When hafnium metal is heated in an atmosphere of chlorine gas, the product of the reaction is found to contain 62.2% hf by mass and 37.4% cl by mass. what is the empirical formula for this compound?

Clearly the percentages are weights. 
For Hafnium metal it’s 62.2%, grams per mole for Hafnium = 178.5 grams / moles 
For Chlorine gas it's 37.4%, grams per mole for Chlorine = 35.45 grams / moles 
Calculating the molar masses,  
Hafnium = 62.2 g / 178.5 grams / moles = 0.35 mol
 Chlorine = 37.4 g / 35.45 grams / moles = 1.05 mol
 Divide by the smallest to get the ratios,
 0.35 mol Hf / 0.35 mol = 1 mol Hf
 1.05 mol Cl / 0.35 mol = 3 mol Cl
 So the emperical formula would be HfCl3

MnSO₄.7H₂O

Explanation:

To solve this question, we need to convert the mass of the dehydrated MnSO₄. The difference between mass of the hydrate and dehydrated compound is the mass of water. With the mass we can find the moles of water and the formula of the hydrate:

Moles MnSO₄ -Molar mass: 151g/mol-:

17.51g * (1mol / 151g) = 0.116 moles

Moles H₂O -Molar mass: 18g/mol-:

32.14g-17.51g = 14.63g * (1mol / 18g) = 0.813 moles

The ratio of moles MnSO₄: Moles H₂O represent the amount of water molecules in the hydrate:

0.813mol / 0.116mol = 7 molecules of water.

The hydrate formula is:

MnSO₄.7H₂O