Question

Ferrous sulfate 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: $\mathrm{Fe}=55.85$; $\mathrm{S}=32.00$; $\mathrm{O}=16.00$)

Answer 1

$$\begin{gathered} 10\mathrm{ppm}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{Fe}(\mathrm{in}\mathrm{gm})}{100\times 1000}\times {10}^6 \\ \mathrm{Mass}\mathrm{of}\mathrm{Fe}=1\mathrm{gm} \\ \mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{FeSO}}_4.7{\mathrm{H}}_2\mathrm{O}=278\mathrm{gm} \\ \mathrm{Mass}\mathrm{of}\mathrm{Fe}\mathrm{is}56\mathrm{gm}\mathrm{in}1\mathrm{mole} \\ \mathrm{So}1\mathrm{gm}\mathrm{in}\frac{1}{56}\mathrm{moles} \\ \mathrm{Therefore},\frac{1}{56}\times 278=4.96\mathrm{gm} \end{gathered}$$

Hence, ferrous sulfate 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 4.96 gm.