Question

Find the empirical formula of the compound containing 54.57% carbon and 9.09% hydrogen and the rest being oxygen.

• A. $C{H}_{2}O$
• B. $CHO$
• C. $C_2{H}_{4}O$
• D. $C_3{H}_4{O}_2$

Answer 1

The correct option is C $C_2{H}_{4}O$

A compound contains 54.57% carbon and 9.09% hydrogen and the rest being oxygen.

100 g of compound will contain 54.57 g of C, 9.09 g of H and $100-54.57-9.09=36.34$ g of O.

The atomic masses of C, H and O are 12 g/mol, 1 g/mol and 16 g/mol respectively.

Number of moles of $C=\frac{54.57g}{12g/mol}=4.55mol$.

Number of moles of $H=\frac{9.09g}{1g/mol}=9.09mol$.

Number of moles of $O=\frac{36.34g}{16g/mol}=2.27mol$.

The mole ratio $C:H:O=4.55:9.09:2.27=2:4:1$

The empirical formula of the compound is $C_2{H}_{4}O$.