Question

Consider a sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) at 0.959 atm and 298 K. Upon combusting the entire sample in oxygen, you collect a mixture of gaseous carbon dioxide and water vapor at 1.208 atm and 375 K. This mixture has a density of 1.1128 g/L and occupies a volume five times as large as that of the pure hydrocarbon. Determine the molecular formula of the hydrocarbon.

Answer 1

Answer:

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!