Determine the molecular formula of an oxide of iron in which Chapter 1: Some Basic Concepts of Chemistry Chemistry Class 11 solutions are developed for assisting understudies with working on their score and increase knowledge of the subjects. Question 8: Question 8: Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is 159.69 g mol-1. is solved by our expert teachers. You can get ncert solutions and notes for class 11 chapter 1 absolutely free. NCERT Solutions for class 11 Chemistry Chapter 1: Some Basic Concepts of Chemistry is very essencial for getting good marks in CBSE Board examinations
Answer
Answer
Mass % of iron = 69.9 % [Given]
Mass % of oxygen = 30.1 % [Given]
Element |
Atomic mass |
Mass % |
Mass % / atomic mass |
Fe |
55.85 |
69.9 |
69.9/55.85 = 1.25 |
O |
16.00 |
30.1 |
30.1/16.00 = 1.88 |
Fe : O = 1.25 : 1.88
Convert in simple ratio we get
Fe : O = 2 : 3
The empirical formula of the iron oxide is Fe2O3.
N = Molar mass / Empirical mass …(1)
Empirical formula mass of Fe2O3 = [2×55.85 + 3×16.00] g = 159.70 g
Given that Molar mass of Fe2O3 = 159.69 g
Plug the values in equation (1), we get
N = 159.69/159.70 = 1
Molecular formula = Empirical formula × n = Fe2O3 × 1 = Fe2O3
Hence molecular formula is also Fe2O3
Molecular formula of a compound is obtained by multiplying the empirical formula with n. Thus, the empirical formula of the given oxide is Fe2O 3 and n is 1.
Hence, the molecular formula of the oxide is Fe2O3.
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