what is the empirical formula of a compound that is 62.0% C, 10.4% H, and 27.5% O by mass?

C3H6O

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

1. Assume we have a 100g sample of the compound.
2. Convert the mass percentages to grams. Based on the given percentages:
- Carbon (C) = 62.0g
- Hydrogen (H) = 10.4g
- Oxygen (O) = 27.5g

3. Convert the masses of each element to moles using their molar masses:
- Molar mass of C = 12.01 g/mol, so moles of C = 62.0g / 12.01 g/mol = 5.16 mol
- Molar mass of H = 1.01 g/mol, so moles of H = 10.4g / 1.01 g/mol = 10.30 mol
- Molar mass of O = 16.00 g/mol, so moles of O = 27.5g / 16.00 g/mol = 1.72 mol

4. Divide each of the mole values obtained in the previous step by the smallest number of moles to obtain a whole-number ratio:
- Carbon (C) = 5.16 mol / 1.72 mol = 3
- Hydrogen (H) = 10.30 mol / 1.72 mol = 6
- Oxygen (O) = 1.72 mol / 1.72 mol = 1

Therefore, the empirical formula of the compound is C3H6O.

To determine the empirical formula of a compound, you need to find the simplest whole-number ratio of the elements present in the compound. Here's how you can do it:

1. Assume you have a 100g sample of the compound, which allows you to convert the mass percentages to grams. So, for a 100g sample:
- Carbon (C): 62.0g
- Hydrogen (H): 10.4g
- Oxygen (O): 27.5g

2. Convert the mass of each element to moles using their respective atomic masses (C: 12.01 g/mol, H: 1.01 g/mol, O: 16.00 g/mol).
- Carbon (C): 62.0g / 12.01 g/mol = 5.165 moles
- Hydrogen (H): 10.4g / 1.01 g/mol = 10.297 moles
- Oxygen (O): 27.5g / 16.00 g/mol = 1.719 moles

3. Find the mole ratio by dividing the number of moles of each element by the smallest number of moles, which, in this case, is oxygen.
- Carbon (C): 5.165 moles / 1.719 moles = 3 moles
- Hydrogen (H): 10.297 moles / 1.719 moles = 6 moles
- Oxygen (O): 1.719 moles / 1.719 moles = 1 mole

4. Simplify the ratio obtained to the smallest whole numbers. In this case, we have:
- Carbon (C): 3
- Hydrogen (H): 6
- Oxygen (O): 1

5. Write the empirical formula using the obtained ratios: C3H6O

Therefore, the empirical formula of the compound is C3H6O.

Take a 100 g sample. This will give you

62.0 g C
10.4 g H
27.5 g O

Now determine moles of each.
62.0/atomic mass C = ??
10.4/atomic mass H = ??
27.5/atomic mass O = ??

Now you want to determine the mole ratios of these three to each other in small whole numbers. The easy way to do this is to divide the smallest number by itself; obviously, that gives 1.000 for that element. Then divide all the other values by that same small number and round to whole numbers. That should provide the empirical formula.