# Calaculate the amount of Elemental Zinc from 200g of Anhydrous Zinc (Given is the molecular wieght)?

Pharmacy Calculations. Can't find a simple formula or any formula for that matter.

I do not need advanced Chemestry definitions just a simple explaination.

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The Actual Question is : A patient is currently taking 220 mg of anhydrous zinc sulfate. To receive the equivalent amount of elemental zinc, how many milligrams of zinc sulfate heptahydrate (•7 H2O) would the patient have to take? (Molecular weights: zinc 65, ZnSO4 161, H2O 18)

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Sorry for the repeats, Technical diffficulties

Relevance

Does the question specify the type of anhydrous zinc? Because ''Anhydrous zinc'' means ''zinc without water'', i.e. elemental zinc. Also, what is the molecular weight given?

[EDIT] Thanks. I'll attempt to explain it as clear and concise as possible.

Determine the moles of zinc in 220 mg of zinc sulfate:

moles zinc = 0.22 g / 161 g-mol = 0.001366 mol

Now, we multiply this amount by the molar mass of zinc sulfate heptahydrate (287 g-mol) to determine the mass required for an equivalent amount of zinc:

mass required = (0.001366 mol)(287 g-mol) = 0.39 g ≈ 0.4 g = 400 mg (rounded to one significant figure).

If you require any more clarification, let it be known.

• Elemental Zinc

• Anonymous
7 years ago