14) A lawn fertilizer is rated as 7.04% nitrogen, meaning 7.04 g of N in 100 g of fertilizer. The nitrogen is present in the form of urea, (NH2)2CO. How many grams of urea are present in 100 g of the fertilizer to supply the rated amount of nitrogen?
If that is 7.04% N (and not N2), you have 7.04g N in 100 g of the fertilizer. To convert to g urea just multiply by the chemical factor (which usually isn't taught in schools now).
7.04 x (molar mass urea/2*atomic mass N) = ?
To determine the number of grams of urea present in 100 g of the fertilizer, we need to calculate the amount of nitrogen in 100 g of the fertilizer.
Given that the fertilizer is rated as 7.04% nitrogen, this means that 7.04 g of nitrogen is present in 100 g of the fertilizer.
Since the nitrogen in the fertilizer is present in the form of urea, (NH2)2CO, we can calculate the molar mass of urea to relate it to nitrogen.
Molar mass of N = 14.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol
To calculate the molar mass of urea:
2 moles of N (from NH2) = 2 * 14.01 g/mol = 28.02 g/mol
4 moles of H = 4 * 1.01 g/mol = 4.04 g/mol
1 mole of C = 12.01 g/mol
1 mole of O = 16.00 g/mol
Total molar mass of urea = 28.02 g/mol + 4.04 g/mol + 12.01 g/mol + 16.00 g/mol = 60.07 g/mol
Now, we can set up the following proportion to find the number of grams of urea:
(7.04 g of N) / (100 g of fertilizer) = (x g of urea) / (60.07 g of urea)
Cross-multiplying, we get:
x g of urea = (7.04 g of N) * (60.07 g of urea) / (100 g of fertilizer)
Calculating this, we find:
x g of urea = 4.213 g
Therefore, approximately 4.213 grams of urea are present in 100 g of the fertilizer to supply the rated amount of nitrogen.