The Kjeldahl procedure was used to analyze 256 mL of a

solution containing 37.9 mg protein/mL The liberated NH3 was

collected in 5.00 mL of 0.033 6 M HCl, and the remaining acid

required 6.34 mL of 0.010 M NaOH for complete titration. What is

the weight percent of nitrogen in the protein?

Weight percent is one of the parameters used to express the concentration of a solution. It is expressed as, the weight percent of a solute

$=\frac{\mathrm{weight}\mathrm{of}\mathrm{solute}}{\mathrm{weight}\mathrm{of}\mathrm{solution}}\times 100\%$

volume of sample solution$256\mathrm{\mu L}=256\times {10}^{‐6}\mathrm{L}=0.256\mathrm{ml}$

mass of protein in the sample $=37.9\mathrm{mg}/\mathrm{mL}$

the volume of $\mathrm{HCI}$consumed by ${\mathrm{NH}}_3=5.00\mathrm{mL}$

Molarity of $\mathrm{HCI}=0.0336\mathrm{M}$

Molarity of $\mathrm{NaOH}=0.010\mathrm{M}$

The volume of $\mathrm{NaOH}$consumed by $\mathrm{HCI}=6.34\mathrm{mL}$

37.9 mg of protein is digested in a solution of $256\mathrm{\mu L}$and liberates ${\mathrm{NH}}_3$. ${\mathrm{NH}}_3$distilled into

whose concentration is $0.0336\mathrm{M}$. Unreacted $\mathrm{HCI}$requires $6.34\mathrm{mL}$of $0.010\mathrm{MNaOH}$.

Quantity of HCl before reaction is $\left(5.00\mathrm{mL}\times 0.0336\mathrm{mmol}/\mathrm{mL}\right)=0.168\mathrm{mmoL}$.

The quantity of $\mathrm{NaOH}$required for titration of unreacted $\mathrm{HCI}$is

$\left(6.34\mathrm{mL}\times 0.010\mathrm{Mmol}/\mathrm{mL}\right)=0.0634\mathrm{mmol}$

.

The difference between two quantities gives quantity of ${\mathrm{NH}}_3$produced by digestion of protein solution which distilled into $\mathrm{HCI}$which is given as

$$\begin{gathered} 0.168\mathrm{mmol}-0.0634\mathrm{mmol} \\ =0.1046\mathrm{mmol} \end{gathered}$$

As we know one mol of Nitrogen in protein furnishes one mol of ${\mathrm{NH}}_3$, the protein must contain $0.1046\mathrm{mmol}$$\left(0.1046\mathrm{mmol}\right)\left(14.0007\frac{\mathrm{mgN}}{\mathrm{mmol}}\right)=1.465\mathrm{mgN}$of Nitrogen and it is given in milligrams as,

Given that in $256\mu \mathrm{L}\left(=0.256\mathrm{mL}\right)$solution $37.9\mathrm{mg}$of protein is present per $m L$ of the solution. Thus 0.256 mL of a solution sample contains $9.702\mathrm{mg}$protein. The weight percent of Nitrogen in protein is,

$\frac{1.465\mathrm{mgN}}{9.702\mathrm{mg}\mathrm{protein}}=15.1\%$