Answer to Question #182953 in Organic Chemistry for smart ben

Question

A 3.1g sample of unknown organic gas molecule compose of carbon, hydrogen and oxygen undergoes complete combustion to produce 4.4g CO2 and 2.7g H2O. Calculate the empirical and molecular formula of the gas if the molar mass is 62g/mol.

Answer 1

Amounts of CO₂ and H₂O in moles are:

$n(CO_2)=\frac{4.4g}{44g/mol}=0.1mol$

$n(H_2O)=\frac{2.7g}{18g/mol}=0.15mol$

The mass of O₂ can also be determined directly since the total mass of reactants is always equal to the total mass of products:

$m(O_2)=4.4+2.7-3.1=4.0g$

The amount of O₂ in moles:

$n(O_2)=\frac{4.0g}{32g/mol}=0.125mol$

Therefore, the O₂ : CO₂ : H₂O molar ratio is 0.125 : 0.1 : 0.15, which equals 5 : 4 : 6 as the lowest whole number ratio.

Now the overall equation can be written:

nC_xH_yO_z + 5O₂ --> 4CO₂ + 6H₂O

For the equation to become balanced, there should be 4 carbon atoms, 12 hydrogen atoms, and 4 oxygen atoms on the left side, in other words, x : y : z = 4 : 12 : 4. The lowest whole number ratio of this (1 : 3 : 1) determines the empirical formula, which is CH₃O. To calculate the molecular formula, the molar masses of the empirical formula unit and the unknown compound should be compared:

$M(CH_3O)=12+3\times1+16=31g/mol$ , which is 2 times less than the actual molar mass (62 g/mol). Therefore, the subscripts of the empirical formula should be multiplied by 2 giving C₂H₆O₂.

Answer:

Empirical formula CH₃O

Molecular formula C₂H₆O₂