1. Copyright Sautter 2003

2. EMPIRICAL FORMULAE An empirical formula is the simplest formula for a compound. For example, H 2 O 2 can be reduced to a simpler formula, HO. Similarly, N 2 O 4 can be reduced to NO 2. Both HO and NO 2 are empirical formulae. Some formulae cannot be reduced and are the empirical formula. An example is H 2 O which cannot be reduced to lower terms. Empirical formulae for compounds can be determined from experimental information such as percent composition data.

3. The molecular formula is always a whole number multiple of the empirical formula

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 CH 4. One mole of CH 4 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 CH 4 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 %

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.

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 NO 2.5 Empirical formula requires the lowest “whole number ratio” Multiplying the formula through by 2 to obtain whole numbers we get N 2 O 5 (dinitrogen pentaoxide)

7. Empirical Formula Calculations Using Percent Composition Problem: What is the empirical formula for a compound consisting of 37.5 % carbon, 12.5 % hydrogen and oxygen? Solution: We will assume a 100 g sample of the compound. It then consists of 37.5 g C, 12.5 g H and 50.0 g O (100 g sample – 37.5g C – 12.5 g H) Moles C = 37.5 g / 12.0 g per mole = 3.125 Moles H = 12.5 g / 1.00 g per mole = 12.5 Moles O = 50.0 g / 16.0 g per mole = 3.125 Divide each by the smallest mole value Moles C / moles O = 3.125 / 3.125 = 1.00 Moles H / moles O = 12.5 / 3.125 = 4.00 The empirical formula is CH 4 O or CH 3 OH (methyl alcohol)

8. Empirical Formula Calculations Steps for finding empirical formulae: (1) If percent composition data is given assume a 100 gram sample (2) Convert grams to moles for each element (3) Divide the lowest mole value into each of the other mole values (4) If all the ratios are not whole numbers multiply through to obtain whole number values for each ratio (5) Write the empirical formula using the whole number ratios as subscripts

9. Empirical Formula Calculations From Combustion (Burning) Data Problem: A compound containing only C, H and O is burned in O 2 to obtain CO 2 and H 2 O. A 0.5438 g sample gives 1.039 g CO 2 and 0.6369 g H 2 O. Find its empirical formula. Solution: All the carbon is contained in the CO 2. Carbon dioxide is 27.3 % C (12.0 / 44.0) so 0.273 x 1.039 g gives 0.283 grams of C The hydrogen is contained in the water which is 11.1 % H (2 / 18) so.111 x 0.6369 g gives 0.0708 g H The rest of the sample is O (0.5438 – 0.283 – 0.0709) or 0.190 grams of O Moles C = 0.283 / 12 = 0.0236 Moles H = 0.0708 / 1.00 = 0.0708 Moles O = 0.190 / 16 = 0.0119 Moles C / moles O = 0.0236 / 0.0119 = 1.98 ~ 2 Moles H / moles O = 0.0708 / 0.0119 = 5.95 ~ 6 Empirical Formula equals C 2 H 6 O or C 2 H 5 OH (ethyl alcohol)

10. Finding the Molecular Formula from the Empirical Formula and Molar Mass Problem: The empirical formula for polystyrene is CH. It has a molecule mass of 104. What is its molecular formula? Solution: Since the molecular formula is always some whole number multiple of the empirical formula, the molecular mass is always some whole number multiple of the empirical formula mass. The empirical formula mass is 13 (12 for C + 1 for H) Dividing 104 by 13 we get 8. Polystyrene must contain 8 CH units and have a molecular formula of C 8 H 8