Law of Multiple Proportions

Lesson Progress

0% Complete

Topic Content:

Law of Multiple Proportions
Verification of the Law

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 (Cu₂O) and copper(II) oxide (CuO) respectively. Reduce the oxide to copper by using hydrogen gas as described in the experiment on the 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 I	Sample II
	Cu₂O	CuO
Mass of oxide	3.04 g	1.91 g
Mass of Cu residue	2.55 g	1.38 g
Mass of oxygen removed from the oxide	0.49 g	0.53 g

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

In sample I:

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

\( \frac{2.55g}{0.49g} \: \times \: \frac{1}{1} \\ \scriptsize = 5.20\: g \)

In sample II:

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

\( \frac{1.38g}{0.53g} \: \times \: \frac{1}{1} \\ \scriptsize = 2.6\:g \)

The mass of copper that has combined with a fixed mass (1.0 g) of oxygen in samples 1 and 2 are in a simple ratio of 5.20:2.60, or 2:1 . Hence, the Law of multiple proportions is obeyed.

Example 8.4.1:

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

(I) In oxide A, 57.1 g of oxygen combines with 42.9 g of carbon.
(II) In oxide B, 72.7 g of O₂ combines with 27.3 g of carbon.

Show that the results illustrate the law of multiple proportions.

Solution:

Sample	Oxide A	Oxide B
Mass of oxygen	57.1 g	72.7 g
Mass of carbon	42.9 g	27.3 g
Using 1g as a fixed mass	\( \scriptsize\frac{57.1\:g}{42.9\:g} \: \times \: \frac{1}{1} \)	\(\scriptsize \frac{72.7\:g}{27.3\:g} \: \times \: \frac{1}{1} \)
	\(\scriptsize = 1.33 \: g\)	\(\scriptsize = 2.66\: g\)

= 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

2. Two samples of Copper(II) Oxide were prepared by

(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 contains 4 g of Copper
(ii) 5.60 g of Copper(II) Oxide contains 4.98 g of Copper

Show that these results illustrate the Law of Constant Composition

3. (a) Define the Law of conservation of mass

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

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. 3.14 g of Iron(II) Sulphide contains 0.12 g of Iron.
ii. 4.16 g of Iron(II) Sulphide contains 0.15 g of Iron.

Show that these results illustrate the Law of Constant Composition.