Question

1. calculate the empirical formula of a compound that is 67.6% hg, 10.8% s, and 21.6% o.

Explanation:

Step1: Assume 100g of the compound

If we assume 100g of the compound, then the masses of each element are: \(m_{Hg}=67.6g\), \(m_{S} = 10.8g\), \(m_{O}=21.6g\)

Step2: Calculate the number of moles of each element

The molar mass of \(Hg\) is \(M_{Hg}=200.59g/mol\), the molar mass of \(S\) is \(M_{S}=32.07g/mol\), and the molar mass of \(O\) is \(M_{O} = 16.00g/mol\)
The number of moles of \(Hg\): \(n_{Hg}=\frac{m_{Hg}}{M_{Hg}}=\frac{67.6g}{200.59g/mol}\approx0.337mol\)
The number of moles of \(S\): \(n_{S}=\frac{m_{S}}{M_{S}}=\frac{10.8g}{32.07g/mol}\approx0.337mol\)
The number of moles of \(O\): \(n_{O}=\frac{m_{O}}{M_{O}}=\frac{21.6g}{16.00g/mol}=1.35mol\)

Step3: Find the mole ratio

Divide each number of moles by the smallest number of moles (\(n = 0.337mol\))
For \(Hg\): \(\frac{n_{Hg}}{0.337mol}=\frac{0.337mol}{0.337mol}=1\)
For \(S\): \(\frac{n_{S}}{0.337mol}=\frac{0.337mol}{0.337mol}=1\)
For \(O\): \(\frac{n_{O}}{0.337mol}=\frac{1.35mol}{0.337mol}\approx4\)

Answer:

The empirical formula of the compound is \(HgSO_{4}\)