Question

A 17.11 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 21.71 g CO2 and 5.926 g H2O are produced. In a separate experiment, the molar mass is found to be 104.1 g/mol. Determine the empirical formula and the molecular formula of the organic compound. Enter the elements in the order C, H, O empirical formula

Answer 1

Answer: The empirical formula and the molecular formula of the organic compound is
`CHO` and
`C_4H_4O_4` respectively.

Step-by-step explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

`C_xH_yO_z+O_2\rightarrow CO_2+H_2O`

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of
`CO_2` = 21.71 g

Mass of
`H_2O`= 5.926 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 21.71 g of carbon dioxide, =
`(12)/(44)* 21.71=5.921g` of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 5.926 g of water, =
`(2)/(18)* 5.926=0.658g` of hydrogen will be contained.

Mass of oxygen in the compound = (17.11) - (5.921+0.658) = 10.53 g

Mass of C = 5.921 g

Mass of H = 0.658 g

Mass of O = 10.53 g

Step 1 : convert given masses into moles.

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (5.921g)/(12g/mole)=0.493moles`

Moles of H=
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (0.658g)/(1g/mole)=0.658moles`

Mass of O=
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (10.53g)/(16g/mole)=0.658moles`

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

For C =
`(0.493)/(0.493)=1`

For H =
`(0.658)/(0.493)=1`

For O=
`(0.658)/(0.493)=1`

The ratio of C : H: O = 1: 1: 1

Hence the empirical formula is
`CHO`.

empirical mass of CHO = 12(1) + 1(1) + 1 (16) = 29

Molecular mass = 104.1 g/mol

`n=\frac{\text {Molecular mass}}{\text {Equivalent mass}}=(104.1)/(29)=4`

Thus molecular formula =
`n* {\text {Empirical formula}}=4* CHO=C_4H_4O_4`