Question

Complete combustion of 5.90 g of a hydrocarbon produced 18.8 g of CO2 and 6.75 g of H2O.
What is the empirical formula for the hydrocarbon?

Answer 1

First we have to find moles of C:
Molar mass of CO2:
12*1+16*2 = 44g/mol
(18.8 g CO2) / (44.00964 g CO2/mol) x (1 mol C/ 1 mol CO2) =0.427 mol C 
Molar mass of H2O:
2*1+16 = 18g/mol
As there is 2 moles of H in H2O,
So,

(6.75 g H2O) / (18.01532 g H2O/mol) x (2 mol H / 1 mol H2O) = 0.74mol H 

Divide both number of moles by the smaller number of moles: 
As Smaaler no moles is 0.427:
So,
Dividing both number os moles by 0.427 :
(0.427 mol C) / 0.427 = 1.000 
(0.74 mol H) / 0.427 = 1.733 

To achieve integer coefficients, multiply by 2, then round to the nearest whole numbers to find the empirical formula: 
C = 1 * 2 = 2
H = 1.733 * 2 =3.466
So , the empirical formula is C2H3

Answer 2

Answer: The empirical formula for the hydrocarbon is $CH_2$

Explanation:

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

$C_xH_y+O_2\rightarrow CO_2+H_2O$

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

We are given:

Mass of $CO_2=18.8g$

Mass of $H_2O=6.75g$

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

So, in 18.8 g of carbon dioxide, $\frac{12}{44}\times 18.8=5.12g$ of carbon will be contained.

• For calculating the mass of hydrogen:

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

So, in 6.75 g of water, $\frac{2}{18}\times 6.75=0.75g$ 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 =$\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5.12g}{12g/mole}=0.426moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.75g}{1g/mole}=0.75moles$

• 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 0.426 moles.

For Carbon = $\frac{0.426}{0.426}=1$

For Hydrogen = $\frac{0.75}{0.426}=1.76\approx 2$

• Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 2

Hence, the empirical formula for the given hydrocarbon is $C_1H_{2}=CH_2$