Law of Multiple Proportions: Solved Numericals
The Law of Multiple Proportions is a cornerstone of basic chemistry and stoichiometry. In examinations such as Class 11 school boards, NEET, JEE Main, and CUET-UG, numerical questions test whether students can interpret analytical mass data correctly. Many students make avoidable errors because they compare element percentages directly without fixing a reference mass first. Learning the systematic calculation method ensures quick, reliable problem solving in exams.
Statement of the Law of Multiple Proportions
Formulated by John Dalton, the Law of Multiple Proportions states: When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in a ratio of small whole numbers.
In simple words, consider elements A and B forming two compounds. When the mass of element A is held constant (such as 1.00 g), the masses of element B reacting with it form a simple integer ratio like 1:2, 2:3, or 1:3. For instance, carbon and oxygen form carbon monoxide (CO) and carbon dioxide (CO2). For a fixed 12.01 g of carbon, oxygen combines in masses of 16.00 g and 32.00 g, giving a simple ratio of 16.00 : 32.00 = 1 : 2.
Core Mathematical Relationship:
For elements A and B forming Compound 1 and Compound 2:
Mass of B per gram of A in Compound 1: R1 = m(B, 1) / m(A, 1) [g B / g A]
Mass of B per gram of A in Compound 2: R2 = m(B, 2) / m(A, 2) [g B / g A]
Comparison ratio: R1 : R2 = a : b (where a and b are small integers, unitless).
Step-by-Step Problem Solving Method
Follow these four steps to solve any multiple proportions numerical:
Step 1: List the mass of each element. If percentages are given, assume a 100.0 g total so percentages equal grams directly.
Step 2: Fix the mass of one element (element A) to a standard reference value, usually 1.00 g.
Step 3: Calculate the mass of the second element (element B) combining with that fixed mass in each compound.
Step 4: Divide the calculated masses of B by the smallest value to obtain a ratio of small integers.
Question: Sulfur and oxygen combine to form two binary oxides. Analysis shows that Oxide A contains 50.00% sulfur and 50.00% oxygen by mass, while Oxide B contains 40.00% sulfur and 60.00% oxygen by mass. Prove that these results illustrate the Law of Multiple Proportions.
Primary Solution:
Take 100.00 g of each oxide sample.
In Oxide A: Mass of sulfur = 50.00 g; mass of oxygen = 50.00 g.
In Oxide B: Mass of sulfur = 40.00 g; mass of oxygen = 60.00 g.
Fix the mass of sulfur to 1.000 g in both oxides:
For Oxide A: Mass of oxygen combining with 1.000 g sulfur = 50.00 g / 50.00 g = 1.000 g.
For Oxide B: Mass of oxygen combining with 1.000 g sulfur = 60.00 g / 40.00 g = 1.500 g.
Compare oxygen masses combining with 1.000 g sulfur:
Ratio = 1.000 g : 1.500 g = 1.000 : 1.500.
Multiply both values by 2 to convert into integers: 1.000 * 2 = 2; 1.500 * 2 = 3.
The oxygen mass ratio is 2 : 3, which is a ratio of small whole numbers.
Double-Check Route: Fix sulfur to 100.00 g. In Oxide A, oxygen = (50.00 / 50.00) * 100.00 = 100.00 g. In Oxide B, oxygen = (60.00 / 40.00) * 100.00 = 150.00 g. Ratio = 100.00 : 150.00 = 2 : 3. Both routes yield identical results.
Answer: The masses of oxygen combining with a fixed mass of sulfur are in the ratio 2 : 3 (unitless ratio).
Question: Two oxides of copper were reduced in a hydrogen stream. 1.428 g of the first oxide gave 1.270 g of copper metal. 1.591 g of the second oxide gave 1.270 g of copper metal. Demonstrate that these data agree with the Law of Multiple Proportions.
Primary Solution:
Find the mass of oxygen in each sample:
First oxide: Mass of oxygen = 1.428 g - 1.270 g = 0.158 g.
Second oxide: Mass of oxygen = 1.591 g - 1.270 g = 0.321 g.
The mass of copper in both samples is already fixed at 1.270 g.
Compare the masses of oxygen directly for this fixed 1.270 g of copper:
Ratio = 0.158 g : 0.321 g.
Divide both numbers by 0.158: 0.158 / 0.158 = 1.000; 0.321 / 0.158 = 2.031 ≈ 2.
The ratio of oxygen masses is 1 : 2, satisfying the law.
Double-Check Route: Find mass of oxygen per 1.000 g copper: First oxide = 0.158 / 1.270 = 0.1244 g O/g Cu. Second oxide = 0.321 / 1.270 = 0.2528 g O/g Cu. Ratio = 0.1244 : 0.2528 = 1.000 : 2.032 ≈ 1 : 2. Both calculation paths verify the law.
Answer: The masses of oxygen combining with a fixed mass of copper are in the ratio 1 : 2 (unitless ratio).
Common Mistakes in Law of Multiple Proportions Numericals
1. Comparing different elements: The law applies only when the same two elements form different compounds (such as CO and CO2). It does not apply to compounds with different elements (such as CO2 and SO2) or ternary compounds with three elements (such as H2SO3 and H2SO4).
2. Comparing percentages without fixing one element: Students often directly divide percentages of the second element. For example, dividing 50% by 60% gives 5 : 6, which is wrong because the sulfur amount also changed. You must fix one element mass first.
3. Confusing with Definite Proportions: Definite Proportions applies to one compound from different sources. Multiple Proportions applies to different compounds formed by the same pair of elements.
4. Rounding decimals prematurely: Do not round intermediate numbers too early during division, as this can turn a clean ratio like 1.5 into an incorrect integer.
| Exam | Question Format | Key Skills Evaluated |
|---|---|---|
| CBSE & State Boards Class 11 | Theory definitions and step-by-step numerical questions | Stating Dalton's law accurately, setting up mass tables, and simplifying ratios |
| NEET | Objective multiple-choice questions | Rapidly identifying valid pairs of compounds and computing simple ratios mentally |
| JEE Main | Multiple-choice and numerical answer questions | Multi-step problems connecting mass ratios with empirical formulas and stoichiometry |
| CUET-UG | Objective concept questions and data matching | Distinguishing chemical combination laws and calculating simple integer proportions |
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: CBSE Class 11, ISC Class 11, State Boards, NEET, JEE Main, CUET-UG
- Topic: Some Basic Concepts of Chemistry — Laws of Chemical Combination
- Difficulty: School foundational to competitive entry level
Previous-year CSIR-NET questions and timed quizzes to check what you just read.
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