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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.

Respuesta :

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:

[tex]\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}[/tex]    

[tex]\text{Moles of} HCl={0.09552\times 50.00}{1000}=0.0047moles[/tex]

[tex]\text{Moles of} NaOH={0.05992\times 37.84}{1000}=0.0023moles[/tex]

[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]  

According to stoichiometry :

1 mole of [tex]NaOH[/tex] require 1 mole of [tex]HCl[/tex]

Thus 0.0023 moles of [tex]NaOH[/tex] will require=[tex]\frac{1}{1}\times 0.0023=0.0023moles[/tex]  of [tex]HCl[/tex]

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

[tex]NH_3+HCl\rightarrow NH_4Cl[/tex]

1 mole of HCl uses = 1 mole of ammonia

Thus 0.0024 moles uses = [tex]\frac{1}{1}\times 0.0024=0.0024moles[/tex] of ammonia

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

17 g of ammonia contains = 14 g of Nitrogen

Thus 0.0408 g of ammonia contains = [tex]\frac{14}{17}\times 0.0408=0.034 g[/tex] of Nitrogen

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

Thus 0.034 g of Nitrogen is present in =[tex]\frac{100}{17.45}\times 0.034=0.195g[/tex] of protein

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

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

mahima6824soni

If the proteins in grains average 17.54% w/w N, the % w/w protein in the sample is 15.2 %.

What is Kjeldahl analysis?

It is a process of converting nitrogen present in any compound into ammonium sulfate.

Step1: calculating the moles by the formula of molarity

molarity × volume

M = 0.09552, V =  50.00 mL =0.05 L

Moles of HCl

[tex]0.09552 \times 0.05 = 0.0047\;mol\\[/tex]

Moles of NaOH

[tex]0.05992 \times 0.03784 = 0.0023 \;mol[/tex]

Moles of HCl used is  0.0047 - 0.0023 = 0.0024

Step2: calculating the mass of ammonia

moles × molar mass

[tex]0.024 \times 17\;g/mol = 0.0408\;g[/tex]

Now, 17 grams of ammonia have 14 grams of nitrogen

so, 0.0408 g ammonia have [tex]\dfrac{14}{17} \times 0.0408 = 0.034[/tex]

In 100 g of protein, there is 17.45 g of nitrogen

So, 0.034 g N have [tex]\dfrac{100}{17.45} \times 0.034 = 0.195 g.\\[/tex]

Step3: calculating the % w/w

[tex]\dfrac{0.195}{1.2846} \times 100 = 15.2 \%[/tex]

Thus, the % w/w is 15.2 %

Learn more about Kjeldahl analysis

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