IBDP Chemistry -Structure 1.5 Ideal gases - IB Style Questions For SL Paper 2 -FA 2025
▶️ Answer/Explanation
Detailed solution
(a)
• \(C\): \(9.49 / 44.01 = 0.216 \text{ mol}\). Mass \(C = 2.59 \text{ g}\).
• \(H\): \(2 \times (5.18 / 18.02) = 0.575 \text{ mol}\). Mass \(H = 0.581 \text{ g}\).
• \(O\): \(4.32 – (2.59 + 0.581) = 1.15 \text{ g}\). Moles \(O = 1.15 / 16.00 = 0.0719 \text{ mol}\).
• Ratio \(C:H:O = 0.216 : 0.575 : 0.0719\). Divide by \(0.0719\) \(\rightarrow\) \(3 : 8 : 1\).
• Empirical formula: \(\text{C}_3\text{H}_8\text{O}\).
(b)
(i) \(n = \frac{PV}{RT} = \frac{1.00 \times 10^5 \times 55.7 \times 10^{-6}}{8.31 \times 373} = \mathbf{0.00180 \text{ mol}}\).
(ii) \(M = \frac{m}{n} = \frac{0.108}{0.00180} = \mathbf{60.0 \text{ g mol}^{-1}}\).