What is the empirical formula of a compound that contains 50.0% carbon, 6.7% hydrogen, and 43.3% oxygen by mass?

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What is the empirical formula of a compound that contains 50.0% carbon, 6.7% hydrogen, and 43.3% oxygen by mass?

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Answer 1 · 2016-04-25T15:16:16

$C_{3} H_{5} O_{2}$ $C_3H_5O_2$

Explanation:

We assume $100$ $100$ $g$ $g$ of unknown compound.

There are thus $\frac{50.0 \cdot g}{12.011 \cdot g \cdot m o l^{- 1}}$ $(50.0*g)/(12.011*g*mol^-1)$ $=$ $=$ $4.163$ $4.163$ $m o l \cdot C$ $mol*C$ .

And $\frac{6.7 \cdot g}{1.00794 \cdot g \cdot m o l^{- 1}}$ $(6.7*g)/(1.00794*g*mol^-1)$ $=$ $=$ $6.65$ $6.65$ $m o l \cdot H$ $mol*H$ .

And $\frac{43.3 \cdot g}{15.999 \cdot g \cdot m o l^{- 1}}$ $(43.3*g)/(15.999*g*mol^-1)$ $=$ $=$ $2.71$ $2.71$ $m o l \cdot O$ $mol*O$ .

We divide thru by the lowest molar quantity, that of oxygen to get a formula of $C_{1.54} H_{2.45} O_{1}$ $C_(1.54)H_(2.45)O_(1)$ , which we double to give a whole number EMPIRICAL formula of $C_{3} H_{5} O_{2}$ $C_3H_5O_2$ .

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What is the empirical formula of a compound that contains 50.0% carbon, 6.7% hydrogen, and 43.3% oxygen by mass?

1 Answer

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