper, as well as a brown crystalline solid with a mass ratio of 1.116 g chlorine to 1 g copper. These ratios by themselves may not seem particularly interesting or informative; however, if we take a ratio of these ratios, we obtain a useful and possibly surprising result: a small, whole-number ratio.


104. Answer: -
$$
\frac{\frac{1.116 \mathrm{~g} \mathrm{Cl}}{1 \mathrm{~g} \mathrm{Cu}}}{\frac{0.558 \mathrm{~g} \mathrm{Cl}}{1 \mathrm{~g} \mathrm{Cu}}}=\frac{2}{1}
$$

This 2-to-1 ratio means that the brown compound has twice the amount of chlorine per amount of copper as the green compound.

This can be explained by atomic theory if the copper-to-chlorine ratio in the brown compound is 1 copper atom to 2 chlorine atoms, and the ratio in the green compound is 1 copper atom to 1 chlorine atom. The ratio of chlorine atoms (and thus the ratio of their masses) is therefore 2 to 1 (Figure 2.5).

FIGURE 2.5 Compared to the copper chlorine compound in (a), where copper is represented by brown spheres and chlorine by green spheres, the copper chlorine compound in (b) has twice as many chlorine atoms per copper atom. (credit a: modification of work by "Benjah-bmm27"/Wikimedia Commons; credit b: modification of work by "Walkerma"/Wikimedia Commons)


106. Laws of Definite and Multiple Proportions -
A sample of compound A (a clear, colorless gas) is analyzed and found to contain 4.27 g carbon and 5.69 g oxygen. A sample of compound B (also a clear, colorless gas) is analyzed and found to contain 5.19 g carbon and 13.84 g oxygen. Are these data an example of the law of definite proportions, the law of multiple proportions, or neither? What do these data tell you about substances A and B?


107. Solution -
In compound A , the mass ratio of oxygen to carbon is:

$$
\frac{1.33 \mathrm{~g} \mathrm{O}}{1 \mathrm{~g} \mathrm{C}}
$$

In compound $B$, the mass ratio of oxygen to carbon is:

$$
\frac{2.67 \mathrm{~g} \mathrm{O}}{1 \mathrm{~g} \mathrm{C}}
$$

The ratio of these ratios is:

$$
\frac{\frac{1.33 \mathrm{~g} \mathrm{O}}{1 \mathrm{~g} \mathrm{C}}}{\frac{2.67 \mathrm{~g} \mathrm{O}}{1 \mathrm{~g} \mathrm{C}}}=\frac{1}{2}
$$

This supports the law of multiple proportions. This means that A and B are different compounds, with A having one-half as much oxygen per amount of carbon (or twice as much carbon per amount of oxygen) as B. A possible pair of compounds that would fit this relationship would be $\mathrm{A}=\mathrm{CO}$ and $\mathrm{B}=\mathrm{CO}_{2}$.


108. Check Your Learning -
A sample of compound $X$ (a clear, colorless, combustible liquid with a noticeable odor) is analyzed and found to contain 14.13 g carbon and 2.96 g hydrogen. A sample of compound Y (a clear, colorless, combustible liquid with a noticeable odor that is slightly different from X's odor) is analyzed and found to contain 19.91 g carbon and 3.34 g hydrogen. Are these data an example of the law of definite proportions, the law of multiple proportions, or neither? What do these data tell you about substances X and Y ?


109. Answer: -
In compound $X$, the mass ratio of carbon to hydrogen is $\frac{14.13 \mathrm{~g} \mathrm{C}}{2.96 \mathrm{~g} \mathrm{H}}$. In compound $Y$, the mass ratio of carbon to
number ratio supports the law of multiple proportions. This means that $X$ and $Y$ are different compounds.


110. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Outline milestones in the development of modern atomic theory
- Summarize and interpret the results of the experiments of Thomson, Millikan, and Rutherford
- Describe the three subatomic particles that compose atoms
- Define isotopes and give examples for several elements

If matter is composed of atoms, what are atoms composed of? Are they the smallest particles, or is there something smaller? In the late 1800 s , a number of scientists interested in questions like these investigated the electrical discharges that could be produced in low-pressure gases, with the most significant discovery made by English physicist J. J. Thomson using a cathode ray tube. This apparatus consisted of a sealed glass tube from which almost all the air had been removed; the tube contained two metal electrodes. When high voltage was applied across the electrodes, a visible beam called a cathode ray appeared between them. This beam was deflected toward the positive charge and away from the negative charge, and was produced in the same way with identical properties when different metals were used for the electrodes. In similar experiments, the ray was simultaneously deflected by an applied magnetic field, and measurements of the extent of deflection and the magnetic field strength allowed Thomson to calculate the charge-to-mass ratio of the cathode ray particles. The results of these measurements indicated that these particles were much lighter than atoms (Figure 2.6).

FIGURE 2.6 (a) J. J. Thomson produced a visible beam in a cathode ray tube. (b) This is an early cathode ray tube, invented in 1897 by Ferdinand Braun. (c) In the cathode ray, the beam (shown in yellow) comes from the cathode and is accelerated past the anode toward a fluorescent scale at the end of the tube. Simultaneous deflections by applied electric and magnetic fields permitted Thomson to calculate the mass-to-charge ratio of the particles composing the cathode ray. (credit a: modification of work by Nobel Foundation; credit b: modification of work by Eugen Nesper; credit c: modification of work by "Kurzon"/Wikimedia Commons)


110. LEARNING OBJECTIVES -
Based on his observations, here is what Thomson proposed and why: The particles are attracted by positive (+) charges and repelled by negative (-) charges, so they must be negatively charged (like charges repel and unlike charges attract); they are less massive than atoms and indistinguishable, regardless of the source material, so they must be fundamental, subatomic constituents of all atoms. Although controversial at the time, Thomson's idea was gradually accepted, and his cathode ray particle is what we now call an electron, a negatively charged, subatomic particle with a mass more than one thousand-times less that of an atom. The term "electron" was coined in 1891 by Irish physicist George Stoney, from "electric ion."


111. LINK TO LEARNING -
Click here (http://openstax.org/l/16JJThomson) to hear Thomson describe his discovery in his own voice.

In 1909, more information about the electron was uncovered by American physicist Robert A. Millikan via his "oil drop" experiments. Millikan created microscopic oil droplets, which could be electrically charged by friction as they formed or by using X-rays. These droplets initially fell due to gravity, but their downward progress could be slowed or even reversed by an electric field lower in the apparatus. By adjusting the electric field strength and making careful measurements and appropriate calculations, Millikan was able to determine the charge on individual drops (Figure 2.7).

FIGURE 2.7 Millikan's experiment measured the charge of individual oil drops. The tabulated data are examples of a few possible values.

Looking at the charge data that Millikan gathered, you may have recognized that the charge of an oil droplet is always a multiple of a specific charge, $1.6 \times 10^{-19} \mathrm{C}$. Millikan concluded that this value must therefore be a fundamental charge-the charge of a single electron-with his measured charges due to an excess of one electron ( 1 times $1.6 \times 10^{-19} \mathrm{C}$ ), two electrons ( 2 times $1.6 \times 10^{-19} \mathrm{C}$ ), three electrons ( 3 times $1.6 \times 10^{-19} \mathrm{C}$ ), and so on, on a given oil droplet. Since the charge of an electron was now known due to Millikan's research, and the charge-to-mass ratio was already known due to Thomson's research ( $1.759 \times 10^{11} \mathrm{C} / \mathrm{kg}$ ), it only required a simple calculation to determine the mass of the electron as well.

$$
\text { Mass of electron }=1.602 \times 10^{-19} \mathrm{C} \times \frac{1 \mathrm{~kg}}{1.759 \times 10^{11} \mathrm{C}}=9.107 \times 10^{-31} \mathrm{~kg}
$$


111. LINK TO LEARNING -
Scientists had now established that the atom was not indivisible as Dalton had believed, and due to the work of Thomson, Millikan, and others, the charge and mass of the negative, subatomic particles-the electrons-were known. However, the positively charged part of an atom was not yet well understood. In 1904, Thomson proposed the "plum pudding" model of atoms, which described a positively charged mass with an equal amount of negative charge in the form of electrons embedded in it, since all atoms are electrically neutral. A competing model had been proposed in 1903 by Hantaro Nagaoka, who postulated a Saturn-like atom, consisting of a positively charged sphere surrounded by a halo of electrons (Figure 2.8).

FIGURE 2.8 (a) Thomson suggested that atoms resembled plum pudding, an English dessert consisting of moist cake with embedded raisins ("plums"). (b) Nagaoka proposed that atoms resembled the planet Saturn, with a ring of
electrons surrounding a positive "planet." (credit a: modification of work by "Man vyi"/Wikimedia Commons; credit b: modification of work by "NASA"/Wikimedia Commons)


111. LINK TO LEARNING -
The next major development in understanding the atom came from Ernest Rutherford, a physicist from New Zealand who largely spent his scientific career in Canada and England. He performed a series of experiments using a beam of high-speed, positively charged alpha particles ( $\alpha$ particles) that were produced by the radioactive decay of radium; $\alpha$ particles consist of two protons and two neutrons (you will learn more about radioactive decay in the chapter on nuclear chemistry). Rutherford and his colleagues Hans Geiger (later famous for the Geiger counter) and Ernest Marsden aimed a beam of $\alpha$ particles, the source of which was embedded in a lead block to absorb most of the radiation, at a very thin piece of gold foil and examined the resultant scattering of the $\alpha$ particles using a luminescent screen that glowed briefly where hit by an $\alpha$ particle.
What did they discover? Most particles passed right through the foil without being deflected at all. However, some were diverted slightly, and a very small number were deflected almost straight back toward the source (Figure 2.9). Rutherford described finding these results: "It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you., ${ }^{\underline{1}}$

FIGURE 2.9 Geiger and Rutherford fired $\alpha$ particles at a piece of gold foil and detected where those particles went, as shown in this schematic diagram of their experiment. Most of the particles passed straight through the foil, but a few were deflected slightly and a very small number were significantly deflected.

