Analysis of a compound shows that it contains 7.0 g nitrogen and 16.0 g oxygen. What is its empirical formula?

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Analysis of a compound shows that it contains 7.0 g nitrogen and 16.0 g oxygen. What is its empirical formula?

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Answer 1 · 2016-10-23T14:59:22

$N O_{2}$ $NO_2$

Explanation:

The empirical formula is the simplest whole number that defines constituent atoms in a species.

Using the given masses, and the atomic masses of each constituent, we define an empirical formula this way:

$Moles of nitrogen$ $"Moles of nitrogen"$ $=$ $=$ $\frac{7.0 \cdot g}{7.0 \cdot g + 16 \cdot g} \times \frac{1}{14.01 \cdot g \cdot m o l^{- 1}} = 0.0217 \cdot m o l$ $(7.0*g)/(7.0*g+16*g)xx1/(14.01*g*mol^-1)=0.0217*mol$ .

$Moles of oxygen$ $"Moles of oxygen"$ $=$ $=$ $\frac{16.0 \cdot g}{7.0 \cdot g + 16 \cdot g} \times \frac{1}{15.999 \cdot g \cdot m o l^{- 1}} = 0.0435 \cdot m o l$ $(16.0*g)/(7.0*g+16*g)xx1/(15.999*g*mol^-1)=0.0435*mol$ .

And thus we have the molar quantities of nitrogen, and oxygen respectively. To access the empirical formula, we simply divide thru by the LOWEST molar quantity, that of nitrogen:

$N : \frac{0.0217 \cdot m o l}{0.0217 \cdot m o l} = 1$ $N:(0.0217*mol)/(0.0217*mol)=1$

$O : \frac{0.0435 \cdot m o l}{0.0217 \cdot m o l} = 2$ $O:(0.0435*mol)/(0.0217*mol)=2$

And thus the empirical formula is $N O_{2}$ $NO_2$ .

Note that to find the molecular formula we need an estimate of molecular mass.

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Analysis of a compound shows that it contains 7.0 g nitrogen and 16.0 g oxygen. What is its empirical formula?

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