What is the empirical chemical formula of a compound that is 69.9% of iron and 30.0% oxygen?

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What is the empirical chemical formula of a compound that is 69.9% of iron and 30.0% oxygen?

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Answer 1 · 2016-04-24T16:44:07

$F e_{2} O_{3}$ $Fe_2O_3$

Explanation:

We assume there are $100 \cdot g$ $100*g$ of compound, and thus there are:

$(i)$ $(i)$ $\frac{69.9 \cdot g}{55.85 \cdot g \cdot m o l^{- 1}}$ $(69.9*g)/(55.85*g*mol^-1)$ $=$ $=$ $1.25 \cdot m o l \cdot F e$ $1.25*mol*Fe$ , and

$(i i)$ $(ii)$ $\frac{30.0 \cdot g}{15.999 \cdot g \cdot m o l^{- 1}}$ $(30.0*g)/(15.999*g*mol^-1)$ $=$ $=$ $1.88 \cdot m o l \cdot O$ $1.88*mol*O$ .

If we divide thru by the lowest molar quantity we get a formula of $F e O_{1.5}$ $FeO_(1.5)$ . Because, by definition the empirical formula is the simplest WHOLE number ratio, we double this result to give $F e_{2} O_{3}$ $Fe_2O_3$ .

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What is the empirical chemical formula of a compound that is 69.9% of iron and 30.0% oxygen?

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