Mole OL Notes.
Uploaded by
Uzair siddiquiMole OL Notes.
Uploaded by
Uzair siddiquiBy Kamal Ahmad (03004112681)
LGS, BSS, CBS
Topic: Mole (O level)
Relative atomic mass (Ar)
“It is mass of an element as compared to carbon-12”. e.g. (H=1), (O=16), (Na=23) etc.
Note: Ar of each element is given in the periodic table. The bigger number with symbol of an element is its Ar
value while the smaller number with symbol is proton number or atomic number.
Relative molecular mass (Mr)
“It is mass of a molecule of element/compound as compared to carbon-12”. e.g. (H2O=1+1+16=18),
(CO2=12+16+16=44) etc.
Note: Mr can be calculated by adding Ar values of atoms in a compound
What is difference between empirical and molecular formula of a compound
Look at this table carefully
Compound Molecular formula Empirical formula Greatest divisor (n)
Ethane C2H6 CH3 2
Hydrogen peroxide H2O2 HO 2
Ammonia NH3 NH3 1
Molecular formula
“It tells us actual number of atoms present in one molecule of a compound”
Empirical formula
“It tells us simplest ratio between elements of a compound”
How to convert empirical formula into molecular formula?
For this we have following relation,
Molecular formula = empirical formula × n
So first we have to find “n” by the following formula,
𝑴
n= Empirical mass can be calculated by adding
𝑬𝒎𝒑𝒊𝒓𝒊𝒄𝒂𝒍 𝒎𝒂𝒔𝒔
Ar values of atoms of empirical formula
Note: Mr of compound is always given in question but empirical mass has to be calculated by using empirical
formula.
Finding mole in “empirical formula calculations”
Mole is a quantity which can be calculated mathematically as,
%
Mole = Or Very important formula
1
By Kamal Ahmad (03004112681)
LGS, BSS, CBS
How to calculate empirical formula of a compound?
Remember that to calculate empirical formula we need percentage of elements/mass of elements
Steps of finding empirical formula
Write down the symbols of elements in a table
Write down the mass or percentage of each element
Write down the Ar value of each element from periodic table
Find mole by dividing percentage/mass of each element by its A r
Divide all mole values by the smallest mole value (this will give you lower numbers or subscripts of
elements)
Example: Solved Question
In a compound percentages are C = 37.5%, H = 12.5%, O = 50%. Find its empirical formula.
Element C H O
% or mass 37.5 12.5 50
Ar 12 1 16
Mole 37.5 12.5 50
= 3.125 = 12.5 = 3.125
12 1 16
Subscript (divide by 3.125 12.5 3.125
smallest mole) =1 =4 =1
3.125 3.125 3.125
So empirical formula of the compound is CH4O
Finding percentage of an element in a compound
Formula
% of element = × 100
Note: If there are two atoms then multiply its atomic mass by 2 in numerator and so on.
Example: Solved Question
Find percentage of N in NH3 and NH4NO3.
Solution:
In NH3
%N= × 100
×
2
By Kamal Ahmad (03004112681)
LGS, BSS, CBS
= × 100
= 82.35 %
In NH4NO3
×
%N= × 100
= × 100
= 35 %
So NH3 has greater percentage of N.
Molar Calculations/Stoichiometry
Memorize these formulae
Important formulae for molar calculations
Mole = (for element like Na, Mg, Al, Fe) or (for molecular element and compound like H 2, HCl, H2O, NaCl)
How to solve questions of molar calculations?
We need a balanced chemical equation (if it is not balanced already, balance it)
Any one substance (reactant or product) would be given in “question statement” and then we can find
any other substance (reactant or product) present in balanced chemical equation.
Steps
1. Find moles of given (or known) substance by using a suitable formula (as given above).
2. Find moles of the target substance (which we want to find) by ratio method in which we do
comparison of coefficients (of substances given in balance chemical equation) and use the mole of
given substance. Use “x” for the mole of target substance.
Example:
Mg + 2HCl → MgCl2 + H2 (balanced chemical equation)
If no. of moles of Mg is 5 then calculate no. of moles of HCl by ratio method as given below,
3
By Kamal Ahmad (03004112681)
LGS, BSS, CBS
So moles of HCl are 10
3. If required, convert the moles of target substance into any other quantity (like mass, concentration
or volume) by using suitable formula (as given above)
Example: Solved Question
Limiting & Excessive reactant
The reactant which has been taken in a relatively less amount and therefore consumed earlier in a reaction, is
called limiting reactant. The other reactant which has been taken in a relatively more amount is called
excessive reactant. Some amount of excessive reactant is left unreacted at the end of reaction.
How to find limiting & excessive reactant in a question?
For this, we need a balanced chemical equation
Steps
Find moles of both reactants.
Divide mole of each reactant by its coefficient (given in balanced chemical equation)
The resultant Smaller value is for limiting reactant and larger value is for excessive reactant
4
By Kamal Ahmad (03004112681)
LGS, BSS, CBS
Example: Solved Question
So Fe is in excess or the excessive reactant while Cu+2 is the limiting reactant.
Note: To find moles of product, in such type of question, always use the moles of limiting reactant.
Percentage yield (of product)
Amount of product, formed in a chemical reaction, is called yield.
There are two types of yield i.e theoretical yield and actual yield.
The amount of product, we calculate in the question, is called the “theoretical yield” while the amount of
product, that is given in the question, is called the “Actual yield”.
Remember actual yield is always less than the theoretical yield. In fact, actual yield is the amount of product
that we obtain in lab experiment after carrying out a chemical reaction. So, some of the amount of product is
wasted (not obtained) during separation of the product from mixture i.e crystallization or filtration of the
product.
Formula to find % yield is,
% Yield = × 100 Very important formula
5
By Kamal Ahmad (03004112681)
LGS, BSS, CBS
Example: Solved Question
Percentage purity (of reactant)
% Purity = × 100 Just memorize this formula
What is the concept of Mole?
Mole is collection of 6 × 1023 particles of a substance. “Particle” means atom (atomic element e.g. Fe, Mg, Li)
or molecule (molecular element e.g. H2, O2, Cl2 /covalent compound e.g. H2O, CO2, NH3) or ion (ionic
compound e.g. NaCl, MgO, LiF).
Examples
1 mole of Fe = 6 × 1023 atoms of Fe
1 mole of H2 = 6 × 1023 molecules of H2
1 mole of H2O = 6 × 1023 molecules of H2O
1 mole of NaCl = 6 × 1023 ions of Na+1 and 6 × 1023 ions of Cl-1
6
By Kamal Ahmad (03004112681)
LGS, BSS, CBS
Mole & Energy (for exothermic and endothermic reaction)
Steps
1. Find moles of the given substance
2. Multiply the moles by energy per one mole of the substance
Example: Solved Question
All important formulae of Mole chapter
(After memorizing these formulae, do practice to solve questions of mole )
1- Molecular formula = empirical formula × n
2- n=
3- % of element = × 100
7- % Yield = × 100
8- % Purity = × 100
9- Energy = Mole × Energy per one mole (For exo & endothermic reaction)
10- No. of atoms or molecules = Mole × (6 × 1023) (Here 6 × 1023 is called Avogadro’s number)