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A pure compound contains $2.4 \mathrm{~g}$ of $C$, $1.2 \times 10^{23}$ atoms of $\mathrm{H}, 0.2$ moles of oxygen atoms. Its empirical formula is
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The correct answer is:
$\mathrm{CHO}$
Moles of carbon $=\frac{w}{M}=\frac{2.4}{12}=0.2 \mathrm{~mol}$
Moles of $\mathrm{H}$ in $1.2 \times 10^{23}$ atoms of $\mathrm{H}=\frac{1.2 \times 10^{23}}{6 \times 10^{23}}$
$=0.2 \mathrm{~mol}$
Moles of oxygen atoms $=0.2$ moles
Now, simplest ratio of the moles of $\mathrm{C}, \mathrm{H}$ and $\mathrm{O}$ is as follows
$\begin{aligned}
&=\mathrm{C}: \mathrm{H}: \mathrm{O} \\
&=0.2: 0.2: 0.2=\mathrm{CHO}
\end{aligned}$
Thus, the empirical formula of the compound is CHO.
Moles of $\mathrm{H}$ in $1.2 \times 10^{23}$ atoms of $\mathrm{H}=\frac{1.2 \times 10^{23}}{6 \times 10^{23}}$
$=0.2 \mathrm{~mol}$
Moles of oxygen atoms $=0.2$ moles
Now, simplest ratio of the moles of $\mathrm{C}, \mathrm{H}$ and $\mathrm{O}$ is as follows
$\begin{aligned}
&=\mathrm{C}: \mathrm{H}: \mathrm{O} \\
&=0.2: 0.2: 0.2=\mathrm{CHO}
\end{aligned}$
Thus, the empirical formula of the compound is CHO.
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