Complete combustion of 2.90g of a hydrocarbon produced 9.32g of CO2 and 3.18g of H2O. What is the empirical formula for the hydrocarbon?

Answer;

= C3H5

Explanation and solution;

1 mole of CO2 contains 44 g, of which 12 g are carbon

Thus, mass of carbon in 9.32 g will be;

(12/44) * 9.32 g = 2.542 g

Mass of Hydrogen in 3.18 g of water;

= (2/18) * 3.18 g = 0.353 g

we then find the number of moles;

Moles of carbon; 2.542 / 12 = 0.2118 moles

Moles of Hydrogen = 0.353 moles

The ratios of C; H;

= 1 : 0.353 / 0.2118

= 1 : 5/3

= 3: 5

Therefore; the empirical formula of the hydrogen carbon is; C3H5

Mass of CO₂ = 9.32g

Molar mass of CO₂ = 44 g/mol

Mass of H₂O = 3.18 g

Molar mass of H₂O = 18 g/mol

Moles = mass / molar mass

9.32 g CO₂ x (1 mol CO₂ / 44 g CO₂) = 0.2118 mol CO₂

Every CO₂ molecule has 1 Carbon atom, therefore 0.2118 mol of CO₂ will have 0.2118 moles of C

3.18 g H₂O x (1 mol H₂O / 18 g H₂O) = 0.177 mol H₂O

In every H₂O molecule there are 2 atoms of H therefore 0.177 mol of H₂O will have 2 x 0.177 or 0.354 moles of H

Now the ratio of C : H = 0.2118 : 0.354

To get the whole number we divide both numbers in the ratio by the lowest number.

C : H

= (0.2118/0.2118) : (0.354 / 0.2118)

= 1:1.67

Since we cannot round, we multiply by 3 to clear the fraction:

C = 1 x 3 = 3

H = 1.67 x 3 = 5

Thus the empirical formula is C₃H₅.