Question

An iron compound yielded C = 64.4%, H = 5.5% and Fe = 29.9%. The simple formula of the compound is:

• A. $C_{10}{H}_{10}Fe$
• B. $C_9{H}_{10}Fe$
• C. $C_{10}{H}_{9}Fe$
• D. $C_9{H}_{9}Fe$

Answer 1

Answer: (A)

$C_{10}{H}_{10}Fe$

An iron compound yielded C = 64.4%, H = 5.5% and Fe = 29.9%.
100 g of compound will contain 64.4 g C, 5.5 g, H and 29.9 g Fe.
The atomic masses of C, H and Fe are 12 g/mol, 1 g/mol and 55.8 g/mol respectively.
The number of moles of C $=\frac{\text{64.4 g}}{\text{12 g/mol}}=\text{5.37 mol}$
The number of moles of H $=\frac{\text{5.5 g}}{\text{1 g/mol}}=\text{5.5 mol}$
The number of moles of Fe $=\frac{\text{29.9 g}}{\text{55.8 g/mol}}=\text{0.54 mol}$
The mole ratio C : H : Fe $=5.37:5.5:0.54s\simeq 10:10:1$
The simple formula of the compound is $C_{10}{H}_{10}Fe$