When 2.52 g of hydrocarbon with a relative hydrogen vapor density of 42 were burned, 7.92 g of carbon

When 2.52 g of hydrocarbon with a relative hydrogen vapor density of 42 were burned, 7.92 g of carbon dioxide and 3.24 g of water were formed. Determine the molecular formula of the hydrocarbon.

1. Let’s compose the reaction equation, in which x is the number of carbon atoms, y is the number of hydrogen atoms. Some coefficients will be in the form of expressions:

CxHy + (x + y / 4) O2 = xCO2 + (y / 2) H2O;

2. Find the molecular weight of the unknown hydrocarbon:

2 * 42 = 84 g / mol;

3. Let’s find the amount of substance in the unknown hydrocarbon:

2.52 / 84 = 0.03 mol;

4. Let’s find the amount of substance in carbon dioxide:

7.92 / 44 = 0.18 mol;

5. Let’s compose the proportion and find the value of x:

0.03 / 1 = 0.18 / x;

0.03x = 0.18;

x = 0.18 / 0.03

x = 6;

6. Find the amount of matter near the water:

3.24 / 18 = 0.18 mol;

7. Let’s compose the proportion and find the value of y:

0.03 / 1 = 0.18 / (y / 2);

0.03 = 0.36 / y;

y = 0.36 / 0.03;

y = 12;

8. The unknown hydrocarbon has the chemical formula C6H12. This is hexene. Its molecular weight is 84, which corresponds to the calculation in the second step.

Answer: The molecular formula of the hydrocarbon is C6H12.