# When burning 0.29 g of hydrocarbon, 0.45 g of water and 0.448 l were obtained. The formula of the substance?

Let’s represent the hydrocarbon formula as CxHy. Let’s write down the reaction scheme and arrange the coefficients:
2СхНу + (4х + y) / 2О2 = 2хСО2 + уН2О
Let us determine the amount of carbon dioxide substance: n (CO2) = V (CO2) / Vm, where V (CO2) is the volume of CO2,
Vm = 22.4 l / mol – molar volume
n (CO2) = 0.448 / 22.4 = 0.02 mol
Next, we find the amount of water substance: n (H2O) = m (H2O) / M (H2O), where m (H2O) is the mass of water, M (H2O) = 18 g / mol is the molar mass of water.
n (H2O) = 0.45 / 18 = 0.025 mol.
This means that 2 * 0.01 = 0.02 mol of carbon and 2 * 0.025 = 0.05 mol of hydrogen burns out, that is, a hydrocarbon burns out, in which carbon and hydrogen are related as 0.02: 0.05 (or 2: 5) – C2H5.
This is an elementary unit in butane C4H10 and its isomers.
Let’s check our assumptions. M (C4H10) = 58 g / mol is the molar mass of butane and its isomers.
Then the amount of butane substance n (C4H10) = m (C4H10) / M (C4H10), where m (C4H10) is the mass of hydrocarbon according to the problem statement
n (C4H10) = 0.29 / 58 = 0.05 mol. According to the reaction equation, from such an amount of substance, x times more CO2 is obtained – that is, 0.05 * 4 = 0.02 mol and y / 2 times more water, that is, 0.05 * 10/2 = 0.025 mol.
Our assumptions are correct. The hydrocarbon has the formula C4H10

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