Here is what Rutherford deduced: Because most of the fast-moving $\alpha$ particles passed through the gold atoms undeflected, they must have traveled through essentially empty space inside the atom. Alpha particles are positively charged, so deflections arose when they encountered another positive charge (like charges repel each other). Since like charges repel one another, the few positively charged $\alpha$ particles that changed paths abruptly must have hit, or closely approached, another body that also had a highly concentrated, positive charge. Since the deflections occurred a small fraction of the time, this charge only occupied a small amount of the space in the gold foil. Analyzing a series of such experiments in detail, Rutherford drew two conclusions:


111. LINK TO LEARNING -
1. The volume occupied by an atom must consist of a large amount of empty space.
2. A small, relatively heavy, positively charged body, the nucleus, must be at the center of each atom.


112. LINK TO LEARNING -
View this simulation (http://openstax.org/l/16Rutherford) of the Rutherford gold foil experiment. Adjust the slit width to produce a narrower or broader beam of $\alpha$ particles to see how that affects the scattering pattern.

[^0]This analysis led Rutherford to propose a model in which an atom consists of a very small, positively charged nucleus, in which most of the mass of the atom is concentrated, surrounded by the negatively charged electrons, so that the atom is electrically neutral (Figure 2.10). After many more experiments, Rutherford also discovered that the nuclei of other elements contain the hydrogen nucleus as a "building block," and he named this more fundamental particle the proton, the positively charged, subatomic particle found in the nucleus. With one addition, which you will learn next, this nuclear model of the atom, proposed over a century ago, is still used today.

FIGURE 2.10 The $\alpha$ particles are deflected only when they collide with or pass close to the much heavier, positively charged gold nucleus. Because the nucleus is very small compared to the size of an atom, very few $\alpha$ particles are deflected. Most pass through the relatively large region occupied by electrons, which are too light to deflect the rapidly moving particles.


113. LINK TO LEARNING -
The Rutherford Scattering simulation (http://openstax.org/1/16PhetScatter) allows you to investigate the differences between a "plum pudding" atom and a Rutherford atom by firing $\alpha$ particles at each type of atom.

Another important finding was the discovery of isotopes. During the early 1900s, scientists identified several substances that appeared to be new elements, isolating them from radioactive ores. For example, a "new element" produced by the radioactive decay of thorium was initially given the name mesothorium. However, a more detailed analysis showed that mesothorium was chemically identical to radium (another decay product), despite having a different atomic mass. This result, along with similar findings for other elements, led the English chemist Frederick Soddy to realize that an element could have types of atoms with different masses that were chemically indistinguishable. These different types are called isotopes-atoms of the same element that differ in mass. Soddy was awarded the Nobel Prize in Chemistry in 1921 for this discovery.

One puzzle remained: The nucleus was known to contain almost all of the mass of an atom, with the number of protons only providing half, or less, of that mass. Different proposals were made to explain what constituted the remaining mass, including the existence of neutral particles in the nucleus. As you might expect, detecting uncharged particles is very challenging, and it was not until 1932 that James Chadwick found evidence of neutrons, uncharged, subatomic particles with a mass approximately the same as that of protons. The existence of the neutron also explained isotopes: They differ in mass because they have different numbers of
neutrons, but they are chemically identical because they have the same number of protons. This will be explained in more detail later.


114. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Write and interpret symbols that depict the atomic number, mass number, and charge of an atom or ion
- Define the atomic mass unit and average atomic mass
- Calculate average atomic mass and isotopic abundance

The development of modern atomic theory revealed much about the inner structure of atoms. It was learned that an atom contains a very small nucleus composed of positively charged protons and uncharged neutrons, surrounded by a much larger volume of space containing negatively charged electrons. The nucleus contains the majority of an atom's mass because protons and neutrons are much heavier than electrons, whereas electrons occupy almost all of an atom's volume. The diameter of an atom is on the order of $10^{-10} \mathrm{~m}$, whereas the diameter of the nucleus is roughly $10^{-15} \mathrm{~m}$-about 100,000 times smaller. For a perspective about their relative sizes, consider this: If the nucleus were the size of a blueberry, the atom would be about the size of a football stadium (Figure 2.11).

FIGURE 2.11 If an atom could be expanded to the size of a football stadium, the nucleus would be the size of a single blueberry. (credit middle: modification of work by "babyknight"/Wikimedia Commons; credit right: modification of work by Paxson Woelber)


114. LEARNING OBJECTIVES -
Atoms-and the protons, neutrons, and electrons that compose them-are extremely small. For example, a carbon atom weighs less than $2 \times 10^{-23} \mathrm{~g}$, and an electron has a charge of less than $2 \times 10^{-19} \mathrm{C}$ (coulomb). When describing the properties of tiny objects such as atoms, we use appropriately small units of measure, such as the atomic mass unit (amu) and the fundamental unit of charge (e). The amu was originally defined based on hydrogen, the lightest element, then later in terms of oxygen. Since 1961, it has been defined with regard to the most abundant isotope of carbon, atoms of which are assigned masses of exactly 12 amu . (This isotope is known as "carbon-12" as will be discussed later in this module.) Thus, one amu is exactly $\frac{1}{12}$ of the mass of one carbon-12 atom: $1 \mathrm{amu}=1.6605 \times 10^{-24} \mathrm{~g}$. (The Dalton (Da) and the unified atomic mass unit (u) are alternative units that are equivalent to the amu.) The fundamental unit of charge (also called the elementary charge) equals the magnitude of the charge of an electron (e) with $\mathrm{e}=1.602 \times 10^{-19} \mathrm{C}$.


114. LEARNING OBJECTIVES -
A proton has a mass of 1.0073 amu and a charge of $1+$. A neutron is a slightly heavier particle with a mass 1.0087 amu and a charge of zero; as its name suggests, it is neutral. The electron has a charge of 1 - and is a much lighter particle with a mass of about 0.00055 amu (it would take about 1800 electrons to equal the mass of one proton). The properties of these fundamental particles are summarized in Table 2.2. (An observant student might notice that the sum of an atom's subatomic particles does not equal the atom's actual mass: The total mass of six protons, six neutrons, and six electrons is 12.0993 amu , slightly larger than 12.00 amu . This "missing" mass is known as the mass defect, and you will learn about it in the chapter on nuclear chemistry.)

Properties of Subatomic Particles

| Name | Location | Charge (C) | Unit Charge | Mass (amu) | Mass (g) |
| :---: | :---: | :---: | :---: | :---: | :---: |
| electron | outside nucleus | $-1.602 \times 10^{-19}$ | $1-$ | 0.00055 | $0.00091 \times 10^{-24}$ |
| proton | nucleus | $1.602 \times 10^{-19}$ | $1+$ | 1.00727 | $1.67262 \times 10^{-24}$ |
| neutron | nucleus | 0 | 0 | 1.00866 | $1.67493 \times 10^{-24}$ |

TABLE 2.2


114. LEARNING OBJECTIVES -
TABLE 2.2

The number of protons in the nucleus of an atom is its atomic number (Z). This is the defining trait of an element: Its value determines the identity of the atom. For example, any atom that contains six protons is the element carbon and has the atomic number 6, regardless of how many neutrons or electrons it may have. A neutral atom must contain the same number of positive and negative charges, so the number of protons equals the number of electrons. Therefore, the atomic number also indicates the number of electrons in an atom. The total number of protons and neutrons in an atom is called its mass number (A). The number of neutrons is therefore the difference between the mass number and the atomic number: $\mathrm{A}-\mathrm{Z}=$ number of neutrons.

$$
\begin{aligned}
\text { atomic number }(\mathrm{Z}) & =\text { number of protons } \\
\text { mass number }(\mathrm{A}) & =\text { number of protons+ number of neutrons } \\
\mathrm{A}-\mathrm{Z} & =\text { number of neutrons }
\end{aligned}
$$

Atoms are electrically neutral if they contain the same number of positively charged protons and negatively charged electrons. When the numbers of these subatomic particles are not equal, the atom is electrically charged and is called an ion. The charge of an atom is defined as follows:

Atomic charge $=$ number of protons - number of electrons
As will be discussed in more detail, atoms (and molecules) typically acquire charge by gaining or losing electrons. An atom that gains one or more electrons will exhibit a negative charge and is called an anion. Positively charged atoms called cations are formed when an atom loses one or more electrons. For example, a neutral sodium atom $(Z=11)$ has 11 electrons. If this atom loses one electron, it will become a cation with a 1+ charge ( $11-10=1+$ ). A neutral oxygen atom $(Z=8)$ has eight electrons, and if it gains two electrons it will become an anion with a $2-$ charge ( $8-10=2-$ ).


116. Composition of an Atom -
Iodine is an essential trace element in our diet; it is needed to produce thyroid hormone. Insufficient iodine in the diet can lead to the development of a goiter, an enlargement of the thyroid gland (Figure 2.12).

FIGURE 2.12 (a) Insufficient iodine in the diet can cause an enlargement of the thyroid gland called a goiter. (b) The addition of small amounts of iodine to salt, which prevents the formation of goiters, has helped eliminate this concern in the US where salt consumption is high. (credit a: modification of work by "Almazi"/Wikimedia Commons; credit b: modification of work by Mike Mozart)

The addition of small amounts of iodine to table salt (iodized salt) has essentially eliminated this health concern in the United States, but as much as $40 \%$ of the world's population is still at risk of iodine deficiency. The iodine atoms are added as anions, and each has a 1-charge and a mass number of 127. Determine the numbers of protons, neutrons, and electrons in one of these iodine anions.


117. Solution -
The atomic number of iodine (53) tells us that a neutral iodine atom contains 53 protons in its nucleus and 53 electrons outside its nucleus. Because the sum of the numbers of protons and neutrons equals the mass number, 127 , the number of neutrons is $74(127-53=74)$. Since the iodine is added as a 1 - anion, the number of electrons is 54 [53-(1-)=54].


118. Check Your Learning -
An ion of platinum has a mass number of 195 and contains 74 electrons. How many protons and neutrons does it contain, and what is its charge?


119. Answer: -
78 protons; 117 neutrons; charge is 4+


120. Chemical Symbols -
A chemical symbol is an abbreviation that we use to indicate an element or an atom of an element. For example, the symbol for mercury is Hg (Figure 2.13). We use the same symbol to indicate one atom of mercury (microscopic domain) or to label a container of many atoms of the element mercury (macroscopic domain).

