Question

An organic compound of molar mass greater than $100$ contains only $C,H$ and $N$. The percentage of $C$ is $6$ times that of $H$ while the sum of percentages of $C$ and $H$ is $1.5$ times that of $N$. What is the least molar mass of the compound?

• A. $210\text{g}$
• B. $140\text{g}$
• C. $105\text{g}$
• D. $175\text{g}$

Answer 1

The correct option is B $140\text{g}$

Let us assume $\text{percent of}Hasx\%$,
$\therefore \text{Percent of}C\text{is}6x\%$
As per question $\%C+\%H=1.5\left(\%N\right)$
$\therefore \text{Percent of}N\text{is}\frac{14}{3}x\%$
Assuming $100$ g of the total compound,
Mass of $H$ is $x$ g $\Rightarrow$ moles of $H$ is $x\text{mol}$
Mass of $C$ is $6x$ g $\Rightarrow$ moles of $C$ is $\frac{x}{2}\text{mol}$
Mass of $N$ is $\frac{14x}{3}$ g $\Rightarrow$ moles of $N$ is $\frac{x}{3}\text{mol}$
Taking simplest mole ratio of all three i.e. $C:H:N$ comes out to be $\frac{x}{2}:x:\frac{x}{3}$.
On simplifying it comes out to be $3:6:2$
Empirical formula: $C_3{H}_6{N}_2$
Empirical mass = $70\text{g}$
It is given in question that molar mass of compound comes out to be greater than $100$ hence we need to multiply empirical formula by $2$ for getting molecular formula having minimum molar mass greater than $100$.
Therefore, Molecular formula becomes $C_6{H}_{12}{N}_4$ whose molar mass is $140$ i.e., greater than $100$.