Problem 123 - General Chemistry

Question

A \(0.608-\mathrm{g}\) sample of fertilizer contained nitrogen as ammonium sulfate, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\). It was analyzed for nitrogen by heating with sodium hydroxide. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(s)+2 \mathrm{NaOH}(a q) \longrightarrow\) $$ \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{NH}_{3}(g) $$ The ammonia was collected in \(46.3 \mathrm{~mL}\) of \(0.213 \mathrm{M} \mathrm{HCl}\) (hydrochloric acid), with which it reacted. $$ \mathrm{NH}_{3}(g)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q) $$ This solution was titrated for excess hydrochloric acid with \(44.3 \mathrm{~mL}\) of \(0.128 \mathrm{M} \mathrm{NaOH}\). $$ \mathrm{NaOH}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ What is the percentage of nitrogen in the fertilizer?

Step-by-Step Solution

Answer

The percentage of nitrogen in the fertilizer is approximately 9.65%.

Step 1: Calculate moles of HCl initially present

Calculate the moles of HCl initially present in the 46.3 mL solution:\[\text{Moles of HCl} = 0.0463 \text{ L} \times 0.213 \text{ M} = 0.0098619 \text{ moles}\]

Step 2: Calculate moles of NaOH used for titration

Calculate the moles of NaOH used in the titration:\[\text{Moles of NaOH} = 0.0443 \text{ L} \times 0.128 \text{ M} = 0.0056704 \text{ moles}\]

Step 3: Calculate moles of excess HCl

Since NaOH reacts with the excess HCl, calculate the moles of excess HCl remained after reaction with ammonia:\[\text{Moles of excess HCl} = \text{Moles of NaOH} = 0.0056704 \text{ moles}\]

Step 4: Determine moles of HCl that reacted with ammonia

Subtract the moles of excess HCl from the total moles of HCl to find the moles of HCl that reacted with ammonia:\[\text{Moles of HCl that reacted with ammonia} = 0.0098619 \text{ moles} - 0.0056704 \text{ moles} = 0.0041915 \text{ moles}\]

Step 5: Calculate moles of ammonia

Ammonia reacts in a 1:1 mole ratio with HCl. Thus, the moles of ammonia generated by the decomposition of ammonium sulfate is the same as the moles of HCl that reacted:\[\text{Moles of } \text{NH}_3 = 0.0041915 \text{ moles}\]

Step 6: Calculate moles of nitrogen

Each mole of \(\text{NH}_3\) contains one mole of nitrogen, so:\[\text{Moles of nitrogen} = 0.0041915 \text{ moles}\]

Step 7: Calculate mass of nitrogen

Convert moles of nitrogen to grams using nitrogen's molar mass \(14.01 \text{ g/mol}\):\[\text{Mass of nitrogen} = 0.0041915 \text{ moles} \times 14.01 \text{ g/mol} = 0.058691115 \text{ g}\]

Step 8: Determine percentage of nitrogen in the fertilizer

Divide the mass of nitrogen by the total mass of the fertilizer sample and multiply by 100 to find the percentage of nitrogen:\[\text{Percentage of nitrogen} = \left(\frac{0.058691115 \text{ g}}{0.608 \text{ g}}\right) \times 100 \approx 9.65\%\]

Key Concepts

Percent Composition

Percent composition is a fundamental concept when analyzing the makeup of a chemical substance. In this context, it refers to the percentage by mass of a particular element within a compound or mixture. To calculate percent composition, divide the mass of the element of interest by the total mass of the compound, then multiply by 100 to convert it to a percentage.
When considering our example, the percent composition was specifically used to determine the nitrogen content of a fertilizer sample. This process helps in understanding how much nitrogen, as a key nutrient, is present relative to the whole fertilizer. Calculating the percent composition provides insight into the efficiency or richness of a fertilizer.
If you know the detailed percentage of each component, such as nitrogen in this case, it is easier to adjust chemical formulations if needed. This makes percent composition a critical aspect of quality control in chemical industries.

Titration Calculation

Titration is a key analytical technique used in chemistry to determine the concentration of an unknown solution. It involves carefully adding a solution of known concentration (the titrant) to a measured amount of the unknown solution until the reaction reaches completion, indicated by a color change or a pH change.
In the given exercise, the titration calculation was crucial to finding out how much hydrochloric acid (HCl) was initially present and how much was used up in the reaction with ammonia. By determining the moles of HCl that reacted, and the moles used in the back-titration with sodium hydroxide (NaOH), we were able to isolate the moles of ammonia produced from the ammonium sulfate.

• First, calculate the total moles of HCl present before the reaction.
• Then, determine the moles of HCl left after back-titration.
• Subtract these to find the moles that reacted with ammonia.
• Finally, relate this to the amount of nitrogen originally in the fertilizer.

Titration is an essential part of chemical analysis because it allows for precise quantification of reactants and their resultant products.

Nitrogen Content Analysis

Analyzing the nitrogen content of a sample is vital in many fields, especially in agriculture, where nitrogen richness determines fertilizer effectiveness. The nitrogen content in a fertilizer can influence plant growth, yields, and environmental effects.
For this exercise, the nitrogen content analysis was carried out via chemical reactions that converted all nitrogen present in the compound to a measurable form of ammonia gas. This conversion is crucial because ammonia reacts in a 1:1 ratio with HCl, allowing us to back-calculate the exact amount of nitrogen present.

• Identifying the total nitrogen through quantifying moles of ammonia is a common technique.
• It involves calculating the mass of nitrogen directly from its molar mass.
• This gives precise nitrogen analysis, paying close attention to stoichiometric relationships.

Such a rigorous analysis ensures efficient nutrient management in agriculture and adherence to environmental guidelines.

Stoichiometry in Reactions

Stoichiometry plays a critical role in chemical calculations, allowing chemists to predict the amounts of substances consumed and produced in a chemical reaction. It is founded on the conservation of mass and the concept of moles in chemistry.
In the original problem, stoichiometry was pivotal in determining the number of moles of ammonia produced. Utilizing the balanced chemical equations, we established the relationships between ammonium sulfate, sodium hydroxide, ammonia, and hydrochloric acid.

• This helped us relate the amount of starting material, ammonium sulfate, to the measurable products, ammonia.
• Each mole of ammonium sulfate was shown to produce two moles of ammonia, fully aligning with stoichiometric coefficients.
• Careful stoichiometric calculations facilitated the conversion of reaction data into practical information, like the percentage of nitrogen.

Mastering stoichiometry ensures accurate predictions in chemical production and analysis, helping to manage resources effectively and efficiently.