Molecular formula of hydrocarbon is: C₂H₆
Explanation:
The combusting of the hydrocarbon:
CxHy + O₂ → CO₂ + H₂O
Using gas law to obtain molar mass of the gas mixture (CO₂ + H₂O):
P/RT = n/V
Where:
P is pressure (1,208 atm)
R is gas constant (0,082 atmL/molK)
T is temperature (375 K)
n/V = 0,0393 mol/L
1,1128 g/L ÷ 0,0393 mol/L = 28,32 g/mol
Thus, average molecular weight is:
28,32 g/mol = 44,01 g/mol X + 18,02 g/mol Y
1 = X + Y
Where X is CO₂ molar percentage and Y is H₂O molar percentage.
Solving:
X = 0,397
Y = 0,603
39,7% H₂O
60,3% CO₂
With this proportion you can obtain ratio CO₂:H₂O thus:
60,3/39,7 = 1,52 So, 2 CO₂: 3H₂O
The moles of the hydrocarbon are:
PV/RT = n
P (0,959 atm)
V ( 1/5 of final volume)
T (298K)
n = 7,85x10⁻³ mol
The moles of CO₂ are:
0,0393 mol × 2 mol CO₂/ 5 mol total =0,01572.
Ratio of CO₂:CxHY =
0,01572 : 7,85x10⁻³ 2 CO₂: 1 CxHY
Doing:
1 CxHy + O₂ → 2 CO₂ + 3 H₂O
By mass balance:
1 C₂H₆ + 7/2 O₂ → 2 CO₂ + 3 H₂O
Thus, molecular formula of hydrocarbon is: C₂H₆
I hope it helps!
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