Chapter 3: Amount of substance

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3.1 Amount of substance and the mole

Avogadro's constant: one mole contains 6.02 x 10^23 particles

The mass of one mole of any element = relative atomic mass, Ar, in grams

E.g. one mole of carbon atoms hass a mass of 12.0g

Molar mass, M: mass per mole of a substance, g mol-1

Calculation: amount of substance (mol) = mass (g) / molar mass (g mol-1)

*same equation as: number of moles = mass (g) / Mr

  • moles ---> molecules   x (6.02 x 10^23)
  • molecules ---> moles   / (6.02 x 10^23)
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3.2 Determination of formulae: molecular and empir

Molecular formula: the number of atoms of each element in a molecule e.g. H2

Used for elements that exist as molecules, H2, N2, O2, F2, Cl2, Br2, I2, P4 and S8

Empirical formula: the simplest whole number ratio of atoms of each element in a compound

Used for elements that don't exist as molecules e.g. metals, some non-metals (C and Si) and ionic compounds

These substances are written as their empirical formula

They exist as giant crystalline structures of atoms or ions

They are too large to be written using the actual number which would be billions

Number would vary depending on the size of the crystals

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3.2 Determination of formulae: relative mass and a

Relative molecular mass, Mr: used for simple molecular substances

Compares the mass of a molecule with the mass of an atom of carbon-12

Mr = sum of all the mass numbers of the element in the molecular formula e.g. H2O = 18

Relative formula mass: used for giant crystalline structures

Compares the mass of a formula unit with the mass of an atom of carbon-12

= sum of all the mass numbers of the elements in the empirical formula e.g. NaCl = 58.5

Analysis: investigating the chemical composition of a substance

EF from mass: 1) convert mass to mol, 2) divide all by smallest answer, 3) write EF from ratio

MF from % analysis:1) convert % mass to mol (n = % m / Mr), 2) calculate EF by dividing to ratio, 3) find relative mass of EF, 4) divide given Mr by relative mass of EF (step 3 answer), 5) write MF by: EF x units in one molecule (step 4 answer)

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3.2 Determination of formulae: water of crystallis

Water of crystallisation: water molecules that are part of the crystalline structure

Hydrated: contains water of crystallisation, anhydrous: doesn't contain water of crystallisation

E.g. CuSO4.5H2O (s) ---> CuSO4 (s) + 5H2O (l)

When water is removed, the crystalline structure is lost (CuSO4 turns from crystals to powder)

Calculating the formula of a hydrated salt from practical results:

  • A: mass of crucible, B: mass of crucible & hydrated, C: mass of crucible & anhydrous
  • Calculate number of moles of anhydrous (C - A) n = m / Mr
  • Calculate number of moles of water (B - C) n = m / Mr
  • Find whole number ratio of salt : water
  • Write down x (molecules of water) in formula
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3.3 Moles and volumes: solutions

Measurements: 1 dm3 = 1000 cm3

Concentration, c (mol dm^-3): the number of moles of solute dissolved in 1 dm3 of solution

n = c x v           number of moles = concentration (mol dm-3) x volume (dm3)

Standard solution: a solution of known concentration, prepared by dissolving an exact mass of solute in a solvent and making the solution up to an exact volume

Calculation for mass required to prepare x cm3 of an x mol dm-3 standard solution

  • work out moles, n = c x (v 1000)
  • work out molar mass, Mr (g mol-1)
  • calculate mass required, m = n x Mr

Mass concentration (g dm^-3) = concentration (mol dm^-3) x Mr

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3.3 Moles and volumes: ideal gas equation

Number of moles = volume / 24         n (mol) = v (dm3) / 24         n (mol) = v (cm3) / 24000

For gases not at RTP, use ideal gas equation: pV = nRT

  • p = pressure (Pa)
  • V = volume (m3)
  • n = amount of gas molecules (mol)
  • R = ideal gas constant = 8.31 J mol-1 K-1
  • T = temperature (K)

Conversions:

  • cm3 ---> m3     x 10^-6
  • dm3 ---> m3     x 10^-3
  • oC ---> K         + 273
  • kPa ---> Pa      x 10^3
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3.4 Reacting quantities: stoichiometry, % yield &

Stoichiometry: the ratio of  the amount, in mol, of each substance in an equation

In a balanced equation, the balancing numbers give the ratio = stoichiometry

Theoretical yield: maximum possible amount of product that can be produced from a complete conversion of reactants to products

Actual yield: the amount of product obtained from a reaction

% yield = (actual yield / theoretical yield) x 100

Limiting reagent: the reagent that is completely used up in the reaction

E.g. 2 H2 + O2 ---> 2H2O

2 mol H2 + 1 mol O2

with equal amounts (1 mol : 1 mol) H2 would be used up first so it the limiting reagent

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3.4 Reacting quantities: atom economy

Atom economy: a measure of how well atoms have been utalised in a chemical reaction

Atoms with a high atom economy are better utalised, as they:

  • produce a large proportion of desired products
  • produce few unwanted products
  • make the best use of natural resources so are more sustainable

Calculation: (worked out from equation, assumes a 100% yield)

(sum of molar masses of desired products / sum of molar masses of all products) x 100

Sustainability(exam question e.g. comment on AE in terms of sustainability)

Ideal chemical processes: no waste, 100% atom economy

  • Low atom economy means lots of waste
  • Use for unwanted / waste products needs to be developed to increase atom economy
  • OR alternative process needs to be developed with high atom economy
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