Question

The active ingredient in iron supplements is iron (II) sulfate, also called ferrous sulfate. One tablet contains 325 mg of iron (II) sulfate.

1. What is the chemical formula for iron (II) sulfate?

2. What is the molar mass of iron (II) sulfate?

3. How many moles of iron (II) sulfate are in 6 tablets?

4. How many molecules of iron (II) sulfate are in 3 tablets?

(HINT: tablets  mass  mol  molecules)

Answer 1

Iron (II) means Iron exists in +2 state and Sulfate exists in -2 state.

So, Chemial formula for Iron (II) Sulfate is FeSO₄

Molar Mass of Iron (II) Sulfate = 1*Molar Mass of Fe + 1*Molar Mass of S + 4*Molar Mass of O = 1*56+1*32+4*16 = 152 g/mol

One Tablet contains 325 mg of Iron (II) Sulfate

Six Tablets will contain 6*325 mg of Iron (II) Sulfate i.e. 1950 mg * 1 g/1000 mg = 1.95 g

Moles = Given Mass/Molar Mass = 1.95 g / 152 g/mol = 0.0128 moles

Number of moles in 3 tablets will be half of number of moles in 6 tablets = 0.0128/2 = 0.0064 moles

One mole contains 6.023 * 10²³ molecules

So, Number of molecules in 3 tablets = 0.0064 * 6.023 * 10²³ molecules = 3.85 * 10²¹ molecules