Acetone is an important chemical solvent and is used at homes as a nail polish remover. Chemical analysis shows that acetone is 62.0%C, 10.4%H, and 27.5%O by mass. A. Determine the empirical formula of acetone. B. In a separate experiment, the molecular mass is found to b 58.1 g/mole. What is the molecular formula of acetone?
To determine the empirical formula of acetone, we need to find the ratio of the elements present in the compound.
A. Determine the empirical formula of acetone:
1. Convert the given percentages of each element to grams:
- 62.0% C: (62.0 g/100g) = 0.62 g C
- 10.4% H: (10.4 g/100g) = 0.104 g H
- 27.5% O: (27.5 g/100g) = 0.275 g O
2. Convert the grams of each element to moles using their atomic masses:
- C: 0.62 g C / 12.01 g/mol = 0.0516 mol C
- H: 0.104 g H / 1.01 g/mol = 0.103 mol H
- O: 0.275 g O / 16.00 g/mol = 0.0172 mol O
3. Divide each mole value by the smallest number of moles calculated to obtain the mole ratio:
- C: 0.0516 mol C / 0.0516 mol C = 1
- H: 0.103 mol H / 0.0516 mol C = 2
- O: 0.0172 mol O / 0.0516 mol C = 0.333 (approximately)
4. Round off the mole ratio to the nearest whole number (if necessary):
- C: 1
- H: 2
- O: 0.333 (approximately 1, when rounded)
Therefore, the empirical formula of acetone is CH₂O.
B. To determine the molecular formula of acetone, we need to know its molecular mass, which is given as 58.1 g/mole.
1. Calculate the empirical formula mass by summing the atomic masses of the elements in the empirical formula:
- C: 1 × 12.01 g/mol = 12.01 g/mol
- H: 2 × 1.01 g/mol = 2.02 g/mol
- O: 1 × 16.00 g/mol = 16.00 g/mol
Empirical formula mass = 12.01 g/mol + 2.02 g/mol + 16.00 g/mol = 30.03 g/mol
2. Divide the given molecular mass by the empirical formula mass to find the whole-number multiple:
- Molecular formula multiple: 58.1 g/mole ÷ 30.03 g/mole = 1.935 (approximately)
3. Round off the molecular formula multiple to the nearest whole number:
- Molecular formula multiple: 2
4. Multiply the subscripts in the empirical formula by the molecular formula multiple:
- Empirical formula: CH₂O
- Molecular formula: C₂H₄O₂
Therefore, the molecular formula of acetone is C₂H₄O₂.
A. To determine the empirical formula of acetone, we need to find the simplest whole number ratio of the elements present in acetone.
Step 1: Convert the mass percentages to grams.
Assuming we have a 100g sample of acetone:
- Mass of carbon (C) = 62.0g
- Mass of hydrogen (H) = 10.4g
- Mass of oxygen (O) = 27.5g
Step 2: Determine the moles of each element.
- Moles of C = (mass of C) / (molar mass of C)
= 62.0g / 12.01g/mol = 5.16 mol
- Moles of H = (mass of H) / (molar mass of H)
= 10.4g / 1.01g/mol = 10.30 mol
- Moles of O = (mass of O) / (molar mass of O)
= 27.5g / 16.00g/mol = 1.72 mol
Step 3: Divide the moles by the smallest number of moles.
Dividing all the moles by the smallest number of moles (1.72 mol), we get:
- C: 5.16 mol / 1.72 mol = 3
- H: 10.30 mol / 1.72 mol = 6
- O: 1.72 mol / 1.72 mol = 1
Therefore, the empirical formula of acetone is CH3O.
B. To determine the molecular formula of acetone, we need to know the molecular mass of acetone and compare it with the empirical formula mass.
Step 1: Calculate the empirical formula mass.
Empirical formula mass = (mass of C in empirical formula) + (mass of H in empirical formula) + (mass of O in empirical formula)
= (12.01g/mol) + (1.01g/mol x 3) + (16.00g/mol)
= 58.08 g/mol
Step 2: Divide the molecular mass by the empirical formula mass.
Molecular formula mass / Empirical formula mass = (58.1 g/mol) / (58.08 g/mol)
≈ 1
Since the molecular formula mass is approximately equal to the empirical formula mass, the empirical formula (CH3O) is also the molecular formula of acetone.