Some Basic Concepts of Chemistry — Part 2: Formulae, stoichiometry and concentration: visual lesson, CBSE Class 11 Chemistry

Lesson

Formulae, reactions and solutions

Part 2 — from the mass of each element to a formula, from a balanced equation to the mass of product, and from a bottle label to a concentration.

In this lesson you will

Calculate percentage composition and derive empirical and molecular formulae
Perform mass-mass calculations from balanced equations
Identify the limiting reagent and find the amount of product
Express and interconvert concentration terms: mass percent, ppm, mole fraction, molarity and molality

Concept

Percentage composition

Mass percent of an element $= \dfrac{\text{mass of the element in one mole of compound}}{\text{molar mass of the compound}} \times 100$ . The percentages of all elements add up to 100.

Table of percentage composition of three compounds
Compound	Molar mass / g mol⁻¹	Mass percent of the main element
Potassium permanganate, $\ce{KMnO4}$	158.0	K 24.7, Mn 34.8, O 40.5
Calcium sulfate, $\ce{CaSO4}$	136.1	Ca 29.4, S 23.6, O 47.0
Iron(III) oxide, $\ce{Fe2O3}$	159.7	Fe 69.9, O 30.1

Worked example

Worked example: percentage composition

Problem. Find the mass percent of each element in ammonium dihydrogen phosphate, $\ce{NH4H2PO4}$ (N 14.0, H 1.0, P 31.0, O 16.0).

Step 1. Molar mass $= 14.0 + 6(1.0) + 31.0 + 4(16.0) = 115.0$ g mol⁻¹.

Step 2. N: $14.0 / 115.0 \times 100 = 12.2$ %; H: $6.0 / 115.0 \times 100 = 5.2$ %; P: $31.0 / 115.0 \times 100 = 27.0$ %; O: $64.0 / 115.0 \times 100 = 55.7$ %.

Check. $12.2 + 5.2 + 27.0 + 55.7 = 100.1$ , equal to 100 within rounding.

Table of the mass percent of each element in ammonium dihydrogen phosphate
Element	Atoms	Mass in 1 mol / g	Percent
N	1	14.0	12.2
H	6	6.0	5.2
P	1	31.0	27.0
O	4	64.0	55.7

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