Some Basic Concepts of Chemistry - Result Question 8
9. The ratio of mass per cent of C and 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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Answer:
Correct Answer: 9. (d)
Solution:
- We can calculate the simplest whole number ratio of C and H from the data given, as
| Element | Telative Mass | Molar mass | Relative mole | Simplest number | Whole tatio |
|---|---|---|---|---|---|
| C | 6 | 12 | $\frac{6}{12}$ = 0.5 | $\frac{0.5}{0.5} $ = 1 | |
| 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 C and 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 $\left[\mathrm{C}_1 \mathrm{H}2 \mathrm{O}{1.5}\right] \times 2=\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_3$
BETA
