What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen?

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What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen?

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Answer 1 · 2016-12-19T15:49:52

$Z n O_{2}$ $ZnO_2$

Explanation:

With all these problems, we assume that there are $100 \cdot g$ $100*g$ of unknown compound, and then we proceed to calculate the empirical formula:

$Moles of zinc$ $"Moles of zinc"$ $=$ $=$ $\frac{67.1 \cdot g}{65.4 \cdot g \cdot m o l^{- 1}} = 1.03 \cdot m o l$ $(67.1*g)/(65.4*g*mol^-1)=1.03*mol$

$Moles of oxygen$ $"Moles of oxygen"$ $=$ $=$ $\frac{32.9 \cdot g}{15.999 \cdot g \cdot m o l^{- 1}} = 2.06 \cdot m o l$ $(32.9*g)/(15.999*g*mol^-1)=2.06*mol$

Given the molar ratios, we divide thru by the lowest molar quantity to get an empirical formula of $Z n O_{2}$ $ZnO_2$ . The examiner has been nasty in that the expectation would have been $Z n O$ $ZnO$ . However, zinc does form a peroxo species composed of $Z n^{2 +}$ $Zn^(2+)$ and $O_{2}^{2 -}$ $O_2^(2-)$ ions.

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What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen?

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