KCET · Chemistry · Some Basic Concepts of Chemistry
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
- A \(\mathrm{C}_{2} \mathrm{HO}\)
- B \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{O}_{2}\)
- C \(\mathrm{CH}_{2} \mathrm{O}\)
- D \(\mathrm{CHO}\)
Answer & Solution
Correct Answer
(D) \(\mathrm{CHO}\)
Step-by-step Solution
Detailed explanation
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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