The combustion of 1.210 g of a mixture of CH4 and C2H6 gives 3.372 g of CO2

I have to determine the composition of the mixture (mastic %)

and

molar fraction of CH4 in the mixture
CH4 + C2H6: CO2 + 2H2O

C2H6 + O2: CO2 + H2O

CH4 + C2H6 + 02: CO2 +H2O

Thanks for the help

This is a problem with two unknowns; therefore, you need two equations and solve them simultaneously. Write and balance the combustion equations.

CH4 + 2O2 ==> CO2 + 2H2O
2C2H6 + 7O2 ==> 4CO2 + 6H2O

Let X = grams CH4
and Y = grams C2H6
----------------------
eqn 1 is X + Y = 1.210

For equation 2 you want to convert X to grams CO2 from the CH4 component and convert Y to grams CO2 from the C2H6 component.
I will use mm for molar mass.
X(mm CO2/mmCH4) + Y(4*mm CO2/2*mm C2H6) = 3.372

Solve those two equations simultaneously for X and Y to give grams each. I don't know what mastic % means; I assume you want % by mass.
%CH4 = (X/1.210)*100 = ?
%C2H6 = (Y/1.210)*100 = ?

I assume you can handle the mole fraction part. Post your work if you get stuck.

Can you give me the result because mine is a negative mass

Thank you

I don't get a negative number. Post your work and I'll find the error.

Thank you

I post my work but there is something wrong

For the CO2: M=44
For CH4 M= 16
For C2H6 M= 30

number of mole of CO2= 3,372/44 = 0,766

Could you explain me why you put: Y(4xmm CO2/2mmC2H6)?

I would like to use

m1/M1 + m2/M2= O,766
and m1+m2= 1,210

but don't know how to manage with C2H6 in relation with CO2

Thank you very much for you attention

You need to re-read my explanation. If you want to convert 1.210 g to mols that's ok but more work than you need to do.

I presume m1 is the mass of CH4 (that's my X) and m2 is the mass of C2H6(that's my Y). Your equation of
m1/M1 + m2/M2 = 0.766 is right but unnecessary.
m1 + m2 = 1.210 is right, also, however, it is not a different equation. The first one is mole = moles and the second one is grams = grams. You can't use both of these. For the second equation you must convert grams CH4 (that's your m1) in the mixture to grams CO2 produced by the CH4. I did that above with X(44/16). That would be m1(44/16) in your scheme.
Then you need to convert grams C2H6 (that's your m2) to grams CO2 produced from the C2H6. That y(88/30) in mine and m2(88/30) in yours. Then those added together is actually g CO2 produced by CH4 + grams CO2 produced by C2H6 = 3.372 g total.
That second equation becomes for me x(44/16) + Y(88/30) = 3.72 and yours is
m1(44/16)+ Y(88/30) = 3.372

Solve those two equations for X and Y for mine or for m1 and m2 for yours.
Post your work if you get stuck. I have effectively answered this twice, once with my symbols and once with yours. It shouldn't be a problem to use X and Y and not m1 and m2 or the reverse but use what you feel familiar with.

You want to convert

thank you for your help. It is very useful

I'll do it and post my work

My result is m1=0,96 gr and the pourcentage is 0,96/1,21O = 79% (for CH4)

For m2 it is the difference: 21% (for C2H6)

but... the result I should found is
75,6% for m1 (for CH4)
and 24,4 % for m2 (for C2H6)

Thank you for your help