The law states that if two elements A and B combine together to form more than one compound, the various masses of B which combine with a fixed mass of A are in a simple whole-number ratio. E.g 1:2, 2:1, 1:3, 3:2, etc.

### Verification of the Law:

Weigh two clean metal boats, weigh them again containing pure copper (I) oxide (Cu_{2}O) and copper(II) oxide (CuO) respectively. Reduce the oxide to copper by using hydrogen gas as described in the experiment on Law of definite proportion.

The mass of copper which combined with a fixed mass of oxygen is found in both samples. The masses are in a simple ratio.

**i.** \( \scriptsize Cu_2O_{(s)} \: + \: H_{2(g)} \: \rightarrow \: 2Cu_{(s)} \: + \: H_2O_{(g)} \)

**ii.** \( \scriptsize CuO_{(s)} \: + \: H_{2(g)} \: \rightarrow \: Cu_{(s)} \: + \: H_2O_{(g)} \)

Sample | Sample ICu _{2}O | Sample IICuO |

Mass of oxide | 3.04g | 1.91g |

Mass of Cu residue | 2.55g | 1.38g |

Mass of oxygen removed from the oxide | 0.49g | 0.53g |

In order to show that these figures conform to the law of multiple proportions, use 1g or 100g as a fixed mass.

**In sample I,**

0.49g of oxygen combined with 2.55g of Cu using 1g as fixed mass.

\( \frac{2.55g}{0.49g} \: \times \: \frac{1}{1} \\ \scriptsize = 5.20g \)**In sample II,**

0.53g of oxygen combined with 1.38g of copper using 1g as fixed mass.

\( \frac{1.38g}{0.53g} \: \times \: \frac{1}{1} \\ \scriptsize = 2.6g \)The mass of copper that have combined with a fixed mass (1.0g) of oxygen in samples 1 and 2 are in simple ratio 5.20, 2.60, or 2:1. Hence, the Law of multiple proportions is obeyed.

Example I:

Carbon combines with oxygen to form two oxides A and B. Analysis of the two oxides show that:

(I) In oxide A, 57.1g of oxygen combines with 42.9g of carbon.

(II) In oxide B, 72.7g of O_{2} combines with 27.3g of carbon.

Show that the results illustrate the law of multiple proportions.

Sample | Oxide A | Oxide B |

Mass of oxygen | 57.1g | 72.7g |

Mass of carbon | 42.9g | 27.3g |

Using 1g as a fixed mass | \( \frac{57.1g}{42.9g} \: \times \: \frac{1}{1} \\ \scriptsize = 1.33g \) | \( \frac{72.7g}{27.3g} \: \times \: \frac{1}{1} \\ \scriptsize = 2.66g \) |

= Ratio 1:2

Since the masses are in simple whole-number ratios, the law of multiple proportions is verified.

**Theory Questions**

**1. **(a) Define the Law of multiple proportions

(b) Name an Oxide you can use for the verification of the Law

View Answer**2. **Two samples of Copper(II) Oxide were prepared byÂ

(i) Heating Copper(II) trioxocarbonate(IV)

(ii) Heating strongly the mixture of Copper(II) tetraoxosulphate(VI) and Sodium hydrogen

Analysis of these two samples shows that: (i) 4.50g of Copper(II) Oxide contain 4g of Copper (ii) 5.60g of Copper(II) Oxide contain 4.98g of Copper

Show that these results illustrates the Law of Constant composition

View Answer**3.** (a) Define the Law of conservation of massÂ

(b) Calculate the mass of Silver Chloride precipitate formed when 20.20g of Silver trioxonitrate(V) reacts with 9.10g of Sodium Chloride to form 14.20g of Sodium trioxonitrate(V)

View Answer**4.** (a) State the Law of Constant composition or definite proportion

(b) How could you prove the Law of constant composition experimentally

(c) Two samples of Iron(II)Sulphide were prepared by:

(i) Heating a mixture of Iron filings and Sulphur

(ii) Passing hydrogen Sulphide gas into a solution of Iron(II) Chloride

Analysis of both samples (i) and (ii) shows that:

i. 3.14g of Iron(II)Sulphide contains 0.12g of Iron

ii. 4.16g of Iron(II)sulphide contains 0.15g of Iron.

Show that these results illustrate the Law of Constant Composition.

View Answer