# Question 0c5e8

Oct 26, 2016

${N}_{2} {H}_{4}$ most likely.

#### Explanation:

First change percent to mass. If there was 100 grams of the sample, there would be 100 grams.

So 87.4 % nitrogen would be 87.4 grams of nitrogen.

To find the moles of nitrogen divide the mass of 87.4 grams by the molar mass of nitrogen 14 grams/ mole This gives

$\frac{87.4}{14} = 6.24$ moles nitrogen.

100 grams - 87.4 grams = 12.6 grams this would be the mass of Hydrogen. To find the moles of Hydrogen divide the mass of 12.6 by the molar mass of hydrogen.

$\frac{12.6}{1} = 12.6$ moles of hydrogen

To find the ratio of Nitrogen to Hydrogen in the empirical formula divide the moles of Hydrogen by the moles of Nitrogen

12.6/6.24 = 2.0 Hydrogen atoms: 1.0 Nitrogen atoms.

${N}_{1} {H}_{2}$ This is not the molecular formula because the empirical formula is not balanced ${N}^{- 3} + 2 \times {H}^{+ 1} = + 1$

It would be helpful if the mass was given for the gas. Calculating the new volume using the combined gas law looks like this.

$\frac{{V}_{1} \times {P}_{1}}{T} _ 1 = {V}_{2} \times {P}_{2} / {T}_{2}$

${V}_{1} = 1 l i t e r s$
${P}_{1} = 710$ torr
${T}_{1} = {373}^{o} K$
${V}_{2} = u n k o w n$
${P}_{2} = 760$torr
${T}_{2} = {273}^{o} C$

$1 \times \frac{710}{373} = {V}_{2} \times \frac{760}{273}$ gives

V_2 = .68 liters.

So the new density at STP is .997 grams/.68 liters. which reduces to

1.47 grams / l liter changing this to the molar volume at STP gives

1.47 x 22.4 = 32.8 grams per mole. Dividing the grams per mole by the grams per empirical formula of $N {H}_{2}$ or 16 grams gives

$\frac{32.8}{16} = 2 e m p e r i c a l f \mathmr{and} \mu l a s \mathmr{and} 2 \times N {H}_{2}$ =${N}_{2} {H}_{4}$