Transcript STANDARD GRADE CHEMISTRY CALCULATIONS

STANDARD GRADE CHEMISTRY
CALCULATIONS
Calculations involving the mole.
1 mole of a solid substance is the formula mass of the
substance in grams. This is known as the gram formula mass (gfm).
The triangle shown below
mass = mass of substance.
You must learn this
in calculations.
n
mass
and be able to apply it
n
gfm
= number of moles
gfm = gram formula mass
can be used to give the following relationships:1.
Mass = number of moles x gram formula mass.
2.
Number of moles =
mass
gram formula mass
Worked example 1.
Calculate the mass of 0.25 moles of butane (C4H10).
Step 1 :- Write the formula for butane
Step 2:- Calculate the gram formula mass
C4H10
(4 x 12) + (10 x 1) = 58g
Step 3:- Using the triangle we have
mass = number of moles x gram formula mass
mass =
0.25
x 58
mass =
14.5 g
Calculations for you to try.
1. Calculate the mass present in 2.5 moles of calcium carbonate
(CaCO3).
mass
n
gfm
gfm = 100g
mass = 2.5 x 100 = 250g
2. Calculate the mass of ammonium sulphate, (NH4) 2SO4,
present in 0.1 mol of ammonium sulphate?
gfm = 132g
Standard Grade Chemistry
mass = 0.1 x 132 = 13.2g
Worked example 2.
Calculate the number of moles in 5.05 g of potassium nitrate, (KNO3).
Step 1 :- Write the formula for potassium nitrate
KNO3
Step 2:- Calculate the gram formula mass
(1 x 39) + (1 x 14) + (3 x 16) = 101g
Step 3:- Using the triangle we have
mass
number of moles =
=
gram formula mass
5.05
101
= 0.05 mole
Calculations for you to try.
1. Calculate the number of moles in 132 g of carbon dioxide, CO2.
mass
n
gfm
gfm = 44g
Number of moles =
132/
44
= 3 moles
2. Calculate the number of moles in 4g of bromine, Br2.
Standard Grade Chemistry
gfm = 160
Number of moles =
4/
160
= 0.025 moles
Mole calculations involving solutions.
The concentration of a solution is measured in moles per litre (mol/l)
The triangle shown below
n
You must learn this
C
n
and be able to apply it
in calculations.
C
= number of moles.
= concentration.
V(l) = volume in litres
V (l)
Remember this is in
litres
can be used to give the following relationships:-
1. number of moles = concentration x volume (in litres).
2.
Standard Grade Chemistry
concentration =
number of moles
volume (in litres)
Worked example 1.
Calculate the number of moles in 200cm3 of 0.5 mol/l sodium hydroxide
solution.
Step 1 :- Change the volume into litres.
0.2 litres
Step 2 :- Using the triangle gives
number of moles = concentration x volume (in litres).
=
0.5
x 0.2
=
0.1 moles
Calculations for you to try.
1. Calculate the number of moles in 50 cm3 of 0.1 mol/l zinc sulphate
solution.
Number of moles
n
C
Volume =
50/
1000
= 0.05 litres
= 0.1 x 0.05
= 0.0005 moles
V (l)
2. Calculate the number of moles in 0.2 litres of 2 mol/l sodium
hydroxide solution
Standard Grade Chemistry
Volume = 0.2 litres
Number of moles
= 0.2 x 2
= 0.4 moles
Calculations involving, concentration, moles and mass.
In this type of calculation both triangles are used.
Worked example 1.
Calculate the mass required to prepare 200cm3 of 0.1 mol/l sodium
hydroxide, (NaOH), solution.
Step 1 :- Calculate the number of moles of sodium hydroxide in
200cm3 of 0.1 mol/l of solution.
n = c x V(l)
n
C
V (l)
=
0.1 x 0.2
=
0.02 moles
Step 2 :- Calculate the mass of NaOH in 0.02 moles.
gfm of NaOH = (1 x 23) + (1 x 16) + (1 x 1) = 40
mass = n
x gfm
= 0.02 x 40
mass
n
gfm
Calculation for you to try.
Calculate the mass of zinc sulphate in 500 cm3 of 0.2 mol/l ZnSO4(aq)
Volume =
Standard Grade Chemistry
= 0.8 g
500/
1000
= 0.5 litres
No. of moles = 0.2 x 0.5 = 0.1
Mass = 0.1 x 161.5
= 16.15 g
Worked example 2.
Calculate the concentration of a solution that contains 7.45 g of
potassium chloride (KCl) in 250cm3 of solution.
Step 1 :- Calculate the number of moles of potassium chloride in
7.45g
mass
Number of moles = gfm
7.45
=
= 0.1
74.5
Step 2 :- Calculate the concentration of the solution.
mass
n
gfm
Concentration =
No. of moles
Volume in litres
=
0.1
0.25
= 0.4 mol/l
Calculation for you to try.
n
C
V (l)
Calculate the concentration of a solution that contains 5.85 g of sodium
Chloride, NaCl, in 200cm3 of solution.
Volume =
Standard Grade Chemistry
200/
1000
= 0.2 litres
No. of moles in 5.85 g of NaCl
=
Concentration =
0.5 mol/l
0.1
/ 0.2
=
5.85/
58.5
= 0.1