Question

A lawn fertilizer is rated as \(6.00 \%\) nitrogen, meaning \(6.00 \mathrm{~g}\) of \(\mathrm{N}\) in \(100 \mathrm{~g}\) of fertilizer. The nitrogen is present in the form of urea, \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO} .\) How many grams of urea are present in \(100 \mathrm{~g}\) of the fertilizer to supply the rated amount of nitrogen?

Short answer

The fertilizer contains 28.57 grams of urea to supply 6.00 g of nitrogen.

Step-by-step solution

Calculate the molar mass of urea

Use the atomic masses of nitrogen (N), hydrogen (H), carbon (C), and oxygen (O) to calculate the molar mass of urea, \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\). The formula for urea is \((NH_2)_2CO\), which means it has 2 nitrogen atoms, 4 hydrogen atoms, 1 carbon atom, and 1 oxygen atom. Molar masses are approximately: N=14 g/mol, H=1 g/mol, C=12 g/mol, O=16 g/mol.

Calculate the mass of nitrogen in urea

Knowing that urea has 2 nitrogen atoms, the mass of nitrogen in one mole of urea can be found by multiplying the amount of nitrogen atoms by the atomic mass of nitrogen. The calculation will give the mass of nitrogen in one mole of urea, which is \(2 \times 14 \, \mathrm{g/mol} \).

Determine the mass of urea corresponding to the required nitrogen content

With a rated content of 6.00 g of \(N\) in 100 g of fertilizer, convert the mass of nitrogen to the mass of urea by using the ratio of the molar mass of urea to the mass of nitrogen in one mole of urea calculated in the previous steps.

Calculate the grams of urea

Using the ratio found in Step 3, multiply the rated nitrogen content by the ratio to find the grams of urea in the fertilizer. This will give us how many grams of urea are required to provide 6.00 g of nitrogen.

Key concepts

Molar Mass Calculation

As we venture into the world of fertilizers in chemistry, understanding the molar mass of a substance, like urea in this instance, is a fundamental skill. Molar mass is the weight of one mole (Avogadro's number, which is approximately 6.022\(\times\)10^23 particles) of a substance, and it is usually expressed in grams per mole (g/mol).

To calculate the molar mass of urea, \( (NH_2)_2CO \), we simply sum up the atomic masses of each atom present in the compound. Here's a breakdown of the calculation:

Percent Composition

Understanding percent composition is essential in agricultural chemistry, especially when analyzing fertilizers like our urea example. Percent composition tells us how much of a certain element is present in a compound by weight. For the lawn fertilizer in our exercise, it's expressed as the percentage of nitrogen present in the product.

With the given information that the fertilizer contains 6.00% nitrogen by weight, we already have a clear indication of its composition. However, to fully grasp what this means in the context of the fertilizer, we must relate it to urea's overall composition. The percent composition of an element in a compound is calculated by dividing the total mass of the element in one mole of the compound by the compound's molar mass, and then multiplying by 100 to convert to a percentage. Therefore, for urea, one would calculate the mass contribution of nitrogen to the total molar mass of urea to establish its percent composition.

Molecular Formula

The molecular formula is a direct representation of the number of atoms of each element found in a molecule of a compound. It provides a visual insight into the composition of substances, such as the urea in our fertilizer example. In this case, the molecular formula of urea is \( (NH_2)_2CO \), indicating the exact number of nitrogen, hydrogen, carbon, and oxygen atoms present in a single molecule.

Using the molecular formula, one can calculate the percent composition and subsequently interpret it to understand how much of each element is required for a certain amount of fertilizer. It also aids in determining how much urea will contribute to the nitrogen content of the fertilizer, ultimately allowing calculation of the amount of urea needed to achieve the desired nitrogen percentage in a 100 g sample of fertilizer.