Question

an iron - containing compound is 20.1% fe, 11.5% s, 63.3% o, and 5.1% h. what is the empirical formula for this compound? answer: a $\ce{fe_{3}so_{19}h_{25}}$ b $\ce{fe_{15}s_{9}o_{50}h}$ c $\ce{fes_{2}o_{22}h_{28}}$ d $\ce{feso_{11}h_{14}}$

Explanation:

Step1: Assume 100g of the compound

If we assume 100g of the compound, then the masses of each element are: \(m_{Fe}=20.1g\), \(m_{S} = 11.5g\), \(m_{O}=63.3g\), \(m_{H}=5.1g\)

Step2: Calculate the number of moles of each element

The molar mass of \(Fe\) is \(M_{Fe}=55.85g/mol\), the molar mass of \(S\) is \(M_{S}=32.07g/mol\), the molar mass of \(O\) is \(M_{O} = 16.00g/mol\), and the molar mass of \(H\) is \(M_{H}=1.01g/mol\)
The number of moles of \(Fe\): \(n_{Fe}=\frac{m_{Fe}}{M_{Fe}}=\frac{20.1g}{55.85g/mol}\approx0.36mol\)
The number of moles of \(S\): \(n_{S}=\frac{m_{S}}{M_{S}}=\frac{11.5g}{32.07g/mol}\approx0.36mol\)
The number of moles of \(O\): \(n_{O}=\frac{m_{O}}{M_{O}}=\frac{63.3g}{16.00g/mol}\approx3.96mol\)
The number of moles of \(H\): \(n_{H}=\frac{m_{H}}{M_{H}}=\frac{5.1g}{1.01g/mol}\approx5.05mol\)

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles (\(n = 0.36mol\))
For \(Fe\): \(\frac{n_{Fe}}{0.36}\approx1\)
For \(S\): \(\frac{n_{S}}{0.36}\approx1\)
For \(O\): \(\frac{n_{O}}{0.36}\approx11\)
For \(H\): \(\frac{n_{H}}{0.36}\approx14\)

Answer:

D. \(FeSO_{11}H_{14}\)