Problem 47

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 \(=51.5 \mathrm{~g} / \mathrm{mol}\)

Step-by-Step Solution

Verified
Answer
The molecular formula for the compounds are: (a) \(\mathrm{C}_{6} \mathrm{H}_{12}\) (b) \(\mathrm{NH}_{2}\mathrm{Cl}\)
1Step 1: (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
2Step 2: (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
3Step 3: (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 \(\mathrm{C}_{6} \mathrm{H}_{12}\).
4Step 4: (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
5Step 5: (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
6Step 6: (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 \(\mathrm{NH}_{2}\mathrm{Cl}\).