FIGURE 2.13 The symbol Hg represents the element mercury regardless of the amount; it could represent one atom of mercury or a large amount of mercury.

The symbols for several common elements and their atoms are listed in Table 2.3. Some symbols are derived from the common name of the element; others are abbreviations of the name in another language. Most symbols have one or two letters, but three-letter symbols have been used to describe some elements that have atomic numbers greater than 112 . To avoid confusion with other notations, only the first letter of a symbol is capitalized. For example, Co is the symbol for the element cobalt, but CO is the notation for the compound carbon monoxide, which contains atoms of the elements carbon (C) and oxygen (O). All known elements and their symbols are in the periodic table in Figure 3.37 (also found in Figure A1).


121. Some Common Elements and Their Symbols -
| Element | Symbol | Element | Symbol |
| :--- | :--- | :--- | :--- |
| aluminum | Al | iron | Fe (from ferrum) |
| bromine | Br | lead | Pb (from plumbum) |
| calcium | Ca | magnesium | Mg |
| carbon | C | mercury | Hg (from hydrargyrum) |
| chlorine | Cl | nitrogen | N |
| chromium | Cr | oxygen | O |
| cobalt | Co | potassium | K (from kalium) |
| copper | Cu (from cuprum) | silicon | Si |
| fluorine | F | silver | Ag (from argentum) |
| gold | Au (from aurum) | sodium | Na (from natrium) |
| helium | He | sulfur | S |

TABLE 2.3

| Element | Symbol | Element | Symbol |
| :--- | :--- | :--- | :--- |
| hydrogen | H | tin | Sn (from stannum) |
| iodine | I | zinc | Zn |

TABLE 2.3

Traditionally, the discoverer (or discoverers) of a new element names the element. However, until the name is recognized by the International Union of Pure and Applied Chemistry (IUPAC), the recommended name of the new element is based on the Latin word(s) for its atomic number. For example, element 106 was called unnilhexium (Unh), element 107 was called unnilseptium (Uns), and element 108 was called unniloctium (Uno) for several years. These elements are now named after scientists (or occasionally locations); for example, element 106 is now known as seaborgium (Sg) in honor of Glenn Seaborg, a Nobel Prize winner who was active in the discovery of several heavy elements. Element 109 was named in honor of Lise Meitner, who discovered nuclear fission, a phenomenon that would have world-changing impacts; Meitner also contributed to the discovery of some major isotopes, discussed immediately below.


122. (2) LINK TO LEARNING -
Visit this site (http://openstax.org/l/16IUPAC) to learn more about IUPAC, the International Union of Pure and Applied Chemistry, and explore its periodic table.


123. Isotopes -
The symbol for a specific isotope of any element is written by placing the mass number as a superscript to the left of the element symbol (Figure 2.14). The atomic number is sometimes written as a subscript preceding the symbol, but since this number defines the element's identity, as does its symbol, it is often omitted. For example, magnesium exists as a mixture of three isotopes, each with an atomic number of 12 and with mass numbers of 24,25 , and 26 , respectively. These isotopes can be identified as ${ }^{24} \mathrm{Mg}$, ${ }^{25} \mathrm{Mg}$, and ${ }^{26} \mathrm{Mg}$. These isotope symbols are read as "element, mass number" and can be symbolized consistent with this reading. For instance, ${ }^{24} \mathrm{Mg}$ is read as "magnesium 24 ," and can be written as "magnesium- 24 " or " $\mathrm{Mg}-24$." ${ }^{25} \mathrm{Mg}$ is read as "magnesium 25," and can be written as "magnesium-25" or "Mg-25." All magnesium atoms have 12 protons in their nucleus. They differ only because a ${ }^{24} \mathrm{Mg}$ atom has 12 neutrons in its nucleus, a ${ }^{25} \mathrm{Mg}$ atom has 13 neutrons, and a ${ }^{26} \mathrm{Mg}$ has 14 neutrons.

FIGURE 2.14 The symbol for an atom indicates the element via its usual two-letter symbol, the mass number as a left superscript, the atomic number as a left subscript (sometimes omitted), and the charge as a right superscript.


123. Isotopes -
Information about the naturally occurring isotopes of elements with atomic numbers 1 through 10 is given in Table 2.4. Note that in addition to standard names and symbols, the isotopes of hydrogen are often referred to using common names and accompanying symbols. Hydrogen- 2 , symbolized ${ }^{2} \mathrm{H}$, is also called deuterium and sometimes symbolized D. Hydrogen-3, symbolized ${ }^{3} \mathrm{H}$, is also called tritium and sometimes symbolized T.

Nuclear Compositions of Atoms of the Very Light Elements


123. Isotopes -
| Element | Symbol | Atomic
 Number | Number of Protons | Number of Neutrons | Mass
 (amu) | \% Natural
 Abundance |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| hydrogen | ${ }_{1}^{1} \mathrm{H}$ (protium) | 1 | 1 | 0 | 1.0078 | 99.989 |
|  | ${ }_{1}^{2} \mathrm{H}$ (deuterium) | 1 | 1 | 1 | 2.0141 | 0.0115 |
|  | ${ }_{1}^{3} \mathrm{H}$ (tritium) | 1 | 1 | 2 | 3.01605 | - (trace) |
| helium | ${ }_{2}^{3} \mathrm{He}$ | 2 | 2 | 1 | 3.01603 | 0.00013 |
|  | ${ }_{2}^{4} \mathrm{He}$ | 2 | 2 | 2 | 4.0026 | 100 |
| lithium | ${ }_{3}^{6} \mathrm{Li}$ | 3 | 3 | 3 | 6.0151 | 7.59 |
|  | ${ }_{3}^{7} \mathrm{Li}$ | 3 | 3 | 4 | 7.0160 | 92.41 |
| beryllium | ${ }_{4}^{9} \mathrm{Be}$ | 4 | 4 | 5 | 9.0122 | 100 |
| boron | ${ }_{5}^{10} \mathrm{~B}$ | 5 | 5 | 5 | 10.0129 | 19.9 |
|  | ${ }_{5}^{11} \mathrm{~B}$ | 5 | 5 | 6 | 11.0093 | 80.1 |


123. Isotopes -
|  | ${ }_{5}^{11} \mathrm{~B}$ | 5 | 5 | 6 | 11.0093 | 80.1 |
| carbon | ${ }_{6}^{12} \mathrm{C}$ | 6 | 6 | 6 | 12.0000 | 98.89 |
|  | ${ }_{6}^{13} \mathrm{C}$ | 6 | 6 | 7 | 13.0034 | 1.11 |
|  | ${ }_{6}^{14} \mathrm{C}$ | 6 | 6 | 8 | 14.0032 | - (trace) |
| nitrogen | ${ }_{7}^{14} \mathrm{~N}$ | 7 | 7 | 7 | 14.0031 | 99.63 |
|  | ${ }_{7}^{15} \mathrm{~N}$ | 7 | 7 | 8 | 15.0001 | 0.37 |
| oxygen | ${ }_{8}^{16} \mathrm{O}$ | 8 | 8 | 8 | 15.9949 | 99.757 |
|  | ${ }_{8}^{17} \mathrm{O}$ | 8 | 8 | 9 | 16.9991 | 0.038 |
|  | ${ }_{8}^{18} \mathrm{O}$ | 8 | 8 | 10 | 17.9992 | 0.205 |
| fluorine | ${ }_{9}^{19} \mathrm{~F}$ | 9 | 9 | 10 | 18.9984 | 100 |
| neon | ${ }_{10}^{20} \mathrm{Ne}$ | 10 | 10 | 10 | 19.9924 | 90.48 |


123. Isotopes -
TABLE 2.4

| Element | Symbol | Atomic
 Number | Number of
 Protons | Number of
 Neutrons | Mass
 (amu) | \% Natural
 Abundance |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
|  | 21
 10 | 10 | 10 | 11 | 20.9938 | 0.27 |
|  | ${ }_{10}^{22} \mathrm{Ne}$ | 10 | 10 | 12 | 21.9914 | 9.25 |

TABLE 2.4


124. LINK TO LEARNING -
Use this Build an Atom simulator (http://openstax.org/l/16PhetAtomBld) to build atoms of the first 10 elements, see which isotopes exist, check nuclear stability, and gain experience with isotope symbols.


125. Atomic Mass -
Because each proton and each neutron contribute approximately one amu to the mass of an atom, and each electron contributes far less, the atomic mass of a single atom is approximately equal to its mass number (a whole number). However, the average masses of atoms of most elements are not whole numbers because most elements exist naturally as mixtures of two or more isotopes.

The mass of an element shown in a periodic table or listed in a table of atomic masses is a weighted, average mass of all the isotopes present in a naturally occurring sample of that element. This is equal to the sum of each individual isotope's mass multiplied by its fractional abundance.

$$
\text { average mass }=\sum_{i}(\text { fractional abundance } \times \text { isotopic mass })_{i}
$$

For example, the element boron is composed of two isotopes: About $19.9 \%$ of all boron atoms are ${ }^{10} \mathrm{~B}$ with a mass of 10.0129 amu , and the remaining $80.1 \%$ are ${ }^{11} \mathrm{~B}$ with a mass of 11.0093 amu . The average atomic mass for boron is calculated to be:

$$
\begin{aligned}
\text { boron average mass } & =(0.199 \times 10.0129 \mathrm{amu})+(0.801 \times 11.0093 \mathrm{amu}) \\
& =1.99 \mathrm{amu}+8.82 \mathrm{amu} \\
& =10.81 \mathrm{amu}
\end{aligned}
$$

It is important to understand that no single boron atom weighs exactly $10.8 \mathrm{amu} ; 10.8 \mathrm{amu}$ is the average mass of all boron atoms, and individual boron atoms weigh either approximately 10 amu or 11 amu .


