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tonio613 tonio613
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Posts: 73
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11 years ago
Combustion of 8.652g of a compound containing C, H, O, and N yields 11.088g of CO2 3.780g of H2O and 3.864g of NO2
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wrote...
11 years ago
this is not an equation question I think itis an empirical formula question

here is the solution

mass of CO2 =11.088g
moles of CO2 =0.252moles
moles of C =0.252moles
mass of C =3.024g

mass of water =3.78g
moles of water =0.21moles
moles of H =0.42moles
mass of H0.42336g

mass of NO2 =3.864g
moles of NO2 =0.084moles
moles of N =0.084moles
mass of N =1.176g

if we assume the compound is C, H and N only we get the O by difference

mass of compound 8.652g
mass of O4.02864g
moles of O0.33572moles

our moles are

C0.252
H0.42
N0.084
O0.33572

or dividing by the smallest

C3
H5
N1
O3.996666667

empirical formula is

C3H5NO4

we don't know if this is the molecular formula but if it was we could write a combustion equation

2C3H5NO4 + 9O2  => 6CO2 +5H2O + 2NO2
wrote...
11 years ago
So you get:

4 C3H5NO4 + 13 O2 Rightwards Arrow 12 CO2 + 10 H2O + 4 NO2
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