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Chapter 12: Problem 66

The amount of nitrogen in an organic substance can be determined by an analytical method called the Kjeldahl method, in which all the nitrogen in the organic substance is converted to ammonia. The ammonia, which is a weak base, can be neutralized with hydrochloric acid, as described by the equation $$ \mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q) $$ If it requires \(27.5\) milliliters of \(0.150 \mathrm{M} \mathrm{HCl}(a q)\) to neutralize all the \(\mathrm{NH}_{3}(g)\) from a 2.25-gram sample of organic material, calculate the mass percentage of nitrogen in the sample.

Short Answer

Expert verified
The mass percentage of nitrogen is approximately 2.57%.

Step by step solution

01

Understanding the Neutralization Reaction

The first step is to recognize the type of reaction involved. The given reaction \( \mathrm{NH}_{3}(aq) + \mathrm{HCl}(aq) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(aq) \) is a neutralization reaction where one mole of ammonia \( \mathrm{NH}_{3} \) reacts with one mole of hydrochloric acid \( \mathrm{HCl} \). Thus, the number of moles of \( \mathrm{HCl} \) used is equivalent to the number of moles of \( \mathrm{NH}_{3} \) present.
02

Calculate Moles of Hydrochloric Acid

To find the moles of \( \mathrm{HCl} \) used, use the concentration and volume of the \( \mathrm{HCl} \) solution:\[ \text{moles of } \mathrm{HCl} = \text{Volume} \times \text{Concentration} = \frac{27.5 \times 10^{-3} \text{ L} \times 0.150 \text{ mol/L}}{1} = 0.004125 \text{ mol} \]
03

Find Moles of Ammonia

Since the reaction between \( \mathrm{NH}_{3} \) and \( \mathrm{HCl} \) is a 1:1 ratio, 0.004125 moles of \( \mathrm{NH}_{3} \) are neutralized by the \( \mathrm{HCl} \). Therefore, there are 0.004125 moles of \( \mathrm{NH}_{3} \).
04

Calculate Mass of Nitrogen

Now, calculate the mass of nitrogen using the molar mass:\[ \text{Molar mass of nitrogen (N)} = 14.01 \text{ g/mol} \]\[ \text{Mass of nitrogen} = 0.004125 \text{ mol} \times 14.01 \text{ g/mol} = 0.0578 \text{ g} \]
05

Determine Mass Percentage of Nitrogen

To find the mass percentage of nitrogen in the sample, use the formula:\[ \text{Mass percentage} = \left( \frac{\text{mass of nitrogen}}{\text{mass of sample}} \right) \times 100% \]\[ \text{Mass percentage} = \left( \frac{0.0578 \text{ g}}{2.25 \text{ g}} \right) \times 100% \approx 2.57% \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Neutralization Reaction
In the context of the Kjeldahl method, a neutralization reaction is essential for determining the amount of nitrogen in an organic material. This type of reaction occurs when an acid and a base react to form a salt and water, essentially cancelling each other鈥檚 pH effects. In this specific case, ammonia (\( NH_3 \)), a weak base, reacts with hydrochloric acid (\( HCl \)), a strong acid, to form ammonium chloride (\( NH_4Cl \)).
  • Ammonia, naturally derived from the decomposition of nitrogen compounds, releases as a gas during the digestion stage of the Kjeldahl analysis.
  • Through neutralization, one mole of ammonia reacts with one mole of hydrochloric acid, highlighting the 1:1 stoichiometric ratio in the reaction equation provided.
  • This stoichiometry is crucial for calculating the amount of nitrogen, as the number of moles of acid used directly correlates to the moles of ammonia, which was initially part of the organic sample.
Overall, the neutralization step allows analytical chemists to accurately determine the nitrogen content by gauging how much \( HCl \) is needed to fully react with the generated \( NH_3 \).
Mass Percentage
Calculating the mass percentage of nitrogen in a sample is a critical step in understanding its nitrogen content. The mass percentage is a representation of concentration, illustrating how much nitrogen is present in proportion to the entire sample. It is calculated using the formula:\[ \text{Mass percentage} = \left( \frac{\text{mass of nitrogen}}{\text{mass of sample}} \right) \times 100% \]
  • Begin by determining the mass of nitrogen produced in the reaction, derived earlier by calculating the moles of ammonia and then converting to grams using the molar mass of nitrogen, which is 14.01 g/mol.
  • In this example, the resultant mass of nitrogen was 0.0578 grams, from a sample weight of 2.25 grams.
  • Using these values, the mass percentage is calculated as approximately 2.57%, indicating the proportion of nitrogen relative to the entire sample.
This calculation is useful in many fields such as agriculture and food science, providing insight into the nitrogen content, which impacts nutritional and growth factors.
Organic Chemistry Analysis
The Kjeldahl method falls under the branch of organic chemistry analysis, which involves determining specific elements in organic compounds, with nitrogen being a focal point in this case. Several components make this analysis methodical and effective:
  • The sample's digestion converts nitrogen into ammonium sulfate, which, through distillation, releases ammonia.
  • The ammonia is trapped by neutralization with an acid, like hydrochloric acid, in a controlled chemical reaction.
  • Quantitative results are expressed through calculations such as determining mass percentage, providing critical insights into the sample's composition.
This method is highly regarded because it accurately measures nitrogen, an essential element in numerous organic substances.
Used globally in laboratories, the Kjeldahl method aids in educational and industrial settings, helping analysts deduce the nitrogen content in samples that range from agricultural fertilizers to protein-laden foods.

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Most popular questions from this chapter

Suppose we have a solution of lead nitrate, \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)\). A solution of \(\mathrm{NaCl}(a q)\) is added slowly until no further precipitation of \(\mathrm{PbCl}_{2}(s)\) occurs. The \(\mathrm{PbCl}_{2}(s)\) precipitate is collected by filtration, dried, and weighed. A total of \(12.79\) grams of \(\mathrm{PbCl}_{2}(s)\) is obtained from \(200.0\) milliliters of the original solution. Calculate the molarity of the \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) solution.

When silicon is heated with an aqueous solution of sodium hydroxide, sodium silicate and hydrogen gas are formed according to $$ \begin{aligned} \mathrm{Si}(s)+2 \mathrm{NaOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \\\ \mathrm{Na}_{2} \mathrm{SiO}_{3}(a q)+2 \mathrm{H}_{2}(g) \end{aligned} $$ How many milliliters of \(6.00 \mathrm{M} \mathrm{NaOH}(a q)\) are required to react with \(12.5\) grams of silicon? How many grams of hydrogen gas will be produced?

What volume of \(0.865 \mathrm{M} \mathrm{CaCl}_{2}(a q)\) is required to precipitate all the \(\mathrm{Ag}^{+}(a q)\) from \(35.0\) milliliters of \(0.500 \mathrm{M} \mathrm{AgNO}_{3}(a q) ?\) How many grams of \(\mathrm{AgCl}(s)\) will precipitate?

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Bromine is obtained commercially from natural brines from wells in Michigan and Arkansas by the reaction described by the equation $$ \mathrm{Cl}_{2}(g)+2 \mathrm{NaBr}(a q) \rightarrow 2 \mathrm{NaCl}(a q)+\mathrm{Br}_{2}(l) $$ If the concentration of \(\operatorname{NaBr}(a q)\) is \(4.00 \times 10^{-3} \mathrm{M}\), how many grams of bromine can be obtained per cubic meter of brine? How many grams of chlorine are required?
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