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Basic concepts: Class 11 Chemistry
Atoms, molecules and ions, relative atomic mass, the mole and Avogadro's number, empirical and molecular formulae, stoichiometry, limiting reactant and percentage yield.
Key points
Matter is made of atoms. An atom is the smallest particle of an element that keeps its chemical identity. A molecule is a group of two or more atoms joined by chemical bonds, such as H₂O or O₂. An ion is an atom or group of atoms that has gained or lost electrons and so carries a charge.
The relative atomic mass (Ar) of an element is the average mass of its atoms, including their isotopes in their natural proportions, compared with one twelfth of the mass of a carbon-12 atom. It has no unit. The relative molecular mass (Mr) is the sum of the relative atomic masses of all the atoms in the formula. For ionic compounds the term relative formula mass is used.
Isotopes are atoms of the same element with the same number of protons and different numbers of neutrons. Chlorine has two common isotopes, Cl-35 (about 75 %) and Cl-37 (about 25 %), so its relative atomic mass is about 35.5. The mass spectrometer is used to find the masses and abundances of isotopes.
The mole is the amount of substance that contains as many elementary particles as there are atoms in exactly 12 g of carbon-12. That number is Avogadro's number, 6.022 × 10²³ per mole. One mole of any substance contains 6.022 × 10²³ particles, whether they are atoms, molecules or ions.
The molar mass of a substance is its mass in grams for one mole. It has the same number as the relative atomic or molecular mass but the unit g/mol. For water, Mr = 18, so the molar mass is 18 g/mol.
The mole links mass, particles and gas volume:
- moles = mass in grams ÷ molar mass
- number of particles = moles × 6.022 × 10²³
- at standard temperature and pressure (0 °C and 1 atm), one mole of any ideal gas occupies 22.4 dm³
The empirical formula shows the simplest whole number ratio of the atoms of each element in a compound. The molecular formula shows the actual number of atoms in one molecule. The molecular formula is a whole number multiple of the empirical formula: molecular formula = (empirical formula)ₙ, where n = molecular mass ÷ empirical formula mass.
To find an empirical formula from percentage composition:
- Assume 100 g of the compound so that each percentage becomes a mass in grams.
- Divide each mass by the relative atomic mass to get moles of each element.
- Divide all the mole values by the smallest one.
- If the results are not whole numbers, multiply them all by a small integer (2, 3 or 4) until they are.
Stoichiometry is the calculation of the amounts of reactants and products from a balanced chemical equation. The coefficients in the equation give the ratio of moles. In the reaction N₂ + 3H₂ → 2NH₃, one mole of nitrogen reacts with three moles of hydrogen to give two moles of ammonia.
The limiting reactant is the reactant that is used up first. It decides how much product can form. The other reactant is in excess. To find it, convert the masses of the reactants into moles, divide each by its coefficient and compare. The smaller value belongs to the limiting reactant.
The theoretical yield is the maximum mass of product predicted by the equation. The actual yield is what is really obtained, and it is usually less because of incomplete reactions, side reactions and losses. The percentage yield is (actual yield ÷ theoretical yield) × 100.
Study tip: Every stoichiometry problem follows the same path: write the balanced equation, convert the given masses to moles, use the mole ratio, then convert back to the unit that the question asks for.
Definitions and flashcards
Say the meaning first, then open the card to check.
Relative atomic mass
The average mass of an atom of an element compared with one twelfth of the mass of a carbon-12 atom.
Isotopes
Atoms of the same element with different numbers of neutrons.
Mole
The amount of substance containing 6.022 × 10²³ elementary particles.
Avogadro's number
6.022 × 10²³, the number of particles in one mole.
Molar mass
The mass of one mole of a substance, in g/mol.
Empirical formula
The simplest whole number ratio of atoms in a compound.
Molecular formula
The actual number of atoms of each element in a molecule.
Limiting reactant
The reactant that is completely used up and limits the amount of product.
Percentage yield
Actual yield divided by theoretical yield, times 100.
