2021 H2 Chem P1 Q10 — Mole Solution

The answer is A (2.00 g of ethyl methanoate).

The greatest number of molecules means the greatest amount in moles.

Ethyl methanoate is \(\text{HCO}_2\text{CH}_2\text{CH}_3\) (\(\text{C}_3\text{H}_6\text{O}_2\)):
\(n_{\text{HCO}_2\text{CH}_2\text{CH}_3} = \frac{2.00}{12.0 \times 3 + 1.0 \times 6 + 16.0 \times 2} = \frac{2.00}{74.0} = 0.0270\) mol

B ✗: \(n_{\text{Br}_2} = \frac{4.00}{79.9 \times 2} = \frac{4.00}{159.8} = 0.0250\) mol, less than 0.0270 mol.

C ✗: \(n_{\text{H}_2} = \frac{550 \text{ cm}^3}{22700 \text{ cm}^3 \text{ mol}^{-1}} = 0.0242\) mol, less than 0.0270 mol.

D ✗: \(n_{\text{H}_2\text{O}} = \frac{1.55 \times 10^{22}}{6.02 \times 10^{23} \text{ mol}^{-1}} = 0.0257\) mol, less than 0.0270 mol.

Order: ethyl methanoate (0.0270) > water (0.0257) > bromine (0.0250) > hydrogen (0.0242), so A contains the most molecules.