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Chapter 4· Stoichiometry
MCQs with Answers Try each one first: tap an option to check it, then read the reasoning and why the others are wrong.
A note about the questions
The wording follows published solved exercises and every answer was re-checked by hand, so the wording or the order of options may differ slightly from your print.
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Set A: Textbook exercise 10 MCQs The multiple-choice questions at the end of Chapter 4 in your book, in the book's option order.
Q4.1 Avogadro's Number and Mole
How many atoms are present in one gram of H₂O?
(a) 1.002 × 10²³ atoms (b) 6.022 × 10²³ atoms (c) 0.334 × 10²³ atoms (d) 2.004 × 10²³ atoms Show answer and reasoning Answer (a) 1.002 × 10²³ atoms
Step by step
n = m ÷ M = 1 ÷ 18 = 0.0556 mol. Molecules = 0.0556 × 6.022 × 10²³ = 3.35 × 10²² (0.334 × 10²³). Atoms = 3 × 3.35 × 10²² = 1.004 × 10²³. Why the others are wrong
(b) 6.022 × 10²³ is the number of molecules in a whole mole (18 g), not in 1 g.(c) 0.334 × 10²³ is the number of molecules; each molecule has three atoms.(d) 2.004 × 10²³ would be six atoms per molecule.Remember
Atoms = molecules × atoms per molecule. H₂O has 3.
Q4.2 Formulas of Ionic Compounds
Which is the correct formula of calcium phosphide?
(a) CaP (b) CaP₂ (c) Ca₂P₃ (d) Ca₃P₂ Show answer and reasoning Answer (d) Ca₃P₂
Step by step
Cation Ca²⁺, anion P³⁻. Cross over: Ca takes 3, P takes 2. Check: +6 − 6 = 0. Why the others are wrong
(a) +2 − 3 = −1: not neutral.(b) +2 − 6 = −4: not neutral.(c) the subscripts are swapped: +4 − 9 = −5.Remember
Charge of one ion becomes the subscript of the other.
Q4.3 Avogadro's Number and Mole
How many atomic mass units (amu) are there in one gram?
(a) 1 amu (b) 10²³ amu (c) 6.022 × 10²³ amu (d) 6.022 × 10²² amu Show answer and reasoning Answer (c) 6.022 × 10²³ amu
Step by step
1 amu = 1.66 × 10⁻²⁴ g. 1 g ÷ 1.66 × 10⁻²⁴ g = 6.022 × 10²³ amu. Why the others are wrong
(a) 1 amu is about the mass of one proton, far smaller than 1 g.(b) close in size, but not the exact value.(d) ten times too small.Remember
Grams to amu: multiply by Avogadro's number.
Q4.4 Empirical and Molecular Formula
Structural formula of 2-hexene is CH₃–CH=CH–(CH₂)₂–CH₃. What will be its empirical formula?
(a) C₂H₂ (b) CH (c) C₆H₁₂ (d) CH₂ Show answer and reasoning Answer (d) CH₂
Step by step
C atoms: 1 + 1 + 1 + 2 + 1 = 6. H atoms: 3 + 1 + 1 + 4 + 3 = 12. C₆H₁₂ ÷ 6 = CH₂. Why the others are wrong
(a) C₂H₂ (ratio 1 : 1) is acetylene's formula.(b) CH is the empirical formula of benzene and acetylene.(c) C₆H₁₂ is the molecular formula, not the simplest ratio.Remember
Empirical = divide all subscripts by their common factor.
Q4.5 Avogadro's Number and Mole
How many moles are there in 25 g of H₂SO₄?
(a) 0.765 moles (b) 0.51 moles (c) 0.255 moles (d) 0.4 moles Show answer and reasoning Answer (c) 0.255 moles
Step by step
M = 98 g mol⁻¹. n = m ÷ M = 25 ÷ 98 = 0.255 mol. Why the others are wrong
(a) three times too big.(b) this is 25 ÷ 49, using half the molar mass.(d) 0.4 mol would be 39.2 g.Remember
Moles = mass ÷ molar mass.
Q4.6 Avogadro's Number and Mole
A necklace has 6 g of diamonds in it. What are the number of carbon atoms in it?
(a) 6.02 × 10²³ (b) 12.04 × 10²³ (c) 1.003 × 10²³ (d) 3.01 × 10²³ Show answer and reasoning Answer (d) 3.01 × 10²³
Step by step
n = 6 ÷ 12 = 0.5 mol. N = n × Nᴀ = 0.5 × 6.02 × 10²³ = 3.01 × 10²³. Why the others are wrong
(a) that is a whole mole, 12 g of carbon.(b) that is two moles, 24 g.(c) that would be 2 g of carbon.Remember
Half a mole = half of Avogadro's number.
Q4.7 Calculations from Equations
What is the mass of Al in 204 g of aluminium oxide, Al₂O₃?
