Question

A compound containing only c, h, and o, was extracted from the bark of the sassafras tree. The combustion of 43.9 mg produced 119 mg of co2 and 24.4 mg of h2o. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas.

Answer 1

The compound contain only carbon, hydrogen and oxygen, let the empirical formula of compound be
`C_(x)H_(y)O_(z)`.

The mass of compound, carbon dioxide and water is 43.9 mg, 119 mg and 24.4 mg respectively . The molar mass of compound, carbon dioxide and water is 162 g/mol, 44.01 g/mol and 18 g/mol respectively.

Converting the mass into number of moles as follows:

`n=(m)/(M)`

Here, m is mass and M is molar mass.

Number of moles of carbon dioxide
`CO_(2)` will be:

`n=(119* 10^(-3))/(44.01 g/mol )=0.0027 mol`

1 mol of
`CO_(2)` has 1 mol of C thus, number of moles of C from 0.0027 mol of
`CO_(2)` will be 0.0027 mol.

Molar mass of C is 12 g/mol thus, mass of C will be:

`m_(C)=0.0027 mol* 12 g/mol=0.0324 g`

Number of moles of water
`H_(2)O` will be:

`n=(24.4* 10^(-3))/(18 g/mol )=0.001356 mol`

1 mol of
`H_(2)O` has 2 mol of H thus, 0.001356 mol of
`H_(2)O` will have
`2* 0.001356 mol=0.002712 mol` mole of H.

Molar mass of H is 1 g/mol thus, mass of H will be:

`m_(H)=0.002712 mol* 1 g/mol=0.002712 g`

Sum of mass of C and H will be:

`m_(C+H)=0.0324+0.002712=0.035112 g`

Mass of compound is 43.9 mg or 0.0439 g thus, mass of oxygen will be:

`m_(O)=m_(compound)-m_(C+H)=(0.0439-0.035112)g=0.008788 g`

Molar mass of oxygen is 16 g/mol thus, number of moles of oxygen will be:

`n=(0.008788 g)/(16 g/mol)=0.00055 mol`

Taking the molar ratio:

C:H:O=0.0027:0.002712:0.00055=5:5:1

Therefore, empirical formula will be
`C_(5)H_(5)O`.

The molar mass of
`C_(5)H_(5)O` is 81 g/mol

The molar mass of given compound is 162 that is 2 times the molar mass calculated above thus, molecular formula will be
`2* C_(5)H_(5)O=C_(10)H_(10)O_(2)`.

Therefore, empirical formula and molecular formula of the compound is
`C_(5)H_(5)O` and
`C_(10)H_(10)O_(2)` respectively.