What is the empirical formula of an unknown compound that contains 60.3% of magnesium and 39.7% of oxygen?

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What is the empirical formula of an unknown compound that contains 60.3% of magnesium and 39.7% of oxygen?

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Answer 1 · 2016-04-26T17:08:45

$M g O$ $MgO$ , magnesium oxide.

Explanation:

For empirical formula problems we assume 100 g of compound, and then divide thru by the ATOMIC weights of each constituent.

There are thus $\frac{60.3 \cdot g}{24.305 \cdot g \cdot m o l^{- 1}}$ $(60.3*g)/(24.305*g*mol^-1$ $=$ $=$ $2.48 \cdot m o l \cdot M g$ $2.48*mol* Mg$ , and

$\frac{39.7 \cdot g}{15.999 \cdot g \cdot m o l^{- 1}}$ $(39.7*g)/(15.999*g*mol^-1$ $=$ $=$ $2.48 \cdot m o l \cdot O$ $2.48*mol*O$ .

There is clearly a $1 : 1$ $1:1$ molar equivalence, and the empirical formula is $M g O$ $MgO$ , as we would anticipate.

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What is the empirical formula of an unknown compound that contains 60.3% of magnesium and 39.7% of oxygen?

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