A 3.00 gram mixture of methane and ethane were allowed to burn in an excess of oxygen. The released CO2 was captured and its mass was determined to be 8.58 grams. How many grams of methane were in the mixture?

Question

australopithecus · Answer

3.00g of CH3, H3C-CH3

Final product being 8.58g of CO2 and H2O

so we have two equations:

CH4 + 2 O2 -> 2 H2O + CO2
2 C2H6 + 7 O2 -> 6 H2O + 4 CO2

we know we have 8.58g of CO2

This question is tricky let me think about it some more, I wish they would have just used a gas chromatograph

australopithecus · Answer

we know that the moles of CO2 is equal to 

8.58g/44.0095g/mol = 0.195mol

anonymous · Answer

assuming x is the mass of methane and y is the mass of CO2 formed in the burning process of the methane:
$$CH_4+2O_2 \rightarrow 2H_2O+CO_2$$
number of moles of methane is: x/16.04
number of moles of CO2 is: y/44.01

$$2C_2H_6+7O_2 \rightarrow 6H_2O+4CO_2$$
number of moles of ethane is: (3-x)/30.07
number of moles of CO2 is:(8.58-y)/44.01

by stochimetric ratios:
$$\frac{ x }{ 16.04 }=\frac{ y }{ 44.01 } and \frac{ 3-x }{ 30.07 }\times2=\frac{ 8.58-y }{ 44.01 }$$
which makes the mass of methane (x) =1.10g

anonymous · Answer

by the way, 16.04 is the molar mass of methane, 30.07 is the molar mass of ethane and 44.01 is the molar mass of CO2

anonymous · Answer

at the end why did you multiply the 3-x/30.07by 2?

anonymous · Answer

because of stoichiometric ratio between the ethane and CO2 2:4 which is 1:2 then there is twice the number of moles of CO2 than ethane.

anonymous · Answer

i can just cross multiply the x and y values and obtain the same values right?

anonymous · Answer

no because x and y refer here to the masses and the stoichiometric ratios refer to the number of moles, this is why you have to divide the masses by the molar mass in order to build the equations.