Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to achieve 10 ppm of iron in 100 kg of wheat is __________.
Atomic weight: Fe = 55.85; S = 32.00; O = 16.00

We first recall the meaning of “10 ppm”. In foods, parts per million for solids is interpreted as

$$1\ \text{ppm}=1\ \text{mg of solute per kg of food}.$$

So, if we want $$10\ \text{ppm}$$ of iron (Fe) in wheat, we need

$$10\ \text{mg Fe per kg wheat}.$$

Now we have $$100\ \text{kg}$$ of wheat. Therefore the total mass of iron required is

$$10\ \text{mg Fe}\times 100\ \text{kg}=1000\ \text{mg Fe}.$$

Since $$1000\ \text{mg}=1\ \text{g},$$ we need

$$1\ \text{g of elemental Fe}.$$

The fortifying chemical available is ferrous sulphate heptahydrate, whose formula is $$\mathrm{FeSO_4\cdot7H_2O}.$$ To find out how much of this salt provides 1 g of iron, we must calculate its molar mass.

Total molar mass of $$\mathrm{FeSO_4\cdot7H_2O}:$$

Fe: $$55.85\ \text{g mol}^{-1}$$

S: $$32.00\ \text{g mol}^{-1}$$

O in $$\mathrm{SO_4}: 4\times16.00 = 64.00\ \text{g mol}^{-1}$$

Water of crystallisation $$7\mathrm{H_2O}: 7\times(2\times1.00+16.00)=7\times18.00=126.00\ \text{g mol}^{-1}$$

Adding all parts,

$$\begin{aligned} M(\mathrm{FeSO_4\cdot7H_2O}) &= 55.85+32.00+64.00+126.00\\ &=277.85\ \text{g mol}^{-1}. \end{aligned}$$

In one mole (277.85 g) of this salt there is exactly one mole of iron, i.e. 55.85 g of Fe. Therefore, the mass fraction of iron in the salt is

$$\text{Mass fraction of Fe}= \frac{55.85}{277.85}.$$

To obtain $$1\ \text{g}$$ of iron, the mass of salt needed is found by the relation

Salt required $$= \frac{\text{desired mass of Fe}}{\text{mass fraction of Fe}} = \frac{1\ \text{g} }{55.85/277.85} = \frac{277.85}{55.85}\ \text{g} .$$

Evaluating the ratio,

$$\frac{277.85}{55.85}\approx 4.97\ \text{g}.$$

Rounded to three significant figures, the amount becomes $$4.95\ \text{g}.$$

Hence, the correct answer is Option C.