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7,901 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CuSO4 from mass 59.54 g | The molar mass of CuSO4 is 159.6 g/mol. A sample of mass 59.54 g therefore contains n = m / M = 59.54 / 159.6 = 0.3731 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,902 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CaCO3 from mass 26.26 g | The molar mass of CaCO3 is 100.1 g/mol. A sample of mass 26.26 g therefore contains n = m / M = 26.26 / 100.1 = 0.2624 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,903 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CO2 from mass 55.38 g | The molar mass of CO2 is 44.01 g/mol. A sample of mass 55.38 g therefore contains n = m / M = 55.38 / 44.01 = 1.258 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,904 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CH4 from mass 98.43 g | The molar mass of CH4 is 16.04 g/mol. A sample of mass 98.43 g therefore contains n = m / M = 98.43 / 16.04 = 6.135 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,905 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of C6H12O6 from mass 1.555 g | The molar mass of C6H12O6 is 180.2 g/mol. A sample of mass 1.555 g therefore contains n = m / M = 1.555 / 180.2 = 0.008634 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,906 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CaCO3 from mass 78.85 g | The molar mass of CaCO3 is 100.1 g/mol. A sample of mass 78.85 g therefore contains n = m / M = 78.85 / 100.1 = 0.7878 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,907 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of H2O from mass 4.82 g | The molar mass of H2O is 18.02 g/mol. A sample of mass 4.82 g therefore contains n = m / M = 4.82 / 18.02 = 0.2675 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,908 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of C6H12O6 from mass 4.688 g | The molar mass of C6H12O6 is 180.2 g/mol. A sample of mass 4.688 g therefore contains n = m / M = 4.688 / 180.2 = 0.02602 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,909 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of NH3 from mass 72.05 g | The molar mass of NH3 is 17.03 g/mol. A sample of mass 72.05 g therefore contains n = m / M = 72.05 / 17.03 = 4.23 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,910 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CH4 from mass 10.27 g | The molar mass of CH4 is 16.04 g/mol. A sample of mass 10.27 g therefore contains n = m / M = 10.27 / 16.04 = 0.6403 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,911 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CuSO4 from mass 35.65 g | The molar mass of CuSO4 is 159.6 g/mol. A sample of mass 35.65 g therefore contains n = m / M = 35.65 / 159.6 = 0.2234 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,912 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of C6H12O6 from mass 1.827 g | The molar mass of C6H12O6 is 180.2 g/mol. A sample of mass 1.827 g therefore contains n = m / M = 1.827 / 180.2 = 0.01014 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,913 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of NH3 from mass 38.8 g | The molar mass of NH3 is 17.03 g/mol. A sample of mass 38.8 g therefore contains n = m / M = 38.8 / 17.03 = 2.278 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,914 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CO2 from mass 55.84 g | The molar mass of CO2 is 44.01 g/mol. A sample of mass 55.84 g therefore contains n = m / M = 55.84 / 44.01 = 1.269 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,915 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CuSO4 from mass 19.24 g | The molar mass of CuSO4 is 159.6 g/mol. A sample of mass 19.24 g therefore contains n = m / M = 19.24 / 159.6 = 0.1205 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,916 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CaCO3 from mass 97.88 g | The molar mass of CaCO3 is 100.1 g/mol. A sample of mass 97.88 g therefore contains n = m / M = 97.88 / 100.1 = 0.9779 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,917 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of H2SO4 from mass 43.26 g | The molar mass of H2SO4 is 98.07 g/mol. A sample of mass 43.26 g therefore contains n = m / M = 43.26 / 98.07 = 0.4411 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,918 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CuSO4 from mass 5.273 g | The molar mass of CuSO4 is 159.6 g/mol. A sample of mass 5.273 g therefore contains n = m / M = 5.273 / 159.6 = 0.03304 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,919 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CaCO3 from mass 91.41 g | The molar mass of CaCO3 is 100.1 g/mol. A sample of mass 91.41 g therefore contains n = m / M = 91.41 / 100.1 = 0.9133 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,920 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of H2O from mass 65.4 g | The molar mass of H2O is 18.02 g/mol. A sample of mass 65.4 g therefore contains n = m / M = 65.4 / 18.02 = 3.63 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,921 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of Fe2O3 from mass 47.26 g | The molar mass of Fe2O3 is 159.7 g/mol. A sample of mass 47.26 g