NAT 5 Chemistry
Calculations
(Answers on page 9)
The cross-over method
The cross-over method for –ite and -ate compounds
Example 1:
Lithium Sulphate
The sulphate ion is a complex ion.
Check the correct formula and ion charge of any complex ion in the table, ‘formulae
of selected ions…’ on page 4 of the SQA data booklet.
SO42-
Sulphate ion is
Place the formula of the ion in brackets (SO4) with the valency of the ion represented
by the charge on the ion i.e. 2.
Elements/complex ions
Lithium
Sulphate ion
1
2
Valency
Write the symbol with the valency number at the top right hand corner of the symbol.
Symbol/valency
Cross-over
Reduce ratio
Chemical formula
Example 2:
Symbols/valency
Cross-over
Reduce ratio
Chemical formula
Li1
(SO4)2
2
2
1
1
Li2SO4
Ammonium Phosphate
(NH4)1
3
3
(PO4)3
1
1
(NH4)3 PO4
Write chemical formulae for the following ten –ite/ate compounds:
1.
2.
2.
4.
5.
Ammonium nitrate
Magnesium phosphate
Iron(iii) sulphate
Calcium nitrate
Aluminium carbonate
6. Calcium hydroxide
7. Potassium sulphate
8. Lithium carbonate
9. Calcium sulphite
10. Magnesium nitrate
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Calculating one mole of a substance – gram formula mass (gfm)
One mole of substance is its formula mass expressed in grams.
Example:
What is the mass of one mole of calcium hydroxide?
One mole of calcium hydroxide (Ca(OH)2)
gfm
= (1 x Ca) + (2 x O) + (2 x H)
= (1 x 40) + (2 x 16) + (2 x 1)
= 40 + 32 + 2
= 74g
What is the mass of one mole of each of the following compounds?
11.
12.
13.
14.
15.
Ammonium nitrate
Magnesium phosphate
Iron(iii) sulphate
Calcium nitrate
Tin(ii) carbonate
16.
17.
18.
19.
20.
Calcium hydroxide
Potassium sulphate
Lithium carbonate
Calcium sulphite
Magnesium nitrate
2
Calculating Mass to Moles and Moles to Mass.
This triangle is very important:
m
n
gfm
Where:
m
n
gfm
= mass of substance (grams).
= number of moles of substance (moles)
= gram formula mass of substance (grams).
Example:
How many moles are in 8g of magnesium oxide (MgO)?
Write down what you know:
Number of moles
Mass of substance
n = ?
m = 8g
Remember you can always work out the gfm of any substance using your data booklet
once you know its name/formula.
gfm of MgO =
=
=
=
(1 x Mg) + (1 x O)
(1 x 24.5) + (1 x 16)
24.5
+ 16
40.5g
Number of moles,
n = _m_
gfm
= _8_
40.5
= 0.196 mols or 0.2 moles
Using the triangle above:
Calculate the mass of:
21. 2 moles of calcium carbonate.
22. 0.1 moles of nitric acid.
Calculate the number of moles of:
23. 10g of calcium carbonate.
24. 10.6g of sodium nitrate.
25. 7.5g of Iron(iii) oxide
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4. Calculating the Concentration of a Solution.
Concentration of a solution is written as:
mol/l or moll-1
This triangle is very important;
n
c
V
Where n = number of moles of substance (moles).
c = concentration of solution (mol/l).
V = volume in litres.
Remember: volume must be in litres.
Example:
Calculate the concentration of one mole of hydrochloric acid dissolved to
make 100cm3 of solution.
Always write down what you are given in the question.
n = 1 mole
c = ?
V = 100cm3 =0.1L
Remember V must be in litres.
C = _n_
V
= _1_
0.1
= 10mol/l
Calculate the concentration in each of the following:
26. 0.4 mol of sodium nitrate in 500cm3 of solution.
27. 3 mol of hydrochloric acid in 2 litres of solution.
Calculate the number of moles in each of the following:
28. 250cm3 of 2 mol/l of lithium nitrate.
29. 500cm3 of 0.1 mol/l potassium iodide.
Calculate the volume of:
30. 6 mol of sulphuric acid are used to prepare a 2 mol/l solution.
4
Calculations based on balanced equations.
Balanced Equations.
A formula equation is balanced if the number and type of atoms on the left hand side
of the equation exactly equal the number and type of atoms on the right hand side of
the equation.
Example:
When hydrogen reacts with chlorine, hydrogen chloride is formed.
Word equation
hydrogen + chlorine → hydrogen chloride
Formula equation
H2 + Cl2 → HCl
Balanced equation
H2 + Cl2 → 2HCl
Balance the following chemical equations:
31.
