Find Percent Composition of Ethylene from Natural Gas Sample

[312213]

Homework Statement

A 5.00g sample of natural gas, containing methane, CH4, and ethylene, C2H4, was burned in excess oxygen, yielding 14.5g CO2 and some H2O as products. What percent of the sample was ethylene?

Homework Equations

Moles and percent composition

The Attempt at a Solution

I don't really understand how to correctly set up the problem. I know that ethylene is about 40% and methane is about 60% by guess-and-checking percents until the two gas more or less result 14.5g CO2 combined, from 5.00g sample, but this way is obviously inefficient. How is it correctly set up to result the correct and accurate answer?

 

[symbolipoint]

Homework Statement

A 5.00g sample of natural gas, containing methane, CH4, and ethylene, C2H4, was burned in excess oxygen, yielding 14.5g CO2 and some H2O as products. What percent of the sample was ethylene?

Homework Equations

Moles and percent composition

The Attempt at a Solution

I don't really understand how to correctly set up the problem. I know that ethylene is about 40% and methane is about 60% by guess-and-checking percents until the two gas more or less result 14.5g CO2 combined, from 5.00g sample, but this way is obviously inefficient. How is it correctly set up to result the correct and accurate answer?

Write the balanced reactions for each combustion - be certain to balance reactants and products correctly. Calculate the molecular weights for methane, ethane(ethylene), and carbon dioxide.

You will have two unknown values; the mass of methane (call it x) and mass of ethylene (call it y). Using these as unknown variables, derive a formula for the resulting mass of carbon dioxide, do so for each gas compound reactant, not including the oxygen reactant (meaning just do this for the methane AND for the ethylene).

Now, you know two things: the expression with x plus the expression with y must equal 14.5 grams, and x+y must equal 5 grams.

If there were a way to construct a table into which I could put values and expressions, and if the typesetting system were easier to use, I would show you the table which I constructed on paper, but into this forum message. Try using the approach described after your quoted message.

 

[symbolipoint]

Homework Statement

A 5.00g sample of natural gas, containing methane, CH4, and ethylene, C2H4, was burned in excess oxygen, yielding 14.5g CO2 and some H2O as products. What percent of the sample was ethylene?

Homework Equations

Moles and percent composition

The Attempt at a Solution

I don't really understand how to correctly set up the problem. I know that ethylene is about 40% and methane is about 60% by guess-and-checking percents until the two gas more or less result 14.5g CO2 combined, from 5.00g sample, but this way is obviously inefficient. How is it correctly set up to result the correct and accurate answer?

Your two reactions should be:

CH₄ + O₂ \rightarrow CO₂+ 2H₂O

CH₂CH₂ + 3O₂ \rightarrow 2CO₂ + 2 waters

 

[312213]

CH₄ + 2O₂ \rightarrow CO₂+ 2H₂O
C₂H₄ + 3O₂ \rightarrow 2CO₂ + 2H₂O

x = CH₄
y = C₂H₄

x + y = 5.00 g

x × (1 mol CH₄/16.0426 g CH₄) × (1 mol CO₂ / 1 mol CH₄) × (44.009 g CO₂ / 1 mol CO₂) \approx 2.74326x

y × (1 mol C₂H₄/28.0536 g C₂H₄) × (2 mol CO₂ / 1 mol C₂H₄) × (44.009 g CO₂ / 1 mol CO₂) \approx 3.137494y

2.74326x + 3.137494y \approx 14.5g CO₂

x + y = 5.00 g \rightarrow y = 5.00 g - x


2.74326x + 3.137494(5.00 g - x) \approx 14.5g CO₂
2.74326x + 15.68747 g - 3.137494x \approx 14.5g CO₂
1.18747 g \approx 0.394234x
3.01209434 \approx x

x + y = 5.00 g
3.01209434 + y = 5.00 g
y \approx 1.988 g

3.01209434 / 5.00 \approx 0.602 \approx 60.2% CH₄ by weight.

1.988 / 5.00 = 0.398 = 39.2% C₂H₄ by weight.

This looks correct. Thank you.

 

[symbolipoint]

312213 you found and corrected my error in the methane reaction: TWO oxygen molecules, not just one oxygen molecule. I trust that the rest of your work is good and will not need to actually check it. In any case, you would not need the actual coefficient of oxygen molecules from the reactants sides in the actual rest of the solution process - but maybe just to check that the reactions are being balanced.