Laws of Chemical Combination

Chemical changes follow some basic laws. There are four main laws that describe the general features of chemical reactions.

1. Law of Conservation of Mass (Matter)

This law states that during ordinary chemical reactions, matter is neither created nor destroyed.

Experiment

• Aim: To verify the law of conservation of mass.
• Reagents: Lead(II) trioxonitrate(V) solution and potassium chloride solution.

Procedure:

Pour some lead(II) trioxonitrate(V) solution into a conical flask. Fill a small test tube with potassium chloride solution and suspend it inside the flask using a string. Ensure the solutions do not mix. Cork the flask with a rubber stopper and weigh the entire setup. Record the mass.

Mix the two solutions by releasing the string and weigh the flask again.

Reaction:

Pb(NO₃)₂(aq) + KCl(aq) → PbCl₂(s) + KNO₃(aq)

(Formation of a white precipitate)

Result:

After mixing, a white precipitate of lead(II) chloride and potassium trioxonitrate(V) forms. The mass before and after the reaction remains the same, showing that mass is conserved.

Conclusion:

The experiment confirms that matter is neither created nor destroyed during a chemical reaction.

2. Law of Definite Proportions (Constant Composition)

This law states that all pure samples of a specific chemical compound contain the same elements combined in the same fixed proportion by mass.

Experiment

• Aim: To verify the law of definite proportions.

Procedure:

Prepare three samples of black copper(II) oxide using different methods:

• Sample A (from copper): Dissolve copper turnings in concentrated trioxonitrate(V) acid. Brown fumes of NO₂ are produced, forming a green copper(II) trioxonitrate(V) solution.

Reaction:

3Cu(s) + 8HNO₃(aq) → 3Cu(NO₃)₂(aq) + 2NO(g) + 4H₂O(l)

Evaporate the solution to dryness and heat the residue to obtain black copper(II) oxide:

2Cu(NO₃)₂(s) → 2CuO(s) + 4NO₂(g) + O₂(g)

• Sample B (from copper(II) sulfate): Add excess sodium hydroxide to copper(II) sulfate solution to form copper(II) hydroxide. Boil the solution to decompose it into copper(II) oxide.

Reactions:

CuSO₄(aq) + 2NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)

Cu(OH)₂(s) → CuO(s) + H₂O(l)

• Sample C (from copper(II) carbonate): Heat copper(II) carbonate strongly to form copper(II) oxide and carbon dioxide gas.

Reaction:

CuCO₃(s) → CuO(s) + CO₂(g)

Analysis of the Samples:

Weigh three clean, dry boats. Add each sample of copper(II) oxide to separate boats and re-weigh. Place them in a combustion tube, pass dry hydrogen gas over them, and heat strongly. Copper metal forms:

CuO(s) + H₂(g) → Cu(s) + H₂O(l)

Weigh the cooled samples to calculate the percentage of copper in each sample.

Results:

Conclusion:

No matter how copper(II) oxide is prepared, the copper content remains nearly the same, confirming the law of definite proportions.

3. Law of Multiple Proportions

This law states that if two elements A and B form more than one compound, the different masses of A that combine with a fixed mass of B are in a simple whole-number ratio.

Experiment

• Aim: To verify the law of multiple proportions.

Procedure:

Weigh two dry boats. Add copper(I) oxide (Cu₂O) to one and copper(II) oxide (CuO) to the other. Weigh again to find the mass of the oxides. Place the boats in a combustion tube, pass dry hydrogen gas, and heat to reduce the oxides to copper. Allow cooling while maintaining hydrogen flow to avoid oxidation. Weigh the copper residues.

Results:

Calculation:

In Cu₂O: 0.65g of oxygen combines with 3.53g of copper.

Thus, 1g of oxygen combines with (3.53 ÷ 0.65) ≈ 5.43g of copper.

In CuO: 0.83g of oxygen combines with 2.25g of copper.

Thus, 1g of oxygen combines with (2.25 ÷ 0.83) ≈ 2.71g of copper.

Ratio of copper in Cu₂O : CuO ≈ 2:1

Conclusion:

The mass of copper combining with a fixed mass of oxygen is in a simple whole number ratio, confirming the law of multiple proportions.