Question

A 1.047 g sample of canned tuna was analyzed by the Kjeldahl method; 24.61 mL of 0.1180 M HCl were required to titrate the liberated ammonia. Calculate the percentage of nitrogen in the sample. [

Answer 1

Question #77235, Chemistry / Other

A 1.047 g sample of canned tuna was analyzed by the Kjeldahl method; 24.61 mL of 0.1180 M HCl were required to titrate the liberated ammonia. Calculate the percentage of nitrogen in the sample. [

Solution:

Chemical equation:

$$\mathrm{NH_3} + \mathrm{HCl} = \mathrm{NH_4Cl}$$ $$\begin{array}{l} n(NH_3) = n(HCl) = c(HCl) \times V(HCl) = 0.1180 \frac{mol}{L} \times 0.02461 \, L = 0.002904 \, mol \\ n(N) = n(NH_3) = 0.002904 \, mol \\ A(N) = 14.0067 \frac{g}{mol} \\ m(N) = 14.0067 \frac{g}{mol} \times 0.002904 \, mol = 0.04068 \, g \\ \%(N) = \frac{0.04068 \, g}{1.047 \, g} \times 100\% = 3.88\% \end{array}$$

Answer:

3.88%

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