Please enable JavaScript.
Coggle requires JavaScript to display documents.
Quantatative Chemistry :star: - Coggle Diagram
Quantatative Chemistry :star:
Relative Formula Mass (Mr)
Mr is the sum of the relative atomic masses (big numbers on periodic table) of the atoms in the numbers shown in the formula
E.G - What is the RFM/Mr of CaCO3: 40 + 12 + (16 * 3)
1 Carbon atom is assigned to having a mass of "12" and alll the other atoms are relative to this
i.e - 1 Hydrogen atom is 1/12th the mass of carbon so has a relative mass of 1
Relative atom mass (Ar) is the average mass of an atom of that element, compared with the mass of 1/12th of an atom of Carbon (12)
Law of conservation of mass
"no atoms are lost or made during a chemical reaction so the mass of the products = the mass of the reactants"
Mass in a chemical reaction always remains constant in a closed system. It cannot be changed but moves from one form to another
Moles
6 x 10
*23
For atoms: 1 mole = Ar in grams
E.G - 1 mole of water will weigh 18g (H
2
O) because 16 + 1 +1 = 18 (Mr of water added up)
Calculating Moles
number of moles = mass of substance (g) / Molar mass (g/mol)
E.G - 2 moles of Cl = 35.5 x 2 = 71. 71 x 2 = 142. as Cl molecule = Cl
2
Making Magnesium Oxide
Method:
Take a piece of Mg and record it's mass using a mass balance
Take a crucible and lid and record it's mass
Place the Mg into the crucible and close lid
Set up apparatus as shown in diagram
Turn Bunsen burner on and heat for 10 mins - every minute remove lid using tongs for 10 seconds
After 10 mins turn off Bunsen burner and leave to cool
Reweigh crucible and lid and calculate mass of oxygen added
Equation : 2Mg + 0
2
-----> 2Mg0
Conclusion: Increase in mass
Added 0
2
into system and what go out must go in therefore increase in mass
Thermal Decomposition of CaCO
3
Method:
Take a piece of limestone and record it's mass using mass balance
Set up apparatus as shown in diagram
Turn on Bunsen using roaring flame heat limestone for 10 mins
Turn off Bunsen and allow rock to cool
Reweigh to obtain final mass
Equation: CaCO
3
-----> CaO + CO
2
Conclusion: Mass decreased as rock not in closed system so CO
2
produced escaped as it's a gas
Work out the mass of aluminium chloride produced when 5.4g aluminium reacts completely with EXCESS chloride
2Al + 3Cl
2
-----> 2AlCl
3
2 moles Al makes 2 moles AlCl
3
or 1 mole Al make 1 mole AlCl
3
, in other words it is a 1:1 conversion ration
Mr of Al = 27
n = m/Ar
5.4/27 + 0.2 moles
Moles of AlCl
3
= 0.2 as 1:1 ratio (ONLY USE BIG NUMBERS AT THIS STAGE. BIG NUMBERS TELL US THE RATIO OF THE MOLECULES
Mass of AlCl
3
= 0.2 x133.5 = 26.7g (mass = Moles *Mr)
Limiting Reactants
The limiting reactant is the one that gets used up first
E.G - 4.8g of Mg reacting with HCl acid, that contains 7.3g of HCl, which reactant is the limiting reactant?
Mg + 2HCl -----> MgCl
2
+ H
2
n = m/Mr
4.8/24 = 0.2
As 1:2 ratio, we need 0.4 moles HCl therefore HCl will run out before. (we don't have 0.4, we have 0.2 HCl)
HCl is the limiting reactant and Mg is in excess (0.2 moles or 2.4g will remain after the reaction is finished)
Balloon Experiment
Add different masses of Sodium Hydrogencarbonate to different balloons
Add 10cm
3
of vinegar to each test tube
Lift the balloons one at a time to allow sodium bicarbonate and vinegar to react
Observe and note the size of the balloons relative to one another
Lower mass = smallest
Last 3 or 4 balloon might appear the same in size despite mass as HCl will be the limiting reactant instead of NaHCO
3
Error and Uncertainty
recorded as a +- value
E.G - The vol of gas produced in an experiment was 83, 71, 78, 61, 82 and 79 cm
3
The anomalous result is 61cm
3
so we can discard that
mean result = 83 +71 + 78 + 81 + 79 / 5 = 79cm
3
(2 sf as figures in question given to 2sf)
The furthest value away from the mean is 71 which is 8cm
3
from the mean value
Therefore the mean result = 79 +- 8cm
3
Percentage Yield
Definition: percentage of the products to reactants
% yield is the mass actually made compared to what could theoretically be made.
