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How many kilograms of a fertilizer made of pure \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) would be required to supply 1 kilogram of nitrogen to the soil?

Short Answer

Expert verified
To supply 1 kilogram of nitrogen, \(\frac{1}{\text{percentage of nitrogen in decimal}}\) kilograms of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) fertilizer would be required.

Step by step solution

01

Determine the molar mass of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\)

Calculate the molar mass by summing the atomic masses of all the atoms in the compound. The molar mass of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) is computed as: 2 * (14.01 for N + 4 * 1.01 for H) + 12.01 for C + 3 * 16.00 for O.
02

Calculate the mass percentage of nitrogen in \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\)

Find the total mass contributed by nitrogen and divide it by the molar mass of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\), then multiply by 100% to get the percentage of nitrogen.
03

Determine the required mass of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) to obtain 1 kg of nitrogen

Use the mass percentage of nitrogen to find out how many kilograms of fertilizer will yield 1 kilogram of nitrogen. This is done by dividing 1 kilogram by the decimal equivalent of the percentage of nitrogen in the fertilizer.

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Key Concepts

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

Molar Mass Calculation
Understanding the molar mass of a substance is crucial for a variety of chemistry problems, including those dealing with stoichiometry. Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). One mole contains Avogadro's number of particles, which is approximately 6.022 x 10^23 particles.

To calculate the molar mass of a compound, you add the atomic masses of all the atoms in the formula. For example, in the compound ammonium carbonate \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\), you first calculate the molar mass of each element present. Nitrogen's molar mass is 14.01 g/mol, hydrogen's is 1.01 g/mol, carbon's is 12.01 g/mol, and oxygen's is 16.00 g/mol. With this information, the molar mass of ammonium carbonate is computed by multiplying the atomic masses by the number of each atom in the formula and adding them all together: \[\text{Molar Mass of } \left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} = 2 \times (14.01 + 4 \times 1.01) + 12.01 + 3 \times 16.00.\] This computation provides the total molar mass necessary to find out how much of the compound is needed to supply a certain amount of an individual element, such as nitrogen.
Mass Percentage Composition
Mass percentage composition is a way of expressing the concentration of an element in a compound. It is calculated by dividing the total mass of the element in one mole of compound by the molar mass of the compound, and then multiplying by 100 to get a percentage.

For the compound ammonium carbonate \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\), you would first determine the mass of nitrogen in one mole of the compound. Since there are two ammonium groups, and each group contains one nitrogen atom, you have two moles of nitrogen. Calculating the total mass of nitrogen as \[2 \times 14.01 \text{ g/mol}\], you then divide this value by the molar mass of the entire compound. Finally, you multiply the result by 100 to find the mass percentage of nitrogen: \[\text{Mass percentage of N} = \left(\frac{2 \times 14.01 \text{ g/mol}}{\text{Molar Mass of } \left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}}\right) \times 100\%.\] This percent tells you how much nitrogen is present compared to the total mass of the compound.
Chemical Fertilizer Analysis
Analyzing chemical fertilizers involves determining the nutrient content to ensure proper nourishment for plants. The key nutrients are typically nitrogen (N), phosphorus (P), and potassium (K), and their concentrations are critical for plant growth.

When analyzing a chemical fertilizer such as ammonium carbonate, the objective may be to determine how much of the fertilizer needs to be applied to the soil to supply a certain amount of nitrogen. You would use the mass percentage composition of nitrogen already calculated and apply it to the amount of nitrogen required.

For instance, if you need to supply 1 kg of nitrogen, you divide this amount by the mass percentage of nitrogen in its decimal form: \[\text{Required mass of } \left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} = \frac{1 \text{ kg}}{\text{Decimal mass percentage of N}}.\] By doing this calculation, you identify the specific mass of the fertilizer necessary to provide the plants with the desired amount of nitrogen.

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Most popular questions from this chapter

Which has a higher percentage of oxygen: morphine \(\left(\mathrm{C}_{17} \mathrm{H}_{19} \mathrm{NO}_{3}\right)\) or heroin \(\left(\mathrm{C}_{21} \mathrm{H}_{23} \mathrm{NO}_{5}\right) ?\)

Some of the acid in acid rain is produced by the following reaction: $$ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) $$ If a falling raindrop weighing \(0.050 \mathrm{~g}\) comes into contact with \(1.0 \mathrm{mg}\) of \(\mathrm{NO}_{2}(g)\), how many milligrams of \(\mathrm{HNO}_{3}\) can be produced?

Give a step-by-step procedure for estimating the number of grams of \(A\) required to completely react with 10 moles of \(B\), given the following information: \(A\) and \(B\) react to form \(A_{5} B_{2}\) \(A\) has a molecular mass of \(100.0 . B\) has a molecular mass of 200.0 There are \(6.02 \times 10^{23}\) molecules of \(A\) in a mole of \(A\). Which of these pieces of information weren't needed?

Phencyclidine ("angel dust") is \(\mathrm{C}_{17} \mathrm{H}_{25} \mathrm{~N}\). A sample suspected of being this illicit drug was found to have a percentage composition of \(84.71 \% \mathrm{C}, 10.42 \% \mathrm{H},\) and \(5.61 \% \mathrm{~N}\). Do these data acceptably match the theoretical data for phencyclidine?

In a report to a supervisor, a chemist described an experiment in the following way: " \(0.0800 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\) decomposed into \(0.0800 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}\) and \(0.0400 \mathrm{~mol}\) of \(\mathrm{O}_{2}\)." Express the chemistry and stoichiometry of this reaction by a conventional chemical equation.

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