therefore contains n = m / M = 47.26 / 159.7 = 0.296 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,922 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CH4 from mass 87.83 g | The molar mass of CH4 is 16.04 g/mol. A sample of mass 87.83 g therefore contains n = m / M = 87.83 / 16.04 = 5.475 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,923 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of H2O from mass 3.22 g | The molar mass of H2O is 18.02 g/mol. A sample of mass 3.22 g therefore contains n = m / M = 3.22 / 18.02 = 0.1787 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,924 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of C6H12O6 from mass 42.83 g | The molar mass of C6H12O6 is 180.2 g/mol. A sample of mass 42.83 g therefore contains n = m / M = 42.83 / 180.2 = 0.2377 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,925 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CO2 from mass 4.856 g | The molar mass of CO2 is 44.01 g/mol. A sample of mass 4.856 g therefore contains n = m / M = 4.856 / 44.01 = 0.1103 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,926 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of H2SO4 from mass 91.9 g | The molar mass of H2SO4 is 98.07 g/mol. A sample of mass 91.9 g therefore contains n = m / M = 91.9 / 98.07 = 0.9371 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,927 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of H2SO4 from mass 94.97 g | The molar mass of H2SO4 is 98.07 g/mol. A sample of mass 94.97 g therefore contains n = m / M = 94.97 / 98.07 = 0.9684 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,928 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of H2SO4 from mass 13.24 g | The molar mass of H2SO4 is 98.07 g/mol. A sample of mass 13.24 g therefore contains n = m / M = 13.24 / 98.07 = 0.135 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,929 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CuSO4 from mass 93.82 g | The molar mass of CuSO4 is 159.6 g/mol. A sample of mass 93.82 g therefore contains n = m / M = 93.82 / 159.6 = 0.5878 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,930 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CH4 from mass 48.45 g | The molar mass of CH4 is 16.04 g/mol. A sample of mass 48.45 g therefore contains n = m / M = 48.45 / 16.04 = 3.02 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,931 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of NH3 from mass 95.47 g | The molar mass of NH3 is 17.03 g/mol. A sample of mass 95.47 g therefore contains n = m / M = 95.47 / 17.03 = 5.606 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,932 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of Fe2O3 from mass 90.25 g | The molar mass of Fe2O3 is 159.7 g/mol. A sample of mass 90.25 g therefore contains n = m / M = 90.25 / 159.7 = 0.5652 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,933 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CH4 from mass 81.87 g | The molar mass of CH4 is 16.04 g/mol. A sample of mass 81.87 g therefore contains n = m / M = 81.87 / 16.04 = 5.103 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,934 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of NH3 from mass 16.57 g | The molar mass of NH3 is 17.03 g/mol. A sample of mass 16.57 g therefore contains n = m / M = 16.57 / 17.03 = 0.9728 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,935 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of H2O from mass 37.85 g | The molar mass of H2O is 18.02 g/mol. A sample of mass 37.85 g therefore contains n = m / M = 37.85 / 18.02 = 2.101 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,936 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of NaCl from mass 11.77 g | The molar mass of NaCl is 58.44 g/mol. A sample of mass 11.77 g therefore contains n = m / M = 11.77 / 58.44 = 0.2014 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,937 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of NH3 from mass 61.17 g | The molar mass of NH3 is 17.03 g/mol. A sample of mass 61.17 g therefore contains n = m / M = 61.17 / 17.03 = 3.592 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,938 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CH4 from mass 87.24 g | The molar mass of CH4 is 16.04 g/mol. A sample of mass 87.24 g therefore contains n = m / M = 87.24 / 16.04 = 5.438 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,939 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of NaCl from mass 31.61 g | The molar mass of NaCl is 58.44 g/mol. A sample of mass 31.61 g therefore contains n = m / M = 31.61 / 58.44 = 0.5409 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,940 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CH4 from mass 76.84 g | The molar mass of CH4 is 16.04 g/mol. A sample of mass 76.84 g therefore contains n = m / M = 76.84 / 16.04 = 4.789 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,941 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of NaCl from mass 93.9 g | The molar mass of NaCl is 58.44 g/mol. A sample of mass 93.9 g therefore contains n = m / M = 93.9 / 58.44 = 1.607 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,942 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of NaCl from mass 12.58 g | The molar mass of NaCl is 58.44 