Chapter 3: Amount of substance
- Created by: hannahc072
- Created on: 28-12-15 21:17
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)
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
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)
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
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
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
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
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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