What is the molecular formula of a compound that is 62.58% carbon, 9.63% hydrogen, 27.79% oxygen?

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Answer 1 · 2016-03-27T16:52:11

$C_6H_11O_2$ is the simplest whole number ration that defines constituent atoms in a species, and is, therefore, the empirical formula.

As is the standard with these problems, we assume $100$ $g$ of compound, and break the percentages up into atoms.

In this mass, there are $(62.58*g)/(12.011*g*mol^-1)$ $C$ $=$ $5.21$ $mol$ $C$ .

And, $(9.63*g)/(1.00794*g*mol^-1)$ $H$ $=$ $9.55$ $mol$ $H$ .

And, $(27.79*g)/(15.999*g*mol^-1)$ $O$ $=$ $1.740$ $mol$ $O$ .

We divide thru, by the smallest molar quantity ( $O$ ), and get (almost) the empirical formula: $C_3H_5.5O$ .

Because the empirical formula is by definition the smallest WHOLE number ratio that describes constituent atoms in a species, we DOUBLE this formula.

$C_6H_11O_2$