127. Calculation of Average Atomic Mass -
A meteorite found in central Indiana contains traces of the noble gas neon picked up from the solar wind during the meteorite's trip through the solar system. Analysis of a sample of the gas showed that it consisted of $91.84 \%{ }^{20} \mathrm{Ne}$ (mass 19.9924 amu ), $0.47 \%{ }^{21} \mathrm{Ne}$ (mass 20.9940 amu ), and $7.69 \%{ }^{22} \mathrm{Ne}$ (mass 21.9914 amu ). What is the average mass of the neon in the solar wind?


128. Solution -
$$
\begin{aligned}
\text { average mass } & =(0.9184 \times 19.9924 \mathrm{amu})+(0.0047 \times 20.9940 \mathrm{amu})+(0.0769 \times 21.9914 \mathrm{amu}) \\
& =(18.36+0.099+1.69) \mathrm{amu} \\
& =20.15 \mathrm{amu}
\end{aligned}
$$

The average mass of a neon atom in the solar wind is 20.15 amu . (The average mass of a terrestrial neon atom is 20.1796 amu . This result demonstrates that we may find slight differences in the natural abundance of
isotopes, depending on their origin.)


129. Check Your Learning -
A sample of magnesium is found to contain $78.70 \%$ of ${ }^{24} \mathrm{Mg}$ atoms (mass 23.98 amu ), $10.13 \%$ of ${ }^{25} \mathrm{Mg}$ atoms (mass 24.99 amu ), and $11.17 \%$ of ${ }^{26} \mathrm{Mg}$ atoms (mass 25.98 amu ). Calculate the average mass of a Mg atom.


130. Answer: -
24.31 amu

We can also do variations of this type of calculation, as shown in the next example.


132. Calculation of Percent Abundance -
Naturally occurring chlorine consists of ${ }^{35} \mathrm{Cl}$ (mass 34.96885 amu ) and ${ }^{37} \mathrm{Cl}$ (mass 36.96590 amu ), with an average mass of 35.453 amu . What is the percent composition of Cl in terms of these two isotopes?


133. Solution -
The average mass of chlorine is the fraction that is ${ }^{35} \mathrm{Cl}$ times the mass of ${ }^{35} \mathrm{Cl}$ plus the fraction that is ${ }^{37} \mathrm{Cl}$ times the mass of ${ }^{37} \mathrm{Cl}$.

$$
\text { average mass }=\left(\text { fraction of }{ }^{35} \mathrm{Cl} \times \text { mass of }{ }^{35} \mathrm{Cl}\right)+\left(\text { fraction of }{ }^{37} \mathrm{Cl} \times \text { mass of }{ }^{37} \mathrm{Cl}\right)
$$

If we let $x$ represent the fraction that is ${ }^{35} \mathrm{Cl}$, then the fraction that is ${ }^{37} \mathrm{Cl}$ is represented by $1.00-x$.
(The fraction that is ${ }^{35} \mathrm{Cl}$ + the fraction that is ${ }^{37} \mathrm{Cl}$ must add up to 1 , so the fraction of ${ }^{37} \mathrm{Cl}$ must equal 1.00 - the fraction of ${ }^{35} \mathrm{Cl}$.)

Substituting this into the average mass equation, we have:

$$
\begin{aligned}
35.453 \mathrm{amu} & =(x \times 34.96885 \mathrm{amu})+[(1.00-x) \times 36.96590 \mathrm{amu}] \\
35.453 & =34.96885 x+36.96590-36.96590 x \\
1.99705 x & =1.513 \\
x & =\frac{1.513}{1.99705}=0.7576
\end{aligned}
$$

So solving yields: $x=0.7576$, which means that $1.00-0.7576=0.2424$. Therefore, chlorine consists of $75.76 \%$ ${ }^{35} \mathrm{Cl}$ and $24.24 \%{ }^{37} \mathrm{Cl}$.


134. Check Your Learning -
Naturally occurring copper consists of ${ }^{63} \mathrm{Cu}$ (mass 62.9296 amu ) and ${ }^{65} \mathrm{Cu}$ (mass 64.9278 amu ), with an average mass of 63.546 amu . What is the percent composition of Cu in terms of these two isotopes?


135. Answer: -
69.15\% Cu-63 and 30.85\% Cu-65


136. LINK TO LEARNING -
Visit this site (http://openstax.org/l/16PhetAtomMass) to make mixtures of the main isotopes of the first 18 elements, gain experience with average atomic mass, and check naturally occurring isotope ratios using the Isotopes and Atomic Mass simulation.

As you will learn, isotopes are important in nature and especially in human understanding of science and medicine. Let's consider just one natural, stable isotope: Oxygen-18, which is noted in the table above and is referred to as one of the environmental isotopes. It is important in paleoclimatology, for example, because scientists can use the ratio between Oxygen-18 and Oxygen-16 in an ice core to determine the temperature of
precipitation over time. Oxygen-18 was also critical to the discovery of metabolic pathways and the mechanisms of enzymes. Mildred Cohn pioneered the usage of these isotopes to act as tracers, so that researchers could follow their path through reactions and gain a better understanding of what is happening. One of her first discoveries provided insight into the phosphorylation of glucose that takes place in mitochondria. And the methods of using isotopes for this research contributed to entire fields of study.


136. LINK TO LEARNING -
The occurrence and natural abundances of isotopes can be experimentally determined using an instrument called a mass spectrometer. Mass spectrometry (MS) is widely used in chemistry, forensics, medicine, environmental science, and many other fields to analyze and help identify the substances in a sample of material. In a typical mass spectrometer (Figure 2.15), the sample is vaporized and exposed to a high-energy electron beam that causes the sample's atoms (or molecules) to become electrically charged, typically by losing one or more electrons. These cations then pass through a (variable) electric or magnetic field that deflects each cation's path to an extent that depends on both its mass and charge (similar to how the path of a large steel ball rolling past a magnet is deflected to a lesser extent that that of a small steel ball). The ions are detected, and a plot of the relative number of ions generated versus their mass-to-charge ratios (a mass spectrum) is made. The height of each vertical feature or peak in a mass spectrum is proportional to the fraction of cations with the specified mass-to-charge ratio. Since its initial use during the development of modern atomic theory, MS has evolved to become a powerful tool for chemical analysis in a wide range of applications.

FIGURE 2.15 Analysis of zirconium in a mass spectrometer produces a mass spectrum with peaks showing the different isotopes of Zr .


137. LINK TO LEARNING -
See an animation (http://openstax.org/l/16MassSpec) that explains mass spectrometry. Watch this video (http://openstax.org/l/16RSChemistry) from the Royal Society for Chemistry for a brief description of the rudiments of mass spectrometry.


138. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Symbolize the composition of molecules using molecular formulas and empirical formulas
- Represent the bonding arrangement of atoms within molecules using structural formulas
- Define the amount unit mole and the related quantity Avogadro's number
- Explain the relation between mass, moles, and numbers of atoms or molecules and perform calculations deriving these quantities from one another


139. Molecular and Empirical Formulas -
A molecular formula is a representation of a molecule that uses chemical symbols to indicate the types of atoms followed by subscripts to show the number of atoms of each type in the molecule. (A subscript is used only when more than one atom of a given type is present.) Molecular formulas are also used as abbreviations for the names of compounds.

The structural formula for a compound gives the same information as its molecular formula (the types and numbers of atoms in the molecule) but also shows how the atoms are connected in the molecule. The structural formula for methane contains symbols for one C atom and four H atoms, indicating the number of atoms in the molecule (Figure 2.16). The lines represent bonds that hold the atoms together. (A chemical bond is an attraction between atoms or ions that holds them together in a molecule or a crystal.) We will discuss chemical bonds and see how to predict the arrangement of atoms in a molecule later. For now, simply know that the lines are an indication of how the atoms are connected in a molecule. A ball-and-stick model shows the geometric arrangement of the atoms with atomic sizes not to scale, and a space-filling model shows the relative sizes of the atoms.

(a)

(b)

(c)

(d)

FIGURE 2.16 A methane molecule can be represented as (a) a molecular formula, (b) a structural formula, (c) a ball-and-stick model, and (d) a space-filling model. Carbon and hydrogen atoms are represented by black and white spheres, respectively.


139. Molecular and Empirical Formulas -
Although many elements consist of discrete, individual atoms, some exist as molecules made up of two or more atoms of the element chemically bonded together. For example, most samples of the elements hydrogen, oxygen, and nitrogen are composed of molecules that contain two atoms each (called diatomic molecules) and thus have the molecular formulas $\mathrm{H}_{2}, \mathrm{O}_{2}$, and $\mathrm{N}_{2}$, respectively. Other elements commonly found as diatomic molecules are fluorine $\left(\mathrm{F}_{2}\right)$, chlorine $\left(\mathrm{Cl}_{2}\right)$, bromine $\left(\mathrm{Br}_{2}\right)$, and iodine $\left(\mathrm{I}_{2}\right)$. The most common form of the element sulfur is composed of molecules that consist of eight atoms of sulfur; its molecular formula is $\mathrm{S}_{8}$ (Figure 2.17).

(a)

(b)

(c)

FIGURE 2.17 A molecule of sulfur is composed of eight sulfur atoms and is therefore written as $\mathrm{S}_{8}$. It can be represented as (a) a structural formula, (b) a ball-and-stick model, and (c) a space-filling model. Sulfur atoms are represented by yellow spheres.

It is important to note that a subscript following a symbol and a number in front of a symbol do not represent the same thing; for example, $\mathrm{H}_{2}$ and 2 H represent distinctly different species. $\mathrm{H}_{2}$ is a molecular formula; it represents a diatomic molecule of hydrogen, consisting of two atoms of the element that are chemically bonded together. The expression 2 H , on the other hand, indicates two separate hydrogen atoms that are not combined as a unit. The expression $2 \mathrm{H}_{2}$ represents two molecules of diatomic hydrogen (Figure 2.18).


139. Molecular and Empirical Formulas -
FIGURE 2.18 The symbols $\mathrm{H}, 2 \mathrm{H}, \mathrm{H}_{2}$, and $2 \mathrm{H}_{2}$ represent very different entities.
Compounds are formed when two or more elements chemically combine, resulting in the formation of bonds. For example, hydrogen and oxygen can react to form water, and sodium and chlorine can react to form table salt. We sometimes describe the composition of these compounds with an empirical formula, which indicates the types of atoms present and the simplest whole-number ratio of the number of atoms (or ions) in the compound. For example, titanium dioxide (used as pigment in white paint and in the thick, white, blocking type of sunscreen) has an empirical formula of $\mathrm{TiO}_{2}$. This identifies the elements titanium ( Ti ) and oxygen (O) as the constituents of titanium dioxide, and indicates the presence of twice as many atoms of the element oxygen as atoms of the element titanium (Figure 2.19).

