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Empirical Formulas and Molecular Formulas

Empirical Formulas and Molecular Formulas. Empirical Formulas. shows the smallest whole-number ratio of the atoms in the compound CH only means it is a 1:1 ratio between carbon and hydrogen, NOT the actual number of atoms in the compound

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Empirical Formulas and Molecular Formulas

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  1. Empirical Formulas and Molecular Formulas

  2. Empirical Formulas • shows the smallest whole-number ratio of the atoms in the compound • CH only means it is a 1:1 ratio between carbon and hydrogen, NOT the actual number of atoms in the compound • could be C2H2 (ecetylene), or C6H6 (benzene), or C8H8 (styrene)

  3. Determining Empirical Formula of a Compound • What is the empirical formula of a compound that’s mass is 25.9% N and 74.1% O? • % is not really a unit of measurement we can use in chemistry at times • if there was 100g of this compound then 25.9g would be nitrogen and 74.1g oxygen • change to moles in order to convert the % to a # • 25.9g N x 1 mol N = 1.85 mol N 14.0 g N • 74.1g O x 1 mol O = 4.63 mol O 16.0 g O

  4. What is the whole number ratio? • a formula is not just the ratio of atoms, it is also the ratio of moles • this leaves a ratio of N1.85O4.63 which is not a a possible formula/whole number • therefore, divide each by the lowest quantity (in this case, 1.85) • now is N1O2.5 • again, still not a whole number, but if you were to double it, it would be N2O5 • the empirical formula is N2O5

  5. Example #2 • Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. • Assume 100 g so… • 38.67 g C x 1mol C = 3.220 mole C 12.01 g C • 16.22 g H x 1mol H = 16.09 mole H 1.01 g H • 45.11 g N x 1mol N = 3.219 mole N 14.01 g N

  6. 3.220 mole C • 16.09 mole H • 3.219 mole N • C1H5N1is the empirical formula • C3.22H16.09N3.219 If we divide all of these by the smallest number (3.22), it will give us whole numbers and therefore the empirical formula

  7. Example #3 (notice this is already in grams and not a percentage) • A compound is found to contain 16g of C and 4g of H. What is the empirical formula? • 16 g C x 1mol C = 1.33 mole C 12.01 g C • 4 g H x 1mol H = 4 mole H 1.01 g H • 1.33 is too far off from 1 • divide both by 1.33 • 1.33/1.33 = 1 • 4/1.33 = 3 • empirical formula is CH3

  8. Comparing Empirical and Molecular Formulas

  9. Molecular Formulas • a molecular formula is the same as, or a multiple of, the empirical formula • it is based on the actual number of atoms of each type in the compound, NOT just the ratio • the following all have same empirical formula (CH2O at a 1-2-1 ratio) but different molecular formulas • C2H4O2 is the molecular formula of ethanoic acid • CH2O is the molecular formula of methanol • C6H12O6 is the molecular formula of glucose

  10. Determine the molecular formula from the empirical formula and the experimental molar mass • divide the experimental molar mass by the mass of one mole of the empirical formula • this results in the multiplier to convert to the molecular formula (this sounds confusing but it really isn’t) • ex: calculate the molecular formula of a compound whose molar mass is found to be 60.0g and has an empirical formula of CH4N • experimental molar mass of ? --- 60.0 g empirical molar mass of CH4N ---- 30.0 g • 2 (CH4N) = C2H8N2 = 2

  11. Determining the empirical formula and then molecular formula of a compound • A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known to be 98.96 g. What is its molecular formula? • need to calculate the empirical formula first • assume 100 g so… • 71.65 g Cl x 1mol Cl = 2.02 mole Cl 35.5 g Cl • 24.27 g C x 1mol C = 2.02 mole C 12.0 g C • 4.07 g H x 1mol H = 4.07 mole H 1.0 g H

  12. Cl2.02C2.02H4.07 • divide by lowest (2.02 mol ) • ClCH2 is the empirical formula • if ClCH2 was the molecular formula, this would give a mass of 48.5 g • however, recall the problem asked for the molecular formula of a compound with a molar mass of 98.96 g which is twice that of 48.5 g • therefore, Cl2C2H4 is the molecular formula

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