Chapter Exam

Chapter-4 Formulae and Stoichiometry — Online MCQ Test

Q: What is the molar mass of calcium carbonate (CaCO₃)? [Ca = 40, C = 12, O = 16]

Answer: B. 100 g/mol — Molar mass = 40 + 12 + (16 × 3) = 40 + 12 + 48 = 100 g/mol.

Q: How many moles are present in 11 g of carbon dioxide (CO₂)? [C = 12, O = 16]

Answer: B. 0.25 mol — Molar mass of CO₂ = 44 g/mol. Number of moles = 11/44 = 0.25 mol.

Q: In the reaction 2H₂ + O₂ → 2H₂O, how many moles of water are produced from 3 moles of H₂?

Answer: C. 3.0 mol — From the equation, 2 mol H₂ produces 2 mol H₂O (1:1 ratio). So 3 mol H₂ produces 3 mol H₂O.

Q: A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen. What is its empirical formula? [C = 12, H = 1, O = 16]

Answer: A. CH₂O — Mole ratio: C = 40/12 = 3.33, H = 6.7/1 = 6.7, O = 53.3/16 = 3.33. Simplest ratio = 1:2:1, giving CH₂O.

CHEMISTRY · Grade 10 · IGCSE cambridge

Practice Chapter-4 Formulae and Stoichiometry with a free chapter-wise online MCQ test for IGCSE cambridge Grade 10 CHEMISTRY. This chapter covers: Writing formulae and balanced equations the mole concept and calculating reacting masses and yields.. AI-generated questions from basic to board-exam level, with instant results and explanations.

Questions

20m

Time Limit

Attempts Left

10 random questions from this chapter (mixed difficulty)
Questions you've seen before won't repeat until the pool resets
You have 20 minutes — exam auto-submits when time is up
Maximum 3 attempts per chapter
Results and explanations shown immediately after submission

New here? Register free — includes 3 free chapter exams.

Chapter-4 Formulae and Stoichiometry — Important Questions & Answers (FAQ)

Frequently asked questions from IGCSE cambridge Grade 10 CHEMISTRY — Chapter-4 Formulae and Stoichiometry, with answers and explanations. These are sample questions; the exam has its own separate question set.

What is the molar mass of calcium carbonate (CaCO₃)? [Ca = 40, C = 12, O = 16]

A. 68 g/mol
B. 100 g/mol ✓
C. 120 g/mol
D. 128 g/mol

Answer: B. 100 g/mol
Molar mass = 40 + 12 + (16 × 3) = 40 + 12 + 48 = 100 g/mol.

How many moles are present in 11 g of carbon dioxide (CO₂)? [C = 12, O = 16]

A. 0.15 mol
B. 0.25 mol ✓
C. 0.35 mol
D. 0.50 mol

Answer: B. 0.25 mol
Molar mass of CO₂ = 44 g/mol. Number of moles = 11/44 = 0.25 mol.

In the reaction 2H₂ + O₂ → 2H₂O, how many moles of water are produced from 3 moles of H₂?

A. 1.0 mol
B. 1.5 mol
C. 3.0 mol ✓
D. 6.0 mol

Answer: C. 3.0 mol
From the equation, 2 mol H₂ produces 2 mol H₂O (1:1 ratio). So 3 mol H₂ produces 3 mol H₂O.

A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen. What is its empirical formula? [C = 12, H = 1, O = 16]

A. CH₂O ✓
B. C₂H₄O₂
C. CH₄O
D. C₃H₆O₃

Answer: A. CH₂O
Mole ratio: C = 40/12 = 3.33, H = 6.7/1 = 6.7, O = 53.3/16 = 3.33. Simplest ratio = 1:2:1, giving CH₂O.

In a limiting reactant problem, 5 g of calcium reacts with 10 g of chlorine to form CaCl₂. Which is the limiting reactant? [Ca = 40, Cl = 35.5]

A. Calcium, because it has less mass
B. Chlorine, because it requires more moles
C. Calcium, because it produces fewer moles of product ✓
D. Chlorine, because it has a higher atomic mass

Answer: C. Calcium, because it produces fewer moles of product
Ca: 5/40 = 0.125 mol. Cl₂: 10/71 = 0.141 mol. Reaction: Ca + Cl₂ → CaCl₂ requires 1:1 ratio. Calcium produces only 0.125 mol product (limiting), chlorine can produce 0.141 mol.

← Previous Chapter

Chapter-3 Chemical Bonding

Next Chapter →

Chapter-5 Electrochemistry