Question

What is the empirical formula of a compound composed of 3.25% hydrogen ( H ), 19.36% carbon ( C ), and 77.39% oxygen ( O ) by mass? Insert subscripts as needed.

Answer 1

Answer : The empirical of the compound is, $C_1H_2O_3$

Solution : Given,

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 19.36 g

Mass of H = 3.25 g

Mass of O = 77.39 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 = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{19.36g}{12g/mole}=1.613moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{3.25g}{1g/mole}=3.25moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{77.39g}{16g/mole}=4.837moles$

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

For C = $\frac{1.613}{1.613}=1$

For H = $\frac{3.25}{1.613}=2.01\approx 2$

For o = $\frac{4.837}{1.613}=2.99\approx 3$

The ratio of C : H : O = 1 : 2 : 3

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

The Empirical formula = $C_1H_2O_3$

Therefore, the empirical of the compound is, $C_1H_2O_3$