Analysis of a sample of a compound indicates that is has 1.04 g K, 0.70 g Cr, and 0.86 g O.What is the empirical mass of the said compound in g/mol?Use:K = 39 amuCr = 52 amuO = 16 amu

To find the empirical mass of the compound, we first need to find the moles of each element in the compound.

1. For Potassium (K): Given mass = 1.04 g Molar mass = 39 g/mol Moles of K = Given mass / Molar mass = 1.04 g / 39 g/mol = 0.0267 mol

2. For Chromium (Cr): Given mass = 0.70 g Molar mass = 52 g/mol Moles of Cr = Given mass / Molar mass = 0.70 g / 52 g/mol = 0.0135 mol

3. For Oxygen (O): Given mass = 0.86 g Molar mass = 16 g/mol Moles of O = Given mass / Molar mass = 0.86 g / 16 g/mol = 0.05375 mol

Next, we divide each mole quantity by the smallest number of moles calculated (which is for Cr) to get the mole ratio:

1. Ratio of K to Cr = 0.0267 mol / 0.0135 mol = 1.98 ≈ 2
2. Ratio of Cr to Cr = 0.0135 mol / 0.0135 mol = 1
3. Ratio of O to Cr = 0.05375 mol / 0.0135 mol = 3.98 ≈ 4

So, the empirical formula of the compound is K2CrO4.

Finally, we calculate the empirical mass by adding the molar masses of all the atoms in the empirical formula:

Empirical mass = (2 * Molar mass of K) + (1 * Molar mass of Cr) + (4 * Molar mass of O) = (2 * 39 g/mol) + (1 * 52 g/mol) + (4 * 16 g/mol) = 78 g/mol + 52 g/mol + 64 g/mol = 194 g/mol

So, the empirical mass of the compound is 194 g/mol.