Chemistry C3 Test Revision

Front 1

Relative Formula Mass

Back 1

Mass of a compound molecule, calculated by adding all of the Ar's (relative atomic masses) for the component atoms (Mr/RFM/RMM)

Front 2

what is the relative atomic mass of H2O?

Back 2

H2O= (2x1)+16=18
H=1
O=16

Front 3

Mr of Al2 (so4)3

Back 3

(so4)3= (2x27)+(32x3)+(16x4x3)=342
Al=27
S=32
0=16

Front 4

Mr of citric acid C6H8O7

Back 4

C6H8O7= (12x6)+(1x8)+(16x7)=192
C=12
H=1
O=16

Front 5

Moles& Avogadro's number

Back 5

Atoms are too small to count/weigh individually, so chemists group them together. 6.02x1023 is the number of atoms called the mole.

Front 6

Moles equation

Back 6

Moles= Mass/Molar mass

Front 7

how many moles in 73g of HCl

Back 7

Moles=Mass/Molar Mass
73/(1+35.5)
= 2 moles

Front 8

196g of H2So4
moles=?

Back 8

Moles=Mass/Molar mass
196/(2+32.1+64)
=2 moles

Front 9

2 moles of iron
what is the mass?

Back 9

Mass=Moles x Molar mass
2 x 55.8
Mass= 111.6 g

Front 10

0.1 moles of S8
Mass=?

Back 10

Mass=moles x molar mass
0.1 x (32.1 x 8)
=25.6 g

Front 11

molar volume of gases

Back 11

when working with gases, chemists calculate how many moles of gas are present from the volume of the gas. therefore 1 mole of gas at RTP (Room Temperature Pressure) has a volume of 24dm3.
Room temperature= 25 degrees
Room pressure= 1 atm

Front 12

Moles in gases equation

Back 12

Moles = volume of gas in dm3/24dm3

1dm3= 1000cm3

cm3/1000=dm

Front 13

24,000cm3 of Cl
Moles=?

Back 13

2.4/24
0.1 moles

Front 14

Mg +H2So4 = MgSo4+ H2
How many moles of Mg?

Back 14

Mg=24.3
H2=2
S=32.1
O4= 64
Moles of Mg= mass/molar mass
=2.4/24.3
=0.1 moles of Mg

Front 15

Hydrated

Back 15

Forms a complex with water

Front 16

Empirical formula or EF

Back 16

shows the simplist whole number molar ratio of atoms in a molecule

Front 17

EF of Fe2O3

Back 17

Fe:O
2:3
EF = Fe2O3

Front 18

EF of Al2O6

Back 18

Al:O
2:6
=1:3
AlO3

Front 19

An experiment shows that 64g of conmpound contains 48g of C and 16g of H. Calculate the empirical formula.

Back 19

C H
Mass= 48 16
Molar mass= 12 1
Moles= 4 16
Ratio= 1:4
EF= CH4

Front 20

Molecular Formula

Back 20

Shopws the actual number of atoms in a compound and is a simple multiple of the Empirical formula
for example:

EF= CH3
MF= CH3 x 2
MF= C2H6

Front 21

Find the MF of a compound whose EF is CH2 and has a mass of 84g

Back 21

Mass of EF=
(12 x 1)+(1 x 2) = 14g
N= the amount to multiply the EF by
N= molar mass/empirical mass
= 84/14
=6
6 x CH2
C6H12

Front 22

N= the amount to multiply by
what is the equation?

Back 22

N= molar mass/empirical mass

Front 23

Find the MF of CH2

Back 23

Mass of EF= 912 x 1) + (1 x 2) =14g
N= molar mass/empirical formula
=28/14
=2
MF+

Front 24

% yield equation

Back 24

% yield = obtained yield/ theoretical yield X 100

Front 25

The theoretical yield of X is 12.5g
Actual yield= 8.25g

Back 25

8.25/12.5 x 100
= 66%

Front 26

Reversible reactions

Back 26

Some reactions can move in both directions. They are known as reversible reactions.

Front 27

Example of an irreversible reaction

Back 27

C+ O2 = CO2

Front 28

Example of a reversible reaction

Back 28

CuSo4.5H20 = CuSo4 + 5H2O
hydrated copper sulfate complex (blue crystals) = Anhydrous copper sulfate (white)
It goes white because the water has been removed.

Front 29

Dynamic equilibrium

Back 29

In a reversible reaction, the forward and back reactions are happening at the same time. The forward reaction always goes first, but as soon as products are formed, the back reaction begins. Eventually, a steady state will be reached and the forward and backwards reaction are happening at he same time. This is Dynamic Equilibrium.

Front 30

How does the position of the equilibrium point tell us the proportion of the products and reactants.

Back 30

EQM point to the right = more products
EQM point to the left = more reactants

Front 31

Kinetic theory of matter

Back 31

all particles above absolute zero have kinetic energy

Front 32

Collision theory

Back 32

In order for a reaction to happen, particles must collide.

Front 33

4 ways to change the equilibrium point

Back 33

1. a higher temperature
2. particle size (smaller particles= greater surface area)
3. Change the concentration
4. A catalyst will make the reaction meet its equilibrium much faster, with less energy

Front 34

The Haber Process

Back 34

Air is liquefied under pressure
Nitrogen is separated out by fractional distillation
Hydrogen is extracted from Methane
Nitrogen and hydrogen are combined to make Ammonia
High temperature, pressure and catalyst is needed.
Used for explosives, dye making, medicines and fertilisers.

Front 35

Why is an iron catalyst used in the Haber process?

Back 35

The iron catalyst is used in the Haber process because it increases the rate of reaction and speeds up both the forward and backwards reaction up.

Front 36

If the temperature increases what will happen to the yield?

Back 36

The yield would decrease because a low temperature favours the exothermic reaction

Front 37

why is a high pressure not used?

Back 37

A high pressure is not used because it is too expensive and dangerous.

Front 38

Le Chatalier's Principle

Back 38

If a dynamic equilibrium is disturbed by the conditions, the reaction will move to counteract the change.