Answer:
[tex]C_{4}H_{10}O[/tex]
Explanation:
Hello!
In this case, since the empirical formula of compounds whose by-mass compositions are known, is computed by first computing the moles considering those compositions as actual masses:
[tex]n_C=\frac{64.82g}{12.01g/mol}=5.40mol\\\\n_H=\frac{13.62g}{1.01g/mol}=13.5mol\\\\n_O=\frac{21.58g}{16g/mol}=1.35mol[/tex]
Next, we divide each moles by the fewest ones, in this case those of oxygen to obtain the subscript in the empirical formula:
[tex]C=\frac{5.40}{1.35}= 4\\\\H=\frac{13.5}{1.35}=10\\\\O=\frac{1.35}{1.35}=1[/tex]
Thus, we obtain:
[tex]C_{4}H_{10}O[/tex]
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