Iron and sulfur comparison

Two beakers sit on a lab bench. One holds grey iron filings stirred through yellow sulfur powder, cold and untouched. The other holds the exact same materials after a Bunsen burner turned them into a single black, glassy mass. Nothing was added and nothing was removed — yet a magnet obeys one beaker and ignores the other, and dilute hydrochloric acid produces a harmless "pop" gas from one and a rotten-egg stench from the other. This is the iron-and-sulfur activity, the signature demonstration that separates a mixture from a compound in the CBSE Class 8 chapter on the nature of matter. It is not just a memorable experiment — it hides a strict numerical rule, the law of constant proportions, which fixes exactly how much iron and how much sulfur combine to make iron sulfide, down to the gram.

The core insight

Cold iron filings and sulfur powder stay a separable, magnetic mixture with unchanged properties, while heating forces them to combine chemically in a fixed mass ratio into iron sulfide, a non-magnetic compound with entirely new properties.

The core idea

Sample A (unheated) is a mixture. The grey iron and yellow sulfur remain two separate substances, simply stirred together. A magnet pulls the iron filings straight out, leaving the sulfur behind — a purely physical separation that changes nothing about either substance. Because the iron is still ordinary iron, it reacts with dilute hydrochloric acid exactly as pure iron does, releasing hydrogen gas.

Sample B (heated) is a compound. Strong heating supplies enough energy to break the "just sitting next to each other" arrangement and force a genuine chemical reaction. The product, iron sulfide, is a single uniform black solid — no grey or yellow specks survive, because the iron and sulfur atoms are now chemically bonded, not merely mixed. That bonding also changes the chemistry: iron sulfide reacts with dilute hydrochloric acid to release hydrogen sulfide gas, not hydrogen.

The hidden number. What is easy to miss in the demonstration is that this combination does not happen in just any ratio. However much iron and sulfur you start with, only a fixed proportion of each actually combines to form iron sulfide — any extra of either reactant is simply left over, unreacted. Using the approximate atomic masses iron $= 56$ and sulfur $= 32$ (as used at this level), the compound iron sulfide, formula $\text{FeS}$ , always contains iron and sulfur in the mass ratio

\frac{\text{mass of iron}}{\text{mass of sulfur}} = \frac{56}{32} = \frac{7}{4}. \qquad (1)

where: the numbers $56$ and $32$ are the atomic masses (in atomic mass units) of one iron atom and one sulfur atom respectively, and the simplified ratio $7:4$ is the fixed mass ratio in which iron and sulfur combine, for any sample size, to form iron sulfide. This is the numerical face of the same idea the magnet test shows physically: a compound has one exact composition, while a mixture can be blended in any ratio you like.

Figure

Two beakers side by side, one showing separate grey iron specks and yellow sulfur specks that a magnet pulls apart, the other showing a single uniform black solid that a magnet cannot affect

Derivation AWhy iron and sulfur always combine in a fixed 7 to 4 mass ratio

The goal is to show, using nothing beyond the atomic masses already known at this level and simple ratio arithmetic, that a compound like iron sulfide is not "roughly iron plus roughly sulfur" but has one exact composition by mass — which is why, when the reactants are not supplied in that exact ratio, some of one reactant is inevitably left over unreacted after heating.

Step 1. Fix the building block. One formula unit of iron sulfide, $\text{FeS}$ , is made of exactly one iron atom bonded to exactly one sulfur atom — that is what the formula $\text{FeS}$ means. This 1-to-1 atom count is fixed by the chemistry of the bond and does not change no matter how much material you start with.

Step 2. Convert the atom count to a mass ratio. Since one iron atom has (approximate) atomic mass $56$ and one sulfur atom has atomic mass $32$ , one iron atom is $56$ times as heavy as a hydrogen-mass unit and one sulfur atom is $32$ times as heavy, so the mass contributed by iron to every formula unit of $\text{FeS}$ compares to the mass contributed by sulfur as $56 : 32$ .

Step 3. Simplify the ratio. Dividing both numbers by their common factor $8$ gives $56/8 = 7$ and $32/8 = 4$ , so the fixed mass ratio is $\text{iron} : \text{sulfur} = 7 : 4$ . Because this ratio comes only from the fixed 1-to-1 atom count and the fixed atomic masses, it is exactly the same whether you make one gram of iron sulfide or one kilogram.

