Some Basic Concepts of Chemistry 2 Question 9
9. The ratio of mass per cent of $\mathrm{C}$ and $\mathrm{H}$ of an organic compound $\left(\mathrm{C}{x} \mathrm{H}{y} \mathrm{O}{z}\right)$ is $6: 1$. If one molecule of the above compound $\left(\mathrm{C}{x} \mathrm{H}{y} \mathrm{O}{z}\right)$ contains half as much oxygen as required to burn one molecule of compound $\mathrm{C}{x} \mathrm{H}{y}$ completely to $\mathrm{CO}{2}$ and $\mathrm{H}{2} \mathrm{O}$. The empirical formula of compound $\mathrm{C}{x} \mathrm{H}{y} \mathrm{O}_{z}$ is
(2018 Main)
(a) $\mathrm{C}{3} \mathrm{H}{6} \mathrm{O}_{3}$
(b) $\mathrm{C}{2} \mathrm{H}{4} \mathrm{O}$
(c) $\mathrm{C}{3} \mathrm{H}{4} \mathrm{O}_{2}$
(d) $\mathrm{C}{2} \mathrm{H}{4} \mathrm{O}_{3}$
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Solution:
- We can calculate the simplest whole number ratio of $\mathrm{C}$ and $\mathrm{H}$ from the data given, as
| Element | Relative mass |
Molar mass |
Relative mole |
Simplest whole number ratio |
|---|---|---|---|---|
| $\mathrm{C}$ | 6 | 12 | $\frac{6}{12}=0.5$ | $\frac{0.5}{0.5}=1$ |
| $\mathrm{H}$ | 1 | 1 | $\frac{1}{1}=1$ | $\frac{1}{0.5}=2$ |
Alternatively this ratio can also be calculated directly in the terms of $x$ and $y$ as
$\frac{12 x}{y}=\frac{6}{1}($ given and molar mass of $\mathrm{C}=12, \mathrm{H}=1$ )
Now, after calculating this ratio look for condition 2 given in the question i.e. quantity of oxygen is half of the quantity required to burn one molecule of compound $\mathrm{C}{x} \mathrm{H}{y}$ completely to $\mathrm{CO}{2}$ and $\mathrm{H}{2} \mathrm{O}$. We can calculate number of oxygen atoms from this as consider the equation.
$$ \mathrm{C}{x} \mathrm{H}{y}+\left[x+\frac{y}{4}\right] \mathrm{O}{2} \longrightarrow x \mathrm{CO}{2}+\frac{y}{2} \mathrm{H}_{2} \mathrm{O} $$
Number of oxygen atoms required $=2 \times\left[x+\frac{y}{4}\right]=\left[2 x+\frac{y}{2}\right]$
Now given, $\quad z=\frac{1}{2}\left[2 x+\frac{y}{2}\right]=\left[x+\frac{y}{4}\right]$
Here we consider $x$ and $y$ as simplest ratios for $\mathrm{C}$ and $\mathrm{H}$ so now putting the values of $x$ and $y$ in the above equation.
$$ z=\left[x+\frac{y}{4}\right]=\left[1+\frac{2}{4}\right]=1.5 $$
Thus, the simplest ratio figures for $x, y$ and $z$ are $x=1, y=2$ and $z=1.5$
Now, put these values in the formula given i.e. $\mathrm{C}{x} \mathrm{H}{y} \mathrm{O}{z}=\mathrm{C}{1} \mathrm{H}{2} \mathrm{O}{1.5}$
So, empirical formula will be $\left[\mathrm{C}{1} \mathrm{H}{2} \mathrm{O}{1.5}\right] \times 2=\mathrm{C}{2} \mathrm{H}{4} \mathrm{O}{3}$
