empirical formula of a barium hydroxide hydrate if a sample consists of 3.70 g of barium hydroxide and 2.64 g water.

mols Ba(OH)2 = grams/molar mass

mols H2O = grams/molar mass

Find the ratio of H2O molecules to 1 molecule Ba(OH)2.
Post your work if you get stuck.

To determine the empirical formula of a compound, we need to find the ratios of the elements present. In this case, we have the mass of barium hydroxide (Ba(OH)₂) and water (H₂O).

1. Start by finding the number of moles of each element present in the compound.

Molar mass of Ba(OH)₂:
- Ba: 137.33 g/mol
- O: 16.00 g/mol (there are two oxygen atoms)
- H: 1.01 g/mol (there are two hydroxide groups)

Molar mass of H₂O:
- H: 1.01 g/mol (there are two hydrogen atoms)
- O: 16.00 g/mol

Number of moles of Ba(OH)₂ = Mass / Molar mass
Number of moles of Ba(OH)₂ = 3.70 g / (137.33 g/mol + 2 * (16.00 g/mol + 1.01 g/mol))
Number of moles of Ba(OH)₂ ≈ 0.0125 mol

Number of moles of H₂O = Mass / Molar mass
Number of moles of H₂O = 2.64 g / (2 * (1.01 g/mol) + 16.00 g/mol)
Number of moles of H₂O ≈ 0.146 mol

2. Determine the mole ratio between the elements.

Divide the number of moles of each element by the smallest number of moles.

Number of moles of Ba(OH)₂ = 0.0125 mol / 0.0125 mol = 1
Number of moles of H₂O = 0.146 mol / 0.0125 mol ≈ 11.7 (round to the nearest whole number)

3. Write the empirical formula using the mole ratios.

The empirical formula of the barium hydroxide hydrate is Ba(OH)₂·11.7H₂O.