FIGURE 2.19 (a) The white compound titanium dioxide provides effective protection from the sun. (b) A crystal of titanium dioxide, $\mathrm{TiO}_{2}$, contains titanium and oxygen in a ratio of 1 to 2 . The titanium atoms are gray and the oxygen atoms are red. (credit a: modification of work by "osseous"/Flickr)

As discussed previously, we can describe a compound with a molecular formula, in which the subscripts indicate the actual numbers of atoms of each element in a molecule of the compound. In many cases, the molecular formula of a substance is derived from experimental determination of both its empirical formula and its molecular mass (the sum of atomic masses for all atoms composing the molecule). For example, it can be determined experimentally that benzene contains two elements, carbon (C) and hydrogen (H), and that for every carbon atom in benzene, there is one hydrogen atom. Thus, the empirical formula is CH. An experimental determination of the molecular mass reveals that a molecule of benzene contains six carbon atoms and six hydrogen atoms, so the molecular formula for benzene is $\mathrm{C}_{6} \mathrm{H}_{6}$ (Figure 2.20).

(a)

(b)

(c)

(d)


139. Molecular and Empirical Formulas -
(a)

(b)

(c)

(d)

FIGURE 2.20 Benzene, $\mathrm{C}_{6} \mathrm{H}_{6}$, is produced during oil refining and has many industrial uses. A benzene molecule can be represented as (a) a structural formula, (b) a ball-and-stick model, and (c) a space-filling model. (d) Benzene is a clear liquid. (credit d: modification of work by Sahar Atwa)

If we know a compound's formula, we can easily determine the empirical formula. (This is somewhat of an academic exercise; the reverse chronology is generally followed in actual practice.) For example, the molecular formula for acetic acid, the component that gives vinegar its sharp taste, is $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}$. This formula indicates that a molecule of acetic acid (Figure 2.21) contains two carbon atoms, four hydrogen atoms, and two oxygen atoms. The ratio of atoms is $2: 4: 2$. Dividing by the lowest common denominator (2) gives the simplest, whole-number ratio of atoms, 1:2:1, so the empirical formula is $\mathrm{CH}_{2} \mathrm{O}$. Note that a molecular formula is always a wholenumber multiple of an empirical formula.

(a)

(b)

(c)

FIGURE 2.21 (a) Vinegar contains acetic acid, $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}$, which has an empirical formula of $\mathrm{CH}_{2} \mathrm{O}$. It can be represented as (b) a structural formula and (c) as a ball-and-stick model. (credit a: modification of work by "HomeSpot HQ"/Flickr)


141. Empirical and Molecular Formulas -
Molecules of glucose (blood sugar) contain 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. What are the molecular and empirical formulas of glucose?


142. Solution -
The molecular formula is $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$ because one molecule actually contains $6 \mathrm{C}, 12 \mathrm{H}$, and 6 O atoms. The simplest whole-number ratio of C to H to O atoms in glucose is $1: 2: 1$, so the empirical formula is $\mathrm{CH}_{2} \mathrm{O}$.


143. Check Your Learning -
A molecule of metaldehyde (a pesticide used for snails and slugs) contains 8 carbon atoms, 16 hydrogen atoms, and 4 oxygen atoms. What are the molecular and empirical formulas of metaldehyde?


144. Answer: -
Molecular formula, $\mathrm{C}_{8} \mathrm{H}_{16} \mathrm{O}_{4}$; empirical formula, $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}$


145. LINK TO LEARNING -
You can explore molecule building (http://openstax.org/l/16molbuilding) using an online simulation.


147. Lee Cronin -
What is it that chemists do? According to Lee Cronin (Figure 2.22), chemists make very complicated molecules by "chopping up" small molecules and "reverse engineering" them. He wonders if we could "make a really cool universal chemistry set" by what he calls "app-ing" chemistry. Could we "app" chemistry?

In a 2012 TED talk, Lee describes one fascinating possibility: combining a collection of chemical "inks" with a 3D printer capable of fabricating a reaction apparatus (tiny test tubes, beakers, and the like) to fashion a "universal toolkit of chemistry." This toolkit could be used to create custom-tailored drugs to fight a new superbug or to "print" medicine personally configured to your genetic makeup, environment, and health situation. Says Cronin, "What Apple did for music, I'd like to do for the discovery and distribution of prescription drugs." ${ }^{2}$ View his full talk (http://openstax.org/l/16LeeCronin) at the TED website.

FIGURE 2.22 Chemist Lee Cronin has been named one of the UK's 10 most inspirational scientists. The youngest chair at the University of Glasgow, Lee runs a large research group, collaborates with many scientists worldwide, has published over 250 papers in top scientific journals, and has given more than 150 invited talks. His research focuses on complex chemical systems and their potential to transform technology, but also branches into nanoscience, solar fuels, synthetic biology, and even artificial life and evolution. (credit: image courtesy of Lee Cronin)

It is important to be aware that it may be possible for the same atoms to be arranged in different ways: Compounds with the same molecular formula may have different atom-to-atom bonding and therefore different structures. For example, could there be another compound with the same formula as acetic acid, $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}$ ? And if so, what would be the structure of its molecules?


147. Lee Cronin -
If you predict that another compound with the formula $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}$ could exist, then you demonstrated good chemical insight and are correct. Two C atoms, four H atoms, and two O atoms can also be arranged to form a methyl formate, which is used in manufacturing, as an insecticide, and for quick-drying finishes. Methyl formate molecules have one of the oxygen atoms between the two carbon atoms, differing from the arrangement in acetic acid molecules. Acetic acid and methyl formate are examples of isomers-compounds with the same chemical formula but different molecular structures (Figure 2.23). Note that this small difference in the arrangement of the atoms has a major effect on their respective chemical properties. You would certainly not want to use a solution of methyl formate as a substitute for a solution of acetic acid (vinegar) when you make salad dressing.

Acetic acid $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}$
(a)

Methyl formate $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}$
(b)

FIGURE 2.23 Molecules of (a) acetic acid and methyl formate (b) are structural isomers; they have the same formula $\left(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\right)$ but different structures (and therefore different chemical properties).

Many types of isomers exist (Figure 2.24). Acetic acid and methyl formate are structural isomers, compounds in which the molecules differ in how the atoms are connected to each other. There are also various types of spatial isomers, in which the relative orientations of the atoms in space can be different. For example, the compound carvone (found in caraway seeds, spearmint, and mandarin orange peels) consists of two isomers that are mirror images of each other. $S$-(+)-carvone smells like caraway, and $R$-(-)-carvone smells like spearmint.

(+)-Carvone $\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{O}$


147. Lee Cronin -
(+)-Carvone $\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{O}$

FIGURE 2.24 Molecules of carvone are spatial isomers; they only differ in the relative orientations of the atoms in space. (credit bottom left: modification of work by "Miansari66"/Wikimedia Commons; credit bottom right: modification of work by Forest \& Kim Starr)


148. LINK TO LEARNING -
Select this link (http://openstax.org/l/16isomers) to view an explanation of isomers, spatial isomers, and why they have different smells (select the video titled "Mirror Molecule: Carvone").


149. The Mole -
The identity of a substance is defined not only by the types of atoms or ions it contains, but by the quantity of each type of atom or ion. For example, water, $\mathrm{H}_{2} \mathrm{O}$, and hydrogen peroxide, $\mathrm{H}_{2} \mathrm{O}_{2}$, are alike in that their respective molecules are composed of hydrogen and oxygen atoms. However, because a hydrogen peroxide molecule contains two oxygen atoms, as opposed to the water molecule, which has only one, the two
substances exhibit very different properties. Today, we possess sophisticated instruments that allow the direct measurement of these defining microscopic traits; however, the same traits were originally derived from the measurement of macroscopic properties (the masses and volumes of bulk quantities of matter) using relatively simple tools (balances and volumetric glassware). This experimental approach required the introduction of a new unit for amount of substances, the mole, which remains indispensable in modern chemical science.

The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a sample of matter. One Latin connotation for the word "mole" is "large mass" or "bulk," which is consistent with its use as the name for this unit. The mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental property, number of atoms, molecules, and so forth. A mole of substance is that amount in which there are $6.02214076 \times$ $10^{23}$ discrete entities (atoms or molecules). This large number is a fundamental constant known as Avogadro's number ( $\mathbf{N}_{\mathbf{A}}$ ) or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly reported with an explicit unit of "per mole," a conveniently rounded version being $6.022 \times 10^{23} / \mathrm{mol}$.


149. The Mole -
Consistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol) (see Figure 2.25).

FIGURE 2.25 Each sample contains $6.022 \times 10^{23}$ atoms -1.00 mol of atoms. From left to right (top row): 65.4 g zinc, 12.0 g carbon, 24.3 g magnesium, and 63.5 g copper. From left to right (bottom row): 32.1 g sulfur, 28.1 g silicon, 207 g lead, and 118.7 g tin. (credit: modification of work by Mark Ott)

The molar mass of any substance is numerically equivalent to its atomic or formula weight in amu. Per the amu definition, a single ${ }^{12} \mathrm{C}$ atom weighs 12 amu (its atomic mass is 12 amu ). A mole of ${ }^{12} \mathrm{C}$ atoms weighs 12 g (its molar mass is $12 \mathrm{~g} / \mathrm{mol}$ ). This relationship holds for all elements, since their atomic masses are measured relative to that of the amu-reference substance, ${ }^{12}$ C. Extending this principle, the molar mass of a compound in grams is likewise numerically equivalent to its formula mass in amu (Figure 2.26).


