Question

Determine the empirical formula of a compound containing 48.38 grams of carbon, 6.74 grams of hydrogen, and 53.5 grams of oxygen​

Answer 1

Answer: The empirical formula of the compound becomes $CH_2O$

Explanation:

The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

We are given:

Mass of C = 48.38 g

Mass of H = 6.74 g

Mass of O = 53.5 g

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

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

• Step 1: Converting the given masses into moles.

Molar mass of C = 12 g/mol

Molar mass of H = 1 g/mol

Molar mass of O = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of C}=\frac{48.38g}{12g/mol}=3.023 mol$

$\text{Moles of H}=\frac{6.74g}{1g/mol}=6.74 mol$

$\text{Moles of O}=\frac{53.5g}{1g/mol}=3.34 mol$

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

Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 3.023 moles

$\text{Mole fraction of C}=\frac{3.023}{3.023}=1$

$\text{Mole fraction of H}=\frac{6.74}{3.023}=2.23\approx 2$

$\text{Mole fraction of O}=\frac{3.34}{3.023}=1.105\approx 1$

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

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

Hence, the empirical formula of the compound becomes $CH_2O$