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 mass 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)

For K: $\frac{1.263}{0.633} \approx 2$
For S: $\frac{0.633}{0.633} = 1$
For O: $\frac{1.894}{0.633} \approx 3$

Answer:

d) $\ce{K2SO3}$