Question

A compound contains only carbon, hydrogen, and oxygen. combustion of 11.75 mg of the compound yields 17.61 mg co2 and 4.81 mg h2o. the molar mass of the compound is 176.1 g/mol. what are the empirical and molecular formulas of the compound? (type your answer using the format co2 for co2.)

Answer 1

Number of moles is defined as the ratio of given mass in g to the molar mass.

First, convert the given mass of carbon dioxide in mg to g:

1 mg = 0.001 g

17.61 mg = 0.01761 g

Number of moles of carbon dioxide =
`(0.01761 g)/(44.01 g/mol)`

=
`0.0004001 mol`

Mass of carbon = number of moles of carbon dioxide \times molar mass of carbon

=
`0.0004001 mol* 12.011 g/mol`

=
`0.004806 g`

Number of moles of water=
`(0.00481 g)/(18 g/mol)`

=
`2.672* 10^(-4)`

Since, water contains two hydrogen atoms. Thus,

Moles of hydrogen =
`2* 2.672* 10^(-4)`

=
`5.34* 10^(-4)`

Mass of hydrogen =
`5.34* 10^(-4)* * 1.008 g/mol`

=
`5.34* 10^(-4) g`

Mass of oxygen =
`0.001175-(5.38* 10^(-4)g+0.004806 g)`

=
`0.006405 g`

Number of moles of oxygen =
`(0.006405 g)/(15.999 g/mol)`

=
`0.000400`

Now,

`C_(0.0004001)  H_(0.000534)  O_(0.000400)`

Divide the smallest number to get the whole number,

`C_{(0.0004001)/(0.000400)}  H_{(0.000534)/(0.000400)}  O_{(0.000400)/(0.000400)}`

we get,

`C_(1)  H_(1.33)  O_(1)`

Now, multiply all the subscript by 3 to get the whole number,

`C_(3)     H_(4)      O_(3)` (empirical fomula)

Molar mass of the compound =
`3* 12.011 g/mol+4* 1.008 g/mol+3* 15.999 g/mol`

=
`88.062 g/mol`

Divide given molar mass of the compound with the molar mass of the compound.

=
`(176.1 g/mol)/(88.062 g/mol)`

=
`1.999\simeq 2`

Thus, multiply the subscripts of empirical formula by 2 to get the molecular formula, we get:

`C_(6)H_(8)O_(6)`

Hence, empirical formula is
`C_(3)H_(4)O_(3)` and molecular formula is
`C_(6)H_(8)O_(6)`