Question

A compound contains only carbon, hydrogen, and oxygen. combustion of 9.612 mg of the compound yields 14.41 mg co2 and 3.93 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

Formula used for determining the number of moles:

number of moles =
`(mass)/(molar mass)` -(1)

Mass of
`CO_2` = 14.41 mg (given)

Mass of
`H_2O` = 3.93 mg (given)

Number of moles of
`C` from
`CO_2` =
`(14.41 * 10^(-3) g)/(44 g/mol)`

=
`0.3275* 10^(-3) mol`

Number of moles of
`H` from
`H_2O` =
`(2 * 3.93 * 10^(-3) g)/(18 g/mol)`

=
`0.437* 10^(-3) mol`

Mass of carbon =
`0.3275* 10^(-3)  mol * 12 g /mol`

=
`3.93 * 10^(-3)  g`

Mass of hydrogen =
`0.437  * 10^(-3) mol * 1 g /mol`

=
`0.437 * 10^(-3)  g`

Mass of oxygen = total mass of the compound - mass of carbon- mass of hydrogen

=
`9.612 * 10^(-3) g-  3.93 * 10^(-3) g - 0.437 * 10^(-3)  g`

=
`5.245 * 10^(-3)g`

Now, number of moles of oxygen =
`(5.245 * 10^(-3) g)/(16 g/mol)`

=
`0.328 * 10^(-3)g`

To identify the empirical formula; divide the number of moles of carbon, hydrogen and oxygen with least number of moles.

Thus,

`C_{(0.3275* 10^(-3) )/(0.3275* 10^(-3) )}`

`H_{(0.437* 10^(-3) )/(0.3275* 10^(-3) )}`

`O_{(0.328* 10^(-3) )/(0.3275* 10^(-3) )}`

`C_(1)H_(1.33)O_(1.00)`

Now, multiply the numbers with 3 to get the hydrogen number in whole number, we get

The empirical formula =
`C_(3)H_(4)O_(3)`

The empirical weight =
`3* 12 g/mol+4* 1 g/mol+3* 16 g/mol`

=
`88 g/mol`

Divide the molecular weight with empirical weight to find the factor between empirical formula and molecular formula , we get

`(molecular weight)/(empirical weight)` =
`(176.1 g/mol)/(88 g/mol)`

=
`2`

Now, multiply the numbers of carbon, hydrogen and oxygen with 2 to get molecular formula, we get

Number of carbon = 6

Number of hydrogen = 8

Number of oxygen = 6

Thus, molecular formula is
`C_(6)H_(8)O_(6)`.