Module 2 - Amount Of Substance

Question 1

What is a mole?

Answer 1

The amount of any substance containing as many elementary particles as there are carbon atoms in exactly 12g of the carbon-12 isotope. (6.02 x10^23 particles).

Question 2

What is the Avogadro constant?

Answer 2

The Avogadro constant is 6.02 x10^23 mol^-1, the number of particles in each mole of carbon-12.

Question 3

What is molar mass? (Mr)

Answer 3

The mass per mole of a substance, in units of g mol^-1.

Question 4

Molar equations:

Answer 4

mol = mass/Mr
Mr = mass/mol
mass = mol x Mr

Question 5

Empirical formula:

Answer 5

The empirical formula is the simplest whole-number ratio of atoms of each element in a compound.

Question 6

Concentration and volume equations:

Answer 6

mol = conc x vol (dm^3)
mol = conc x vol (cm^3)/1000
vol (dm^3) = mol/conc
vol (cm^3) = (mol x 1000)/conc
conc = mol/vol (dm^3)
conc = (mol x 1000) /vol (cm^3)

Question 7

Gas volume equations:

Answer 7

mol = vol (dm^3)/24
mol = vol (cm^3)/24 000
vol (dm^3) = mol x 24
vol (cm^3) = mol x 24 000

Question 8

Molar gas volume:

Answer 8

The molar gas volume is the volume per mole of gas molecules at a stated temperature and pressure.

Question 9

Room Temperature Pressure…

Answer 9

• RTP is about 20*C and 101 kPa (1 atm) pressure.
• At RTP, 1 mole of gas molecules has a volume of approximately 24.0 dm^3 = 24 000 cm^3.
• Therefore, at RTP, the molar gas volume = 24.0 dm^3 mol^-1.

Question 10

Ideal gas equation:

Answer 10

pV = nRT

pressure (Pa) x volume (m^3) = mol x gas constant (8.31 Jmol^-1K^-1) x temperature (K)

Question 11

Assumptions for ideal gas rule…

Answer 11

• random motion.
• elastic collisions.
• negligible size.
• no intermolecular forces.

Question 12

Unit conversions:

Answer 12

cm^3 to m^3 = x 10^-6
dm^3 to m^3 = x 10^-3
*C to K = + 273
kPa to Pa = x 10^3

Question 13

Example Question 1:
The solubility of a substance is a measure of how much of it will dissolve in a given solvent. At 20*C, the solubility of calcium hydroxide in water is 0.0233 mol/dm^-3. Calculate the maximum amount, in grams, that will dissolve in 250 cm^3.

Answer 13

Ca(OH)2,    0.0233 mol/dm^-3,   250 cm^3  
mol = 0.0233 x 250/1000
mol = 5.825 x 10^-3
mass = mol x Mr
Mr (Ca(OH)2) = 40.1 + (2 x 16) + (1 x 2)
mass = 5.825 x10^-3 x 74.1
= 0.432g

Question 14

Example Question 1:

Explanation…

Answer 14

1. ) Find the moles of calcium hydroxide.
2. ) Find the Mr value.
3. ) Find the mass of Ca(OH)2.

Question 15

Example Question 2:
A student prepares a solution of calcium hydroxide, but thinks that the sample has been contaminated. She weighs out 0.50g of the impure calcium hydroxide and dissolves it in 500 cm^3 of distilled water. To check its purity, she titrates it against a solution of 0.04 mol/dm^-3 nitric acid, HNO3. 25cm^3 of the calcium hydroxide requires 12.45 cm^3 of 0.04 mol/dm^-3 hydrochloric acid to be completely neutralised. Calculate the percentage purity of the calcium hydroxide.

Answer 15

Ca(OH)2,   25cm^3
HCl,    12.45cm^3,   0.04mol/dm^-3
HNO3,  12.45cm^3,  0.04mol/dm^-3
500 cm^3 water
mol = conc x vol/1000
mol = 0.04 x 12.45/1000
mol = 4.98 x10^-4 mol
Ca(OH)2 aq + 2HCl aq
= mol Ca(OH)2 = (4.98 x10^-4)/2
= mol Ca(OH)2 = 2.49 x10^-4 mol
500cm^3 water/25cm^3 = 20
therefore 20 x bigger = 2.49 x10^-4 mol x 20
=4.98 x10^-3 mol
Mr(Ca(OH)2) = 74.1
mass = 4.98 x10^-3 x 74.1
mass = 0.369g
purity % = (0.369g/0.5g) x 100
purity % = 73.8 %

Question 16

Example Question 2:

Explanation…

Answer 16

1. ) Find the moles of nitric acid used in the titration.
2. ) Find the moles of calcium hydroxide in 25 cm^3 of solution.
3. ) Find the moles of calcium hydroxide in the original solution.
4. ) Find the mass of calcium hydroxide in the original solution.
5. ) Find the percentage purity of the calcium hydroxide.