Question

What is the empirical formula of a compound that contains 49.4% K, 20.3% S, and 30.3% by mass? A) KSO3 B) K2SO3 C) KSO2 D) KSO E) K2SO4

Answer 1

Answer: $K_2SO_3$.

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of K = 49.4 g

Mass of S = 20.3 g

Mass of O = 30.3 g

Step 1 : convert given masses into moles.

Moles of K=$\frac{\text{ given mass of K}}{\text{ molar mass of K}}= \frac{49.4g}{40g/mole}=1.23moles$

Moles of S= \frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{20.3g}{32g/mole}=0.63moles[/tex]

Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{30.3g}{16g/mole}=1.89moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For K = $\frac{1.23}{0.63}=2$

For S =$\frac{0.63}{0.63}=1$

For O =$\frac{1.89}{0.63}=3$

The ratio of K: S:O = 2: 1: 3

Hence the empirical formula is $K_2SO_3$.