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}}= (40.00g)/(12g/mole)=3.33moles`

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

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (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 =
`(3.33)/(3.33)=1`

For H =
`(6.71)/(3.33)=2.01\approx 2`

For O =
`(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}}=(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=(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.