g/mol. A sample of mass 12.58 g therefore contains n = m / M = 12.58 / 58.44 = 0.2152 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,943 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of H2O from mass 33.25 g | The molar mass of H2O is 18.02 g/mol. A sample of mass 33.25 g therefore contains n = m / M = 33.25 / 18.02 = 1.845 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,944 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of C6H12O6 from mass 93.19 g | The molar mass of C6H12O6 is 180.2 g/mol. A sample of mass 93.19 g therefore contains n = m / M = 93.19 / 180.2 = 0.5173 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,945 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of H2SO4 from mass 44.01 g | The molar mass of H2SO4 is 98.07 g/mol. A sample of mass 44.01 g therefore contains n = m / M = 44.01 / 98.07 = 0.4488 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,946 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of H2O from mass 25.94 g | The molar mass of H2O is 18.02 g/mol. A sample of mass 25.94 g therefore contains n = m / M = 25.94 / 18.02 = 1.44 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,947 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of H2SO4 from mass 48.09 g | The molar mass of H2SO4 is 98.07 g/mol. A sample of mass 48.09 g therefore contains n = m / M = 48.09 / 98.07 = 0.4904 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,948 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CH4 from mass 30.43 g | The molar mass of CH4 is 16.04 g/mol. A sample of mass 30.43 g therefore contains n = m / M = 30.43 / 16.04 = 1.897 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,949 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of H2SO4 from mass 68.5 g | The molar mass of H2SO4 is 98.07 g/mol. A sample of mass 68.5 g therefore contains n = m / M = 68.5 / 98.07 = 0.6985 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,950 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of NH3 from mass 37.3 g | The molar mass of NH3 is 17.03 g/mol. A sample of mass 37.3 g therefore contains n = m / M = 37.3 / 17.03 = 2.19 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,951 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of C6H12O6 from mass 60.12 g | The molar mass of C6H12O6 is 180.2 g/mol. A sample of mass 60.12 g therefore contains n = m / M = 60.12 / 180.2 = 0.3337 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,952 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of NH3 from mass 17.16 g | The molar mass of NH3 is 17.03 g/mol. A sample of mass 17.16 g therefore contains n = m / M = 17.16 / 17.03 = 1.007 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,953 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of C6H12O6 from mass 53.48 g | The molar mass of C6H12O6 is 180.2 g/mol. A sample of mass 53.48 g therefore contains n = m / M = 53.48 / 180.2 = 0.2969 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,954 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of NH3 from mass 0.7101 g | The molar mass of NH3 is 17.03 g/mol. A sample of mass 0.7101 g therefore contains n = m / M = 0.7101 / 17.03 = 0.04169 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,955 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CO2 from mass 96.95 g | The molar mass of CO2 is 44.01 g/mol. A sample of mass 96.95 g therefore contains n = m / M = 96.95 / 44.01 = 2.203 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,956 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of H2SO4 from mass 33.37 g | The molar mass of H2SO4 is 98.07 g/mol. A sample of mass 33.37 g therefore contains n = m / M = 33.37 / 98.07 = 0.3402 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,957 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of NH3 from mass 21.89 g | The molar mass of NH3 is 17.03 g/mol. A sample of mass 21.89 g therefore contains n = m / M = 21.89 / 17.03 = 1.286 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,958 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of NaCl from mass 70.18 g | The molar mass of NaCl is 58.44 g/mol. A sample of mass 70.18 g therefore contains n = m / M = 70.18 / 58.44 = 1.201 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,959 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CuSO4 from mass 5.73 g | The molar mass of CuSO4 is 159.6 g/mol. A sample of mass 5.73 g therefore contains n = m / M = 5.73 / 159.6 = 0.0359 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,960 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of C6H12O6 from mass 39.5 g | The molar mass of C6H12O6 is 180.2 g/mol. A sample of mass 39.5 g therefore contains n = m / M = 39.5 / 180.2 = 0.2193 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,961 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of NaCl from mass 94.66 g | The molar mass of NaCl is 58.44 g/mol. A sample of mass 94.66 g therefore contains n = m / M = 94.66 / 58.44 = 1.62 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,962 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CO2 from mass 92.94 g | The molar mass of CO2 is 44.01 g/mol. A sample of mass 92.94 g therefore contains n = m / M = 92.94 / 44.01 = 2.112 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,963 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CaCO3 