Chapter 3 Amount of substance

Question 1

amount of substance n

Answer 1

The quantity whose unit of the mole, used as a means of counting any species such as atoms, ions and molecules.
n=m/M amount=mass/molar mass

Question 2

Mole

Answer 2

The amount of any substance containing as many elementary particles as there are carbon
atoms in exactly 12 g of the carbon-12 isotope, that is, 6.02 × 10^23 particles.

Question 3

Avogadro constant NA

Answer 3

The number of atoms per mole of the carbon-12 isotope. (6.02 × 1023 mol–1)

Question 4

Molar mass

Answer 4

M The mass per mole of a substance, in units of g mol–1.

Question 5

Molecular formula

Answer 5

molecular formula A formula that shows the number and type of atoms of each element present in a molecule.

Question 6

Empirical formula

Answer 6

The formula that shows the simplest whole-number ratio of atoms of each element present in a compound.

Question 7

Relative molecular mass

Answer 7

Mr The weighted mean mass of a molecule of a compound compared with one-twelfth of the mass of an atom of carbon-12.

Question 8

Relative formula mass

Answer 8

The weighted mean mass of the formula unit of a compound compared with one-twelfth of the mass of an atom of carbon-12.

Question 9

Hydrated

Answer 9

A crystalline compound containing water molecules.

Question 10

Water of crystallisation

Answer 10

Water molecules that are bonded into a crystalline structure of a compound.

Question 11

Anhydrous

Answer 11

Containing no water molecules.

Question 12

Concentration.

Answer 12

The amount of solute, in moles, dissolved in 1dm3 (1000 cm3) of solution.

Question 13

Standard solution

Answer 13

n A solution of known concentration.

Question 14

Molar gas volume

Answer 14

Vm The volume per mole of gas molecules at a stated temperature and pressure
At RTP molar gas volume is 24.0dm3 mol-1

n=volume/molar gas volume

Question 15

RTP

Answer 15

Room temp 20 degrees Celsius

Pressure 101kPa

Question 16

Ideal gas

Answer 16

Molecules have: 
Random motion 
Elastic collisions 
Negligible size 
No intermolecular forces 

Equation
pV=nRT
R ideal gas constant = 8.314
V=m^3 p=Pa temp=K

Question 17

Conversions
Cm^3 to m^3
Dm3 to m3 
C to K
KPa to Pa

Answer 17

X10^-6
X10^-3
+273
X10^3

Question 18

Stoichiometry

Answer 18

The ratio of the amount, in moles, of each substance in a chemical equation (essentially the ratio of the balancing numbers).

Question 19

Chemists use balanced equations to find

Answer 19

• the quantities of reactants required to prepare a required quantity of a product
• the quantities of products that should be formed from certain quantities of reactants

Question 20

Theoretical yield

Answer 20

The yield resulting from complete conversion of reactants into products.

Question 21

Theoretical yield is difficult to achieve

Answer 21

• the reaction may not have gone to completion
• other reactions may have taken place with the main reactions
• purification of the product may result in loss of some product

Question 22

Actual yield

Answer 22

The amount of product obtained from a reaction.

Question 23

Percentage yield

Answer 23

% yield = actual amount,inmol,of product / theoretical amount,inmol,of product x100

Question 24

Limiting reagent

Answer 24

The reactant that is not in excess, which will be used up first and stop the reaction.

Question 25

Atom economy

Answer 25

(Sum of molar masses of desired products)/(sum of molar masses of all products) x100%.
Measure of how well atoms have been utilised
take account of balancing numbers

Question 26

Reactions with high atom economies

Answer 26

• Produce a large proportion of desired products and few unwanted waste products
• are important for sustainability as they make the best use of natural resources

Question 27

To work out mass concentration

Answer 27

e.g conc is 0.250 mol dm^-3 then do 0.250 x molar mass to get grams per dm^3

Question 28

moles of a solution

Answer 28

n=cV

Question 29

other way of showing concentration

Answer 29

e.g 0.250 mol dm-3 in gym-3

use m= n x M