Formula sheet
| Name | Formula | What it means |
|---|---|---|
| Moles from mass | n = m / M | m in grams and M in g/mol. |
| Number of particles | N = n × 6.022 × 10²³ | N is the number of atoms, molecules or ions. |
| Gas volume at STP | V = n × 22.4 dm³ | For an ideal gas at 0 °C and 1 atm. |
| Percentage yield | (actual yield / theoretical yield) × 100 | Both yields in the same unit. |
| Molecular formula | n = molecular mass / empirical formula mass | The molecular formula is (empirical formula)ₙ. |
Short questions with model answers
These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.
Define the mole and state Avogadro's number. (2 marks)
The mole is the amount of substance that contains as many elementary particles as there are atoms in 12 g of carbon-12. Avogadro's number is 6.022 × 10²³ per mole.
Why is the relative atomic mass of chlorine not a whole number? (3 marks)
Chlorine has two isotopes, Cl-35 and Cl-37, in the approximate ratio 3 to 1. The relative atomic mass is the weighted average of their masses, which is about 35.5.
What is the difference between the empirical formula and the molecular formula? (2 marks)
The empirical formula gives the simplest whole number ratio of atoms in the compound. The molecular formula gives the actual numbers of atoms in one molecule. For glucose the empirical formula is CH₂O and the molecular formula is C₆H₁₂O₆.
What is a limiting reactant? Why is it important? (3 marks)
It is the reactant that is completely consumed first in a reaction. The reaction stops when it is used up, so it decides the maximum amount of product that can form. The other reactants are in excess.
Give three reasons why the actual yield of a reaction is less than the theoretical yield. (3 marks)
The reaction may not go to completion, side reactions may form other products, and some product may be lost during separation, filtering and transfer.
Numericals with worked steps
Calculate the number of moles and the number of molecules in 36 g of water. (H = 1, O = 16)
- Molar mass of H₂O = 2 × 1 + 16 = 18 g/mol.
- Moles = 36 ÷ 18 = 2 mol.
- Molecules = 2 × 6.022 × 10²³ = 1.204 × 10²⁴.
Answer: 2 mol and 1.204 × 10²⁴ molecules
A compound contains 40.0 % carbon, 6.7 % hydrogen and 53.3 % oxygen. Find its empirical formula. (C = 12, H = 1, O = 16)
- In 100 g: C = 40.0 g, H = 6.7 g, O = 53.3 g.
- Moles: C = 40.0/12 = 3.33, H = 6.7/1 = 6.7, O = 53.3/16 = 3.33.
- Divide by the smallest (3.33): C = 1, H = 2, O = 1.
Answer: CH₂O
The empirical formula of a compound is CH₂O and its molecular mass is 180. Find the molecular formula.
- Empirical formula mass = 12 + 2 + 16 = 30.
- n = 180 ÷ 30 = 6.
- Molecular formula = (CH₂O)₆.
Answer: C₆H₁₂O₆
28 g of nitrogen reacts with 8 g of hydrogen in N₂ + 3H₂ → 2NH₃. Find the limiting reactant and the mass of ammonia formed. (N = 14, H = 1)
- Moles of N₂ = 28 ÷ 28 = 1 mol. Moles of H₂ = 8 ÷ 2 = 4 mol.
- Divide by coefficients: N₂ = 1/1 = 1 and H₂ = 4/3 = 1.33.
- N₂ has the smaller value, so nitrogen is the limiting reactant.
- 1 mol N₂ gives 2 mol NH₃. Mass = 2 × 17 = 34 g.
Answer: Nitrogen is limiting and 34 g of ammonia is formed
In a reaction the theoretical yield of a product is 25.0 g and 20.0 g is obtained. Find the percentage yield.
- Percentage yield = (20.0 ÷ 25.0) × 100.
- = 80 %.
Answer: 80 %
Common mistakes
- Using the mass in grams directly in a mole ratio. Always convert to moles first.
- Forgetting that Mr has no unit and molar mass has the unit g/mol.
- Writing the empirical formula with decimals. Multiply the mole ratio until all values are whole numbers.
- Choosing the limiting reactant from the smaller mass instead of the smaller moles divided by the coefficient.
- Applying 22.4 dm³ to a gas at room temperature. That value is for standard temperature and pressure.