Step 4. Read the consequence for the classroom activity. Mixing iron and sulfur in any ratio other than $7:4$ by mass and heating completely means only the pairs matching that ratio combine into $\text{FeS}$ ; whatever is in excess is left over, unreacted, inside the black solid. This is the law of constant proportions: a pure compound always has its elements combined in the same fixed mass ratio, unlike a mixture, which can be blended in any proportion at all.

Worked example

Question: $5.6\ \text{g}$ of iron filings and $4.0\ \text{g}$ of sulfur powder are mixed and heated strongly until the reaction is complete. How much iron sulfide forms, and how much of which reactant, if any, is left over unreacted?

Step 1. Find how much sulfur $5.6\ \text{g}$ of iron actually needs. The fixed combining ratio is iron to sulfur $= 7:4$ , so the sulfur required is $\dfrac{4}{7} \times 5.6\ \text{g} = 3.2\ \text{g}$ .

Step 2. Compare with the sulfur available. We were given $4.0\ \text{g}$ of sulfur, but only $3.2\ \text{g}$ is needed to react completely with all $5.6\ \text{g}$ of the iron. Since $4.0\ \text{g} > 3.2\ \text{g}$ , sulfur is in excess and iron is the reactant that runs out first.

Step 3. Find the leftover sulfur and the mass of iron sulfide formed. Leftover sulfur $= 4.0 - 3.2 = 0.8\ \text{g}$ , unreacted and still yellow inside the black mass. The iron sulfide formed uses all the iron plus the sulfur that reacted with it: $5.6\ \text{g} + 3.2\ \text{g} = 8.8\ \text{g}$ .

Step 4. Check the total mass balance. Everything we started with must still be accounted for: $8.8\ \text{g}\ (\text{FeS}) + 0.8\ \text{g}\ (\text{leftover S}) = 9.6\ \text{g}$ , which exactly equals the starting total, $5.6\ \text{g} + 4.0\ \text{g} = 9.6\ \text{g}$ . Mass is conserved. Answer: $8.8\ \text{g}$ of iron sulfide forms, with $0.8\ \text{g}$ of unreacted sulfur left over; no iron remains unreacted.

Common traps

Thinking any amount of iron and any amount of sulfur heated together must fully combine — only the amounts matching the fixed $7:4$ mass ratio react; the rest is left over unreacted, still detectable by the magnet (leftover iron) or as a slight yellow tinge or acid-test clue (leftover sulfur, though it is trapped inside the solid).
Forgetting to check which reactant is in excess before calculating leftover mass — always find how much of one reactant is needed for all of the other, then compare with what was actually supplied.
Assuming the mass of iron sulfide formed equals the total starting mass — it only equals the total starting mass when neither reactant is left over; otherwise, leftover mass must be subtracted first.
Confusing this fixed-ratio idea with the mixture, Sample A — a mixture of iron and sulfur can be blended in any ratio (10% iron, 90% iron, anything) because nothing has chemically combined; only the compound, iron sulfide, is locked to $7:4$ .

Ask the tutor: Want to try a leftover-reactant problem with your own starting masses of iron and sulfur and check your answer step by step?

Why does heating change the mass ratio rule at all — isn't it the same iron and sulfur either way?

It is the same atoms, but heating changes how they are arranged. In the cold mixture, iron and sulfur atoms simply sit next to each other in whatever proportion you poured them — no bond links them, so no fixed ratio applies. Heating supplies the energy for one iron atom to bond one sulfur atom, forming $\text{FeS}$ . Once that bond forms, the fixed atomic masses of iron ( $56$ ) and sulfur ( $32$ ) lock the mass ratio at $7:4$ for every bonded pair, whatever the original mixture's proportions were.

If I mix iron and sulfur in exactly the 7 to 4 ratio, will heating use up every last atom?

In an ideal, complete reaction, yes — mixing exactly $7$ parts iron to $4$ parts sulfur by mass and heating fully should leave no measurable iron or sulfur behind, only iron sulfide. In a real school lab, tiny amounts might remain if the heating is uneven or too brief, but the $7:4$ ratio is, in principle, the point where both reactants are used up together with nothing wasted.

Does the leftover reactant in the black solid still behave like plain iron or plain sulfur?

Yes. Anything left unreacted keeps its original element's properties, since it never actually combined chemically — it is just physically trapped within the solid mass of iron sulfide. Leftover iron filings would still be pulled out by a strong magnet, and leftover sulfur would still show its characteristic reaction if enough acid reached it. Only the iron and sulfur that actually bonded together, in the $7:4$ ratio, have become the new substance with its own distinct properties.

Iron and sulfur comparison

The core idea

Worked example

Common traps

Before this

Where this leads