Question

When 3.539 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 12.15 grams of CO2 and 1.990 grams of H2O were produced.

In a separate experiment, the molar mass of the compound was found to be 128.2 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Enter the elements in the order presented in the question.

empirical formula =
molecular formula =

Answer 1

Answer: The empirical formula and molecular formula for the given hydrocarbon is,
`CH` and
`C_(10)H_(10)` respectively.

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' and 'y' are the subscripts of Carbon and hydrogen respectively.

We are given:

Mass of
`CO_2=12.15g`

Mass of
`H_2O=1.990g`

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 12.15 g of carbon dioxide,
`(12)/(44)* 12.15=3.314g` of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 1.990 g of water,
`(2)/(18)* 1.990=0.221g` of hydrogen will be contained.

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =
`\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=(3.314g)/(12g/mole)=0.276moles`

Moles of Hydrogen =
`\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=(0.221g)/(1g/mole)=0.221moles`

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.221 moles.

For Carbon =
`(0.276)/(0.221)=1.25\approx 1`

For Hydrogen =
`(0.221)/(0.221)=1`

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

Hence, the empirical formula for the given compound is
`C_1H_1=CH`

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

`n=\frac{\text{molecular mass}}{\text{empirical mass}}`

We are given:

Mass of molecular formula = 128.2 g/mol

Mass of empirical formula =
`CH=12+1=13g/mol</p><p>`

Putting values in above equation, we get:

`n=(128.2g/mol)/(13g/mol)=9.86\approx 10`

Multiplying this valency by the subscript of every element of empirical formula, we get:

`C_(n)H_(n)=C_(10)H_(10)`

Thus, the molecular formula for the given organic compound is
`C_(10)H_(10)`