What is the empirical formula of a compound that contains 30.4% nitrogen and 69.6% oxygen by mass?

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What is the empirical formula of a compound that contains 30.4% nitrogen and 69.6% oxygen by mass?

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Answer 1 · 2017-05-27T20:51:30

$N O_{2}$ $NO_2$ ..........

Explanation:

AS is typical in these problems, we ASSUME an $100 \cdot g$ $100*g$ mass of unknown compound, and we divide thru by the atomic mass:

$Moles of nitrogen = \frac{30.4 \cdot g}{14.01 \cdot g \cdot m o l^{- 1}} = 2.17 \cdot m o l .$ $"Moles of nitrogen"=(30.4*g)/(14.01*g*mol^-1)=2.17*mol.$

$Moles of oxygen = \frac{69.6 \cdot g}{15.999 \cdot g \cdot m o l^{- 1}} = 4.35 \cdot m o l .$ $"Moles of oxygen"=(69.6*g)/(15.999*g*mol^-1)=4.35*mol.$

And we divide thru by the SMALLEST molar quantity to get an empirical formula of $N O_{2}$ $NO_2$ ...........

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What is the empirical formula of a compound that contains 30.4% nitrogen and 69.6% oxygen by mass?

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