Question

When 3.915 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 12.74 grams of CO2 and 3.913 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 54.09 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Answer 1

Answer: The empirical formula and the molecular formula of the hydrocarbon is
`C_2H_3` and
`C_4H_6` respectivley.

Step-by-step explanation:

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

`C_xH_y+O_2\rightarrow CO_2+H_2O`

where, 'x', 'y' are the subscripts of Carbon, hydrogen

We are given:

Mass of
`CO_2` = 12.74 g

Mass of
`H_2O`= 3.913 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 12.74 g of carbon dioxide, =
`(12)/(44)* 12.74=3.474g` of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 3.913 g of water, =
`(2)/(18)* 3.913=0.435g` of hydrogen will be contained.

Mass of C = 3.474 g

Mass of H = 0.435 g

Step 1 : convert given masses into moles.

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (3.474g)/(12g/mole)=0.289moles`

Moles of H=
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (0.435g)/(1g/mole)=0.435moles`

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

For C =
`(0.289)/(0.289)=1`

For H =
`(0.435)/(0.289)=1.5`

The ratio of C : H = 1: 1.5

The whole number ratio will be = 2: 3

Hence the empirical formula is
`C_2H_3`.

The empirical weight of
`C_2H_3` = 2(12.01)+3(1.008)= 27.04 g.

The molecular weight = 54.09 g/mole

Now we have to calculate the molecular formula.

`n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=(54.09)/(27.04)=2`

The molecular formula will be=
`2* C_2H_3=C_4H_6`