4.4 Moles I

Question 1

Define Relative formula mass (RFM)

Answer 1

The mass of a formula unit of a substance.

Question 2

How is RFM calculated?

Answer 2

-Adding the RAMs of the atoms or ions

Question 3

Define percentage by mass

Answer 3

Mass of a substance that comes from a particular element

Question 4

How is % by mass calculated?

Answer 4

-from the formula for the substance, using the RFM
-%X = mass of X/RFM x 100

Question 5

Define a mole

Answer 5

The unit of amount of a substance

Question 6

What does a mole represent?

Answer 6

-Represents how many particles of a substance there are
-Taken into account that particles of different substances have different masses

Question 7

How do you calculate moles?

Answer 7

Moles= Mass(g)/RFM

Question 8

Define empirical formula

Answer 8

The EF of a substance contains the simplest mole ratio of elements.

Question 9

RFM of H20??

Answer 9

RFM of H20= (2x1) + 16= 18

Question 10

RFM of Ba(OH)2?

Answer 10

RFM= 137+(2x(16+1))= 171

Question 11

Moles in 11g of CO2??

Answer 11

Moles CO2= 11/44= 0.25 mol

Question 12

Percentage of C in C3H8?

Answer 12

%C = 3x12/(3x12)+8 x 100= 81.8%

Question 12

Calculate the mass of 2.5 moles of H20?

Answer 12

Mass H20= moles x RFM
= 2.5x18= 45g

Question 13

How do you calculate EF?

Answer 13

Empirical formula can be calculated from the masses (or percentages) of each element:
1. Convert the masses (or percentages) into moles.
2. Cancel down the mole ratio.
3. Write the simplest mole ratio into a formula.

Question 14

Calculate the EF of a substance containing 6.2g P AND 4.8g O

Answer 14

MASS: 6.2g P 4.8g O
Moles: 0.2 P 0.3 O
Ratio: 2 P 3 P
P2O3

Question 15

Define molecular formula

Answer 15

The molecular formula of a substance contains the actual numbers of atoms in a molecule.
It’s like empirical formula, but uncancelled:

Question 16

How is Molecular formula calculated?

Answer 16

-Molecular formula from empirical formula & RFM
-Find uncancelled multiple of empirical formula
-Multiply to get correct RFM

Question 17

Calculate the molecular formula if the EF is CH2 and RFM is 70

Answer 17

RFM of CH2= 14
MULTIPLE= 70/14= 5
molecular formula= 5xCH2= C5H10

Question 18

Water of Crystallisation

Answer 18

1. Calculate the mass of water lost.
   Mass of water = 16.7 g — 9.5 g = 7.2 g
2. Draw a “mass, moles, ratio” table and stick the numbers in.
   Be careful to use the final mass of solid for the salt, not the original mass of crystals.
   MgCl2:H2O
   Mass: 9.5 7.2
   Moles: 0.1 0.4 (whole-number ratio 1:4)
   Ratio: 1 4 → x = 4
   So the formula of hydrated magnesium chloride crystals is MgCl2*4H2O

Question 19

CuSO4*5H2O(s) → CuSO4(s) + 5H2O(g)

Answer 19

Hydrated copper sulphate → Anhydrous copper sulphate + Water (vapour)

Question 20

What is water of crystallisation calculated from?

Answer 20

Water of crystallisation is calculated from the masses of salt and water produced on heating.

Question 21

What do the balancing numbers in a balanced equation for a reaction tell us?

Answer 21

• How many moles of each reactant are needed.
• How many moles of each product are made.
  e.g. N2 + 3H2 –> 2NH3

Question 22

Calculate the mass of N2 needed to react with 1.2 g of H2

Answer 22

1. Convert mass of H2 into moles of H2.
2. Use the mole ratio from the balancing numbers.
3. Convert moles of N2 into mass of N2.
   1N2 : 3H2
   mole ratio- 1 : 3
   actual moles- 0.2 : 0.6
   (remember moles= mass/Mr)
   -Mass N2= 0.2x28= 5.6g
   -Mass H2= 1.2/2= 0.6g

Question 23

If there isn’t enough substance than is needed to do a reaction..

Answer 23

-If there isn’t enough, we say that substance is limiting.

Question 24

If there is more than enough substance than is needed to do a reaction..

Answer 24

-If there is more than enough, we say that substance is in excess.

Question 25

What happens if we mix 10 g of Mg and 10 g of HCl?

Answer 25

-Mg + 2HCl → MgCl2 + H2
-Start with Mg, calculate needed HCl, compare to available HCl.
-1 Mg : 2HCl
-moles-0.416 (x2): 0.833
->mol Mg 10/24:mass HCL-> 0.833 x 36.5= 30.4
-10g Mg needed to react with 30.4g of HCl, but we only have 10g of HCl
-HCl limiting (isn’t enough) & Mg in excess

Question 26

How is percentage yield calculated?

Answer 26

% yield= experimental/theoretical x 100

Question 27

Calculate the percentage yield if 10 g of Al produces 0.96 g of H2 in an experiment

Answer 27

1. Find or invent a balanced equation for the reaction:
   2Al + 6HCl → 2AlCl3 + 3H2
2. Calculate the theoretical mass of H2 that 10 g of Al should have produced:
   2AL : 3H2
   mole ratio- 2:3
   actual moles- 0.37:0.55
   mass H2= 1.11g
3. Calculate the percentage yield using the experimental mass from the question
   % yield= 0.96g/1.11g x 100= 86.4%

Question 28

How can the formulae of metal oxides be found experimentally?

Answer 28

The formulae of metal oxides can be found experimentally by reacting a metal with oxygen and recording the mass changes

Question 29

What is the method for finding the formulae of metal oxides?

Answer 29

1.Weigh crucible and lid
2.Place magnesium ribbon in crucible, replace lid, and reweigh
3.Calculate magnesium mass: (mass of crucible + lid + magnesium) - (mass of crucible + lid)
4.Heat crucible with lid on until magnesium burns (lid prevents magnesium oxide escaping)
5.Lift lid occasionally to allow air in
6.Stop heating when no further reaction observed (ensures all Mg reacted)
7.Allow to cool and reweigh
8.Repeat heating, cooling, and reweighing until two consecutive masses are the same (ensures accurate results)
9.Calculate mass of magnesium oxide formed: (mass of crucible + lid + magnesium oxide) - (mass of crucible + lid)