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8. What is the percent by mass of carbon in sucrose, C12H22O11? 9. What is the empirical formula of a compound that is 21.2% nitrogen, 8.1% hydrogen, 24.3% sulfur, and 48.4% oxygen by mass? solve everything step by step cancelation to cancelation gen chem
Accepted Answer
Problem 8:
1. Find the molar mass of sucrose (C12H22O11):
Carbon (C): 12.01 g/mol
12 = 144.12 g/mol
Hydrogen (H): 1.01 g/mol
22 = 22.22 g/mol
Oxygen (O): 16.00 g/mol
11 = 176.00 g/mol
Total molar mass: 144.12 + 22.22 + 176.00 = 342.34 g/mol
2. Calculate the mass of carbon in one mole of sucrose:
Mass of carbon: 144.12 g/mol
3. Calculate the percent by mass of carbon:
(Mass of carbon / Molar mass of sucrose)
100%
(144.12 g/mol / 342.34 g/mol)
100% = 42.11%
Problem 9:
1. Assume a 100 g sample of the compound. This makes the percentages equal to grams:
Nitrogen (N): 21.2 g
Hydrogen (H): 8.1 g
Sulfur (S): 24.3 g
Oxygen (O): 48.4 g
2. Convert grams to moles using the molar mass of each element:
N: 21.2 g / 14.01 g/mol = 1.51 mol
H: 8.1 g / 1.01 g/mol = 8.02 mol
S: 24.3 g / 32.07 g/mol = 0.76 mol
O: 48.4 g / 16.00 g/mol = 3.03 mol
3. Divide each mole value by the smallest mole value (0.76 mol) to find the mole ratio:
N: 1.51 mol / 0.76 mol ≈ 2
H: 8.02 mol / 0.76 mol ≈ 11
S: 0.76 mol / 0.76 mol = 1
O: 3.03 mol / 0.76 mol ≈ 4
4. The empirical formula is N2H11SO4.
1. Find the molar mass of sucrose (C12H22O11):
Carbon (C): 12.01 g/mol
12 = 144.12 g/mol
Hydrogen (H): 1.01 g/mol
22 = 22.22 g/mol
Oxygen (O): 16.00 g/mol
11 = 176.00 g/mol
Total molar mass: 144.12 + 22.22 + 176.00 = 342.34 g/mol
2. Calculate the mass of carbon in one mole of sucrose:
Mass of carbon: 144.12 g/mol
3. Calculate the percent by mass of carbon:
(Mass of carbon / Molar mass of sucrose)
100%
(144.12 g/mol / 342.34 g/mol)
100% = 42.11%
Problem 9:
1. Assume a 100 g sample of the compound. This makes the percentages equal to grams:
Nitrogen (N): 21.2 g
Hydrogen (H): 8.1 g
Sulfur (S): 24.3 g
Oxygen (O): 48.4 g
2. Convert grams to moles using the molar mass of each element:
N: 21.2 g / 14.01 g/mol = 1.51 mol
H: 8.1 g / 1.01 g/mol = 8.02 mol
S: 24.3 g / 32.07 g/mol = 0.76 mol
O: 48.4 g / 16.00 g/mol = 3.03 mol
3. Divide each mole value by the smallest mole value (0.76 mol) to find the mole ratio:
N: 1.51 mol / 0.76 mol ≈ 2
H: 8.02 mol / 0.76 mol ≈ 11
S: 0.76 mol / 0.76 mol = 1
O: 3.03 mol / 0.76 mol ≈ 4
4. The empirical formula is N2H11SO4.
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