from mass 23.76 g | The molar mass of CaCO3 is 100.1 g/mol. A sample of mass 23.76 g therefore contains n = m / M = 23.76 / 100.1 = 0.2374 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,964 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of NaCl from mass 63.37 g | The molar mass of NaCl is 58.44 g/mol. A sample of mass 63.37 g therefore contains n = m / M = 63.37 / 58.44 = 1.084 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,965 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CaCO3 from mass 49.52 g | The molar mass of CaCO3 is 100.1 g/mol. A sample of mass 49.52 g therefore contains n = m / M = 49.52 / 100.1 = 0.4947 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,966 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of CH4 from mass 77.26 g | The molar mass of CH4 is 16.04 g/mol. A sample of mass 77.26 g therefore contains n = m / M = 77.26 / 16.04 = 4.816 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,967 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of Fe2O3 from mass 64.29 g | The molar mass of Fe2O3 is 159.7 g/mol. A sample of mass 64.29 g therefore contains n = m / M = 64.29 / 159.7 = 0.4026 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,968 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of C6H12O6 from mass 61.79 g | The molar mass of C6H12O6 is 180.2 g/mol. A sample of mass 61.79 g therefore contains n = m / M = 61.79 / 180.2 = 0.343 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,969 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of Fe2O3 from mass 32.17 g | The molar mass of Fe2O3 is 159.7 g/mol. A sample of mass 32.17 g therefore contains n = m / M = 32.17 / 159.7 = 0.2014 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,970 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of NH3 from mass 55.46 g | The molar mass of NH3 is 17.03 g/mol. A sample of mass 55.46 g therefore contains n = m / M = 55.46 / 17.03 = 3.256 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,971 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of NaCl from mass 18.18 g | The molar mass of NaCl is 58.44 g/mol. A sample of mass 18.18 g therefore contains n = m / M = 18.18 / 58.44 = 0.311 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,972 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of H2SO4 from mass 17.11 g | The molar mass of H2SO4 is 98.07 g/mol. A sample of mass 17.11 g therefore contains n = m / M = 17.11 / 98.07 = 0.1745 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,973 | chemistry | stoichiometry | mole_concept | 3 | worked_example | Moles of NH3 from mass 14.08 g | The molar mass of NH3 is 17.03 g/mol. A sample of mass 14.08 g therefore contains n = m / M = 14.08 / 17.03 = 0.8267 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant. | n = m / M | atomic masses; chemical formulas | Convert between mass and moles for a pure compound. |
7,974 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=4.297 mol, V=6.225 L, T=240.6 K | For an ideal gas, P V = n R T. With n = 4.297 mol, V = 6.225 L, T = 240.6 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 13.63 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,975 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=2.517 mol, V=30.94 L, T=395.7 K | For an ideal gas, P V = n R T. With n = 2.517 mol, V = 30.94 L, T = 395.7 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 2.641 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,976 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=2.249 mol, V=45.85 L, T=221.9 K | For an ideal gas, P V = n R T. With n = 2.249 mol, V = 45.85 L, T = 221.9 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 0.8929 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,977 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=2.65 mol, V=18.48 L, T=454.7 K | For an ideal gas, P V = n R T. With n = 2.65 mol, V = 18.48 L, T = 454.7 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 5.35 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,978 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=3 mol, V=22.98 L, T=303.3 K | For an ideal gas, P V = n R T. With n = 3 mol, V = 22.98 L, T = 303.3 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 3.25 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,979 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=1.354 mol, V=18.62 L, T=292.1 K | For an ideal gas, P V = n R T. With n = 1.354 mol, V = 18.62 L, T = 292.1 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 1.742 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,980 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=1.082 mol, V=15.23 L, T=509 K | For an ideal gas, P V = n R T. With n = 1.082 mol, V = 15.23 L, T = 509 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 2.967 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,981 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=4.37 mol, V=42.73 L, T=492.7 K | For an ideal gas, P V = n R T. With n = 4.37 mol, V = 42.73 L, T = 492.7 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 4.135 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,982 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=1.486 mol, V=42.28 L, T=533.3 K | For an ideal gas, P V = n R T. With n = 1.486 mol, V = 42.28 L, T = 533.3 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 