Question

What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{HCO}_{2}\), molar mass \(=90.0 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\), molar mass \(=88 \mathrm{~g} / \mathrm{mol}\)

Short answer

The molecular formulas for the given compounds are: (a) \(\mathrm{H_{2}C_{2}O_{4}\) (b) \(\mathrm{C_{4}H_{8}O_{2}\)

Step-by-step solution

Calculate the molar mass of the empirical formulas

For each given empirical formula, calculate its molar mass using the atomic masses of the elements present in the empirical formula. (a) For the empirical formula \(\mathrm{HCO}_{2}\): Molar mass of H: 1 g/mol Molar mass of C: 12 g/mol Molar mass of O: 16 g/mol Molar mass = (1)(1) + (1)(12) + (2)(16) = 1 + 12 + 32 = 45 g/mol (b) For the empirical formula \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\): Molar mass of C: 12 g/mol Molar mass of H: 1 g/mol Molar mass of O: 16 g/mol Molar mass = (2)(12) + (4)(1) + (1)(16) = 24 + 4 + 16 = 44 g/mol

Calculate the ratio between the molar mass of the compound and the empirical formula

For each given compound, divide the molar mass of the compound by the molar mass of its empirical formula. (a) Ratio for \(\mathrm{HCO}_{2}\) compound: \(\frac{90.0 \ \mathrm{g/mol}}{45 \ \mathrm{g/mol}} = 2\) (b) Ratio for \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) compound: \(\frac{88 \ \mathrm{g/mol}}{44 \ \mathrm{g/mol}} = 2\)

Multiply the empirical formula by the ratio to find the molecular formula

For each compound, multiply the coefficients of the elements in the empirical formula by the ratio calculated in step 2. (a) Molecular formula for \(\mathrm{HCO}_{2}\) compound: \(\mathrm{(H_{1}C_{1}O_{2})} \times 2 = \mathrm{H_{2}C_{2}O_{4}}\) (b) Molecular formula for \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) compound: \(\mathrm{(C_{2}H_{4}O_{1})} \times 2 = \mathrm{C_{4}H_{8}O_{2}}\) Therefore, the molecular formulas for the given compounds are as follows: (a) \(\mathrm{H_{2}C_{2}O_{4}}\) (b) \(\mathrm{C_{4}H_{8}O_{2}}\)