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1.2Unit 1. Structure, Bonding and Introduction to Organic Chemistry

Amount of Substance — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

5 min read 12 parts Lesson 2 of 36 1 quiz

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1.2 Amount of Substance — IAL EDEXCEL LINEAR (CHEMISTRY) notes: Carry out calculations of moles, concentration, reacting masses, gas volumes, yield and atom economy.Amount of substance — the number of particles in a sample, measured in moles (mol).Main equations — n = mass / mola…

  • Other equations
  • Molar mass
  • Mass concentration
  • Molar concentration
  • Unit conversion
  • Significant figures

Part 1 of 10Amount of substance equations

Amount of substance
  • Amount of substance the number of particles in a sample, measured in moles (mol).
Main equations
  • Main equations n = mass / molar mass and n = concentration x volume.
Other equations
  • Other equations n = number of particles / Avogadro constant, and n = PV/RT for gases.
Molar mass
  • Molar mass the mass per mole of a substance, in g mol-1.

Part 2 of 10Concentration calculations

Mass concentration
  • Mass concentration mass of solute (g) / volume of solution (dm3), usually in g dm-3.
  • Other units are g cm-3, kg m-3 and mg/ml for medicines.
Molar concentration
  • Molar concentration moles of solute / volume of solution (dm3), in mol dm-3.
  • Moles come from mass / molar mass.
  • Multiply mol dm-3 by M to give g dm-3.
Unit conversion
  • Unit conversion divide cm3 by 1000 to get dm3.
Dilution
  • Dilution moles stay constant, so c1V1 = c2V2.
Significant figures
  • Significant figures give answers to the same number of significant figures as the least precise data value.

Part 3 of 10Parts per million (mass)

Parts per million (mass)
ppm
  • ppm used for very low concentrations such as pollutants in water.
Definition
  • 1 ppm is 1 mg dissolved in 1 dm3 (1 kg) of water, which is 1 part in 106.
Conversion
  • 1 mg = 10-6 kg, so mass in mg = ppm x volume in dm3.

Part 4 of 10Atmospheric gas concentration (ppmv)

ppmv
  • ppmv concentration by volume, where 1 ppmv is 1 cm3 of gas in 106 cm3 (1000 dm3) of air.
Equation
  • Equation concentration in ppm = x 106.
    • (volume of gas / volume of air)
  • Both volumes must be in the same units.
Gas identity
  • Gas identity the identity of the gas does not affect the calculation.

Part 5 of 10Reacting mass calculations

1
Write the balanced equation.
2
Calculate the moles of the known substance from mass / Mr.
3
Use the mole ratio from the equation to find moles of the other substance.
4
Convert to mass using mass = moles x Mr.
Units
  • Units any mass unit (g, kg, tonnes) can be used if consistent, because masses stay in proportion to the equation.
Deducing an equation
  • Deducing an equation convert each mass to moles and divide by the smallest value to give the ratio, then use it as the coefficients.
Limiting reagent
  • Limiting reagent when amounts of two reactants are given, compare the mole ratios.
  • The reactant that runs out first sets the product amount and the other is in excess.

Part 6 of 10Molar gas volume and reacting volumes of gases

1Equal volumes
  • Equal volumes all gases occupy the same volume under the same conditions.
2r.t.p.
  • 293 K (20 °C) and 101.3 kPa, where one mole of gas occupies 24.0 dm3.
3s.t.p.
  • 273 K (0 °C) and 101.3 kPa, where one mole of gas occupies 22.4 dm3.
4Moles of gas
  • Moles of gas volume (dm3) / molar gas volume (dm3 mol-1).
5Gas volumes from moles
  • Gas volumes from moles volume = moles x molar gas volume.
6Gas volumes from other volumes
  • Gas volumes from other volumes volume ratios equal mole ratios at the same temperature and pressure.
  • Propane burns as C3H8 + 5O2 → 3CO2 + 4H2O(g), so 1 volume of propane gives 7 volumes of gas.
  • 1 volume N2 + 3 volumes H2 → 2 volumes NH3.

Part 7 of 10Kinetic theory and ideal gases

Kinetic theory and ideal gases
Kinetic theory assumptions
  • Gas molecules move very fast and randomly.
  • Molecules have almost no volume.
  • There are no intermolecular forces of attraction or repulsion.
  • Collisions are elastic, so no kinetic energy is lost.
  • Temperature is related to the average kinetic energy of the molecules.
Ideal gas
  • Ideal gas one that follows these assumptions.
Real gases
  • Real gases do not fit exactly but can come very close.

Part 8 of 10The ideal gas equation

Equation
  • PV = nRT.
    What the symbols mean: p = pressure (Pa) · V = volume (m3) · n = amount of gas (mol) · R = 8.31 J mol−1 K−1 · T = temperature (K = °C + 273)
  • P in pascals, V in m3, n in mol, R = 8.31 J K-1 mol-1 and T in kelvin.
Unit conversions
  • Unit conversions kPa x 1000 = Pa, cm3 x 10-6 = m3, dm3 x 10-3 = m3 and °C + 273 = K.
Rearrangements
  • V = nRT/P to find a volume, and n = PV/RT to find moles.
  • Then molar mass = mass / n.
Skill
  • Skill be able to rearrange the equation for any quantity and check the units.

Part 9 of 10Percentage yield

Percentage yield
Equation
  • Equation percentage yield = actual yield / theoretical yield x 100.
Actual yield
  • Actual yield the moles or mass of product obtained experimentally.
Theoretical yield
  • Theoretical yield the moles or mass from a reacting mass calculation.
Reasons for a lower yield
  • Reasons for a lower yield other reactions take place at the same time, the reaction does not go to completion,
    • or product is lost during separation and purification.
Yield above 100%
  • Yield above 100% an error, caused by wet crystals, impurities or including the mass of the filter paper or container.

Part 10 of 10Atom economy

Atom economy
  • Atom economy shows how many of the atoms used become the desired product, with the rest wasted.
Equation
  • Mr of desired product / sum of Mr of all reactants x 100.
  • It can also be calculated using masses.
Addition reactions
  • Addition reactions have 100% atom economy because there is only one product.
  • CH2=CH2 + Br2 → CH2BrCH2Br.
Comparing routes
  • Comparing routes hydration of ethene to ethanol has a higher atom economy than substitution of bromoethane, which makes NaBr as waste.
Working
  • Working show the calculation even for an addition reaction when asked.

Key terms

TermMeaning
ConcentrationThe amount of solute dissolved in a given volume of solution, in mol dm−3 or g dm−3.
Molar gas volumeThe volume occupied by one mole of gas, 24.0 dm3 at r.t.p. and 22.4 dm3 at s.t.p.
Limiting reagentThe reactant that is used up first and so sets the amount of product.
Percentage yieldActual yield divided by theoretical yield, multiplied by 100.
Atom economyThe mass of desired product divided by the total mass of reactants, multiplied by 100.
ppmParts per million, a concentration of 1 part in 10⁶.
Ideal gasA gas that obeys the kinetic theory assumptions exactly.

Avoid these mistakes

1
Using cm³ directly in c = n/V — convert to dm³ first.
2
Using 24.0 dm³ at s.t.p. — s.t.p. uses 22.4 dm³ and r.t.p. uses 24.0 dm³.
3
Using °C or kPa in PV = nRT — convert to K and Pa.
4
Calculating atom economy from only one reactant — divide by the sum of the Mr of all reactants.
5
Using the amount of the excess reagent to find the product — use the limiting reagent.

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