find the percentage by mass of nitrogen present in (NH4)2SO4
[2*atomic mass N/molar mass (NH4)2SO4]*100 = ?
To find the percentage by mass of nitrogen (N) in (NH4)2SO4 (ammonium sulfate), you need to calculate the molar mass of nitrogen and the molar mass of (NH4)2SO4.
1. Calculate the molar mass of N:
The atomic mass of nitrogen (N) is approximately 14.01 g/mol.
2. Calculate the molar mass of (NH4)2SO4:
(NH4)2SO4 consists of two ammonium ions (NH4+) and one sulfate ion (SO4^2-).
Ammonium ion (NH4+):
The atomic mass of nitrogen (N) is 14.01 g/mol.
The atomic mass of hydrogen (H) is approximately 1.01 g/mol.
So, the molar mass of NH4+ = (14.01 g/mol) + 4 * (1.01 g/mol) = 18.04 g/mol.
Sulfate ion (SO4^2-):
The atomic mass of sulfur (S) is approximately 32.07 g/mol.
The atomic mass of oxygen (O) is approximately 16.00 g/mol.
So, the molar mass of SO4^2- = (32.07 g/mol) + 4 * (16.00 g/mol) = 96.06 g/mol.
Now, calculate the molar mass of (NH4)2SO4:
Molar mass of (NH4)2SO4 = 2 * (molar mass of NH4+) + molar mass of SO4^2-
= 2 * 18.04 g/mol + 96.06 g/mol
= 132.14 g/mol.
3. Calculate the mass of nitrogen (N) in (NH4)2SO4:
In (NH4)2SO4, there are 2 nitrogen atoms.
Mass of N = (molar mass of N) * (number of nitrogen atoms)
= 14.01 g/mol * 2
= 28.02 g.
4. Calculate the percentage by mass of nitrogen (N) in (NH4)2SO4:
Percentage by mass of N = (mass of N / molar mass of (NH4)2SO4) * 100
= (28.02 g / 132.14 g/mol) * 100
≈ 21.21%.
Therefore, the percentage by mass of nitrogen present in (NH4)2SO4 is approximately 21.21%.