An element forms two oxides of 2.9 g and 2.25 g of these oxides each was found to contain 1.12 L of $O_{2}$ at STP. What is the simple ratio if it obeys law of multiple proportions?

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Solution

According to gram molecular volume law, 22.4 L of all gases at STP weigh equal to their molecular weights denoted in grams. Hence, the weight of 1.12 L of oxygen at STP

= $\frac{32 \times 1.12}{22.4}$ = 1.6g

In the first oxide:
Weight of oxide = 2.9 g; weight of O = 1.6 g
$∴$ Weight of the element = 2.9 - 1.6 = 1.300 g
In the other oxide:
Weight of oxide = 2.250 g; weight of O = 1.600 g
Weight of element = 2.25- 1.6 = 0.650 g

Thus, the weights of the element that combine with the same weight of oxygen in its two oxides are in the ratio of 1.3 : 0.65 or 2 : 1 a simple ratio. Hence, it illustrates the law of multiple proportions.

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An element forms two oxides of 2.9 g and 2.25 g of these oxides each was found to contain 1.12 L of O2 at STP. What is the simple ratio if it obeys law of multiple proportions?

An element forms two oxides of 2.9 g and 2.25 g of these oxides each was found to contain 1.12 L of $O_{2}$ at STP. What is the simple ratio if it obeys law of multiple proportions?