Question

8. The protein in a 1.2846-g sample of an oat cereal is determined by a Kjeldahl analysis. The sample is digested with H2SO4, the resulting solution made basic with NaOH, and the NH3 distilled into 50.00 mL of 0.09552 M HCl. The excess HCl is back titrated using 37.84 mL of 0.05992 M NaOH. Given that the proteins in grains average 17.54% w/w N, report the %w/w protein in the sample.

Answer 1

Answer: The %w/w protein in the sample is 15.2 %

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}$    

$\text{Moles of} HCl={0.09552\times 50.00}{1000}=0.0047moles$

$\text{Moles of} NaOH={0.05992\times 37.84}{1000}=0.0023moles$

$HCl+NaOH\rightarrow NaCl+H_2O$  

According to stoichiometry :

1 mole of $NaOH$ require 1 mole of $HCl$

Thus 0.0023 moles of $NaOH$ will require=$\frac{1}{1}\times 0.0023=0.0023moles$  of $HCl$

moles of HCl used = (0.0047-0.0023) = 0.0024

$NH_3+HCl\rightarrow NH_4Cl$

1 mole of HCl uses = 1 mole of ammonia

Thus 0.0024 moles uses = $\frac{1}{1}\times 0.0024=0.0024moles$ of ammonia

Mass of ammonia= $moles\times {\text {Molar mass}}=0.0024\times 17g/mol=0.0408g$

17 g of ammonia contains = 14 g of Nitrogen

Thus 0.0408 g of ammonia contains = $\frac{14}{17}\times 0.0408=0.034 g$ of Nitrogen

Now 17.45 g of Nitrogen is present in = 100 g of protein

Thus 0.034 g of Nitrogen is present in =$\frac{100}{17.45}\times 0.034=0.195g$ of protein

Now % w/w of protein = $\frac{0.195}{1.2846}\times 100=15.2\%$

Thus %w/w protein in the sample is 15.2%