149. The Mole -
FIGURE 2.26 Each sample contains $6.02 \times 10^{23}$ molecules or formula units-1.00 mol of the compound or element. Clock-wise from the upper left: 130.2 g of $\mathrm{C}_{8} \mathrm{H}_{17} \mathrm{OH}$ (1-octanol, formula mass 130.2 amu ), 454.4 g of $\mathrm{HgI}_{2}$ (mercury(II) iodide, formula mass 454.4 amu ), 32.0 g of $\mathrm{CH}_{3} \mathrm{OH}$ (methanol, formula mass 32.0 amu ) and 256.5 g of $\mathrm{S}_{8}$ (sulfur, formula mass 256.5 amu ). (credit: Sahar Atwa)

| Element | Average Atomic Mass (amu) | Molar Mass (g/mol) | Atoms/Mole |
| :---: | :---: | :---: | :---: |
| C | 12.01 | 12.01 | $6.022 \times 10^{23}$ |
| H | 1.008 | 1.008 | $6.022 \times 10^{23}$ |
| O | 16.00 | 16.00 | $6.022 \times 10^{23}$ |
| Na | 22.99 | 22.99 | $6.022 \times 10^{23}$ |
| Cl | 35.45 | 35.45 | $6.022 \times 10^{23}$ |

While atomic mass and molar mass are numerically equivalent, keep in mind that they are vastly different in terms of scale, as represented by the vast difference in the magnitudes of their respective units (amu versus g). To appreciate the enormity of the mole, consider a small drop of water weighing about 0.03 g (see Figure 2.27). Although this represents just a tiny fraction of 1 mole of water $(\sim 18 \mathrm{~g})$, it contains more water molecules than can be clearly imagined. If the molecules were distributed equally among the roughly seven billion people on earth, each person would receive more than 100 billion molecules.


149. The Mole -
FIGURE 2.27 The number of molecules in a single droplet of water is roughly 100 billion times greater than the number of people on earth. (credit: "tanakawho"/Wikimedia commons)


150. LINK TO LEARNING -
The mole is used in chemistry to represent $6.022 \times 10^{23}$ of something, but it can be difficult to conceptualize such a large number. Watch this video (http://openstax.org/l/16molevideo) and then complete the "Think" questions that follow. Explore more about the mole by reviewing the information under "Dig Deeper."

The relationships between formula mass, the mole, and Avogadro's number can be applied to compute various quantities that describe the composition of substances and compounds. For example, if we know the mass and chemical composition of a substance, we can determine the number of moles and calculate number of atoms or molecules in the sample. Likewise, if we know the number of moles of a substance, we can derive the number of atoms or molecules and calculate the substance's mass.


152. Deriving Moles from Grams for an Element -
According to nutritional guidelines from the US Department of Agriculture, the estimated average requirement for dietary potassium is 4.7 g . What is the estimated average requirement of potassium in moles?


153. Solution -
The mass of $K$ is provided, and the corresponding amount of $K$ in moles is requested. Referring to the periodic table, the atomic mass of K is 39.10 amu , and so its molar mass is $39.10 \mathrm{~g} / \mathrm{mol}$. The given mass of $\mathrm{K}(4.7 \mathrm{~g})$ is a bit more than one-tenth the molar mass ( 39.10 g ), so a reasonable "ballpark" estimate of the number of moles would be slightly greater than 0.1 mol .

The molar amount of a substance may be calculated by dividing its mass (g) by its molar mass (g/mol):

The factor-label method supports this mathematical approach since the unit "g" cancels and the answer has units of "mol:"

$$
4.7 \mathrm{gK}\left(\frac{\mathrm{~mol} \mathrm{~K}}{39.10 \mathrm{~g} \mathrm{~K}}\right)=0.12 \mathrm{~mol} \mathrm{~K}
$$

The calculated magnitude $(0.12 \mathrm{~mol} \mathrm{~K})$ is consistent with our ballpark expectation, since it is a bit greater than 0.1 mol.


154. Check Your Learning -
Beryllium is a light metal used to fabricate transparent X-ray windows for medical imaging instruments. How many moles of Be are in a thin-foil window weighing 3.24 g ?


155. Answer: -
0.360 mol


157. Deriving Grams from Moles for an Element -
A liter of air contains $9.2 \times 10^{-4} \mathrm{~mol}$ argon. What is the mass of Ar in a liter of air?


158. Solution -
The molar amount of Ar is provided and must be used to derive the corresponding mass in grams. Since the amount of Ar is less than 1 mole, the mass will be less than the mass of 1 mole of Ar , approximately 40 g . The molar amount in question is approximately one-one thousandth $\left(\sim 10^{-3}\right)$ of a mole, and so the corresponding mass should be roughly one-one thousandth of the molar mass ( $\sim 0.04 \mathrm{~g}$ ):

In this case, logic dictates (and the factor-label method supports) multiplying the provided amount (mol) by the molar mass (g/mol):

$$
9.2 \times 10^{-4} \text { mol Ar }\left(\frac{39.95 \mathrm{~g} \mathrm{Ar}}{\text { mol Ar }}\right)=0.037 \mathrm{~g} \mathrm{Ar}
$$

The result is in agreement with our expectations, around 0.04 g Ar .
Check Your Learning
What is the mass of 2.561 mol of gold?


159. Answer: -
504.4 g


161. Deriving Number of Atoms from Mass for an Element -
Copper is commonly used to fabricate electrical wire (Figure 2.28). How many copper atoms are in 5.00 g of copper wire?

FIGURE 2.28 Copper wire is composed of many, many atoms of Cu. (credit: Emilian Robert Vicol)


162. Solution -
The number of Cu atoms in the wire may be conveniently derived from its mass by a two-step computation: first calculating the molar amount of Cu , and then using Avogadro's number ( $N_{A}$ ) to convert this molar amount to number of Cu atoms:

Considering that the provided sample mass ( 5.00 g ) is a little less than one-tenth the mass of 1 mole of $\mathrm{Cu}(\sim 64$ g ), a reasonable estimate for the number of atoms in the sample would be on the order of one-tenth $N_{A}$, or approximately $10^{22} \mathrm{Cu}$ atoms. Carrying out the two-step computation yields:

$$
5.00 \mathrm{~g} \mathrm{Cu}\left(\frac{\mathrm{molCu}}{63.55-\mathrm{g} \mathrm{Cu}}\right)\left(\frac{6.022 \times 10^{23} \mathrm{Cu} \text { atoms }}{\mathrm{molCu}}\right)=4.74 \times 10^{22} \mathrm{Cu} \text { atoms }
$$

The factor-label method yields the desired cancellation of units, and the computed result is on the order of $10^{22}$ as expected.


163. Check Your Learning -
A prospector panning for gold in a river collects 15.00 g of pure gold. How many Au atoms are in this quantity of gold?


164. Answer: -
$4.586 \times 10^{22} \mathrm{Au}$ atoms


166. Deriving Moles from Grams for a Compound -
Our bodies synthesize protein from amino acids. One of these amino acids is glycine, which has the molecular formula $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}_{2} \mathrm{~N}$. How many moles of glycine molecules are contained in 28.35 g of glycine?


167. Solution -
We can derive the number of moles of a compound from its mass following the same procedure we used for an element in Example 2.7:

The molar mass of glycine is required for this calculation, and it is computed in the same fashion as its molecular mass. One mole of glycine, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}_{2} \mathrm{~N}$, contains 2 moles of carbon, 5 moles of hydrogen, 2 moles of oxygen, and 1 mole of nitrogen:

| Element | Quantity (mol element/ mol compound) |  | Molar mass (g/mol element) |  | Subtotal (g/mol compound) |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| C | 2 | $\times$ | 12.01 | $=$ | 24.02 |  |
| H | 5 | $\times$ | 1.008 | $=$ | 5.040 |  |
| 0 | 2 | $\times$ | 16.00 | $=$ | 32.00 |  |
| N | 1 | $\times$ | 14.007 | $=$ | 14.007 |  |
| Molecular mass (g/mol compound) |  |  |  |  | 75.07 |  |

The provided mass of glycine ( $\sim 28 \mathrm{~g}$ ) is a bit more than one-third the molar mass ( $\sim 75 \mathrm{~g} / \mathrm{mol}$ ), so we would expect the computed result to be a bit greater than one-third of a mole ( $\sim 0.33 \mathrm{~mol}$ ). Dividing the compound's mass by its molar mass yields:

$$
28.35 \text { golycine- }\left(\frac{\text { mol glycine }}{75.07 \text {-g glyeine }}\right)=0.378 \text { mol glycine }
$$

This result is consistent with our rough estimate.


168. Check Your Learning -
How many moles of sucrose, $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$, are in a 25 -g sample of sucrose?


169. Answer: -
0.073 mol


171. Deriving Grams from Moles for a Compound -
Vitamin C is a covalent compound with the molecular formula $\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}$. The recommended daily dietary allowance of vitamin $C$ for children aged $4-8$ years is $1.42 \times 10^{-4} \mathrm{~mol}$. What is the mass of this allowance in grams?


172. Solution -
As for elements, the mass of a compound can be derived from its molar amount as shown:

The molar mass for this compound is computed to be $176.124 \mathrm{~g} / \mathrm{mol}$. The given number of moles is a very small fraction of a mole ( $\sim 10^{-4}$ or one-ten thousandth); therefore, we would expect the corresponding mass to be about one-ten thousandth of the molar mass ( $\sim 0.02 \mathrm{~g}$ ). Performing the calculation, we get:

$$
1.42 \times 10^{-4} \text { mol vitamin } \mathrm{C}\left(\frac{176.124 \mathrm{~g} \text { vitamin } \mathrm{C}}{\text { mot vitamin } \mathrm{C}}\right)=0.0250 \mathrm{~g} \text { vitamin } \mathrm{C}
$$

This is consistent with the anticipated result.
Check Your Learning
What is the mass of 0.443 mol of hydrazine, $\mathrm{N}_{2} \mathrm{H}_{4}$ ?


173. Answer: -
14.2 g


175. Deriving the Number of Atoms and Molecules from the Mass of a Compound -
A packet of an artificial sweetener contains 40.0 mg of saccharin $\left(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S}\right)$, which has the structural formula:

Given that saccharin has a molar mass of $183.18 \mathrm{~g} / \mathrm{mol}$, how many saccharin molecules are in a 40.0-mg ( $0.0400-\mathrm{g}$ ) sample of saccharin? How many carbon atoms are in the same sample?


