Question

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)

Answer 1

Hint: We are given that the ferrous sulphate heptahydrate is used to fortify foods with iron. Also, we are given atomic weights of elements in gram as: Fe: \[55.85\] ; S: \[32.00\]; O: \[16.00\]. We have to find out the amount of salt required to achieve iron of \[10\] ppm in wheat of \[100\]kg. We will be using following formulae to find out

Formula used : Following formulae are useful for solving this question.
\[ppm = \dfrac{{{w_1}}}{{{w_2}}} \times {10^6}\]
\[moles = \dfrac{w}{M}\]
\[{w_1} = \]moles \[ \times \]atomic mass
Where \[{w_1},{w_2},w,M\]are the weight of an element (here, iron), weight of the object (here, wheat), weight of another object (here, salt), weight of compound (here, ferrous sulphate heptahydrate)

Complete Step by Step Solution:

Let us assume the weight of salt required be \[w\]g
We are given a compound ferrous sulphate heptahydrate, whose chemical formula is \[FeS{O_4}.7{H_2}O\]. On substituting the atomic masses of the elements in the formula of compound we get the mass of the compound as \[277.85\]g.

Weight of Iron (Fe) is given by \[\dfrac{w}{M} \times 55.85\]. On substituting the values, we get weight of Iron as \[\dfrac{w}{{277.85}} \times 55.85\]g, which we will represent by \[{w_1}\]

Now, we are given the weight of wheat as \[100\]kg which is equal to \[{10^5}\]g. This we will represent by \[{w_2}\]
Also, we are given ppm of iron as \[10\]ppm
Now by using formula \[ppm = \dfrac{{{w_1}}}{{{w_2}}} \times {10^6}\], we get
\[
  10 = \dfrac{{\dfrac{w}{{277.85}} \times 55.85}}{{{{10}^{^5}}}} \times {10^6} \\
  10 = \dfrac{w}{{277.85}} \times 55.85 \times 10 \\
  1 = \dfrac{w}{{277.85}} \times 55.85 \\
  w = \dfrac{{277.85}}{{55.85}} \\
 \]

On further solving, we get the weight of salt as
\[w = 4.97\]g

Note: Students may make mistakes while calculating the mass of the compound. They should properly substitute the mass in \[FeS{O_4}.7{H_2}O\]i.e., for this compound they should calculate as \[55.85 + 32 + 16 \times 4 + 7(1 \times 2 + 16) = 277.85\]g.