Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt

Q: Why do we include the mass of water ($7\text{H}_2\text{O}$) in the molar mass?

Because the salt is hydrated. When you weigh $x$ grams of the salt on a scale, the water molecules are part of that weight. If you ignore them, your calculation for the iron fraction will be completely wrong.

Q: Can I use the mass of wheat in kg directly in the ppm formula?

No! The ppm formula requires the mass of the solute and the mass of the solution (or solvent, if solute is negligible) to be in the *same units*. Since we are calculating the mass of iron in grams, the wheat must also be in grams.

Q: Why is the mass of the solution taken as just the mass of the wheat?

Technically, the mass of the solution is the mass of wheat + mass of iron. However, 100 kg is $10^5$ grams, and the iron is just a few grams. Adding them makes a negligible difference, so we approximate the solution mass as the wheat mass.

Q: What does 10 ppm actually mean physically?

It means '10 parts per million'. For every 1 million grams (which is 1000 kg) of wheat, there are exactly 10 grams of iron.

Q: How do I avoid calculation errors with such large numbers?

Always keep the powers of 10 separate and simplify them first. In this problem, $10^6$ in the numerator and $10^5$ in the denominator neatly cancel out to leave just a factor of 10, making the final division much easier.

Question

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

Sigma Percentile · Accepted Answer

### The Challenge of Fortification

Imagine you are tasked with fortifying a massive $100 	ext{ kg}$ sack of wheat with iron. The goal is to achieve a concentration of exactly $10 	ext{ ppm}$ (parts per million).

To do this, we are given a specific salt: Ferrous sulphate heptahydrate, chemically written as $	ext{FeSO}_4 \cdot 7	ext{H}_2	ext{O}$.

Our mission is to find out exactly how many grams of this salt, let's call it $x$, we need to mix into the wheat.

### Decoding the Salt: Molar Mass

Before we can extract the iron, we need to understand the salt itself. The first step in any stoichiometry problem involving a compound is calculating its molar mass.

We must account for every single atom in $	ext{FeSO}_4 \cdot 7	ext{H}_2	ext{O}$, including the water of crystallization.

$$M = 55.85 + 32.00 + 4(16.00) + 7(2(1.00) + 16.00)$$

Adding these up, we get a total molar mass of $277.85 	ext{ g/mol}$.

**This means that in every $277.85 	ext{ g}$ of this salt, there is exactly $55.85 	ext{ g}$ of pure iron.**

### Extracting the Iron Content

Now, we don't have $277.85 	ext{ g}$ of the salt; we have an unknown amount, $x$ grams.

We need to find the mass of iron hidden within this $x$ grams. We can use a simple mass fraction.

$$	ext{Mass of Fe} = \left( \frac{55.85}{277.85} ight) x$$

This expression represents the exact amount of iron that will actually go into fortifying our wheat.

### The PPM Master Equation

Now, let's bring in the concept of ppm. Parts per million is a unit of concentration that tells us how many parts of solute exist in one million parts of the solution.

The formula is:

$$	ext{ppm} = \frac{	ext{Mass of solute}}{	ext{Mass of solution}} 	imes 10^6$$

Here, our solute is the iron we just extracted, and our solution is effectively the $100 	ext{ kg}$ of wheat.

**Crucially, the masses must be in the same units!** We must convert the $100 	ext{ kg}$ of wheat into grams, which gives us $100 	imes 10^3 	ext{ g}$, or $10^5 	ext{ g}$.

Let's substitute everything into our master equation:

$$10 = \frac{\frac{55.85x}{277.85}}{10^5} 	imes 10^6$$

### Final Calculation

This equation might look a bit messy, but it simplifies beautifully.

Look at the powers of $10$. We have $10^6$ in the numerator and $10^5$ in the denominator. They cancel out to leave just a factor of $10$.

$$10 = \frac{55.85x 	imes 10}{277.85}$$

We can divide both sides by $10$, leaving us with:

$$1 = \frac{55.85x}{277.85}$$

Now, simply solve for $x$:

$$x = \frac{277.85}{55.85}$$

$$x = 4.97 	ext{ g}$$

And there we have it! We need exactly $4.97 	ext{ g}$ of Ferrous sulphate heptahydrate to perfectly fortify our $100 	ext{ kg}$ sack of wheat.

Animated Solution for Chemistry - Basic Concepts in Chemistry: 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 : $Fe = 55.85$ ; $S = 32.00$ ; $O = 16.00$

Enter Numerical Value:

Visualized Solution

$P r o b l e m S e t u p$

$M o l a r M a sso f S a l t$

$M a sso f I r o nin xg o f S a l t$

$P P M F or m u l a$

$S u b s t i t u t i o n$

$S im pl i f i c a t i o n$

$F ina l C a l c u l a t i o n$

$C o n ce pt u a l R e f l ec t i o n$

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

The Challenge of Fortification

Decoding the Salt

Extracting the Iron Content

The PPM Master Equation

Final Calculation

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Visualized Solution

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