1.538 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,983 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=0.6042 mol, V=35.27 L, T=432.2 K | For an ideal gas, P V = n R T. With n = 0.6042 mol, V = 35.27 L, T = 432.2 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 0.6077 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,984 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=4.9 mol, V=13.37 L, T=383.8 K | For an ideal gas, P V = n R T. With n = 4.9 mol, V = 13.37 L, T = 383.8 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 11.54 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,985 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=4.239 mol, V=2.314 L, T=441.2 K | For an ideal gas, P V = n R T. With n = 4.239 mol, V = 2.314 L, T = 441.2 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 66.32 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,986 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=1.165 mol, V=1.999 L, T=404.2 K | For an ideal gas, P V = n R T. With n = 1.165 mol, V = 1.999 L, T = 404.2 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 19.32 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,987 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=0.9842 mol, V=21.35 L, T=411.9 K | For an ideal gas, P V = n R T. With n = 0.9842 mol, V = 21.35 L, T = 411.9 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 1.558 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,988 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=4.37 mol, V=10.45 L, T=227.5 K | For an ideal gas, P V = n R T. With n = 4.37 mol, V = 10.45 L, T = 227.5 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 7.805 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,989 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=4.129 mol, V=30.24 L, T=408.5 K | For an ideal gas, P V = n R T. With n = 4.129 mol, V = 30.24 L, T = 408.5 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 4.576 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,990 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=0.9193 mol, V=25.89 L, T=574.5 K | For an ideal gas, P V = n R T. With n = 0.9193 mol, V = 25.89 L, T = 574.5 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 1.674 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,991 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=2.535 mol, V=40.7 L, T=555.8 K | For an ideal gas, P V = n R T. With n = 2.535 mol, V = 40.7 L, T = 555.8 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 2.841 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,992 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=2.789 mol, V=9.117 L, T=341.1 K | For an ideal gas, P V = n R T. With n = 2.789 mol, V = 9.117 L, T = 341.1 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 8.559 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,993 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=2.721 mol, V=14.62 L, T=278.5 K | For an ideal gas, P V = n R T. With n = 2.721 mol, V = 14.62 L, T = 278.5 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 4.254 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,994 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=2.779 mol, V=34.37 L, T=273.6 K | For an ideal gas, P V = n R T. With n = 2.779 mol, V = 34.37 L, T = 273.6 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 1.815 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,995 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=1.574 mol, V=34.73 L, T=248.9 K | For an ideal gas, P V = n R T. With n = 1.574 mol, V = 34.73 L, T = 248.9 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 0.9257 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,996 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=2.049 mol, V=11.03 L, T=532.7 K | For an ideal gas, P V = n R T. With n = 2.049 mol, V = 11.03 L, T = 532.7 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 8.12 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,997 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=1.644 mol, V=40.7 L, T=290.6 K | For an ideal gas, P V = n R T. With n = 1.644 mol, V = 40.7 L, T = 290.6 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 0.963 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,998 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=3.923 mol, V=25.84 L, T=220 K | For an ideal gas, P V = n R T. With n = 3.923 mol, V = 25.84 L, T = 220 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 2.741 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
7,999 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=4.319 mol, V=34.44 L, T=342.5 K | For an ideal gas, P V = n R T. With n = 4.319 mol, V = 34.44 L, T = 342.5 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 3.525 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
8,000 | chemistry | gases | ideal_gas_law | 4 | worked_example | Ideal-gas pressure for n=0.9481 mol, V=19.39 L, T=311.7 K | For an ideal gas, P V = n R T. With n = 0.9481 mol, V = 19.39 L, T = 311.7 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 1.251 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points. | P V = n R T | mole_concept | Apply the ideal-gas law to compute pressure, volume, or temperature. |
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