Chapter 17: Problem 12
About \(0.5 \mathrm{~g}\) of an organic compound containing nitrogen on \(\mathrm{Kjeldahlising}\) requires \(29 \mathrm{~mL}\) of N/5 \(\mathrm{H}_{2} \mathrm{SO}_{4}\) for complete neutralization of ammonia. The percentage of nitrogen in the compound is (a) \(34.3 \%\) (b) \(16.2 \%\) (c) \(21.6 \%\) (d) \(14.8 \%\)
Short Answer
Step by step solution
Calculate moles of Hâ‚‚SOâ‚„
Determine moles of NH₃
Convert moles of NH₃ to grams of Nitrogen
Compute the percentage of nitrogen
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Nitrogen Percentage Calculation
To calculate the nitrogen percentage, you need to:
- Determine the moles of \( \text{NH}_3 \). In our case, this matches the moles of \( \text{H}_2\text{SO}_4 \) used since they react 1:1.
- Convert the moles of \( \text{NH}_3 \) to grams of nitrogen by using the atomic weight of nitrogen.
- Take the grams of nitrogen obtained and divide it by the weight of the original compound.
- Multiply by 100 to convert this value into a percentage, which in this exercise is approximately 16.24%.
Stoichiometry
In the neutralization step detailed in our solution:
- The reaction between \( \text{NH}_3 \) and \( \text{H}_2\text{SO}_4 \) follows a 1:1 molar ratio.
- As a result, the moles of \( \text{NH}_3 \) are equivalent to the moles of \( \text{H}_2\text{SO}_4 \) used.
Chemical Reactions
The key reaction in the exercise is the neutralization of ammonia:
- Ammonia reacts with sulfuric acid (\( \text{H}_2\text{SO}_4 \)) in a linear, straightforward 1:1 ratio.
- This reaction changes the ammonia into an ammonium sulfate compound, ultimately indicating the nitrogen content in a measurable form.
Acid-Base Titration
Here’s how it works in our exercise:
- The sulfuric acid has a known normality, and its volume is recorded as it neutralizes the ammonia.
- Completion of the reaction indicates that a specific amount of \( \text{NH}_3 \) was present, allowing us to back-calculate to the moles of nitrogen.