(a) 26 g (b) 27 g (c) 54 g (d) 108 g Show answer and reasoning Answer (d) 108 g
Step by step
n(Al₂O₃) = 204 ÷ 102 = 2 mol. n(Al) = 2 × 2 = 4 mol. m(Al) = 4 × 27 = 108 g. Why the others are wrong
(a) 26 g is not even one mole of Al.(b) 27 g is one mole of Al.(c) 54 g is the Al in 102 g (one mole) of Al₂O₃, not in 204 g.Remember
102 g of Al₂O₃ holds 54 g of Al; double it.
Q4.8 Calculations from Equations
Which one of the following compounds will have the highest percentage of the mass of nitrogen?
(a) CO(NH₂)₂ (b) N₂H₄ (c) NH₃ (d) NH₂OH Show answer and reasoning Answer (b) N₂H₄
Step by step
N₂H₄: 28 ÷ 32 × 100 = 87.5 %. NH₃: 14 ÷ 17 × 100 = 82.4 %. CO(NH₂)₂: 28 ÷ 60 × 100 = 46.7 %. NH₂OH: 14 ÷ 33 × 100 = 42.4 %. Why the others are wrong
(a) urea is only 46.7 % N (the C and O add mass).(c) NH₃ is 82.4 %: close, but lower.(d) NH₂OH is 42.4 %, the lowest.Remember
Fewest non-nitrogen mass units per N wins: N₂H₄ carries only 4 g of H with 28 g of N.
Q4.9 Calculations from Equations
When one mole of each of the following compounds is reacted with oxygen, which will produce the maximum amount of CO₂?
(a) Carbon (b) Diamond (c) Ethane (C₂H₆) (d) Methane (CH₄) Show answer and reasoning Answer (c) Ethane (C₂H₆)
Step by step
C + O₂ → CO₂ (1 mol). CH₄ + 2O₂ → CO₂ + 2H₂O (1 mol). 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O: 2 mol CO₂ per mol ethane. Why the others are wrong
(a) 1 mol C gives 1 mol CO₂.(b) diamond is carbon: 1 mol CO₂.(d) methane has one C atom: 1 mol CO₂.Remember
Count carbons per formula.
Q4.10 Calculations from Equations
What mass of 95% CaCO₃ will be required to neutralize 50 cm³ of 0.5 M HCl solution?
(a) 9.5 g (b) 1.25 g (c) 1.32 g (d) 1.45 g Show answer and reasoning Answer (c) 1.32 g
Step by step
n(HCl) = M × V = 0.5 × 0.050 = 0.025 mol. n(CaCO₃) = 0.025 ÷ 2 = 0.0125 mol. Pure mass = 0.0125 × 100 = 1.25 g. Impure sample = 1.25 × 100 ÷ 95 = 1.32 g. Why the others are wrong
(a) far too much.(b) 1.25 g is the mass of pure CaCO₃; the 95 % sample must weigh more.(d) this does not follow from the mole ratio.Remember
Impure sample is always heavier than the pure mass.
Set B: More MCQs from this chapter 14 MCQs Extra questions on the same ideas, of the kind set in board papers and class tests.
QB1 Chemical Formula
Which of these is represented by a formula unit rather than a molecular formula?
Show answer and reasoning Answer (b) NaCl
Step by step
Ionic compounds: formula units. Covalent molecules: molecular formulas. Why the others are wrong
(a) water exists as molecules.(c) methane is a molecule.(d) ozone is a molecule of three O atoms.Remember
Ionic = formula unit.
QB2 Empirical and Molecular Formula
The empirical formula of hydrogen peroxide (H₂O₂) is:
Show answer and reasoning Answer (a) HO
Step by step
Divide both subscripts by 2. Why the others are wrong
(b) that is the molecular formula.(c) that is water, a different compound.(d) the ratio is 1 : 1, not 1 : 2.Remember
Simplest whole-number ratio.
QB3 Empirical and Molecular Formula
Which pair has the same empirical formula?
(a) H₂O and H₂O₂ (b) C₂H₂ and C₆H₆ (c) CO and CO₂ (d) NH₃ and N₂H₄ Show answer and reasoning Answer (b) C₂H₂ and C₆H₆
Step by step
C₂H₂ ÷ 2 = CH. C₆H₆ ÷ 6 = CH. Why the others are wrong
(a) H₂O and HO.(c) CO and CO₂.(d) NH₃ and NH₂.Remember
Reduce both and compare.
QB4 Empirical and Molecular Formula
A compound has empirical formula CH₂O and molar mass 60 g mol⁻¹. Its molecular formula is:
(a) C₂H₄O₂ (b) CH₂O (c) C₃H₆O₃ (d) C₆H₁₂O₆ Show answer and reasoning Answer (a) C₂H₄O₂
Step by step
EF mass = 30. n = 60 ÷ 30 = 2. Molecular formula = C₂H₄O₂. Why the others are wrong
(b) that would need molar mass 30.(c) that needs molar mass 90.(d) that is glucose, molar mass 180.Remember
n = molar mass ÷ empirical formula mass.
