Question

A 0.2 mole of an alkane on complete combustion gave 26.4g of $C{O_2}$. The molecular weight of alkane is
1)16
2)30
3)44
4)58

Answer 1

Hint: At first, calculate the mass of $C{O_2}$ produce by 1 mole of an alkane on dividing the given mass of $C{O_2}$ by number of moles of alkane. Then, calculate the molar mass of $C{O_2}$ and the number of moles of $C{O_2}$ is calculated on dividing mass of $C{O_2}$ by molar mass of $C{O_2}$. The number of moles of $C{O_2}$ is equal to the number of atoms of carbon in $C{O_2}$ which gives us the idea of an alkane and then molar mass or molecular weight of alkane is calculated.

Complete Step by step answer:
An alkane is an acyclic saturated hydrocarbon. In other words, an alkane consists of hydrogen and carbon atoms arranged in a tree structure in which all the carbon-carbon and carbon-hydrogen bonds are single. Alkanes have the general formula of ${C_n}{H_{2n + 2}}$. We are given that 0.2 mole of an alkane on complete combustion i.e,, total quantity of oxygen is used up to produce 26.4g of $C{O_2}$.

From unitary method, if 0.2 mole of an alkane produces = 26.4g of $C{O_2}$
1 mole of an alkane will produce = $\dfrac{{26.4}}{{0.2}}$
$\Rightarrow$ 1 mole of an alkane will produce = 132g of $C{O_2}$
Now, molar mass of $C{O_2}$ = $1 \times molar{\text{ }}or{\text{ }}atomic{\text{ }}mass{\text{ }}of{\text{ }}C + $$2 \times molar{\text{ }}mass{\text{ }}of{\text{ }}O$
$\Rightarrow$ Molar mass of $C{O_2}$ = $1 \times 12 + 2 \times 16$
$\Rightarrow$ Molar mass of $C{O_2}$ = $12 + 32$
$\Rightarrow$ Molar mass of $C{O_2}$ = 44g
Now, we’ll calculate number of moles of $C{O_2}$ = $\dfrac{{mass{\text{ }}of{\text{ C}}{{\text{O}}_2}}}{{Molar{\text{ }}mass{\text{ }}of{\text{ C}}{{\text{O}}_2}}}$
$\Rightarrow$ Number of moles of $C{O_2}$ = $\dfrac{{132}}{{44}}$
$\Rightarrow$ Number of moles of $C{O_2}$ = 3

Hence, 3 moles of $C{O_2}$ will be produced on complete combustion which means 3 atoms of Carbon is present in the compound of $C{O_2}$. Thus, the number of carbon atoms present in the given alkane is 3 as the number of carbon atoms in alkane would be the same as the atoms present in $C{O_2}$ since it is a case of complete combustion.
Therefore, the given alkane is a propane with formula ${C_3}{H_8}$. So, the molecular weight of ${C_3}{H_8}$ is $3 \times 12 + 8 \times 1$ i.e., $36 + 8 = 44g$

Therefore, option 3 is correct.

Note: The number of atoms of carbon in $C{O_2}$ or Carbon dioxide is equal to the number of atoms of carbon in the given alkene as the alkane undergoes through complete combustion due to which all the carbon atoms in alkane combined with oxygen atoms to give carbon dioxide and also remember the unitary method.