Pharmacy PEBC Practice Exam

A patient is currently taking 220 mg of anhydrous zinc sulphate. To receive the equivalent amount of elemental zinc, how many milligrams of zinc sulphate heptahydrate (•7 H20) would the patient require? (Molecular weights: zinc 65, ZnSO4 161, H20 18)

1. 123 mg

2. 220 mg

3. 300 mg

4. 392 mg

The correct answer is: 392 mg

To determine how many milligrams of zinc sulfate heptahydrate would provide the same amount of elemental zinc as 220 mg of anhydrous zinc sulfate, it's essential to first calculate the amount of elemental zinc present in the 220 mg of the anhydrous form. Anhydrous zinc sulfate has a molecular weight of 161 g/mol. Of this, zinc (which has a molecular weight of 65 g/mol) represents approximately 40.37% of the molecular weight of the anhydrous form. This means that for every 161 mg of anhydrous zinc sulfate, 65 mg is elemental zinc. To find the elemental zinc in 220 mg of anhydrous zinc sulfate: 1. Calculate the zinc content: \[ \text{Elemental zinc} = \left( \frac{65 \, \text{mg}}{161 \, \text{mg}} \right) \times 220 \, \text{mg} = 86.83 \, \text{mg} \] Now, we need to find out how much zinc sulfate heptahydrate is needed to obtain the same amount of elemental zinc. Zinc sulfate heptahydrate (Zn