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S1.4Structure 1 — Models of the particulate nature of matter

Counting particles by mass: the mole — IB Chemistry SL Revision Notes & Topical Questions

5 min read 11 parts Lesson 4 of 6 1 quiz

📚 IB Chemistry SL Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Lesson summary

S1.4 Counting particles by mass: the mole — IB CHEMISTRY SL notes: Use the mole, molar mass, empirical and molecular formulae, concentration and Avogadro's law to link mass, amount, particles and volume.Mole = the SI unit of amount of substance, containing as many specified parti…

  • Avogadro constant
  • State the particle
  • Formula units
  • Number of particles
  • What the symbols mean
  • Relative atomic mass Ar

Part 1 of 9The mole and the Avogadro constant

The mole and the Avogadro constant
Mole
Mole the SI unit of amount of substance, containing as many specified particles as there are atoms in exactly 12.00 g of carbon-12.
Avogadro constant
NA = 6.02 × 1023 mol−1, the number of particles per mole of atoms, molecules, ions or electrons.
State the particle
  • 1 mol of O2 contains 2 mol of O atoms.
Formula units
  • Formula units ionic compounds have formula units, so 1 mol of CaCl2 contains 6.02 × 1023 formula units but 3 mol of ions.
Number of particles
  • N = n × NA.
    What the symbols mean: N = number of particles · n = amount (mol) · NA = Avogadro constant (6.02 × 1023 mol−1)

Part 2 of 9Relative and molar masses

Relative atomic mass Ar
Relative atomic mass Ar the weighted mean mass of an element's atoms (all isotopes) relative to one twelfth of a carbon-12 atom, with no units.
Relative formula mass Mr
  • Relative formula mass Mr the sum of the Ar values of all atoms in the formula, with no units, used for ionic compounds as well as molecules.
Molar mass M
  • Molar mass M the mass of one mole in g mol−1, numerically equal to Ar or Mr.
Amount from mass
  • Amount from mass n = m/M.
    What the symbols mean: n = amount of substance (mol) · m = mass (g) · M or Mr = molar mass (g mol−1, numerically equal to Mr)
Percentage composition
  • % by mass of an element = (number of atoms × Ar ÷ Mr) × 100.

Part 3 of 9Empirical and molecular formulae

Empirical formula
Empirical formula the simplest whole-number ratio of atoms of each element in a compound, found from mass or percentage composition.
Molecular formula
Molecular formula the actual number of atoms of each element in one molecule, a whole-number multiple of the empirical formula.
Ionic compounds
  • Ionic compounds their formula is always the empirical formula.
1
Write the mass or percentage of each element.
2
Divide each by its relative atomic mass to get moles.
3
Divide all mole values by the smallest.
4
Multiply to reach whole numbers if needed, e.g. ×2 for 1.5 and ×3 for 1.33.
5
Write the empirical formula from the ratio.
6
Divide Mr by the empirical formula mass and multiply the subscripts by this whole number.
1Empirical versus molecular
  • Empirical versus molecular ethene C2H4 has empirical formula CH2, glucose C6H12O6 has CH2O, and hydrogen peroxide H2O2 has HO.
2Rounding
  • Rounding round only when values are very close to whole numbers.

Part 4 of 9Molar concentration

Concentration
Concentration the amount of solute per unit volume of solution, usually in mol dm−3.
Formula
  • Formula c = n/V with V in dm3, so n = cV.
Volume conversion
  • Volume conversion divide cm3 by 1000 to get dm3.
Notation
  • Notation square brackets, e.g. [HCl], denote concentration in mol dm−3.
Mass concentration
  • Mass concentration g dm−3 equals c × M, and dividing g dm−3 by M gives mol dm−3.
Concentrated and dilute
  • Concentrated and dilute a concentrated solution has a large amount of solute per unit volume and a dilute solution has a small amount.

Part 5 of 9Standard solutions and dilution

  • Standard solution
    • one whose concentration is accurately known.
1
Weigh the solute accurately.
2
Dissolve it in a beaker with some solvent.
3
Transfer to a volumetric flask with rinsings.
4
Make up to the mark and invert to mix.
  • Dilution
    • the amount of solute does not change, so c1V1 = c2V2.

Part 6 of 9Parts per million

Parts per million
Use
  • Use expresses very low concentrations such as pollutants in water or air.
Definition
  • 1 ppm is 1 g of solute per 106 g of solution.
In water
  • In water this is about 1 mg per dm3 .
    • (1 mg in 1 kg of water)

Part 7 of 9Avogadro's law

1Statement
Statement equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
2Consequence
  • Consequence gas volume is proportional to amount, so mole ratios equal volume ratios.
3Molar volume at STP
  • 1 mol of any gas occupies 22.7 dm3 at 273 K and 100 kPa.
4Reacting volumes
  • Reacting volumes the volume ratio of gaseous reactants and products equals the coefficient ratio, e.g. N2 + 3H2 → 2NH3 is 1 : 3 : 2.
5Conditions
  • Conditions all gases must be measured at the same temperature and pressure.

Part 8 of 9Limiting reactant with gas volumes

Method
  • Method divide each gas volume by its coefficient, and the smallest value is the limiting reactant.
Liquids
  • Liquids water at room temperature does not count in the final gas volume.
Remaining gas
  • Remaining gas the final volume includes products and any excess reactant.

Part 9 of 9More from the specification

1Molar mass formula triangle
  • Visual aid linking mass (m), molar mass (M), and amount in moles (n) through the relation m = n x M.
2Calculating constituent atoms present
  • Determining the number of individual atoms present by multiplying moles of compound by elemental subscript and Avogadro's constant (6.022 x 1023 mol-1.
3Limiting reactant identification
  • Calculating mole-to-coefficient ratios from balanced chemical equations;
    • the reactant giving the lowest value limits total yield.
4Large-scale concentration conversions
  • Converting millions of litres or cubic metres into molar amounts to establish parts-per-million (ppm) specifications.
5Oxygen
  • A gas that relights a glowing splint.
6Weighted average
  • An average that allows for the abundance of each isotope.

Key terms

TermMeaning
MoleThe SI unit of amount of substance, containing 6.02 × 1023 elementary entities.
Avogadro constantThe number of particles per mole, 6.02 × 1023 mol−1.
Molar massThe mass of one mole of a substance in g mol−1.
Empirical formulaThe simplest whole-number ratio of atoms of each element in a compound.
Molecular formulaThe actual number of atoms of each element in one molecule.
Molar concentrationThe amount of solute in moles per cubic decimetre of solution.
Avogadro's lawEqual volumes of gases at the same T and p contain equal numbers of particles.

Avoid these mistakes

1
Using cm3 in c = n/V — convert the volume to dm3 first.
2
Forgetting the number of atoms per molecule — 1 mol of H2 contains 2 mol of H atoms.
3
Giving Mr or Ar units — they are ratios, but molar mass has units g mol−1.
4
Applying gas volume ratios to liquids or solids — only gases at the same T and p follow them.

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