Question

determine the empirical formula of a compound that contains 49.4% k, 20.3% s and 30.3% o.
a) kso₂ b) kso₃ c) k₂so₄ d) k₂so₃ e) k₂s₂o₅

Explanation:

Step1: Assume 100g of the compound.

So, we have 49.4g of K, 20.3g of S, and 30.3g of O.

Step2: Convert masses to moles.

Molar mass of K: $39.10\ g/mol$, moles of K: $\frac{49.4\ g}{39.10\ g/mol} \approx 1.263\ mol$.
Molar mass of S: $32.07\ g/mol$, moles of S: $\frac{20.3\ g}{32.07\ g/mol} \approx 0.633\ mol$.
Molar mass of O: $16.00\ g/mol$, moles of O: $\frac{30.3\ g}{16.00\ g/mol} \approx 1.894\ mol$.

Step3: Divide by the smallest number of moles (0.633 mol) to get mole ratios.

For K: $\frac{1.263}{0.633} \approx 2$.
For S: $\frac{0.633}{0.633} = 1$.
For O: $\frac{1.894}{0.633} \approx 3$. Wait, no, wait—wait, recalculating O: 1.894 / 0.633 ≈ 3? Wait, no, 0.6333=1.899, close to 1.894. Wait, but wait, maybe I made a mistake. Wait, let's check again. Wait, 49.4/39.1 ≈ 1.263, 20.3/32.07≈0.633, 30.3/16≈1.89375. Then divide by 0.633: K: 1.263/0.633≈2, S:1, O:1.89375/0.633≈3? Wait, but 0.6333=1.899, which is close to 1.89375. But wait, the options: d is K₂SO₃ (K₂SO₃ has K:2, S:1, O:3), c is K₂SO₄ (O:4). Wait, maybe my calculation for O is wrong? Wait, 30.3/16=1.89375. 1.89375/0.633≈3.0 (since 0.6333=1.899). So the ratio is K:2, S:1, O:3, which is K₂SO₃ (option d). Wait, but let's check the mass percentages. Let's calculate the mass percentage of K₂SO₃: K:239.1=78.2, S:32.07, O:3*16=48. Total mass:78.2+32.07+48=158.27. %K:78.2/158.27≈49.4%, %S:32.07/158.27≈20.3%, %O:48/158.27≈30.3%. Yes! So that matches. So the empirical formula is K₂SO₃, which is option d.

Answer:

d) $\ce{K2SO3}$