Calculating Empirical Formulas Directions: Use the sample problems on your Empirical Formula Resource sheets to solve the following problems. Set up your solutions like on the resource sheets. 1. A compound is found to have the following mass composition: 92.3% Carbon; 7.70% Hydrogen. Calculate the empirical formula. 100 g compound contains 92.3 g C 7.7 g H 92.3 g 1 mol 7.70 g 1 mol No. of Moles of Atoms 12.01 g 1.01 g 7.685 mol 7.624 mol Relative No. of Moles = 1 = 1 7.624 mol 7.624 mol Empirical Formula = CH 2. A sulfur compound contains 50.0% sulfur and 50.0% oxygen. What is the empirical formula? 100 g compound contains 50.0 g S 50.0 g O 50.0 g 1 mol 50.0 g 1 mol No. of Moles of Atoms 32.06 g 16.00 g 1.56 mol 3.125 mol Relative No. of Moles = 1 = 2 1.56 mol 1.56 mol 2 Empirical Formula = SO 2 3. A compound is found to contain 39.95% carbon; 6.69% hydrogen; and 53.36% oxygen. Calculate the empirical formula. 100 g compound contains 35.95 g C 6.69 g H 53.36 g O 39.95 g 1 mol No. of Moles of 6.69 g 1 mol 53.36 g 1 mol Atoms 12.01 g 1.01 g 16.00 g 3.33 mol 6.62 mol 3.34 mol Relative No. of Moles = 1 = 2 = 1 3.33 mol 3.33 mol 3.33 mol Empirical Formula = CH 2O 4. If a compound is found to contain 30.4% nitrogen and 69.6% oxygen, what is the empirical formula? 100 g compound contains 30.43 g N 69.6 g O No. of Moles of 30.4 g 1 mol 69.6 g 1 mol Atoms 14.01 g 16.00 g 2.17 mol 4.35 mol Relative No. of Moles = 1 = 2 2.17 mol 2.17 mol Empirical Formula = NO 2 The following are the results of analyses of compounds. In each case calculate the empirical formula. 5. 75.0% Carbon; 25.0% Hydrogen 6. 27.3% Carbon; 72.7% Oxygen 100 g has 75.0 g C 25.0 g H 100 g has 27.3 g C 72.7 g O # Moles of 75.0 g 1 mol 25.0 g 1 mol # Moles 27.3 g 1 mol 72.7g 1 mol Atoms 12.01 g 1.01 g of Atoms 12.01 g 16.00 g Relative 6.24 mol 24.75 mol Relative 2.27 mol 4.54 mol = 1 = 4 = 1 = 2 # Moles 6.24 mol 6.24 mol # Moles 2.27 mol 2.27 mol Empirical Formula = CH 4 Empirical Formula = CO 2 7. 92.6% Mercury; 7.40% Oxygen 8. 79.69% Copper; 20.31% Sulfur 100 g has 92.6 g Hg 7.40 g O 100 g has 79.69 g Cu 20.31 g S # Moles 92.6 g 1 mol 7.4 g 1 mol # Moles 79.69 g 1 mol 20.31 g 1 mol of Atoms 200.59 g 16.00 g of Atoms 63.55 g 32.06 g Relative 0.462 mol 0.463 mol Relative 1.25 mol 0.633 mol # Moles = 1 = 1 = 2 = 1 0.462 mol 0.462 mol # Moles 0.633 mol 0.633 mol Empirical Formula = HgO Empirical Formula = Cu 2S 9. 2.04% Hydrogen; 32.65% Sulfur; 65.31% Oxygen 100 g compound contains 2.04 g H 32.65 g S 65.31 g O 2.04 g 1 mol 32.65 g 1 mol 65.31 g 1 mol No. of Moles of Atoms 1.01 g 32.06 g 16.00 g 2.02 mol 1.02 mol 4.08 mol Relative No. of Moles = 2 = 1 = 4 1.02 mol 1.02 mol 1.02 mol Empirical Formula = H 2SO4 10. A compound consists of: 65.45% Carbon; 5.492% Hydrogen; and 29.06% Oxygen. Calculate its empirical formula. If its correct molecular weight is 110, calculate its correct molecular formula. 100 g compound contains 65.45 g C 5.492 g H 29.06 g O 65.45 g 1 mol 5.492 g 1 mol 29.06 g 1 mol No. of Moles of Atoms 12.01 g 1.01 g 16.00 g 5.45 mol 5.44 mol 1.82 mol Relative No. of Moles = 3 = 3 = 1 1.82 mol 1.82 mol 1.82 mol Empirical Formula = C 3H3O Simplest Weight = 55.06 55 Molecular Weight = 110 Mult. Factor = 110 / 55 = 2 Molecular Formula = 2 x [C 3H3O] = C 6H6O2 .