An element X forms two distinct gaseous oxides, Oxide A and Oxide B. Quantitative analysis… — JAMB UTME

An element $X$ forms two distinct gaseous oxides, Oxide A and Oxide B. Quantitative analysis reveals that $14.0\text{ g}$ of $X$ combines with $16.0\text{ g}$ of oxygen in Oxide A, whereas $14.0\text{ g}$ of $X$ combines with $32.0\text{ g}$ of oxygen in Oxide B. What is the ratio of the masses of oxygen combining with a fixed mass of element $X$ , and which law of chemical combination does this illustrate?

$1:2$ ; Law of Multiple ProportionsAnswer
B
$1:2$ ; Law of Definite Proportions
C
$7:8$ ; Law of Conservation of Mass
D
$2:1$ ; Law of Combining Volumes

Answer

The ratio of oxygen masses is

1:2

, which illustrates the Law of Multiple Proportions.

For a constant mass of element

X

(

14.0\text{ g}

), the mass of oxygen in Oxide A (

16.0\text{ g}

) and Oxide B (

32.0\text{ g}

) forms a simple whole-number ratio of

16:32 = 1:2

. This directly satisfies John Dalton's Law of Multiple Proportions.

Step-by-Step Solution

Identify the fixed mass of element

X

in both compounds

Mass of element

X = 14.0\text{ g}

in both Oxide A and Oxide B

To apply the Law of Multiple Proportions, the mass of one element must be held constant

Determine the masses of oxygen combined with the fixed mass of element

X

Oxide A has

16.0\text{ g}

of oxygen, Oxide B has

32.0\text{ g}

of oxygen

These are the given masses of oxygen reacting with

14.0\text{ g}

X

Calculate the simple ratio between the masses of oxygen

\frac{16.0}{32.0} = \frac{1}{2} \text{ or } 1:2

Dividing both masses by the common factor

16.0\text{ g}

yields a simple whole-number ratio

Match the observed phenomenon to the appropriate chemical law

Law of Multiple Proportions

When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers

Key Concept

Law of Multiple Proportions

Estimated Time:1m 30s

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