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3.51 Allicin is the compound responsible for the characteristic smell of garlic. an analysis of the compound gives the following percent composition by mass: c: 44.4 percent, h: 6.21 percent, s: 39.5 percent, o: 9.86 percent. Calculate its empirical formula. What is its molecular formula given that its molar mass is about 162 g? solve it accurately 4 decimal point always hydrogen is 1.008
Accepted Answer
Here's how to determine the empirical and molecular formulas of allicin:
1. Empirical Formula
Assume a 100g sample: This makes the percentages directly equal to grams. So, we have:
44.4 g C
6.21 g H
39.5 g S
9.86 g O
Convert grams to moles: Divide each mass by the respective element's molar mass:
C: 44.4 g / 12.01 g/mol = 3.698 mol
H: 6.21 g / 1.008 g/mol = 6.163 mol
S: 39.5 g / 32.06 g/mol = 1.232 mol
O: 9.86 g / 16.00 g/mol = 0.6163 mol
Find the smallest whole-number ratio: Divide each mole value by the smallest mole value (0.6163 mol):
C: 3.698 mol / 0.6163 mol ≈ 6
H: 6.163 mol / 0.6163 mol ≈ 10
S: 1.232 mol / 0.6163 mol ≈ 2
O: 0.6163 mol / 0.6163 mol ≈ 1
The empirical formula is C₆H₁₀S₂O.
2. Molecular Formula
Calculate the empirical formula mass: (6
12.01) + (10
1.008) + (2
32.06) + (1
16.00) = 162.24 g/mol
Determine the ratio between the molecular mass and the empirical formula mass: 162 g/mol / 162.24 g/mol ≈ 1
Since the ratio is approximately 1, the molecular formula is the same as the empirical formula: C₆H₁₀S₂O.
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