176. Solution -
The number of molecules in a given mass of compound is computed by first deriving the number of moles, as demonstrated in Example 2.10, and then multiplying by Avogadro's number:

Using the provided mass and molar mass for saccharin yields:

$$
\begin{aligned}
& 0.0400-{ }_{8}^{8} \mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S}-\left(\frac{\mathrm{molC}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S}}{183.18-\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S}}\right)\left(\frac{6.022 \times 10^{23} \mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S} \text { molecules }}{1-\mathrm{mol}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S}}\right) \\
& =1.31 \times 10^{20} \mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S} \text { molecules }
\end{aligned}
$$

The compound's formula shows that each molecule contains seven carbon atoms, and so the number of C atoms in the provided sample is:

$$
1.31 \times 10^{20} \mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S} \text { molecules }\left(\frac{7 \mathrm{C} \text { atoms }}{1 \mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S} \text { molecule }}\right)=9.17 \times 10^{20} \mathrm{C} \text { atoms }
$$


177. Check Your Learning -
How many $\mathrm{C}_{4} \mathrm{H}_{10}$ molecules are contained in 9.213 g of this compound? How many hydrogen atoms?


178. Answer: -
$9.545 \times 10^{22}$ molecules $\mathrm{C}_{4} \mathrm{H}_{10} ; 9.545 \times 10^{23}$ atoms H


180. Counting Neurotransmitter Molecules in the Brain -
The brain is the control center of the central nervous system (Figure 2.29). It sends and receives signals to and from muscles and other internal organs to monitor and control their functions; it processes stimuli detected by sensory organs to guide interactions with the external world; and it houses the complex physiological processes that give rise to our intellect and emotions. The broad field of neuroscience spans all aspects of the structure and function of the central nervous system, including research on the anatomy and physiology of the brain. Great progress has been made in brain research over the past few decades, and the BRAIN Initiative, a federal initiative announced in 2013, aims to accelerate and capitalize on these advances through the concerted efforts of various industrial, academic, and government agencies (see BRAIN Initiative website (https://openstax.org/l/16BRAINInit) for details).

FIGURE 2.29 (a) A typical human brain weighs about 1.5 kg and occupies a volume of roughly 1.1 L. (b) Information is transmitted in brain tissue and throughout the central nervous system by specialized cells called neurons (micrograph shows cells at $1600 \times$ magnification).

Specialized cells called neurons transmit information between different parts of the central nervous system by way of electrical and chemical signals. Chemical signaling occurs at the interface between different neurons when one of the cells releases molecules (called neurotransmitters) that diffuse across the small gap between the cells (called the synapse) and bind to the surface of the other cell. These neurotransmitter molecules are stored in small intracellular structures called vesicles that fuse to the cell membrane and then break open to release their contents when the neuron is appropriately stimulated. This process is called exocytosis (see Figure 2.30). One neurotransmitter that has been very extensively studied is dopamine, $\mathrm{C}_{8} \mathrm{H}_{11} \mathrm{NO}_{2}$. Dopamine is involved in various neurological processes that impact a wide variety of human behaviors. Dysfunctions in the dopamine systems of the brain underlie serious neurological diseases such as Parkinson's and schizophrenia.


180. Counting Neurotransmitter Molecules in the Brain -
FIGURE 2.30 (a) Chemical signals are transmitted from neurons to other cells by the release of neurotransmitter molecules into the small gaps (synapses) between the cells. (b) Dopamine, $\mathrm{C}_{8} \mathrm{H}_{11} \mathrm{NO}_{2}$, is a neurotransmitter involved in a number of neurological processes.

One important aspect of the complex processes related to dopamine signaling is the number of neurotransmitter molecules released during exocytosis. Since this number is a central factor in determining neurological response (and subsequent human thought and action), it is important to know how this number changes with certain controlled stimulations, such as the administration of drugs. It is also important to understand the mechanism responsible for any changes in the number of neurotransmitter molecules released-for example, some dysfunction in exocytosis, a change in the number of vesicles in the neuron, or a change in the number of neurotransmitter molecules in each vesicle.

Significant progress has been made recently in directly measuring the number of dopamine molecules stored in individual vesicles and the amount actually released when the vesicle undergoes exocytosis. Using miniaturized probes that can selectively detect dopamine molecules in very small amounts, scientists have determined that the vesicles of a certain type of mouse brain neuron contain an average of 30,000 dopamine molecules per vesicle (about $5 \times 10^{-20} \mathrm{~mol}$ or 50 zmol ). Analysis of these neurons from mice subjected to various drug therapies shows significant changes in the average number of dopamine molecules contained in individual vesicles, increasing or decreasing by up to three-fold, depending on the specific drug used. These studies also indicate that not all of the dopamine in a given vesicle is released during exocytosis, suggesting that it may be possible to regulate the fraction released using pharmaceutical therapies. ${ }^{-3}$

[^1]


181. Key Terms -
alpha particle ( $\alpha$ particle) positively charged particle consisting of two protons and two neutrons
anion negatively charged atom or molecule (contains more electrons than protons)
atomic mass average mass of atoms of an element, expressed in amu
atomic mass unit (amu) (also, unified atomic mass unit, $u$, or Dalton, Da) unit of mass equal to $\frac{1}{12}$ of the mass of a ${ }^{12} \mathrm{C}$ atom
atomic number (Z) number of protons in the nucleus of an atom
cation positively charged atom or molecule (contains fewer electrons than protons)
chemical symbol one-, two-, or three-letter abbreviation used to represent an element or its atoms
Dalton (Da) alternative unit equivalent to the atomic mass unit
Dalton's atomic theory set of postulates that established the fundamental properties of atoms
electron negatively charged, subatomic particle of relatively low mass located outside the nucleus
empirical formula formula showing the composition of a compound given as the simplest whole-number ratio of atoms
fundamental unit of charge (also called the elementary charge) equals the magnitude of the charge of an electron (e) with $\mathrm{e}=1.602 \times 10^{-19} \mathrm{C}$
ion electrically charged atom or molecule (contains unequal numbers of protons and electrons)
isomers compounds with the same chemical formula but different structures
isotopes atoms that contain the same number of protons but different numbers of neutrons
law of constant composition (also, law of definite proportions) all samples of a pure compound contain the same elements in the same proportions by mass
law of definite proportions (also, law of constant composition) all samples of a pure compound contain the same elements in the same proportions by mass
law of multiple proportions when two elements react to form more than one compound, a fixed mass of one element will react with masses of the other element in a ratio of small whole numbers
mass number (A) sum of the numbers of neutrons and protons in the nucleus of an atom
molecular formula formula indicating the composition of a molecule of a compound and giving the actual number of atoms of each element in a molecule of the compound.
neutron uncharged, subatomic particle located in the nucleus
nucleus massive, positively charged center of an atom made up of protons and neutrons


181. Key Terms -
neutron uncharged, subatomic particle located in the nucleus
nucleus massive, positively charged center of an atom made up of protons and neutrons
proton positively charged, subatomic particle located in the nucleus
spatial isomers compounds in which the relative orientations of the atoms in space differ
structural formula shows the atoms in a molecule and how they are connected
structural isomer one of two substances that have the same molecular formula but different physical and chemical properties because their atoms are bonded differently
unified atomic mass unit (u) alternative unit equivalent to the atomic mass unit


182. Key Equations -
average mass $=\sum_{i}(\text { fractional abundance } \times \text { isotopic mass })_{i}$


183. Summary - 183.1. Early Ideas in Atomic Theory
The ancient Greeks proposed that matter consists of extremely small particles called atoms. Dalton postulated that each element has a characteristic type of atom that differs in properties from atoms of all other elements, and that atoms of different elements can combine in fixed, small, wholenumber ratios to form compounds. Samples of a particular compound all have the same elemental
proportions by mass. When two elements form different compounds, a given mass of one element will combine with masses of the other element in a small, whole-number ratio. During any chemical change, atoms are neither created nor destroyed.


183. Summary - 183.2. Evolution of Atomic Theory
Although no one has actually seen the inside of an atom, experiments have demonstrated much about
atomic structure. Thomson's cathode ray tube showed that atoms contain small, negatively charged particles called electrons. Millikan discovered that there is a fundamental electric charge-the charge of an electron. Rutherford's gold foil experiment showed that atoms have a small, dense, positively charged nucleus; the positively charged particles within the nucleus are called protons. Chadwick discovered that the nucleus also contains neutral particles called neutrons. Soddy demonstrated that atoms of the same element can differ in mass; these are called isotopes.


183. Summary - 183.3. Atomic Structure and Symbolism
An atom consists of a small, positively charged nucleus surrounded by electrons. The nucleus contains protons and neutrons; its diameter is about 100,000 times smaller than that of the atom. The mass of one atom is usually expressed in atomic mass units (amu), which is referred to as the atomic mass. An amu is defined as exactly $\frac{1}{12}$ of the mass of a carbon-12 atom and is equal to $1.6605 \times 10^{-24} \mathrm{~g}$.

Protons are relatively heavy particles with a charge of $1+$ and a mass of 1.0073 amu . Neutrons are relatively heavy particles with no charge and a mass of 1.0087 amu . Electrons are light particles with a charge of 1 - and a mass of 0.00055 amu . The number of protons in the nucleus is called the atomic number ( Z ) and is the property that defines an atom's elemental identity. The sum of the numbers of protons and neutrons in the nucleus is called the mass number and, expressed in amu, is
approximately equal to the mass of the atom. An atom is neutral when it contains equal numbers of electrons and protons.

Isotopes of an element are atoms with the same atomic number but different mass numbers; isotopes of an element, therefore, differ from each other only in the number of neutrons within the nucleus. When a naturally occurring element is composed of several isotopes, the atomic mass of the element represents the average of the masses of the isotopes involved. A chemical symbol identifies the atoms in a substance using symbols, which are one-, two-, or three-letter abbreviations for the atoms.


