0.53g of acetanilide was subjected to Kjeldahl determination, and the ammonia produced was collected in 50cm^3 of 0.50M H2SO4. On dilution to 250cm^3, the acid was titrated against a 25cm^3 portion of 0.05M sodium carbonate solution. 13.60g of acid was required for each titration. What is the percentage of nitrogen in acetanilide?
Question
0.53g of acetanilide was subjected to Kjeldahl determination, and the ammonia produced was collected in 50cm^3 of 0.50M H2SO4. On dilution to 250cm^3, the acid was titrated against a 25cm^3 portion of 0.05M sodium carbonate solution. 13.60g of acid was required for each titration. What is the percentage of nitrogen in acetanilide?
Solution
To solve this problem, we need to follow these steps:
-
Calculate the moles of H2SO4 used in the reaction. The molarity of H2SO4 is 0.50M, and the volume used is 50cm^3 (or 0.05L). So, moles of H2SO4 = Molarity * Volume = 0.50 * 0.05 = 0.025 moles.
-
In the Kjeldahl method, nitrogen in the organic compound is converted to ammonia, which reacts with H2SO4 to form (NH4)2SO4. The reaction is: 2NH3 + H2SO4 -> (NH4)2SO4. From the stoichiometry of the reaction, 1 mole of H2SO4 reacts with 2 moles of NH3. So, the moles of NH3 = 2 * moles of H2SO4 = 2 * 0.025 = 0.05 moles.
-
The molar mass of nitrogen (N) is approximately 14g/mol. So, the mass of nitrogen in the ammonia = moles of NH3 * molar mass of N = 0.05 * 14 = 0.7g.
-
The mass of acetanilide used in the experiment is 0.53g. So, the percentage of nitrogen in acetanilide = (mass of N / mass of acetanilide) * 100% = (0.7 / 0.53) * 100% = 132.08%.
However, this value is greater than 100%, which is not possible. There seems to be a mistake in the problem. The mass of the acid required for each titration (13.60g) is not relevant to the calculation and might have caused some confusion. Please check the problem and try again.
Similar Questions
The percentage of nitrogen in ammonia is given by the expressionSelect an answerA 14 × 10017 B 317 × 100 C 14 × 10034 D 334 × 100
Total number of reaction products for which Kjeldahl's method of nitrogen estimation fails is (X). What is the value of (X)?
The titration between potassium hydrogen phthalate and sodium hydroxide is a 1:1 stoichiometry. If the mean titration volume (titre) of sodium hydroxide to reach the endpoint is 21.5 mL against 25 mL of a 7.17 mM standard solution of potassium hydrogen phthalate, what is the concentration of sodium hydroxide (in mM)? Provide your answer to 3 significant figures.Note: this is a hypothetical scenario and the concentrations may not match the previous question or the values used in your experiment.
An analytical chemist is titrating 63.1mL of a 0.3800M solution of acetic acid HCH3CO2 with a 0.8600M solution of KOH. The pKa of acetic acid is 4.70. Calculate the pH of the acid solution after the chemist has added 21.9mL of the KOH solution to it.Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added.Round your answer to 2 decimal places.
Suppose a chemical engineer studying a new catalyst for the Haber reaction finds that 349. liters per second of dinitrogen are consumed when the reaction is run at 222.°C and the dinitrogen is supplied at 0.64atm. Calculate the rate at which ammonia is being produced. Give your answer in kilograms per second. Round your answer to 2 significant digits.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.