Answer :
Sure, let's go through the step-by-step process of calculating the percentage composition of each element in the given compounds.
### Atomic Weights Reference
1. Hydrogen (H): 1.008 g/mol
2. Carbon (C): 12.011 g/mol
3. Oxygen (O): 15.999 g/mol
4. Phosphorus (P): 30.974 g/mol
5. Nitrogen (N): 14.007 g/mol
### a) Acetic Acid (CH₃COOH)
The molecular formula for acetic acid is CH₃COOH.
- Number of Carbon (C) atoms: 2
- Number of Hydrogen (H) atoms: 4
- Number of Oxygen (O) atoms: 2
1. Calculate the molar mass of acetic acid:
- Carbon: 2 12.011 g/mol = 24.022 g/mol
- Hydrogen: 4 1.008 g/mol = 4.032 g/mol
- Oxygen: 2 15.999 g/mol = 31.998 g/mol
- Total molar mass = 24.022 + 4.032 + 31.998 = 60.052 g/mol
2. Calculate the percentage composition:
- %C = (24.022 / 60.052) 100 ≈ 40.00%
- %H = (4.032 / 60.052) 100 ≈ 6.71%
- %O = (31.998 / 60.052) 100 ≈ 53.29%
### b) Phosphoric Acid (HPO)
The given molecular formula for phosphoric acid is HPO.
(Note: The usual molecular formula for phosphoric acid is H₃PO₄, but we'll proceed with HPO as given:)
- Number of Hydrogen (H) atoms: 1
- Number of Phosphorus (P) atoms: 1
- Number of Oxygen (O) atoms: 1
1. Calculate the molar mass of HPO:
- Hydrogen: 1 1.008 g/mol = 1.008 g/mol
- Phosphorus: 1 30.974 g/mol = 30.974 g/mol
- Oxygen: 1 15.999 g/mol = 15.999 g/mol
- Total molar mass = 1.008 + 30.974 + 15.999 = 47.981 g/mol
2. Calculate the percentage composition:
- %H = (1.008 / 47.981) 100 ≈ 2.10%
- %P = (30.974 / 47.981) 100 ≈ 64.56%
- %O = (15.999 / 47.981) 100 ≈ 33.34%
### c) Ammonium Nitrate (NH₄NO₃)
The molecular formula for ammonium nitrate is NH₄NO₃.
- Number of Nitrogen (N) atoms: 2
- Number of Hydrogen (H) atoms: 4
- Number of Oxygen (O) atoms: 3
1. Calculate the molar mass of ammonium nitrate:
- Nitrogen: 2 14.007 g/mol = 28.014 g/mol
- Hydrogen: 4 1.008 g/mol = 4.032 g/mol
- Oxygen: 3 15.999 g/mol = 47.997 g/mol
- Total molar mass = 28.014 + 4.032 + 47.997 = 80.043 g/mol
2. Calculate the percentage composition:
- %N = (28.014 / 80.043) 100 ≈ 35.00%
- %H = (4.032 / 80.043) 100 ≈ 5.04%
- %O = (47.997 / 80.043) 100 ≈ 59.96%
### Summary
- Acetic Acid (CH₃COOH):
- Carbon: 40.00%
- Hydrogen: 6.71%
- Oxygen: 53.29%
- Phosphoric Acid (HPO):
- Hydrogen: 2.10%
- Phosphorus: 64.56%
- Oxygen: 33.34%
- Ammonium Nitrate (NH₄NO₃):
- Nitrogen: 35.00%
- Hydrogen: 5.04%
- Oxygen: 59.96%
### Atomic Weights Reference
1. Hydrogen (H): 1.008 g/mol
2. Carbon (C): 12.011 g/mol
3. Oxygen (O): 15.999 g/mol
4. Phosphorus (P): 30.974 g/mol
5. Nitrogen (N): 14.007 g/mol
### a) Acetic Acid (CH₃COOH)
The molecular formula for acetic acid is CH₃COOH.
- Number of Carbon (C) atoms: 2
- Number of Hydrogen (H) atoms: 4
- Number of Oxygen (O) atoms: 2
1. Calculate the molar mass of acetic acid:
- Carbon: 2 12.011 g/mol = 24.022 g/mol
- Hydrogen: 4 1.008 g/mol = 4.032 g/mol
- Oxygen: 2 15.999 g/mol = 31.998 g/mol
- Total molar mass = 24.022 + 4.032 + 31.998 = 60.052 g/mol
2. Calculate the percentage composition:
- %C = (24.022 / 60.052) 100 ≈ 40.00%
- %H = (4.032 / 60.052) 100 ≈ 6.71%
- %O = (31.998 / 60.052) 100 ≈ 53.29%
### b) Phosphoric Acid (HPO)
The given molecular formula for phosphoric acid is HPO.
(Note: The usual molecular formula for phosphoric acid is H₃PO₄, but we'll proceed with HPO as given:)
- Number of Hydrogen (H) atoms: 1
- Number of Phosphorus (P) atoms: 1
- Number of Oxygen (O) atoms: 1
1. Calculate the molar mass of HPO:
- Hydrogen: 1 1.008 g/mol = 1.008 g/mol
- Phosphorus: 1 30.974 g/mol = 30.974 g/mol
- Oxygen: 1 15.999 g/mol = 15.999 g/mol
- Total molar mass = 1.008 + 30.974 + 15.999 = 47.981 g/mol
2. Calculate the percentage composition:
- %H = (1.008 / 47.981) 100 ≈ 2.10%
- %P = (30.974 / 47.981) 100 ≈ 64.56%
- %O = (15.999 / 47.981) 100 ≈ 33.34%
### c) Ammonium Nitrate (NH₄NO₃)
The molecular formula for ammonium nitrate is NH₄NO₃.
- Number of Nitrogen (N) atoms: 2
- Number of Hydrogen (H) atoms: 4
- Number of Oxygen (O) atoms: 3
1. Calculate the molar mass of ammonium nitrate:
- Nitrogen: 2 14.007 g/mol = 28.014 g/mol
- Hydrogen: 4 1.008 g/mol = 4.032 g/mol
- Oxygen: 3 15.999 g/mol = 47.997 g/mol
- Total molar mass = 28.014 + 4.032 + 47.997 = 80.043 g/mol
2. Calculate the percentage composition:
- %N = (28.014 / 80.043) 100 ≈ 35.00%
- %H = (4.032 / 80.043) 100 ≈ 5.04%
- %O = (47.997 / 80.043) 100 ≈ 59.96%
### Summary
- Acetic Acid (CH₃COOH):
- Carbon: 40.00%
- Hydrogen: 6.71%
- Oxygen: 53.29%
- Phosphoric Acid (HPO):
- Hydrogen: 2.10%
- Phosphorus: 64.56%
- Oxygen: 33.34%
- Ammonium Nitrate (NH₄NO₃):
- Nitrogen: 35.00%
- Hydrogen: 5.04%
- Oxygen: 59.96%
