When 5.144 grams of a hydrocarbon, C,Hy were burned in a combustion analysis app

Question

When 5.144 grams of a hydrocarbon, C,Hy were burned in a combustion analysis apparatus, 17.39 grams of CO, and 3.560 grams of H,0 were produced. In a separate experiment, the molar mass of the compound was found to be 26.04 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon. Enter the elements in the order presented in the question. empirical formula = molecular formula = n Analysis: Empirical and Molecular For... : This is group attempt 2 of 5 Autosaved at 8:24 PM

Explanation / Answer

1)
let in compound number of moles of C and H be x and y respectively

Number of moles of CO2 = mass of CO2 / molar mass CO2
= 17.39/44
= 0.3952

Number of moles of H2O = mass of H2O / molar mass H2O
= 3.56/18
= 0.1978

Since 1 mol of CO2 has 1 mol of C
Number of moles of C in CO2= 0.3952
so, x = 0.3952

Since 1 mol of H2O has 2 mol of H
Number of moles of H = 2*0.1978 = 0.3956

Divide by smallest to get simplest whole number ratio:
C: 0.3952/0.3952 = 1
H: 0.3956/0.3952 = 1


So empirical formula is:CH

2)

Molar mass of CH,
MM = 1*MM(C) + 1*MM(H)
= 1*12.01 + 1*1.008
= 13.018 g/mol

Now we have:
Molar mass = 26.04 g/mol
Empirical formula mass = 13.018 g/mol
Multiplying factor = molar mass / empirical formula mass
= 26.04/13.018
= 2

So molecular formula is:C2H2

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