Why would the yield of a chemical reaction not be 100% 1. Reaction may be reversible
Some reactants may react to give other products
Some of the products may have been lost in handling
Reactants may not have been pure
May be difficult to separate the product wanted from a mixture of products
Percentage yield = actual yield / predicted yield x100
E.G - In a reaction 5g of Mg was added to a sulphuric acid. From this, it is expected that 25g of magnesium sulphate would be made. However, only 18g are actually made
Mg + H
2
SO
4
-----> MgSO
4
+ H
2
O
18/25 x100 = 72%
Atom Economy
Atom economy is the mass of the product you want as a % of the mass of all the reactants
Mr of useful products / Mr of all reactants x100%
E.G - CaCO
3
-----> CaO + CO
2
56/100 x100% = 56%
ONLY in ATOM ECONOMY you COUNT the BIG NUMBER when CALCULATING the MR
E.G 2Mg + O
2
-----> 2MgO
80/80 x100% = 100% therefore no waste product
Atom Economy vs Percentage Yield
Atom Economy is the MASS OF THE PRODUCT YOU WANT AS A % OF THE MASS OF ALL THE REACTANTS THEY COME FROM
Raw materials are scarce and expensive and so must be carefully conserved.
Chemical processes need to produce as little waste as possible, minimise costs, energy use and pollution
Scientist must plan to maximise Atom Economy by:
Choosing the most efficient reaction to make the product to...
Reduce waste and pollution
Percentage Yield is the AMOUNT OF PRODUCT YOU ACTUALLY MAKE AS A % OF THE AMOUNT YOU SHOULD THEORETICALLY MAKE
Scientists must plan to maximise Percentage Yield by:
Using the most efficient reaction conditions and apparatus to...
Reduce energy use, costs and consume raw materials
Percentage Composition
Mass of Element / Mass of Compound x100
E.G - 7.5g of C
6
H
12
O
6
. 0.5g of H + 4g of O a) mass of carbon
b) percentage by mass of each component
Percentage Composition by Relative Mass
More accurate
H: 1%
O: 16 / 79.5 x100 = 20.1%
Cu: 63.5 / 79.5 x100 = 79.9%
Mr of O = 16
Mr of Cu = 63.5
E.G - CuOH
2
-----> Cu + H
2
O
a) 0.5 + 4 = 4.5 so 7.5 - 4.5 =3g
b) Carbon - 3 / 7.5 x100 = 40%
Hydrogen - 0.5 / 7.5 x100 = 6.7%
Oxygen - 4 / 7.5 x100 = 53.3%
Gas Volumes
Ideal Gas Law: 1 mole of any gas would occupy the same volume at standard RTP as any other gas. We can use this fact to calculate the number of moles of any gas we produce
Molar volume = 24dm
3
-----> for 1 mole of any gas at SRTP
1dm
3
= 1000cm
3
Amount of gas formed (mol) = vol of gas at RTP / molar volume at RTP
E.G - 48dm
3
of O
2
= V / Vm - 48 / 24 = 2 moles
When all reactants and products are gas, we can simply use molar ratios to ratio the gas amounts
Not so easy if we have a solid
E.G - What mass of sodium azide would be needed to produce 48 dm
3
of nitrogen gas as SRTP
2NAN
3
-----> 2Na + 3N
2
n = V/Vm = 48/24 = 2 moles
As it is a 3:2 ratio, number of moles of NaN
3
= 1.3 moles
mass = Mr x n = 65 x 1.3 = 86.6g
E.G - H
2
+ Cl
2
-----> 2HCl
If I start with 50cm
3
H
2
then it will react with:
50cm
3
of Cl
2
to form
100cm
3
of HCl
