calculate the percentage of nitrogen in the following fertilizer i NH4NO3 ii NH4/2SO4
To calculate the percentage of nitrogen in a fertilizer, you need to determine the molar mass of nitrogen and divide it by the molar mass of the complete compound. Then, multiply the result by 100 to obtain the percentage.
Let's calculate the percentage of nitrogen in the given fertilizers:
i. NH4NO3 (Ammonium Nitrate):
The molar mass of nitrogen (N) is 14.01 grams/mol, and the molar mass of ammonium nitrate (NH4NO3) is:
(1 x 14.01) + (4 x 1.01) + (1 x 14.01) + (3 x 16.00) = 80.04 grams/mol
Now, divide the molar mass of nitrogen by the molar mass of ammonium nitrate and multiply by 100 to obtain the percentage:
(14.01 / 80.04) x 100 = 17.50%
Therefore, the percentage of nitrogen in NH4NO3 fertilizer is approximately 17.50%.
ii. NH4/2SO4 (Ammonium Sulfate):
In this case, the fertilizer formula is not clear. It seems like there is a typo or missing information. Please provide the correct formula for NH4/2SO4, and I will be happy to help you further.
To calculate the percentage of nitrogen in a fertilizer, you need to know the molar mass of nitrogen and the molar mass of the compound in the fertilizer. Let's calculate the percentage of nitrogen in both given fertilizers:
i. NH4NO3:
The molar mass of nitrogen (N) is 14.01 g/mol, and the molar mass of ammonium nitrate (NH4NO3) can be calculated as follows:
(1 x molar mass of N) + (4 x molar mass of H) + (3 x molar mass of O)
= (1 x 14.01) + (4 x 1.01) + (3 x 16.00)
= 14.01 + 4.04 + 48.00
= 66.05 g/mol
To calculate the percentage of nitrogen in NH4NO3, divide the molar mass of nitrogen by the molar mass of the compound and multiply by 100:
(molar mass of N / molar mass of NH4NO3) x 100
= (14.01 / 66.05) x 100
≈ 21.21%
Therefore, NH4NO3 contains approximately 21.21% nitrogen.
ii. NH4/2SO4:
The molar mass of nitrogen (N) is 14.01 g/mol, and the molar mass of ammonium sulfate ((NH4)2SO4) can be calculated as follows:
(2 x molar mass of N) + (8 x molar mass of H) + (1 x molar mass of S) + (4 x molar mass of O)
= (2 x 14.01) + (8 x 1.01) + (1 x 32.07) + (4 x 16.00)
= 28.02 + 8.08 + 32.07 + 64.00
= 132.17 g/mol
To calculate the percentage of nitrogen in (NH4)2SO4 , divide the molar mass of nitrogen by the molar mass of the compound and multiply by 100:
(molar mass of N / molar mass of (NH4)2SO4) x 100
= (28.02 / 132.17) x 100
≈ 21.21%
Therefore, (NH4)2SO4 contains approximately 21.21% nitrogen as well.
Both NH4NO3 and (NH4)2SO4 fertilizers have approximately 21.21% nitrogen content.
Let's do apples and oranges. If you had 10 lbs apples and 10 lbs oranges what percent apples do you have. That't 50%, of course, How did you get that. %apples = (lbs apples/total weight in lbs)*100 = (10/20)*100 = 50. So you have a chemistry problem and not apples and oranges BUT it works the same way.
%N = (mass N/molar mass NH4NO3)*100 = ? Don't forget there are 2 N atoms in NH4NO3.
Do (NH4)2SO4 the same way.
Post your work if you get stuck. Don't let the word "chemistry" intimidate you.