2020 H2 Chem P1 Q9 — Mole Solution

The answer is D (\(\text{B}_4\text{C}\)).

All the B in boron carbide ends up in \(\text{B}_2\text{O}_3\) and all the C ends up in \(\text{CO}_2\).
\(M_\text{r}(\text{B}_2\text{O}_3) = 2(10.8) + 3(16.0) = 69.6\)
\(n(\text{B}_2\text{O}_3) = 2.50 \div 69.6 = 0.0359\) mol, so \(n(\text{B}) = 2 \times 0.0359 = 0.0718\) mol
\(n(\text{C}) = n(\text{CO}_2) = 0.80 \div 44.0 = 0.0182\) mol
B : C = 0.0718 : 0.0182 = 3.95 : 1, which is 4 : 1, so the empirical formula is \(\text{B}_4\text{C}\).
Check: mass of B = \(0.0718 \times 10.8 = 0.776\) g and mass of C = \(0.0182 \times 12.0 = 0.218\) g, a total of 0.994 g, which matches the 1.00 g sample.

• A is wrong: \(\text{BC}_2\) has B : C = 1 : 2, more C than B, but the data give about four times as many B atoms as C atoms.
• B is wrong: \(\text{BC}_4\) has B : C = 1 : 4, the inverse of the correct ratio.
• C is wrong: \(\text{B}_2\text{C}\) (2 : 1) is obtained if the amount of \(\text{B}_2\text{O}_3\) (0.0359 mol) is used as the amount of B, forgetting that each \(\text{B}_2\text{O}_3\) contains 2 B atoms (0.0359 : 0.0182 = 2 : 1).