When a hydrocarbon weighing 6.4 g was burned, 19.8 g of carbon dioxide and 9 g of water were formed.

When a hydrocarbon weighing 6.4 g was burned, 19.8 g of carbon dioxide and 9 g of water were formed. The relative density of a substance for oxygen is 4. Find the molecular formula.

1. Let us write down the equation of combustion of an unknown hydrocarbon:

CxHy + O2 → xCO2 + 0.5yH2O;

2.Calculate the chemical amount of the generated carbon dioxide:

n (CO2) = m (CO2): M (CO2);

M (CO2) = 12 + 16 * 2 = 44 g / mol;

n (CO2) = 19.8: 44 = 0.45 mol;

3.Let us write down the formula of carbon dioxide according to its composition:

CO2 → C + 2O;

4.determine the amount of carbon:

n (C) = n (CO2) = 0.45 mol;

5.Calculate the chemical amount of the resulting water:

n (H2O) = m (H2O): M (H2O) = 9: 18 = 0.5 mol;

6.Set the amount of hydrogen:

H2O → 2H + O;

n (H) = n (H2O) * 2 = 0.5 * 2 = 1 mol;

7.determine the ratio of carbon to hydrogen in a hydrocarbon molecule:

n (C): n (H) = 0.45: 1 = 1: 2.2222, that is, the empirical formula CH2.2222;

8.Calculate the relative molecular weight of the hydrocarbon:

Mr (CxHy) = DO2 (CxHy) * Mr (O2) = 4 * 32 = 128;

9.Let’s find the ratio of the molecular weight of the hydrocarbon to the molecular weight of the empirical formula:

Mr (CH2.2222) = 12 + 2.2222 = 14.2222;

Mr (CxHy): Mr (CH2.2222) = 128: 14.2222 = 9,

10. Having increased the indices of the empirical formula by 9 times, we write down the molecular formula of the hydrocarbon:

C1 * 9 H2.2222 * 9 = C9H20.

Answer: С9Н20.