183. Summary - 183.4. Chemical Formulas
A molecular formula uses chemical symbols and subscripts to indicate the exact numbers of different atoms in a molecule or compound. An empirical formula gives the simplest, whole-number ratio of atoms in a compound. A structural formula indicates the bonding arrangement of the atoms in the molecule. Ball-and-stick and space-filling models show the geometric arrangement of atoms in a molecule. Isomers are compounds with the same molecular formula but different arrangements of atoms. A convenient amount unit for expressing very large numbers of atoms or molecules is the mole. Experimental measurements have determined the number of entities composing 1 mole of substance to be $6.022 \times 10^{23}$, a quantity called Avogadro's number. The mass in grams of 1 mole of substance is its molar mass.


184. Exercises - 184.1. Early Ideas in Atomic Theory
1. In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres of different elements touch, they are part of a single unit of a compound. The following chemical change represented by these spheres may violate one of the ideas of Dalton's atomic theory. Which one?

2. Which postulate of Dalton's theory is consistent with the following observation concerning the weights of reactants and products? When 100 grams of solid calcium carbonate is heated, 44 grams of carbon dioxide and 56 grams of calcium oxide are produced.
3. Identify the postulate of Dalton's theory that is violated by the following observations: $59.95 \%$ of one sample of titanium dioxide is titanium; $60.10 \%$ of a different sample of titanium dioxide is titanium.
4. Samples of compound $X, Y$, and $Z$ are analyzed, with results shown here.

| Compound | Description | Mass of Carbon | Mass of Hydrogen |
| :---: | :---: | :---: | :---: |
| X | clear, colorless, liquid
 with strong odor | 1.776 g | 0.148 g |
| Y | clear, colorless, liquid
 with strong odor | 1.974 g | 0.329 g |
| Z | clear, colorless, liquid
 with strong odor | 7.812 g | 0.651 g |

Do these data provide example(s) of the law of definite proportions, the law of multiple proportions, neither, or both? What do these data tell you about compounds X, Y, and Z?


184. Exercises - 184.2. Evolution of Atomic Theory
5. The existence of isotopes violates one of the original ideas of Dalton's atomic theory. Which one?
6. How are electrons and protons similar? How are they different?
7. How are protons and neutrons similar? How are they different?
8. Predict and test the behavior of $\alpha$ particles fired at a "plum pudding" model atom.
(a) Predict the paths taken by $\alpha$ particles that are fired at atoms with a Thomson's plum pudding model structure. Explain why you expect the $\alpha$ particles to take these paths.
(b) If $\alpha$ particles of higher energy than those in (a) are fired at plum pudding atoms, predict how their paths will differ from the lower-energy $\alpha$ particle paths. Explain your reasoning.
(c) Now test your predictions from (a) and (b). Open the Rutherford Scattering simulation
(http://openstax.org/l/16PhetScatter) and select the "Plum Pudding Atom" tab. Set "Alpha Particles
Energy" to "min," and select "show traces." Click on the gun to start firing $\alpha$ particles. Does this match your prediction from (a)? If not, explain why the actual path would be that shown in the simulation. Hit the pause button, or "Reset All." Set "Alpha Particles Energy" to "max," and start firing $\alpha$ particles. Does this match your prediction from (b)? If not, explain the effect of increased energy on the actual paths as shown in the simulation.
9. Predict and test the behavior of $\alpha$ particles fired at a Rutherford atom model.
(a) Predict the paths taken by $\alpha$ particles that are fired at atoms with a Rutherford atom model structure. Explain why you expect the $\alpha$ particles to take these paths.
(b) If $\alpha$ particles of higher energy than those in (a) are fired at Rutherford atoms, predict how their paths will differ from the lower-energy $\alpha$ particle paths. Explain your reasoning.
(c) Predict how the paths taken by the $\alpha$ particles will differ if they are fired at Rutherford atoms of elements other than gold. What factor do you expect to cause this difference in paths, and why?


184. Exercises - 184.2. Evolution of Atomic Theory
(c) Predict how the paths taken by the $\alpha$ particles will differ if they are fired at Rutherford atoms of elements other than gold. What factor do you expect to cause this difference in paths, and why?
(d) Now test your predictions from (a), (b), and (c). Open the Rutherford Scattering simulation (http://openstax.org/l/16PhetScatter) and select the "Rutherford Atom" tab. Due to the scale of the simulation, it is best to start with a small nucleus, so select " 20 " for both protons and neutrons, "min" for energy, show traces, and then start firing $\alpha$ particles. Does this match your prediction from (a)? If not, explain why the actual path would be that shown in the simulation. Pause or reset, set energy to "max," and start firing $\alpha$ particles. Does this match your prediction from (b)? If not, explain the effect of increased energy on the actual path as shown in the simulation. Pause or reset, select " 40 " for both protons and neutrons, "min" for energy, show traces, and fire away. Does this match your prediction from (c)? If not, explain why the actual path would be that shown in the simulation. Repeat this with larger numbers of protons and neutrons. What generalization can you make regarding the type of atom and effect on the path of $\alpha$ particles? Be clear and specific.


184. Exercises - 184.3. Atomic Structure and Symbolism
10. In what way are isotopes of a given element always different? In what way(s) are they always the same?
11. Write the symbol for each of the following ions:
(a) the ion with a $1+$ charge, atomic number 55 , and mass number 133
(b) the ion with 54 electrons, 53 protons, and 74 neutrons
(c) the ion with atomic number 15, mass number 31, and a 3 - charge
(d) the ion with 24 electrons, 30 neutrons, and a $3+$ charge
12. Write the symbol for each of the following ions:
(a) the ion with a $3+$ charge, 28 electrons, and a mass number of 71
(b) the ion with 36 electrons, 35 protons, and 45 neutrons
(c) the ion with 86 electrons, 142 neutrons, and a 4+ charge
(d) the ion with a $2+$ charge, atomic number 38 , and mass number 87
13. Open the Build an Atom simulation (http://openstax.org/l/16PhetAtomBld) and click on the Atom icon.
(a) Pick any one of the first 10 elements that you would like to build and state its symbol.
(b) Drag protons, neutrons, and electrons onto the atom template to make an atom of your element. State the numbers of protons, neutrons, and electrons in your atom, as well as the net charge and mass number.
(c) Click on "Net Charge" and "Mass Number," check your answers to (b), and correct, if needed.
(d) Predict whether your atom will be stable or unstable. State your reasoning.
(e) Check the "Stable/Unstable" box. Was your answer to (d) correct? If not, first predict what you can do to make a stable atom of your element, and then do it and see if it works. Explain your reasoning.
14. Open the Build an Atom simulation (http://openstax.org/l/16PhetAtomBld)
(a) Drag protons, neutrons, and electrons onto the atom template to make a neutral atom of Oxygen-16 and give the isotope symbol for this atom.


184. Exercises - 184.3. Atomic Structure and Symbolism
(a) Drag protons, neutrons, and electrons onto the atom template to make a neutral atom of Oxygen-16 and give the isotope symbol for this atom.
(b) Now add two more electrons to make an ion and give the symbol for the ion you have created.
15. Open the Build an Atom simulation (http://openstax.org/l/16PhetAtomBld)
(a) Drag protons, neutrons, and electrons onto the atom template to make a neutral atom of Lithium-6 and give the isotope symbol for this atom.
(b) Now remove one electron to make an ion and give the symbol for the ion you have created.
16. Determine the number of protons, neutrons, and electrons in the following isotopes that are used in medical diagnoses:
(a) atomic number 9, mass number 18, charge of 1-
(b) atomic number 43 , mass number 99 , charge of $7+$
(c) atomic number 53, atomic mass number 131, charge of $1-$
(d) atomic number 81, atomic mass number 201, charge of $1+$
(e) Name the elements in parts (a), (b), (c), and (d).
17. The following are properties of isotopes of two elements that are essential in our diet. Determine the number of protons, neutrons and electrons in each and name them.
(a) atomic number 26 , mass number 58 , charge of $2+$
(b) atomic number 53 , mass number 127, charge of 1-
18. Give the number of protons, electrons, and neutrons in neutral atoms of each of the following isotopes:
(a) ${ }_{5}^{10} \mathrm{~B}$
(b) ${ }_{80}^{199} \mathrm{Hg}$
(c) ${ }_{29}^{63} \mathrm{Cu}$
(d) ${ }_{6}^{13} \mathrm{C}$
(e) ${ }_{34}^{77} \mathrm{Se}$
19. Give the number of protons, electrons, and neutrons in neutral atoms of each of the following isotopes:
(a) ${ }_{3}^{7} \mathrm{Li}$


184. Exercises - 184.3. Atomic Structure and Symbolism
19. Give the number of protons, electrons, and neutrons in neutral atoms of each of the following isotopes:
(a) ${ }_{3}^{7} \mathrm{Li}$
(b) ${ }_{52}^{125} \mathrm{Te}$
(c) ${ }_{47}^{109} \mathrm{Ag}$
(d) ${ }_{7}^{15} \mathrm{~N}$
(e) ${ }_{15}^{31} \mathrm{P}$
20. Click on the site (http://openstax.org/l/16PhetAtomMass) and select the "Mix Isotopes" tab, hide the "Percent Composition" and "Average Atomic Mass" boxes, and then select the element boron.
(a) Write the symbols of the isotopes of boron that are shown as naturally occurring in significant amounts.
(b) Predict the relative amounts (percentages) of these boron isotopes found in nature. Explain the reasoning behind your choice.
(c) Add isotopes to the black box to make a mixture that matches your prediction in (b). You may drag isotopes from their bins or click on "More" and then move the sliders to the appropriate amounts.
(d) Reveal the "Percent Composition" and "Average Atomic Mass" boxes. How well does your mixture match with your prediction? If necessary, adjust the isotope amounts to match your prediction.
(e) Select "Nature's" mix of isotopes and compare it to your prediction. How well does your prediction compare with the naturally occurring mixture? Explain. If necessary, adjust your amounts to make them match "Nature's" amounts as closely as possible.
21. Repeat Exercise 2.20 using an element that has three naturally occurring isotopes.
22. An element has the following natural abundances and isotopic masses: $90.92 \%$ abundance with 19.99 amu, $0.26 \%$ abundance with 20.99 amu , and $8.82 \%$ abundance with 21.99 amu . Calculate the average atomic mass of this element.