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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 4 decimal pointpls hydrogen is 1.008 s is 32.7
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 translate to grams.
Convert grams to moles: Divide each element's mass by its 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) / (32.07 g/mol) = 1.231 mol
O: (9.86 g) / (16.00 g/mol) = 0.616 mol
Find the smallest whole-number ratio: Divide each mole value by the smallest one (0.616 mol):
C: 3.698 mol / 0.616 mol ≈ 6
H: 6.163 mol / 0.616 mol ≈ 10
S: 1.231 mol / 0.616 mol ≈ 2
O: 0.616 mol / 0.616 mol ≈ 1
Empirical Formula: C6H10S2O
2. Molecular Formula
Calculate the empirical formula mass: (6
12.01) + (10
1.008) + (2
32.07) + (1
16.00) = 162.24 g/mol
Determine the ratio between molecular mass and empirical formula mass: 162 g/mol / 162.24 g/mol ≈ 1
Molecular Formula: Since the ratio is approximately 1, the molecular formula is the same as the empirical formula: C6H10S2O.
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