What is the molecular formula for the compound that is made up of 30.45 g N and 69.55 g O, and has a formula mass of 92.02 amu?

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What is the molecular formula for the compound that is made up of 30.45 g N and 69.55 g O, and has a formula mass of 92.02 amu?

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Answer 1 · 2018-04-14T10:43:58

$N_{2} O_{4}$ $"N"_2"O"_4$

Explanation:

Moles of nitrogen

$n_{N} = \frac{30.45 g}{14 g/mol} = 2.175 mol$ $"n"_"N" = "30.45 g"/"14 g/mol" = "2.175 mol"$

Moles of oxygen

$n_{O} = \frac{69.55 g}{16 g/mol} = 4.347 mol$ $"n"_"O" = "69.55 g"/"16 g/mol" = "4.347 mol"$

Ratio of moles of nitrogen and oxygen

$\frac{n_{N}}{n_{O}} = \frac{2.175 mol}{4.347 mol} \approx \frac{1}{2}$ $"n"_"N"/"n"_"O" = "2.175 mol"/"4.347 mol" ≈ 1/2$

∴ Emperical formula ${= NO}_{2}$ $"= NO"_2$

$\frac{Molecular formula mass}{Emperical formula mass} = \frac{92.02 amu}{46.0 amu} \approx 2$ $"Molecular formula mass"/"Emperical formula mass" = "92.02 amu"/"46.0 amu" ≈ 2$

∴ Molecular formula $= 2 \times {NO}_{2} = N_{2} O_{4}$ $= 2 × "NO"_2 = color(Green)("N"_2"O"_4)$

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What is the molecular formula for the compound that is made up of 30.45 g N and 69.55 g O, and has a formula mass of 92.02 amu?

1 Answer

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