Experiments:

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N = number of particles

n = number of moles

m = mass

V = volume

C = concentration

N$_A$ = Avogadro’s constant

$A_{r}$ = relative atomic mass of an element ( bigger number on periodic table )

$M_{r}$ = relative formula mass of a compound

M = molar mass ( same as Mr but it has a unit - g/mol )

V$_m$ = molar volume

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moles

• Mole: the amount of substance that contains the same number of particles (atoms, molecules, ions, electrons) as there are atoms in 12 g of carbon-12.
• Avogadro’s constant: 6.02× 10$^{23}$
• 1 mole = 6.022 × 10²³ particles (atoms, ions, molecules, electrons, or formula units).
• Molar mass (Mr in g mol⁻¹) = mass of 1 mole of a substance.

Key ideas

• Reactions happen in ratios of particles, not mass.
• Moles let us compare substances in reactions.

Why moles are useful

• Chemical equations show ratios of particles, but in labs we measure mass.
• The mole links mass ↔ particles ↔ volume so equations can be used in real life.

Key conversions

• Mass → moles: n = m ÷ Mr
• Moles → particles: N = n × NA
• Particles → moles: n = N ÷ NA

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To calculate number of moles

• n = $\frac{m}{M_r}$ / $\frac{m}{M_r}$ / $\frac{m}{M_r}$

Number of particles

• N = n × N$_A$

Moles from volume of gas ( STP / RTP )

• n = $\frac{V}{V_m}$

Moles from concentration in solution

• n = c × v

Concentration formula

• c = $\frac{n}{V}$

Percentage yield

% yield = $\frac{Actual-yield}{Theoretical-yield}$ × 100

• Percentage composition of element in a compound

  % = (no. of atoms × $A_{r}$)/ Mr × 100

• Empirical mass

  % = (actual yield / theoretical yield) × 100

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Example: 2H₂ + O₂ → 2H₂O

• Ratio H₂:O₂:H₂O = 2:1:2
• Means 2 moles of H₂ react with 1 mole O₂ to form 2 moles H₂O.

How many molecules are in 0.25 mol H₂O?

• N = n × NA = 0.25 × 6.022×10²³ =
• 1.51 × 10²³ molecules

mass

Definitions

• Mass number (A): protons + neutrons in one atom.
• Relative atomic mass (Ar): weighted average of isotopes compared to carbon-12.
• Relative molecular mass (Mr): sum of all Ar values in a molecule.

Formula

• n = $\frac{m}{M_r}$

Example

• Calculate the mass of CCl, formed from 2 moles of Cl in CS₂ + 3Cl₂ → CCl₄ + S₂Cl₂

  1. Find the ratio of Cl₂ to CCl₄ in the balanced equation

     Cl₂ : CCl₄ = 3:1

  2. Work out how many moles of CCl₄ are made from 2 moles of Cl₂

     n = 2 : 0.67

  3. Find the molar mass of CCl₄

     Mr = 12 + (4 × 35.5)

     Mr = 154

  4. Use formula to calculate mass of CCl

     n = $\frac{m}{M_r}$

     m = 0.67 × 154

     m = 103g

concentration

Definitions

• Solute = solid or gas dissolved.
• Solvent = liquid (usually water) that dissolves solute.
• Solution = mixture of solute in solvent.
• Concentration (c) = how many moles of solute are present in 1 dm³ of solution.

Formula

• C = $\frac{n}{V}$

Example

volume of gases

empirical & molecular formula