What is the percentage composition of nitrogen in silver nitrate?
You need to know the formula for silver nitrate which is AgNO3
Then you work out the MR of silver nitrate = 108+14+(16x3) = 170
Then you work out the MR of the nitrogen contained in silver nitrate. There is just one nitrogen atom so it would be 14.
Then you need to work out the percentage nitrogen in silver nitrate like you would doing any percentage.
(MR nitrogen/MR silver nitrate) x 100%
(14/170)x100 = 8.2%
Hope this helps
To find the percentage composition of nitrogen in silver nitrate (AgNO3), you need to calculate the molar mass of nitrogen and the molar mass of silver nitrate.
Step 1: Calculate the molar mass of nitrogen (N)
- The atomic mass of nitrogen (N) is approximately 14.01 g/mol.
Step 2: Calculate the molar mass of silver nitrate (AgNO3)
- The atomic mass of silver (Ag) is approximately 107.87 g/mol.
- The atomic mass of nitrogen (N) is approximately 14.01 g/mol.
- The atomic mass of oxygen (O) is approximately 16.00 g/mol.
- Since silver nitrate (AgNO3) contains 1 silver atom, 1 nitrogen atom, and 3 oxygen atoms, the molar mass of silver nitrate is:
Molar mass of AgNO3 = (1 × atomic mass of Ag) + (1 × atomic mass of N) + (3 × atomic mass of O)
= (1 × 107.87 g/mol) + (1 × 14.01 g/mol) + (3 × 16.00 g/mol)
= 169.87 g/mol + 14.01 g/mol + 48.00 g/mol
= 231.88 g/mol
Step 3: Calculate the percentage of nitrogen in silver nitrate
- The percentage of nitrogen in silver nitrate can be calculated using the formula:
Percentage of nitrogen = (Mass of nitrogen / Mass of silver nitrate) × 100%
- Since the molar mass of nitrogen (N) is 14.01 g/mol and the molar mass of silver nitrate is 231.88 g/mol, the mass of nitrogen in silver nitrate is:
Mass of nitrogen = 14.01 g/mol × 1 mol
= 14.01 g
- Therefore, the percentage of nitrogen in silver nitrate is:
Percentage of nitrogen = (14.01 g / 231.88 g) × 100%
≈ 6.04%
So, the percentage composition of nitrogen in silver nitrate is approximately 6.04%.
To find the percentage composition of nitrogen in silver nitrate, you need to consider the formula of silver nitrate, AgNO3, and determine the molar mass.
The molar mass of AgNO3 can be calculated by adding up the atomic masses of its constituent elements.
The atomic mass of silver (Ag) is approximately 107.87 g/mol, nitrogen (N) is approximately 14.01 g/mol, and oxygen (O) is approximately 16.00 g/mol.
So, the molar mass of AgNO3 can be calculated as follows:
Ag: 1 atom x 107.87 g/mol = 107.87 g/mol
N: 1 atom x 14.01 g/mol = 14.01 g/mol
O: 3 atoms x 16.00 g/mol = 48.00 g/mol
Adding up these values gives:
107.87 + 14.01 + 48.00 = 169.88 g/mol
Next, calculate the percentage composition of nitrogen in silver nitrate:
To find the percentage composition of nitrogen, divide the molar mass of nitrogen by the molar mass of AgNO3 and multiply by 100.
% nitrogen = (molar mass of N / molar mass of AgNO3) x 100
% nitrogen = (14.01 g/mol / 169.88 g/mol) x 100
% nitrogen ≈ 8.24%
Therefore, the percentage composition of nitrogen in silver nitrate is approximately 8.24%.