Question

27 calculate the empirical formula for a compound that contains 67.6% hg, 10.8% s, and 21.6% o.

Explanation:

Step1: Assume 100g of compound

Masses: $m_{\text{Hg}}=67.6\ \text{g}$, $m_{\text{S}}=10.8\ \text{g}$, $m_{\text{O}}=21.6\ \text{g}$

Step2: Calculate moles of each element

Molar masses: $M_{\text{Hg}}=200.59\ \text{g/mol}$, $M_{\text{S}}=32.07\ \text{g/mol}$, $M_{\text{O}}=16.00\ \text{g/mol}$
$n_{\text{Hg}}=\frac{67.6}{200.59}\approx0.337\ \text{mol}$
$n_{\text{S}}=\frac{10.8}{32.07}\approx0.337\ \text{mol}$
$n_{\text{O}}=\frac{21.6}{16.00}=1.35\ \text{mol}$

Step3: Divide by smallest mole value

Smallest $n=0.337\ \text{mol}$
$N_{\text{Hg}}=\frac{0.337}{0.337}=1$
$N_{\text{S}}=\frac{0.337}{0.337}=1$
$N_{\text{O}}=\frac{1.35}{0.337}\approx4$

Answer:

$\text{HgSO}_4$