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.Empirical formula

2.Molecular mass

3.Molecular formula

Expert's answer

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.

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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