A compound contains 63g Mn and 37g O, what is the empirical formula?

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A compound contains 63g Mn and 37g O, what is the empirical formula?

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Answer 1 · 2016-09-19T19:20:21

$M n O_{2}$ $MnO_2$ .

Explanation:

We divide each constituent mass thru by the ATOMIC mass of each element:

$Moles of manganese$ $"Moles of manganese"$ $=$ $=$ $\frac{63 \cdot g}{54.94 \cdot g \cdot m o l^{- 1}}$ $(63*g)/(54.94*g*mol^-1)$ $=$ $=$ $1.15 \cdot m o l$ $1.15*mol$ .

$Moles of oxygen$ $"Moles of oxygen"$ $=$ $=$ $\frac{37 \cdot g}{15.99 \cdot g \cdot m o l^{- 1}}$ $(37*g)/(15.99*g*mol^-1)$ $=$ $=$ $2.31 \cdot m o l$ $2.31*mol$ .

We divide thru by the smallest molar quantity, that of the metal, to give an empirical formula of $M n O_{2}$ $MnO_2$ .

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A compound contains 63g Mn and 37g O, what is the empirical formula?

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