When burning 2.15 g of organic matter, 6.6 g of carbon dioxide and 3.15 g of water were obtained.

When burning 2.15 g of organic matter, 6.6 g of carbon dioxide and 3.15 g of water were obtained. The relative molecular weight of this substance. is equal to 86. Set the formula of the substance.

Let’s write down the solution:
1. Let’s compose the equation:
CxHy + O2 = CO2 + H2O – the combustion reaction of organic matter is accompanied by the release of carbon dioxide and water;
2. Let’s calculate the amount of carbon that is part of organic matter, which confirms the release of CO2.
Formula: n (C) = M. in * m (CO2) / m in * M (CO2);
n (C) = 86 * 6.6 / 2.15 * 44 = 567.6 / 94.6 = 6 mol;
Let’s calculate the amount of hydrogen that is included in the composition of water and organic matter:
Formula: n (H) = M. in * m (H2) / m (in-va) * M (H2O);
n (H) = 86 * 3.15 / 2.15 * 18 = 270.9 / 38.7 = 7 mol;
3. Determine the mass of carbon and hydrogen:
m (C) = 12 * 6 = 72 g;
m (H) = 1 * 7 = 7 g;
4. The sum of the masses is: 72 +7 = 78, which is less than Mr = 86;
5. Suppose that the general formula of the hydrocarbon is: СnH2n + 2 hence, 12n + 2n + 2 = 86;
14n = 84; n = 6;
Formula of substance: С6Н14 – hexane;
Combustion reaction:
2С6Н14 + 19О2 = 12СО2 + 14Н2О.
Answer: C6H14 – organic matter, hexane.