If carbon =62.07%,hydrogen=10.34%, and oxygen=27.59%. Determine its empirical formula, name each of them. Do not consider ones > 6-carbons.

the % given is a % by mass. so they are grams.

convert the mass to mole for each element
divide each mole with the smallest mole
round off the moles to 1 sig figure.

then you can name them. the empirical formula is C3H6O. this formula can be an aldehyde and a ketone.

To determine the empirical formula, we need to find the simplest ratio of atoms in the compound based on the given percentages.

Step 1: Convert the percentages to grams.
- Assume we have 100 grams of the compound.
- Carbon = 62.07 grams
- Hydrogen = 10.34 grams
- Oxygen = 27.59 grams

Step 2: Convert the grams to moles.
- Carbon: moles = grams / molar mass of carbon
- Molar mass of carbon = 12.01 g/mol
- Moles of carbon = 62.07 g / 12.01 g/mol ≈ 5.17 mol

- Hydrogen: moles = grams / molar mass of hydrogen
- Molar mass of hydrogen = 1.01 g/mol
- Moles of hydrogen = 10.34 g / 1.01 g/mol ≈ 10.24 mol

- Oxygen: moles = grams / molar mass of oxygen
- Molar mass of oxygen = 16.00 g/mol
- Moles of oxygen = 27.59 g / 16.00 g/mol ≈ 1.72 mol

Step 3: Divide each result by the smallest number of moles to get the simplest (whole number) ratio. In this case, it is the moles of oxygen.
- Carbon: 5.17 mol / 1.72 mol = 3
- Hydrogen: 10.24 mol / 1.72 mol ≈ 6
- Oxygen: 1.72 mol / 1.72 mol = 1

Step 4: Write the empirical formula using the ratios obtained.
- Carbon = C3
- Hydrogen = H6
- Oxygen = O1

The empirical formula for the compound with carbon = 62.07%, hydrogen = 10.34%, and oxygen = 27.59% is C3H6O, and the names of the elements are carbon, hydrogen, and oxygen, respectively.