Question

What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{CH}_{2}\), molar mass \(=84 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{NH}_{2} \mathrm{Cl}\), molar mass \(=51.5 \mathrm{~g} / \mathrm{mol}\)

Short answer

The molecular formulas for the given compounds are (a) \(\mathrm{C}_{6}\mathrm{H}_{12}\) and (b) \(\mathrm{NH}_{2}\mathrm{Cl}\).

Step-by-step solution

Calculate empirical formula molar mass.

To calculate the empirical formula molar mass, add the molar mass of each element in the empirical formula, so for \(\mathrm{CH}_{2}\), that is C (Carbon - \(12.01 g/mol\)) and H (Hydrogen - \(1.01 g/mol\) each): Empirical formula molar mass = \(12.01 + (2 * 1.01) = 14.03\text{ g/mol}\)

Calculate molecular formula multiplier ratio.

Divide the given molar mass of the molecular formula by the molar mass of the empirical formula, and round to the nearest whole number: Ratio = \(\frac{84}{14.03} = 5.99 \approx 6\)

Determine the molecular formula.

Multiply the empirical formula by the ratio calculated in step 2, this will give the molecular formula. In this case, \(\mathrm{CH}_{2} \times 6 = \mathrm{C}_{6}\mathrm{H}_{12}\). The molecular formula for compound (a) is \(\mathrm{C}_{6}\mathrm{H}_{12}\). b) empirical formula \(\mathrm{NH}_{2} \mathrm{Cl}\), molar mass \(=51.5 \mathrm{~g} / \mathrm{mol}\)

Calculate empirical formula molar mass.

To calculate the empirical formula molar mass, add the molar mass of each element in the empirical formula, so for \(\mathrm{NH}_{2} \mathrm{Cl}\), that is N (Nitrogen - \(14.01 g/mol\)), H (Hydrogen - \(1.01 g/mol\) each), and Cl (Chlorine - \(35.45 g/mol\)): Empirical formula molar mass = \(14.01 + (2 * 1.01) + 35.45 = 51.48\text{ g/mol}\)

Calculate molecular formula multiplier ratio.

Divide the given molar mass of the molecular formula by the molar mass of the empirical formula, and round to the nearest whole number: Ratio = \(\frac{51.5}{51.48} = 1.0004 \approx 1\)

Determine the molecular formula.

Multiply the empirical formula by the ratio calculated in step 2, this will give the molecular formula. In this case, \(\mathrm{NH}_{2} \mathrm{Cl} \times 1 = \mathrm{NH}_{2}\mathrm{Cl}\). The molecular formula for compound (b) is \(\mathrm{NH}_{2}\mathrm{Cl}\).