# Answer to Question #68128 in Inorganic Chemistry for suryanshu

Question #68128
A welding fuel gas containing C and H only. 3.38g Co2 and 0.690g H2O was produced when some`quantities this gas of burn it O2 volume of 10l of this gas was found to be 11.69 .Find the following .
1.Empirical formula
2.Molecular mass
3.Molecular formula
1
2017-05-09T02:15:09-0400
Combustion of fuel gas:
CxHy+(x+y4)O2=xCO2+(y2)H2O
Convert the masses of CO2 and H2O to moles:
n(CO2)=mM=3.3844=0.077 mol
n(H2O)=mM=0.69018=0.038 mol
So, the mol numbers correspond to the coefficients of the combustion reaction:
x=0.077
y2=0.038
y=0.074
O2 coefficient:
x+y4=0.077+0.0744=0.1
The number of moles O2:
n(O2)=VVM=1022.4=0.44 mol
Assume that oxygen and fuel burnt completely, this means that the mole number of O2 is equal to fuel gas. And molar mass of this fuel is:
M(CxHy)=mn=11.690.44=26 gmol
One can note that coefficients x and y are close to 1 and multiplying by 2 will give empirical - CH and molecular formulaC2H2.

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