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3.50 Peroxyacyl mitrate (pan) is one of the components of smog. it is a compound of c, h n, and o. Determine the percent composition of oxygen and the empirical formula from the following percent composition by mass: 19.8 percent c, 2.50 percent h, 11.6 percent C. What is its molecular formula given that its molar mass is about 120 g? solve it accurately 4 decimal point calcu
Accepted Answer
1. Calculate the mass of oxygen:
Since the percentages represent the mass composition of the compound, the mass of oxygen is 100% - 19.8% - 2.50% - 11.6% = 66.1%.
2. Convert percentages to grams:
Assume a 100g sample of PAN. This means we have 19.8 g C, 2.50 g H, 11.6 g N, and 66.1 g O.
3. Convert grams to moles:
Moles of C = 19.8 g / 12.01 g/mol = 1.65 mol C
Moles of H = 2.50 g / 1.01 g/mol = 2.48 mol H
Moles of N = 11.6 g / 14.01 g/mol = 0.828 mol N
Moles of O = 66.1 g / 16.00 g/mol = 4.13 mol O
4. Determine the mole ratio:
Divide each of the mole values by the smallest mole value (0.828 mol N):
C: 1.65 mol / 0.828 mol = 2
H: 2.48 mol / 0.828 mol = 3
N: 0.828 mol / 0.828 mol = 1
O: 4.13 mol / 0.828 mol = 5
5. Empirical Formula:
The empirical formula is C₂H₃NO₅.
6. Determine the molecular formula:
Calculate the empirical formula mass: (2
12.01) + (3
1.01) + 14.01 + (5
16.00) = 120.06 g/mol
Divide the molar mass of PAN by the empirical formula mass: 120 g/mol / 120.06 g/mol ≈ 1
Since the ratio is approximately 1, the molecular formula is the same as the empirical formula: C₂H₃NO₅.
Answer:
The percent composition of oxygen in PAN is 66.1%.
The empirical formula of PAN is C₂H₃NO₅.
The molecular formula of PAN is C₂H₃NO₅.
Since the percentages represent the mass composition of the compound, the mass of oxygen is 100% - 19.8% - 2.50% - 11.6% = 66.1%.
2. Convert percentages to grams:
Assume a 100g sample of PAN. This means we have 19.8 g C, 2.50 g H, 11.6 g N, and 66.1 g O.
3. Convert grams to moles:
Moles of C = 19.8 g / 12.01 g/mol = 1.65 mol C
Moles of H = 2.50 g / 1.01 g/mol = 2.48 mol H
Moles of N = 11.6 g / 14.01 g/mol = 0.828 mol N
Moles of O = 66.1 g / 16.00 g/mol = 4.13 mol O
4. Determine the mole ratio:
Divide each of the mole values by the smallest mole value (0.828 mol N):
C: 1.65 mol / 0.828 mol = 2
H: 2.48 mol / 0.828 mol = 3
N: 0.828 mol / 0.828 mol = 1
O: 4.13 mol / 0.828 mol = 5
5. Empirical Formula:
The empirical formula is C₂H₃NO₅.
6. Determine the molecular formula:
Calculate the empirical formula mass: (2
12.01) + (3
1.01) + 14.01 + (5
16.00) = 120.06 g/mol
Divide the molar mass of PAN by the empirical formula mass: 120 g/mol / 120.06 g/mol ≈ 1
Since the ratio is approximately 1, the molecular formula is the same as the empirical formula: C₂H₃NO₅.
Answer:
The percent composition of oxygen in PAN is 66.1%.
The empirical formula of PAN is C₂H₃NO₅.
The molecular formula of PAN is C₂H₃NO₅.
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