32.
33.
34.
35.
Mg(s) + O2(g) → MgO(s)
NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l)
AgNO3(aq) + BaCl2(aq) → Ba(NO3)2(aq) + AgCl(s)
C3H8(g) + O2(g) → CO2(g) + H2O(l)
C5H12(g) + O2(g) → CO2(g) + H2O(l)
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Calculations based on balanced equations.
Balanced equations indicate the masses of reactants and products involved in a
particular reaction.
Example:
What mass of hydrogen would be produce when 6g of magnesium reacts
completely with dilute hydrochloric acid.
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
*
*
1 mole
1 mole
gfm of Mg
24.5g
_24.5_ x 6 = 6g
24.5
→
→
gfm of H2
2g
_2_ x 6 = 0.5g
24.5
6g of magnesium will produce 0.5g 0f hydrogen gas.
36.
2P(s) +
3H2(g) → 2PH3(g)
Calculate the mass of phosphorus trihydride produced from 7.75g of phosphorus
37.
N2(g) + 3H2(g) → 2NH3(g)
Calculate the mass of hydrogen required to react completely with 56g of
nitrogen.
38.
4Al(s) + 3O2(g) → 2Al2O3(s)
Calculate the mass of oxygen required to react with 2.7g of aluminium.
39.
Mg(s) + CuSO4(aq) → Cu(s) + MgSO4(aq)
Calculate the mass of copper produced in the reaction of 1.2g of
magnesium with excess copper(ii) sulphate.
40.
Mg(s) + Cl2(g) → MgCl2(s)
Calculate the mass of magnesium chloride produced in the reaction of
1.2g of magnesium with excess chlorine.
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Titration Calculations.
V1C1 = V2C2
n1
n2
Example;
In a titration it was found that 20cm3 of potassium hydroxide solution was neutralised
by 15cm3 of dilute sulphuric acid with a concentration of 0.1 mol/l. Calculate the
concentration of potassium hydroxide.
2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l)
n1 = 2
n2 = 1
Remember to write down what you know.
C1
V1
n1
= ?
= 20cm3
= 2
C2
V2
n2
= 0.1 moll-1
= 15cm3
= 1
Note volumes are in same units.
C1V1 = C2V2
n1
n2
C1 = n1 C2V2
n2V1
= _2 x 0.1 x 15_
1 x 20
= 0.15 mol/l
Titration Questions
41.What is the concentration of H2SO4 if 17.3cm3 neutralises 25cm3 of NaOH
(concentration 0.5 mol/l).
42. Sodium hydroxide and nitric acid react as follows;
NaOH + HNO3  NaNO3 + H2O
In a titration, 30cm3 of nitric acid neutralised 20cm3 of 0.2 mol/l sodium
hydroxide. Calculate the concentration of nitric acid.
43. 25cm3 of 0.2 mol/l sodium hydroxide solution was neutralised by 16cm3 of
hydrochloric acid. Calculate the concentration of hydrochloric acid.
7
44. 10cm3 of potassium hydroxide solution was neutralised by 12cm3 of 0.1
mol/l nitric acid. Calculate the concentration of potassium hydroxide
solution.
45. 20cm3 of 0.5 mol/l lithium hydroxide solution was neutralised by 30cm3 of
sulphuric acid. Calculate the concentration of sulphuric acid.
8
Answers
1. NH4NO3
2. Mg3(PO4)2
3. Fe2(SO4)3
4. Ca(NO3)2
5. Al2(CO3)3
6. Ca(OH)2
7. K2SO4
8. Li2CO3
9. CaSO4
10. Mg(NO3)2
11. 80g
12. 193.5g
13. 400g
14. 164g
15. 234g
16. 74g
17. 174g
18. 67g
19. 120g
20. 148.5g
21. 200g
22. 63g
23. 0.1mol
24. 0.12mol
25. 0.05mol
26. 0.8 moll-1
27. 1.5 moll-1
28. 0.5 moll-1
29. 0.05 moll-1
30. 3L
31. 2Mg(s) + O2(g) → 2MgO(s)
32. 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)
33. 2AgNO3(aq) + BaCl2(aq) → Ba(NO3)2(aq) + 2AgCl(s)
34. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
35. C5H12(g) + 8O2(g) → 5CO2(g) + 6H2O(l)
36. 8.5g
37. 12g
38. 2.4g
39. 3.1g
40. 4.7g
41. 0.72moll-1
42. 0.13 moll-1
43. 0.31 moll-1
44. 0.12 moll-1
45. 0.17 moll-1
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