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Empirical Formula

Empirical Formula. The simplest ratio of atoms For example, the empirical formula of H 2 O 2 (hydrogen peroxide) is HO (1 hydrogen atom for every oxygen atom). How to calculate EF. From percentage composition or mass E.g. 49.5g C, 5.2g H, 28.8g N, 16.5g O

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Empirical Formula

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  1. Empirical Formula • The simplest ratio of atoms • For example, the empirical formula of H2O2 (hydrogen peroxide) is HO (1 hydrogen atom for every oxygen atom)

  2. How to calculate EF • From percentage composition or mass E.g. 49.5g C, 5.2g H, 28.8g N, 16.5g O • Arrange the elements in % or mass Done • Divide each by the molar mass of the element to get moles C49.5 H5.2 N28.8 O16.5 12 1 14 16

  3. C 4.121 H 5.159 N 2.056 O 1.031 • Divide everything by the SMALLEST NUMBER 4.121 5.159 2.056 1.031 1.031 1.031 1.031 1.031 = 4 : 5 : 2 : 1 • Write the formula with the simplest whole number ratio C4H5N2O ANSWER

  4. Now try these! • Nutrasweet is 57.14% C, 6.16% H, 9.52% N and 27.18% O. • A compound is 72.2% Mg and 27.8% N • An oxide of nitrogen contains 42.05 g N and 95.95 g of O. • Mercury forms a compound that is 73.9% Hg and 26.1% Cl. • Vitamin C contains 40.92% C, 4.58% H and 54.50% O.

  5. Now try these - ANSWERS • C14H18N2O5 • Mg3N2 • NO2 • HgCl2 • C3H4O3

  6. From EF to molecular formula • The molecular formula is the ACTUAL NUMBER OF EACH ATOM in the compound

  7. How to work out the MF • Calculate the molar mass for the empirical formula. Example: for a compound where there EF is HO and actual molar mass is 34 gmol-1 Molar mass of EF = 1 + 16 = 17 gmol-1 • Divide the molar mass MF ÷ EF Actual MF ÷ EF 34 ÷ 17 gmol-1 = 2 • Multiply the formula by the factor calculated in part 2. HO x 2 = H2O2

  8. Questions – Molecular formula • A compound has an empirical formula of CH2Cl and a molar mass of 99 gmol-1. Calculate the molecular formula. • A compound has 75.46% C, 4.43% H and 20.1% O and a molar mass of 318 gmol-1. Calculate the molecular formula.

  9. Answers – Molecular formula • The EF is CH2Cl. The molar mass for this is 12 + 2 x 1 + 35.5 = 49.5. 99 ÷ 49.5 = 2, therefore the molecular formula is TWICE that of the EF – C2H4Cl2. • The EF is C10H7O2. The molar mass for this is 12 x 10 + 7 x 1 + 2 x 16 = 159. 318 ÷ 159 = 2, therefore the molecular formula is TWICE that of the EF – C20H14O4*.

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