Question

Acetic acid is an organic compound composed of 40.00% C, 6.71% H, and the rest oxygen. If 0.234 mol of acetic acid has a mass of 14.1 g, what are the empirical and molecular formulas of acetic acid?

Answer 1

Answer : The empirical and molecular formulas of acetic acid is, $CH_2O$  and $C_2H_4O_2$ respectively.

Solution :

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 = 40.00 g

Mass of H = 6.71 g

Mass of O = 100 - (40.00+6.71) = 53.29 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 = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40.00g}{12g/mole}=3.33moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.71g}{1g/mole}=6.71moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.29g}{16g/mole}=3.33moles$

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

For C = $\frac{3.33}{3.33}=1$

For H = $\frac{6.71}{3.33}=2.01\approx 2$

For O = $\frac{3.33}{3.33}=1$

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 = $C_1H_2O_1=CH_2O$

The empirical formula weight = 1(12) + 2(1) + 1(16) = 30 gram/eq

Now we have to calculate the molar mass of acetic acid.

$\text{Molar mass of acetic acid}=\frac{\text{Given mass of acetic acid}}{\text{Moles of acetic acid}}=\frac{14.1g}{0.234mol}=60.2g/mol$

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{60.2}{30}=2$

Molecular formula = $(CH_2O)_n=(CH_2O)_2=C_2H_4O_2$

Therefore, the empirical and molecular formulas of acetic acid is, $CH_2O$  and $C_2H_4O_2$ respectively.