Question

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:

9.67%

Explanation:

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Answer 2

Answer:

%w/w protein in the sample is 15.2%

Explanation:

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

Moles of solute = Molarity of the solution × Volume of solution (in L)

Moles of HCI = 0.09552 x 50.001000 = 0.0047moles

Moles of NaOH = 0.05992 x 37.841000 = 0.0023moles

HCI + Na0H -> NaCl + H20

According to stoichiometry:

1 mole of NaOH require 1 mole of HCI

Thus 0.0023 moles of NaOH will require=1/1 × 0.0023 = 0.0023moles of HCI

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

NH3 + HC -> NH4Cl

1 mole of HCI uses = 1 mole of ammonia

Thus 0.0024 moles uses = 1/1× 0.0024 = 0.0024moles of ammonia

Mass of ammonia= moles x Molar mass = 0.0024 x 17g/mol = 0.0408g

We can say that, here, 17 g of ammonia contains = 14 g of Nitrogen

Thus 0.0408 g of ammonia contains =14/17 × 0.0408 = 0.034g of Nitrogen

Hence, 17.45 g of Nitrogen is present in = 100 g of protein

Thus 0.034 g of Nitrogen is present in = 100/17.45 × 0.034 = 0.195g of protein

Now % w/w of protein = (0.195/1.2846) × 100 = 15.2%

Now % w/w of protein = (0.195/1.2846) × 100 = 15.2%Thus %w/w protein in the sample is 15.2%

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