Question

One method of analyzing amino acids is the van Slyke method. The characteristic amino groups(-NH₂) in protein material are allowed to react with nitrous acid, HNO₂, to form N₂ gas. From the volume of the gas, the amount of amino acid can be determined. A 0.0604-g sample of a biological sample containing glycine, CH₂(NH₂)CO₂H, was analyzed by the van Slyke method and yielded 3.70mL of N₂ collected over water at a pressure of 735 torrs and 29 ̊C. What was the percentage of glycine in the sample?

\[{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{(N}}{{\rm{H}}_{\rm{2}}}{\rm{)C}}{{\rm{O}}_{\rm{2}}}{\rm{H+HN}}{{\rm{O}}_{\rm{2}}}\to{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{(OH)C}}{{\rm{O}}_{\rm{2}}}{\rm{H+}}{{\rm{H}}_{\rm{2}}}{\rm{O+}}{{\rm{N}}_{\rm{2}}}\]

Short answer

The percentage of glycine in the sample is 17.22%.

Step-by-step solution

Dalton’s Law

Dalton's law states that the overall pressure exerted in a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases.

Calculation of number of moles

To calculate the percentage of glycine in the sample, we need to find the mass of glycine in the sample.

\[{\rm{masspercentage=}}\frac{{{\rm{massofglycine}}}}{{{\rm{massofsample}}}}{\rm{ \times 100\% }}\]

According to Dalton's law, the total pressure is the sum of the partial pressures of nitrogen and gaseous water.

\[{{\rm{P}}_{\rm{T}}}{\rm{=}}{{\rm{P}}_{{{\rm{N}}_{\rm{2}}}}}{\rm{+}}{{\rm{P}}_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\]

Rearranging the above equation to solve for the pressure of nitrogen gives

\[{{\rm{P}}_{{{\rm{N}}_{\rm{2}}}}}{\rm{=}}{{\rm{P}}_{\rm{T}}}{\rm{-}}{{\rm{P}}_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\]

The pressure of water vapor above a sample of liquid water at \[{\rm{2}}{{\rm{9}}^{\rm{^\circ }}}{\rm{C}}\] is \[{\rm{30 torr}}\],

so\[{{\rm{P}}_{{{\rm{N}}_{\rm{2}}}}}{\rm{=735torr-30torr=705 torr}}\]

It is given that

\begin{aligned}{\rm{P=705torr\times}}\frac{{{\rm{1\;atm}}}}{{{\rm{760torr}}}}{\rm{=0}}{\rm{.938\;atm}}\\{\rm{T=2}}{{\rm{9}}^{\rm{^\circ}}}{\rm{C+273=302\;K}}\\{\rm{V=3}}{\rm{.7\;mL=3}}{\rm{.7\times1}}{{\rm{0}}^{{\rm{-3}}}}{\rm{\;L}}\end{aligned}

The number of moles of\[{{\rm{N}}_2}\]formed is

\begin{aligned}{\rm{n=}}\frac{{{\rm{PV}}}}{{{\rm{RT}}}}\\{\rm{=}}\frac{{{\rm{0}}{\rm{.938\;atm\times3}}{\rm{.7\times1}}{{\rm{0}}^{{\rm{-3}}}}{\rm{\;L}}}}{{\left({{\rm{0}}{\rm{.08206\;L\;atm\;mol\;}}{{\rm{K}}^{{\rm{-1}}}}}\right){\rm{\times302\;K}}}}\\{\rm{=1}}{\rm{.4\times1}}{{\rm{0}}^{{\rm{-4}}}}{\rm{\;mol}}\end{aligned}

Calculation for percentage of Glycine

The balanced reaction is –

\[{\rm{C}}{{\rm{H}}_{\rm{2}}}\left({{\rm{N}}{{\rm{H}}_{\rm{2}}}}\right){\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{H+HN}}{{\rm{O}}_{\rm{2}}}\to{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{(OH)C}}{{\rm{O}}_{\rm{2}}}{\rm{H+}}{{\rm{H}}_{\rm{2}}}{\rm{O+}}{{\rm{N}}_{\rm{2}}}\]

1 mol of glycine gives 1 mol of N₂. The mole ratio is 1:1. The number of moles of glycine reacted to produce1.38×10^-4 mol N₂ is

\begin{aligned}{\rm{1}}{\rm{.38\times1}}{{\rm{0}}^{{\rm{-4}}}}{\rm{\;mol\times}}\frac{{{\rm{1\;molglycine}}}}{{{\rm{1\;mol}}{{\rm{N}}_{\rm{2}}}}}{\rm{=1}}{\rm{.38\times1}}{{\rm{0}}^{{\rm{-4}}}}{\rm{\;mol}}\\{\rm{Molesofglycinereacted=}}\frac{{{\rm{Massofglycine}}}}{{{\rm{Molarmassofglycine}}}}\\{\rm{Massofglycinereacted=}}\frac{{{\rm{Massofglycine}}}}{{{\rm{Molesofglycine}}}}\\{\rm{=}}\frac{{{\rm{1}}{\rm{.38\times1}}{{\rm{0}}^{{\rm{-4}}}}{\rm{\;mol}}}}{{{\rm{75}}{\rm{.07\;g/mol}}}}\\{\rm{=1}}{\rm{.04\times1}}{{\rm{0}}^{{\rm{-2}}}}{\rm{\;g}}\end{aligned}

Mass percentage of glycine in the sample is

\begin{aligned}{\rm{masspercentage=}}\frac{{{\rm{massofglycine}}}}{{{\rm{massofsample}}}}{\rm{\times100\%}}\\{\rm{=}}\frac{{{\rm{1}}{\rm{.04\times1}}{{\rm{0}}^{{\rm{-2}}}}{\rm{\;g}}}}{{{\rm{0}}{\rm{.0604\;g}}}}{\rm{\times100}}\\{\rm{=17}}{\rm{.22\%}}\end{aligned}

Therefore, the percentage of glycine is 17.22%.