When 6.646 grams of a hydrocarbon, C,Hy, were burned in a combustion analysis ap

Question

When 6.646 grams of a hydrocarbon, C,Hy, were burned in a combustion analysis apparatus, 20.85 grams of CO2 and 8.538 grams of H,0 were produced In a separate expe molecular formula of the hydrocarbon. nt, the molar mass of the compound was found to be 70.13 g/mol. Determine the empirical formula and the Enter the elements in the order presented in the question. empirical formula = Combustion analysis can also be performed on compounds containing carbon, hydrogen, and oxygen, as shown in the following example. In this case, the amount of oxygen must be determined from the mass of the original sample, not directly from the tion data

Explanation / Answer

The mass of CO2 obtained in the combustion of CxHy = 20.85 g

i.e. The no. of moles of CO2 obtained = 20.85 g/44 g mol-1 = 0.47386 mol

i.e. The no. of moles of carbon (C) in the given hydrocarbon = 0.47386 mol

i.e. The no. of moles of H2O obtained = 8.538 g/18 g mol-1 = 0.47433

Therefore, the no. of moles hydrogen (H) in the given hydrocarbon = 0.47433*2 = 0.94867 mol

Here, the no. of moles of C is minimum. So, divide both the moles of C and H by the moles of C.

Therefore, the no. of moles of C = 0.47386/0.47386 = 1

And the no. of moles of H = 0.94867/0.47386 ~ 2

Hence, the empirical formula of the required hydrocarbon is CH2.

The molar mass of CH2 = 14 g mol-1

The actual molar mass of the required hydrocarbon ~ 70 g mol-1

i.e. n*CH2 = CxHy

i.e. n*14 = 70

i.e. n = 70/14 = 5

Therefore, the molecular formula of the required hydrocarbon = (CH2)5 = C5H10

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