Question

a compound sample contains 53.7 g fe and 46.3 g s. what is the empirical formula? fe₃s₂ fes₂ fe₂s₃ fes

Explanation:

Step1: Calculate moles of Fe and S

Molar mass of Fe ($M_{Fe}$) = $55.85\ g/mol$, moles of Fe ($n_{Fe}$) = $\frac{53.7\ g}{55.85\ g/mol}\approx0.96\ mol$
Molar mass of S ($M_{S}$) = $32.07\ g/mol$, moles of S ($n_{S}$) = $\frac{46.3\ g}{32.07\ g/mol}\approx1.44\ mol$

Step2: Find mole ratio

Divide each mole value by the smaller mole value (0.96).
For Fe: $\frac{0.96}{0.96} = 1$
For S: $\frac{1.44}{0.96}=1.5$
Multiply by 2 to get whole - number ratio. Fe: $1\times2 = 2$, S: $1.5\times2=3$

Answer:

$Fe_2S_3$