Empirical Formulae

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How to find the empirical formula of a compound from given data. Also calculating the molecular formula.
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Empirical Formulae

  1. 1. Copyright Sautter 2015 1
  2. 2. EMPIRICAL FORMULAE • An empirical formula is the simplest formula for a compound. For example, H2O2 can be reduced to a simpler formula, HO. Similarly, N2O4 can be reduced to NO2. Both HO and NO2 are empirical formulae. • Some formulae cannot be reduced and are the empirical formula. An example is H2O which cannot be reduced to lower terms. • Empirical formulae for compounds can be determined from experimental information such as percent composition data. 2
  3. 3. The molecular formula is always a whole number multiple of the empirical formula 3
  4. 4. PERCENT COMPOSITION • Percent composition refers to the mass percentage of each element contained in a compound. • Percent composition can be determined experimentally or from the formula of a substance. For example, the percentage of carbon and hydrogen contained in methane can be found from its formula CH4. • One mole of CH4 contains 1 mole of carbon and 4 moles of hydrogen. One mole of carbon has a mass of 12.0 grams and 4 moles of hydrogen have a mass of 4.0 grams (4 x 1.0 grams per mole). The mass of one mole of CH4 is 16.0 grams (12.0 + 4.0 = 16.0 g) • % C = (12.0 / 16.0) x 100 % = 75.0 % carbon • % H = (4.0/ 16.0) x 100 % = 25.0 % hydrogen • The total of the percents for each element must = 100 % • 75.0 % + 25.0 % = 100 % 4
  5. 5. PERCENT COMPOSITION • Problem: A compound is found to contain 0.1417 g of nitrogen and 0.4045 g of oxygen. The sample has a mass of 0.5462 g. What is the percent composition of the compound? • Solution: % of X = (grams of X / total mass) x 100 % • % N = (0.1417 / 0.5462) x 100 % = 25.6 % • % O = (0.4045 / 0.5462) x 100 % = 74.1 % • Check : 25.6 % + 74.1 % = 99.7 % ~ 100 % • It is often required to find the empirical formula of a compound from data such as that given in this problem. Remember, empirical formula is the lowest whole number ratio of atoms in a compound. 5
  6. 6. Empirical Formula Calculations • Problem: A compound is found to contain 0.1417 g of nitrogen and 0.4045 g of oxygen. The sample has a mass of 0.5462 g. What is the empirical formula of the compound? • Solution: All formulae are based on moles. First find the moles of each element present. • Moles N = 0.1417 g / 14.0 grams per mole = 0.0101 • Moles O = 0.4045 g / 16.0 grams per mole = 0.0253 • Now divide the smallest mole value into the other mole values • Moles O / moles N = 0.0253moles / 0.0101 = 2.50 • Moles N / moles N is of course 1.00 • The formula at this point is then NO2.5 • Empirical formula requires the lowest “whole number ratio” • Multiplying the formula through by 2 to obtain whole numbers we get N2O5 (dinitrogen pentaoxide) 6
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