Question

60 g of a compound on analysis gave 24g of C ,4 g of H and 32 g of O.The emprical formula of a compound is

Answer 1

its molecular formula is C2H4O2 and its empirical formula is CH2O

Answer 2

Answer : The empirical formula of the compound is, $C_1H_2O_1$ or $CH_2O$

Solution : Given,

Mass of C = 24 g

Mass of H = 4 g

Mass of O = 32 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{24g}{12g/mole}=2moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{4g}{1g/mole}=4moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{32g}{16g/mole}=2moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{2}{2}=1$

For H = $\frac{4}{2}=2$

For O = $\frac{2}{2}=1$

The ratio of C : H : O = 1 : 2 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_1H_2O_1$

Therefore, the empirical formula of the compound is, $C_1H_2O_1$ or $CH_2O$