Question

What is the empirical formula of a compound that is 62.0% c, 10.4% h, and 27.5% o by mass?

Answer 1

C=62.0/12 H=10.4/1 O=27.5/16, C=5.1667/1.71875 H=10.4/1.71875 O=1.71875/1.71875, C=3 H=6 O=1 Empirical formula=C3H6O

Answer 2

Answer: The empirical formula for the given compound is $C_3H_6O$

Explanation:

We are given:

Percentage of C = 62.0 %

Percentage of H = 10.4 %

Percentage of O = 27.5 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 62.0 g

Mass of H = 10.4 g

Mass of O = 27.5 g

To formulate the empirical formula, we need to follow some steps:

• Step 1: Converting the given masses into moles.

Moles of Carbon =$\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{62.0g}{12g/mole}=5.16moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{10.4g}{1g/mole}=10.4moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{27.5g}{16g/mole}=1.72moles$

• Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.72 moles.

For Carbon = $\frac{5.16}{1.72}=3$

For Hydrogen  = $\frac{10.4}{1.72}=6.04\approx 6$

For Oxygen  = $\frac{1.72}{1.72}=1$

• Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_6O$