QB5 Formulas of Ionic Compounds
The formula of aluminium oxide is:
(a) Al₂O₃ (b) Al₂O (c) AlO₃ (d) AlO Show answer and reasoning Answer (a) Al₂O₃
Step by step
Al takes 2, O takes 3. Why the others are wrong
(b) +6 − 2 ≠ 0.(c) +3 − 6 ≠ 0.(d) +3 − 2 ≠ 0.QB6 Formulas of Ionic Compounds
The formula of magnesium nitride is:
(a) MgN (b) Mg₃N₂ (c) Mg₂N₃ (d) MgN₂ Show answer and reasoning Answer (b) Mg₃N₂
Step by step
Cation Mg²⁺, anion N³⁻. Mg takes 3, N takes 2. Why the others are wrong
(a) not neutral.(c) subscripts swapped.(d) not neutral.QB7 Avogadro's Number and Mole
The quantity of a substance containing Avogadro's number of particles is called:
(a) molar mass (b) a mole (c) atomic mass unit (d) formula unit Show answer and reasoning Answer (b) a mole
Step by step
1 mol = 6.022 × 10²³ particles. Why the others are wrong
(a) molar mass is the mass of one mole.(c) amu is a unit of atomic mass.(d) formula unit is the simplest ratio unit of an ionic compound.Remember
Mole is a counting number, like a dozen.
QB8 Avogadro's Number and Mole
The molar mass of hydrogen molecules (H₂) is:
(a) 2.016 g (b) 1.008 g (c) 6.022 g (d) 0.504 g Show answer and reasoning Answer (a) 2.016 g
Step by step
2 × 1.008 = 2.016 g mol⁻¹. Why the others are wrong
(b) that is for hydrogen atoms.(c) confuses with Avogadro's number.(d) half an atom's mass.Remember
Say which particle: atoms or molecules.
QB9 Avogadro's Number and Mole
Which contains the largest number of molecules?
(a) 22 g of CO₂ (b) 16 g of O₂ (c) 2 g of H₂ (d) 9 g of H₂O Show answer and reasoning Answer (c) 2 g of H₂
Step by step
H₂: 1 mol = 6.022 × 10²³ molecules. The other three: 0.5 mol = 3.011 × 10²³ each. Why the others are wrong
(a) 0.5 mol, though it is the heaviest sample.(b) 0.5 mol.(d) 0.5 mol.Remember
Light molecules: a few grams is already a mole.
QB10 Chemical Equations
A chemical equation must obey the law of:
(a) multiple proportions (b) conservation of energy only (c) definite proportions (d) conservation of mass Show answer and reasoning Answer (d) conservation of mass
Step by step
Same number and kind of atoms on both sides. Why the others are wrong
(a) not what balancing checks.(b) balancing counts atoms, not energy.(c) that is about a compound's fixed composition.Remember
Balanced = mass conserved.
QB11 Chemical Equations
The symbol (aq) in a chemical equation stands for:
(a) a solid (b) a liquid (c) dissolved in water (d) a gas Show answer and reasoning Answer (c) dissolved in water
Step by step
(s) solid, (l) liquid, (g) gas, (aq) in water. Why the others are wrong
QB12 Chemical Equations
The balanced equation for copper with concentrated sulphuric acid is Cu + _H₂SO₄ → CuSO₄ + SO₂ + _H₂O. The missing coefficients are:
(a) 2 and 2 (b) 2 and 1 (c) 1 and 1 (d) 1 and 2 Show answer and reasoning Answer (a) 2 and 2
Step by step
S: 2 on the left needs CuSO₄ + SO₂. H: 4 H needs 2H₂O. O: 8 = 4 + 2 + 2. Why the others are wrong
(b) H does not balance.(c) S does not balance.(d) S does not balance.Remember
Balance S first, then H, then check O.
QB13 Calculations from Equations
25 g of CaCO₃ reacts with excess HCl. The mass of CaCl₂ formed is (Ca = 40, Cl = 35.5):
(a) 11.1 g (b) 27.75 g (c) 25 g (d) 55.5 g Show answer and reasoning Answer (b) 27.75 g
Step by step
Ratio 100 g : 111 g. 111 ÷ 100 × 25 = 27.75 g. Why the others are wrong
(a) that would come from 10 g.(c) mass is not carried across unchanged.(d) that would come from 50 g.Remember
Quarter of the reactant gives quarter of 111 g.
QB14 Calculations from Equations
How many moles of O₂ are needed to burn 1.80 mol of ethyl alcohol completely? (C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O)
(a) 1.8 mol (b) 3.6 mol (c) 0.6 mol (d) 5.4 mol Show answer and reasoning Answer (d) 5.4 mol
Step by step
Mole ratio C₂H₅OH : O₂ = 1 : 3. 3 × 1.8 = 5.4 mol. Why the others are wrong
(a) that assumes a 1 : 1 ratio.(b) 3.6 mol is the CO₂ produced.(c) divides instead of multiplying.Remember
Multiply by the coefficient ratio.
