A 0.500 g sample of a compound Containing only antimony and oxygen was found to
contain 0.418 g of antimony and 0.082 g of oxygen. What is the simplest formula for the
compound? (a) Sbo (b) SbO2 (c) Sb.O. (a) Sb2O,
(c) Sb;0. (a) Sb2O, e) Sb2O3
Answers
Explanation:
On analysis, a certain compound was found to contain iodine and oxygen in the ratio 254 g of iodine and 80 g of oxygen. The atomic mass of iodine is 127 and that of oxygen is 16. Which is the formula of the compound?
Answer:
The correct answer is e) Sb₂O₃.
Given:
Total mass of the compound, M = 0.500 g
Mass of antimony, m₁ = 0.418 g
Mass of oxygen, m₂ = 0.082 g
We know that,
Molar mass of antimony, W₁ = 122
Molar mass of oxygen, W₂ = 16
Find:
The simplest formula for the compound.
Solution:
For finding out the formula, we need to find out the their number of moles.
Number of moles of antimony (Sb), n₁ = m₁/W₁
= 0.418/122
= 0.209/61
= 0.00343
Number of moles of oxygen (O), n₂ = m₂/W₂
= 0.082/16
= 0.041/8
= 0.00513
Ratio of coefficients = (0.00343/0.00343) : (0.00513/0.00343)
= 1 : 1.5
= 1 : (15/10)
= 1 : (3/2)
= 2/3
= 2 : 3
It means that antimony and oxygen are present in the ratio 2 : 3.
∴ The simplest formula for the compound = Sb₂O₃
Hence, option e) is the correct answer.
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