'm confused on how to set up the problem for each question..

1.) Calculate the empirical formula of a compound if a 6.21 g sample contains 1.67g of cerium and the remainder iodine

2.) An unknown compound contains 85.64% carbon with the remainder hydrogen. It has a molar mass of 42.08 g/mol. ... I don't know how to find its empirical and molecular formulas?

3.) The molar mass of benzene, an important industrial solvent, is 78.0 g/mol and its empirical formula is CH. What is the molecular formula for benzene?

They are all similar. Lets take on the second.

Assume 100 grams. Then
85.64gC=85.64/12=7.13MoleC
14.36gH=14.36moleH

Now divide them all by the least, and you get CH2 as the empirical formula, and it formula mass is 14
but we know molecular mass is 42.08, so that is three times the empirical formula mass, so the molecular mass must be C3H6. You can work it all with more accurate numbers...

On the third, divide the mole mass by the empirical formula, and you then know the real formual (as in above).

Number one is very much like the second.

To set up the problem and find the empirical formula for each of these questions, follow these steps:

1.) Calculate the empirical formula of the compound if a 6.21 g sample contains 1.67g of cerium and the remainder iodine:
- Determine the mass of each element in the sample. In this case, the mass of cerium is 1.67 g, and the remaining mass is iodine, so it is 6.21 g - 1.67 g = 4.54 g.
- Convert the masses of each element to moles by dividing by their respective atomic masses. To do this, you need to know the atomic masses of cerium and iodine, which you can find on the periodic table. Atomic mass of cerium (Ce) is 140.12 g/mol, and iodine (I) is 126.90 g/mol. So, the number of moles of cerium is 1.67 g / 140.12 g/mol = 0.0119 mol, and the number of moles of iodine is 4.54 g / 126.90 g/mol = 0.0357 mol.
- Divide each number of moles by the smallest value obtained. In this case, the smallest value is 0.0119 mol, so you divide by 0.0119 mol. The ratio is approximately 1:3, so the empirical formula is CeI3.

2.) Find the empirical and molecular formulas of an unknown compound that contains 85.64% carbon with the remainder hydrogen and has a molar mass of 42.08 g/mol:
- Convert the percent composition of carbon to mass. To do this, assume that you have 100 g of the compound, so the mass of carbon is 85.64 g.
- Subtract the mass of carbon from the molar mass of the compound to find the mass of hydrogen. In this case, the mass of hydrogen is 42.08 g - 85.64 g = -43.56 g. However, a negative mass is not possible, so there must be an error in the information given. Check the question or data again to make sure you have the correct information.

3.) Determine the molecular formula for benzene given its empirical formula is CH and its molar mass is 78.0 g/mol:
- Calculate the empirical formula mass by adding the atomic masses of carbon and hydrogen. Carbon has an atomic mass of 12.01 g/mol, and hydrogen has an atomic mass of 1.01 g/mol. So, the empirical formula mass is 12.01 g/mol + 1.01 g/mol = 13.02 g/mol.
- Divide the molar mass of benzene by the empirical formula mass to find the multiple of the empirical formula. In this case, 78.0 g/mol / 13.02 g/mol = 5.98.
- Round the multiple obtained in the previous step to the nearest whole number. In this case, it is approximately 6.
- Multiply the empirical formula by the whole number obtained in the previous step. The molecular formula for benzene is (CH)6, which can be simplified as C6H6.

Remember, these steps are general guidelines and should be adjusted based on the specific details of the problem.