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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 point always hydrogen is 1.008 s is 32.7
Accepted Answer
To determine the empirical formula, we'll follow these steps:
1. Assume a 100g sample: This simplifies the percentage composition to grams. So, we have 44.4g C, 6.21g H, 39.5g S, and 9.86g O.
2. Convert grams to moles: Divide each element's mass by its molar mass:
C: 44.4g / 12.01 g/mol = 3.697 mol
H: 6.21g / 1.008 g/mol = 6.161 mol
S: 39.5g / 32.06 g/mol = 1.232 mol
O: 9.86g / 16.00 g/mol = 0.616 mol
3. Find the mole ratio: Divide each mole value by the smallest mole value (0.616 mol):
C: 3.697 mol / 0.616 mol ≈ 6
H: 6.161 mol / 0.616 mol ≈ 10
S: 1.232 mol / 0.616 mol ≈ 2
O: 0.616 mol / 0.616 mol ≈ 1
4. Empirical formula: The empirical formula is C6H10S2O.
Now, let's determine the molecular formula:
1. Calculate the empirical formula mass: (6
12.01) + (10
1.008) + (2
32.06) + (1
16.00) = 162.24 g/mol.
2. Divide the molar mass by the empirical formula mass: 162 g/mol / 162.24 g/mol ≈ 1
3. Molecular formula: Since the result is approximately 1, the molecular formula is the same as the empirical formula: C6H10S2O.
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