A metal X forms two distinct chlorides. Quantitative analysis shows that in Chloride 1, 5.… — JAMB UTME

A metal $X$ forms two distinct chlorides. Quantitative analysis shows that in Chloride 1, $5.40\text{ g}$ of $X$ combines with $10.65\text{ g}$ of chlorine. In Chloride 2, $3.60\text{ g}$ of $X$ combines with $10.65\text{ g}$ of chlorine. If the empirical formula of Chloride 1 is $XCl_2$ , calculate the subscript value $y$ in the empirical formula $XCl_y$ of Chloride 2.

Answer: 3

Answer

The value of the subscript y is 3.

Applying the Law of Multiple Proportions, when a fixed mass of chlorine (

10.65\text{ g}

) reacts with different masses of metal

X

(

5.40\text{ g}

and

3.60\text{ g}

), the mass ratio of

X

5.40 : 3.60 = 3 : 2

. This implies that for a fixed amount of metal

X

, the ratio of chlorine atoms in Chloride 1 to Chloride 2 is

2 : 3

. Since Chloride 1 is

XCl_2

, Chloride 2 must be

XCl_3

, yielding

y = 3

Step-by-Step Solution

Determine the mass of chlorine per gram of metal X in Chloride 1.

\frac{10.65\text{ g}}{5.40\text{ g}} = 1.9722\text{ g } Cl / \text{g } X

This establishes the baseline quantitative relationship for the formula

XCl_2

Determine the mass of chlorine per gram of metal X in Chloride 2.

\frac{10.65\text{ g}}{3.60\text{ g}} = 2.9583\text{ g } Cl / \text{g } X

This determines the mass of chlorine per unit mass of metal in the second compound.

Calculate the simple multiple proportion ratio between the two compounds.

\frac{2.9583}{1.9722} = 1.5

According to the Law of Multiple Proportions, the masses of chlorine combining with a fixed mass of X stand in a simple whole-number ratio.

Multiply the subscript of chlorine in the first compound by the calculated ratio.

y = 2 \times 1.5 = 3

Since Chloride 1 has 2 chlorine atoms (

XCl_2

), Chloride 2 must have

2 \times 1.5 = 3

chlorine atoms (

XCl_3

Key Concept

Law of Multiple Proportions

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