Question

Combustion analysis of a hydrocarbon produced 33.01 g of CO2 and 6.75 g of H2O.Calculate the empirical formula of the hydrocarbon

Answer 1

Given, 33.019 mol of co₂ of and 6.75g of  h₂o

mol of co₂ = mass/molar mass = 33.01g/44gmol = 0.75 mol

Since 1mol have co₂ 1 mol c

Hence 0.75 mol have = 0.75 mol

Now,

mol of H₂O = 6.75g/18gmol = 0.375 mol

Since 1mol H20 have = 2 mol H

Hence 0.375 mol H20 have = 2x0.375 mol H = 0.75 mol H.

Now

mol of H = [0.75 mol x4] = 3mol

mol of c = [0.75 molx4] = 3 mol

Hence

Empirical Formula = CH.

This is done by dividing the atomic mass by the molecular mass and then multiplying it by the mass of the compound is made. Therefore, the molecular formula for this hydrocarbon is CH3 because it has a ratio of 1 carbon atom to 3 hydrogen atoms. The general formula for saturated hydrocarbons with one ring is CnH2n. Most compounds require information other than the chemical formula to elucidate the structure. However, the CH ratio in the chemical formula can provide information about the chemical structure.

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