A sample containing the amino acid alanine, CH3CH(NH2)COOH, plus inert matter is analyzed by the Kjeldahl method. A 2.00-g sample is digested, the NH3 is distilled and collected in 50.0 mL of 0.150 M H2SO4, and a volume of 9.0 mL of 0.100 M NaOH is required for back-titration. Calculate the percent alanine in the sample.
The correct answer and explanation is :
To determine the percent alanine in the sample using the Kjeldahl method, we need to calculate how much nitrogen (N) originated from alanine and then relate it to the molecular weight of alanine to find its mass fraction in the sample.
Step-by-step Solution:
1. Acid amount (H₂SO₄) added to trap NH₃:
$$
\text{mol H}_2\text{SO}_4 = M \times V = 0.150 \, \text{mol/L} \times 0.0500 \, \text{L} = 0.00750 \, \text{mol}
$$
2. Base (NaOH) used in back titration:
$$
\text{mol NaOH} = 0.100 \, \text{mol/L} \times 0.0090 \, \text{L} = 0.00090 \, \text{mol}
$$
H₂SO₄ reacts with NH₃ in a 1:2 ratio (1 mol H₂SO₄ reacts with 2 mol NH₃). But for titration purposes, we use:
$$
\text{mol H}_2\text{SO}_4 \text{ neutralized by NH}_3 = \text{total mol H}_2\text{SO}_4 – \text{mol NaOH}
$$
$$
= 0.00750 – 0.00090 = 0.00660 \, \text{mol H}_2\text{SO}_4}
$$
Since 1 mol H₂SO₄ traps 2 mol NH₃:
$$
\text{mol NH}_3 = 2 \times 0.00660 = 0.01320 \, \text{mol}
$$
This is the amount of nitrogen in moles (1 mol NH₃ = 1 mol N).
3. Mass of nitrogen:
$$
\text{mass N} = 0.01320 \, \text{mol} \times 14.01 \, \text{g/mol} = 0.1849 \, \text{g}
$$
4. Mass of alanine:
Alanine (CH₃CH(NH₂)COOH) has one N atom per molecule. Molar mass = 89.09 g/mol.
From nitrogen content:
$$
\text{mol alanine} = \text{mol N} = 0.01320 \, \text{mol}
$$
$$
\text{mass alanine} = 0.01320 \times 89.09 = 1.176 \, \text{g}
$$
5. Percent alanine:
$$
\% \text{alanine} = \frac{1.176}{2.00} \times 100\% = \boxed{58.8\%}
$$
Explanation (300 words):
The Kjeldahl method is a classical technique for determining nitrogen content in organic compounds. Since amino acids like alanine contain nitrogen (from the amino group, –NH₂), the amount of nitrogen in a sample can be used to determine the quantity of alanine present.
In the procedure described, a 2.00 g sample containing alanine and inert (non-nitrogen-containing) matter is digested with sulfuric acid, converting nitrogen into ammonium ions (NH₄⁺). These are then released as ammonia (NH₃) by making the solution basic and are trapped in a known volume and concentration of sulfuric acid (H₂SO₄). The remaining (unreacted) acid is then back-titrated with NaOH to determine how much acid was actually used to trap NH₃.
The amount of H₂SO₄ that reacted with NH₃ gives the number of moles of ammonia, and therefore nitrogen. Since alanine contains one nitrogen atom per molecule, the number of moles of nitrogen is equal to the number of moles of alanine. Using alanine’s molar mass (89.09 g/mol), we calculate how much alanine corresponds to that nitrogen amount.
Finally, the mass of alanine is divided by the total sample mass to calculate the percentage of alanine. This allows us to determine how much of the original 2.00 g sample was actually alanine. The final result is that 58.8% of the sample is alanine.