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 formula for the compounds are: (a) \(\mathrm{C}_{6} \mathrm{H}_{12}\) (b) \(\mathrm{NH}_{2}\mathrm{Cl}\)

Step-by-step solution

(a) Step 1: Calculate the empirical formula weight for CH2

Calculate the empirical formula weight by adding the atomic weights of the elements in the empirical formula. For \(\mathrm{CH}_{2}\): Empirical Formula Weight = Atomic Weight of Carbon + (2 × Atomic Weight of Hydrogen) = 12.01 + (2 × 1.01) = 12.01 + 2.02 = 14.03 g/mol

(a) Step 2: Find the ratio between molar mass and empirical formula weight

Divide the molar mass given by the empirical formula weight. Ratio = (Molar Mass) / (Empirical Formula Weight) = 84 / 14.03 = 6

(a) Step 3: Determine the molecular formula

Multiply each element in the empirical formula by the ratio found in Step 2. Molecular Formula = \(\mathrm{C}(6) \mathrm{H}_{2}(6) = \mathrm{C}_{6} \mathrm{H}_{12}\) So, the molecular formula for a compound with an empirical formula of \(\mathrm{CH}_{2}\) and molar mass of 84 g/mol is \(\mathrm{C}_{6} \mathrm{H}_{12}\).

(b) Step 1: Calculate the empirical formula weight for NH2Cl

Calculate the empirical formula weight by adding the atomic weights of the elements in the empirical formula. For \(\mathrm{NH}_{2} \mathrm{Cl}\): Empirical Formula Weight = Atomic Weight of Nitrogen + (2 × Atomic Weight of Hydrogen) + Atomic Weight of Chlorine = 14.01 + (2 × 1.01) + 35.45 = 14.01 + 2.02 + 35.45 = 51.48 g/mol

(b) Step 2: Find the ratio between molar mass and empirical formula weight

Divide the molar mass given by the empirical formula weight. Ratio = (Molar Mass) / (Empirical Formula Weight) = 51.5 / 51.48 ≈ 1

(b) Step 3: Determine the molecular formula

Multiply each element in the empirical formula by the ratio found in Step 2. Since the ratio is approximately 1, the molecular formula is the same as the empirical formula. Molecular Formula = \(\mathrm{NH}_{2}\mathrm{Cl}\) So, the molecular formula for a compound with an empirical formula of \(\mathrm{NH}_{2} \mathrm{Cl}\) and molar mass of 51.5 g/mol is \(\mathrm{NH}_{2}\mathrm{Cl}\).