Question

Complete combustion of 3.20g of a hydrocarbon produced 9.69g of CO2 and 4.96g of H2O. What is the empirical formula for the hydrocarbon?

Answer 1

Let empirical formula for hydrocarbon is CxHy

it will undergo combustion as

CxHy + (x + y/4) O2  ---> xCO2 + (y/2 )H2O

Given that mass of CO2 produced = 9.69 g

So moles of CO2 produced = 9.69 / 44 = 0.22 moles

So moles of carbon present = 0.22 moles

mass of H2O produced = 4.96 g

Moles of H2O produced = mass / molar mass = 4.96 / 18 = 0.28 moles

So moles of H present = 2 X 0.28 = 0.56 moles

Let us divided the moles of each with lowest value of moles

Moles of Carbon = 0.22 / 0.22 = 1 moles

moles of H = 0.56 / 0.22 = 2.55

Multiplying with two to get whole number

the ratio of carbon and hydrogen will be : C:H = 2:5

empirical formula : C2H5

Answer 2

The empirical formula for the hydrocarbon : C₂H₅

Further explanation

Complete combustion of Hydrocarbons with Oxygen will be obtained by CO₂ and H₂O compounds.

If O₂ is insufficient there will be incomplete combustion produced by CO and H and O

Hydrocarbon combustion reactions (specifically alkanes)

$\large {\boxed {\bold {C_nH _ (_2_n _ + _ 2_) + \frac {3n + 1} {2} O_2 ----> nCO_2 + (n + 1) H_2O}}}$

$\rm Mass~A~in~AxBy=\dfrac{x\times atomic~mass~A}{molar~mass~AxBy}\times mass~AxBy$

$\rm mass~B~inAxBy=\dfrac{y\times atomic~mass~B}{molar~mass~AxBy}\times mass~AxBy$

Complete combustion of 3.20g of a hydrocarbon produced 9.69g of CO₂ and 4.96g of H₂O

Then the mass of C in CO₂: (molar mass of CO₂ = 44 g / mol, atomic mass C = 12 g / mol)

$\rm \dfrac{1\times 12}{44}\times 9.69=2.64~grams$

$\rm mol~C=\dfrac{2.64}{12}=0.22$

mass H in H2O (molar mass H2O = 18 g / mol, atomic mass H = 1 g / mol)

$\rm \dfrac{2\times 1}{18}\times 4.96=0.5511~grams$

$\rm mol~H=\dfrac{0.5511}{1}=0.5511$

For example the hydrocarbons: CxHy

x and y are the mole ratio of C and H atoms

the mole ratio of C: H (from CO₂ and H₂O) is the same as the ratio of C and H in hydrocarbon compounds then:

mol C: mol H = 0.22: 0.5511 = 0.4 = 4: 10

So that the compound molecular formula: C₄H₁₀ or can be simplified in the form of the empirical formula C₂H₅

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