The percent composition by mass of an unknown compound with a molecular mass of 60.052 amu is 40.002% C, 6.7135% H, and 53.284% O. Determine the compound's empirical and molecular formulas.

To determine the empirical formula, we need to find the simplest whole number ratio of atoms present in the compound.

Let's assume we have 100 grams of the unknown compound. This means that in 100 grams, we have:

- 40.002 grams of carbon (C)
- 6.7135 grams of hydrogen (H)
- 53.284 grams of oxygen (O)

Next, we need to find the number of moles of each element using their respective atomic masses:

- Carbon (C): 40.002 grams ÷ 12.01 g/mol = 3.333 moles
- Hydrogen (H): 6.7135 grams ÷ 1.008 g/mol = 6.666 moles
- Oxygen (O): 53.284 grams ÷ 16.00 g/mol = 3.3315 moles

Now, we need to find the simplest whole number ratio of these moles. We divide all the mole values by the smallest mole value:

- C: 3.333 moles ÷ 3.3315 moles = 1
- H: 6.666 moles ÷ 3.3315 moles = 2
- O: 3.3315 moles ÷ 3.3315 moles = 1

The empirical formula is then C1H2O1, which can be simplified to CH2O.

To find the molecular formula, we need to determine the molecular mass of the compound. Given that the molecular mass is 60.052 amu, the empirical formula mass (CH2O) is:

(12.01 amu x 1) + (1.008 amu x 2) + (16.00 amu x 1) = 30.026 amu

To determine the molecular formula mass, we divide the molecular mass by the empirical formula mass:

60.052 amu ÷ 30.026 amu = 1.998 ≈ 2

This suggests that the molecular formula is twice the empirical formula, so the compound's molecular formula is C2H4O2.

To determine the compound's empirical formula, we need to find the simplest whole number ratio of the elements present in the compound.

1. Start by assuming we have 100 grams of the compound. This means we have 40.002 grams of carbon, 6.7135 grams of hydrogen, and 53.284 grams of oxygen.

2. Convert the masses to moles using the molar masses of the elements:
- Carbon (C): molar mass = 12.01 g/mol
- Hydrogen (H): molar mass = 1.01 g/mol
- Oxygen (O): molar mass = 16.00 g/mol

Moles of carbon (C) = 40.002 g / 12.01 g/mol = 3.333 mol
Moles of hydrogen (H) = 6.7135 g / 1.01 g/mol = 6.65 mol
Moles of oxygen (O) = 53.284 g / 16.00 g/mol = 3.33 mol

3. Divide each element's moles by the smallest number of moles to get the simplest whole number ratio:
Carbon: 3.333 mol / 3.333 mol = 1
Hydrogen: 6.65 mol / 3.333 mol ≈ 1.99 ≈ 2
Oxygen: 3.33 mol / 3.333 mol ≈ 0.999 ≈ 1

The empirical formula is C1H2O1, which can be simplified to just CH2O.

To determine the compound's molecular formula, we need to find the actual number of atoms of each element in the molecular formula.

4. Calculate the empirical formula mass:
Empirical formula mass (CH2O) = (12.01 g/mol * 1) + (1.01 g/mol * 2) + (16.00 g/mol * 1) = 30.03 g/mol

5. Divide the molecular mass by the empirical formula mass:
Molecular mass = 60.052 g/mol
60.052 g/mol / 30.03 g/mol = 1.999 ≈ 2

This means the molecular formula is a multiple of the empirical formula. Thus, the compound's molecular formula is C2H4O2.

multiply the mass by the percent to find the element masses

divide the element masses by the amu's to find the number of atoms

adjust for fractions by dividing through by the fraction