
Answers to the more advanced level mole concept
calculations
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Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) Online Chemical Calculations
ANSWERS to Part 7
- the more advanced mole Q's
ORIGINAL
mole Q
Quantitative chemistry
calculations This page has the
answers to the more advanced mole based questions from
calculations section 7. for more
advanced chemistry students. Online practice exam chemistry CALCULATIONS and
solved problems for KS4 Science
GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB.
These revision notes and practice questions on how to do chemical
calculations and worked examples should prove useful for the new AQA,
Edexcel and OCR GCSE (9–1) chemistry science courses.
Spotted any careless error?
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GCSE calculation not covered?
Mole
calculations introduction * Molar gas
volume * Advanced
Redox titration Q's
* Non-redox titration Q's
Qa7.1
(a) f. mass Al2O3 = 102, 2
÷ 102 x 3 x 6.02 x 1023 = 3.54 x 1022 oxide ions
(b) f. mass H2 = 2, 3 ÷ 2 x 6.02 x 1023 =
9.03 x 1023 molecules
(c) 1.2 ÷ 24000 x 6.02 x 1023 =
3.01 x 1019 molecules
(d) f. mass Cl2 = 71, 3 ÷ 71 x 6.02 x 1023 =
2.54 x 1022 molecules
(e) Neon exists as single atoms (Ar = 20), 10
÷ 20 x 6.02 x 1023 = 3.01 x 1023 atoms
(f) 2Na + 2H2O
==>
2NaOH + H2,
1 mole sodium gives 0.5 moles hydrogen,
mole Na = 0.2
÷ 23 = 0.008696, so mole H2 = 0.008696 ÷ 2 = 0.004348
so volume H2 = 0.004348 x 24000 =
104.3 cm3
or 0.104 dm3
(g) e.g. Mg + 2HCl ==> MgCl2 + H2
1 mole magnesium gives 1 mole hydrogen, mole Mg = 2
÷ 24.3 = 0.0823
so mole H2 = 0.0823, so
volume H2 = 0.0823 x 24 =
1.975 dm3
(h) both 1 mole of Na2CO3 or NaHCO3 will give 1 mole of CO2
(1) VCO2 = mol Na2CO3 x 24000 = 0.76
÷ 106 x 24000 = 172cm3
(2) VCO2 = mol NaHCO3 x 24000 = 0.76÷ 84
x 24000 = 217cm3
(i) Zn + 2HCl
==> ZnCl2 + H2 , mole H2 = mole HCl
÷ 2
mol HCl = 50 ÷ 1000 x 0.2 = 0.01 mol
so mole H2 = 0.005, VH2 = 0.005 x 24000 =
120cm3
(j) CaCO3 + 2HCl
==>
CaCl2 + H2O + CO2
mole CO2 = mole HCl
÷ 2, mol HCl = 75 ÷ 1000 x 0.05 = 0.00375 mol
so mole CO2 = 0.001875, VCO2 = 0.001875 x 24000 =
45cm3

Above is typical periodic table used in GCSE science-chemistry specifications in
doing chemical calculations,
and I've 'usually' used these values in my exemplar calculations to cover most
syllabuses
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Calculating relative
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Law of Conservation of Mass and simple reacting mass calculations
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Composition by percentage mass of elements
in a compound
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Empirical formula and formula mass of a compound from reacting masses
(easy start, not using moles)
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Reacting mass ratio calculations of reactants and products
from equations
(NOT using
moles) and brief mention of actual percent % yield and theoretical yield,
atom economy
and formula mass determination
-
Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations
(relating reacting masses and formula
mass)
-
Using
moles to calculate empirical formula and deduce molecular formula of a compound/molecule
(starting with reacting masses or % composition)
-
Moles and the molar volume of a gas, Avogadro's Law
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Reacting gas volume
ratios, Avogadro's Law
and Gay-Lussac's Law (ratio of gaseous
reactants-products)
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Molarity, volumes and solution
concentrations (and diagrams of apparatus)
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How to do acid-alkali
titration calculations, diagrams of apparatus, details of procedures
-
Electrolysis products calculations (negative cathode and positive anode products)
-
Other calculations
e.g. % purity, % percentage & theoretical yield, dilution of solutions
(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy
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Energy transfers in physical/chemical changes,
exothermic/endothermic reactions
-
Gas calculations involving PVT relationships,
Boyle's and Charles Laws
-
Radioactivity & half-life calculations including
dating materials
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