Question

A 3.00 g sample of unknown hydrocarbon is prepared for combustion analysis. after the hydrocarbon undergoes complete combustion, 8.80 g of co2 and 5.40 g of h2o are produced. what is the empirical formula of the unknown?

Answer 1

Answer:

CH₃.

Explanation:

• The complete combustion of hydrocarbon in an excess of oxygen produces CO₂ and H₂O according to the general reaction equation:

CₐHₓ + O₂(excess) → b CO₂ + c H₂O,

Where a is the no. of C atoms in the unknown hydrocarbon.

x is the no. of H atoms in the unknown hydrocarbon.

b is the no. of moles of CO₂ produced from the combustion of unknown hydrocarbon.

c is no. of moles of H₂O produced from the combustion of unknown hydrocarbon.

• We can calculate b and c using the relation:

n = mass / molar mass.

• n of CO₂ = (8.80 g) / (44.0 g/mol) = 0.2 mol.

Every molecule of CO₂ contains 1 C atom, so no. of moles of C atoms in CO₂ and thus in the original hydrocarbon is 0.2 mol.

• n of H₂O = (5.40 g) / (18.0 g/mol) = 0.3 mol.

Every molecule of H₂O contains 2 H atoms, so no. of moles of H atoms in H₂O and thus in the original hydrocarbon is (2 x 0.3) 0.6 mol.

• The ratio of C to H in the unknown hydrocarbon is (0.2): (2 x 0.3) = 0.2: 0.6.

• Dividing by the lowest no. of moles (0.2) that of C.

∴ The ratio of C to H in the unknown hydrocarbon is 1: 3.

So, the empirical formula of the unknown hydrocarbon is CH₃.

Answer 2

Combustion reactions are those who react with oxygen and produce flames. The empirical formula of the hydrocarbon that is undergoing combustion reaction is $\bold{CH_3}$ (methane).

What is combustion reaction?

Combustion reactions are those in which a compound react with oxygen to produce heat as a byproduct.

Given,

Let take the hydrocarbon be x

The reaction will be

$\bold{x + O_2(excess) =b (CO_2) + c H_2O}$

To calculate the number of moles

$\bold{Number\;of \;moles= \dfrac{mass}{molar\;mass}}\\\\\\\bold{Number\;of \;moles\;of \;CO_2= \dfrac{ 8.80 }{44.0 g/mol}=0.2\;mol}\\\\\\\bold{Number\;of \;moles\;of \;H_2O= \dfrac{ 5.40}{18.0 g/mol}=0.3\;mol}$

Since there is 2 molecules of hydrogen in $\bold{H_2O}}$.

So $\bold{(2 \times 0.3)= 0.6 mol}$

Now, the ratio of carbon to hydrogen is  $\bold{\dfrac{0.2}{0.6}= 3 }$

The ratio is 1:3

Thus, the hydrocarbon is $\bold{CH_3}$,  Methane.

Learn more about combustion, here:

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