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Heavy. Heavy. Light. Light. Percent composition of an element in a compound = Mass (g) of element Mass (g) of the compound. Example: A sample contains 32.5g Carbon, 44.0g Oxygen, and 7.5g of Hydrogen. Solution:. % Carbon = 32.5g C / 84.0g total = 0.387 = 38.7% C.

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  1. Heavy Heavy Light Light

  2. Percent composition of an element in a compound = Mass (g) of element Mass (g) of the compound Example: A sample contains 32.5g Carbon, 44.0g Oxygen, and 7.5g of Hydrogen. Solution: % Carbon = 32.5g C / 84.0g total = 0.387 = 38.7% C % Oxygen = 44.0g O / 84.0g total = 0.524 = 52.4% O % Hydrogen = 7.5g H / 84.0g total = 0.089 = 8.9% H 38.7% + 52.4% + 8.9% = 100%

  3. Percent composition of an element in a KNOWN compound 1 x (16.0 g) 2 x (1.0 g) 6 x (1.0 g) 1 x (16.0 g) 2 x (12.0 g) %H = %O = %O = %C = %H = x 100% = 34.7% x 100% = 13.1% x 100% = 52.1% x 100% = 11.1% x 100% = 88.9% 46.0 g 46.0 g 18.0 g 18.0 g 46.0 g C2H6O 52.14% + 13.13% + 34.73% = 100.0% H2O

  4. H2O 2 x (1.0 g) 1 x (16.0 g) %H = %O = x 100% = 11.1% x 100% = 88.9% 18.0 g 18.0 g Is it possible to determine what a substance is by knowing only the mass % Compositions…? In other words, can you determine that the molecule is water by knowing it’s 11.1% H and 88.9% O?

  5. 2 x (1.0 g) 2 x (16.0 g) %H = %O = x 100% = 5.9% x 100% = 94.1% 34.0 g 34.0 g H2O2 How about Hydrogen Peroxide?

  6. A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance An empirical formula shows the simplest whole-number ratio of the atoms in a substance molecular empirical H2O2 H1O1 H2O H2O CH2O C6H12O6 O3 O N2H4 NH2 2.6

  7. An empirical formula shows the simplest whole-number ratio of the atoms in a substance Can also be thought of as the simplest MOLE Ratio of the atoms in a substance empirical H2O 2 moles of H per every 1mole of O CH2O 2 moles of H per every 1mole of O & C 2.6

  8. Empirical Formula Example An unknown sample contains 6.0g of C, 1.5g of H, and 4.0g of O and has a molecular weight of 138g/mol 6.0 g C x (1 mol/12g) = 0.5 mol C 0.25mol = 2 C 1.5 g H x (1mol/1g) = 1.5 mol H 0.25mol = 6 H = 1 O 4.0 g O x (1mol/16g) = 0.25 mol O 0.25mol Empirical formula C2H6O

  9. Molecular Formula M.W. Molecular Formula = Empirical Formula x E.F.W.

  10. Molecular Formula Example An unknown sample contains 6.0g of C, 1.5g of H, and 4.0g of O and has a molecular weight of 138g/mol

  11. Molecular Formula Example An unknown sample contains 6.0g of C, 1.5g of H, and 4.0g of O and has a molecular weight of 138g/mol 6.0 g C x (1 mol/12g) = 0.5 mol C 0.25mol = 2 C 1.5 g H x (1mol/1g) = 1.5 mol H 0.25mol = 6 H = 1 O 4.0 g O x (1mol/16g) = 0.25 mol O 0.25mol Empirical formula C2H6O M.W. Molecular Formula = Empirical Formula x E.F.W. 138 M.F. = C2H6O x = C2H6O x 3 = C6H18O3 46

  12. Formula of a Hydrate Example ~ ~ CuSO4 · X H2O Dish/hydrate 44.00g Dish (only) 20.00g Heat gently Dish/Anhydrous 35.00g ? 15g CuSO4 1 mol X = 0.0934 mol = 1 CuSO4 159.7g 0.0934 mol 9g H2O ? 1 mol = 0.500 mol = 5.3 H2O 5 H2O X 18g 0.0934 mol CuSO4 · 5 H2O

  13. Formula of a Hydrate (For Lab) CuSO4 · X H2O Dish/hydrate 44.00g Dish (only) 20.00g Heat gently Dish/Anhydrous 35.00g ? 15g CuSO4 1 mol X = 0.0934 mol = 1CuSO4 159.7g 0.0934 mol 9g H2O ? 1 mol = 0.500 mol = 5.3H2O X 18g 0.0934 mol CuSO4 · 5.3 H2O

  14. Hydrate Example CaCl2 · ___ H2O Mass of Crucible: 30g Mass of crucible with hydrate: 48.3g Mass of crucible after heating (anhydrate with crucible): 41.1g Mass of anhydrous (CaCl2) = 41.1g – 30g = 11.1g CuCl2 Mass of water (H2O) = 48.3g – 41.1g = 7.2g H2O 11.1g CaCl2 1 mol X = 0.10 mol CaCl2 / 0.10 mol 111.1g = 1 CaCl2 7.2g H2O 1 mol = 0.40 mol H2O / 0.10 mol X 18g = 4 H2O CaCl2 · 4 H2O

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