Question

1. what is the empirical formula for a compound that is 67.6% mercury, 10.8% sulfur, and 21.6 % oxygen?

Explanation:

Step1: Assume 100g sample, get masses

Mass of Hg = 67.6 g, S = 10.8 g, O = 21.6 g

Step2: Convert mass to moles

Molar mass: Hg = 200.59 g/mol, S = 32.07 g/mol, O = 16.00 g/mol
Moles of Hg: $\frac{67.6}{200.59} \approx 0.337$ mol
Moles of S: $\frac{10.8}{32.07} \approx 0.337$ mol
Moles of O: $\frac{21.6}{16.00} = 1.35$ mol

Step3: Divide by smallest mole value

Smallest mole value = 0.337 mol
Ratio of Hg: $\frac{0.337}{0.337} = 1$
Ratio of S: $\frac{0.337}{0.337} = 1$
Ratio of O: $\frac{1.35}{0.337} \approx 4$

Answer:

$\text{HgSO}_4$