Quantitative Chemistry

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Transcript Quantitative Chemistry

Quantitative Chemistry
Foundation tier - You must be able to:
• Find the relative atomic mass (mass number) of an
element
• Calculate the Relative formula mass of a compound
• Calculate the % mass of an element of a compound
Higher tier - You must be able to:
• Work out the Empirical formula (HT)
• Balance equations (HT)
• Use chemical equations to calculate masses of reactants
and products (using scale factors) (HT)
• Calculate % yield (HT)
• Calculate atom economy (HT)
Atomic Number
Atomic number (smaller number):
= number of protons
= number of electrons
Mass Number
Mass Number (larger number)
= protons + neutrons
(also known as Relative Atomic MASS)
Relative formula mass
Calculating relative formula mass
To calculate this we add together the masses of all
the atoms shown in the formula
(N=14; H=1; Na=23; O=16; Mg=24; Ca=40)
Substance
Formula Formula Mass
Ammonia
NH3
Sodium Oxide
Na2O
Magnesium
Hydroxide
Calcium Nitrate
Mg(OH)2
Ca(NO3)2
Calculating Formula Mass
To calculate this we ADD TOGETHER the MASSES
of ALL the ATOMS shown in the FORMULA
(N=14; H=1; Na=23; O=16; Mg=24; Ca=40)
Substance
Formula Formula Mass
Ammonia
NH3
14 + (3x1)=17
Sodium Oxide
Na2O
(2x23) + 16 =62
Mg(OH)2
24+ 2(16+1)=58
Magnesium
Hydroxide
Calcium Nitrate
Ca(NO3)2
40+ 2(14+(3x16))=164
Calculating % mass of an element in a compound
We can use the relative atomic mass of elements and the
relative formula mass of compounds to help us work out the
percentage of an element in a compound.
a.) Calculate % of Magnesium (Mg) in Magnesium oxide (MgO).
Ar of Mg = 24;
Ar of O = 16;
Mr of MgO = 24 + 16 = 40
% Mg = 24/40 x 100 = 60%
b.) Calculate % of Aluminium (Al) in Aluminium bromide (AlBr3)
Ar of Al = 27
Ar of Br = 80
Mr of AlBr3 = 27 + (80x3) = 27 + 240 = 267
% Al = 27/267 x 100 = 10.1%
Calculating Percentage Mass
Calculate the percentage of Oxygen in the
compounds shown below
% O = (Number of atoms of O) x (Atomic Mass of Z) x 100
Formula Mass of the compound
Formula
Atoms
of O
MgO
1
K2 O
1
NaOH
1
SO2
2
Mass
of O
Formula
Mass
%age Oxygen
Calculating Percentage Mass
Calculate the percentage of Oxygen in the
compounds shown below
% O = (Number of atoms of O) x (Atomic Mass of Z) x 100
Formula Mass of the compound
Formula
Atoms
of O
Mass
of O
Formula
Mass
%age Oxygen
MgO
1
16
24+16=40
16/40 x 100 = 40%
K2 O
1
16
(2x39)+16=
94
16/94 x 100 = 17%
NaOH
1
16
23+16+=
40
16/40 x 100 = 40%
SO2
2
16
32+(2x16)
= 64
32/64 x100 = 50%
Empirical formula
1.) Make a column for each
element
2.) Divide the mass or % by its
relative atomic mass
3.) Simplify the ratio by
dividing all answers by the
smallest
4.) Find the simplest whole
number ratio
5. Write the empirical formula
Calculating Empirical formula
If 9g of aluminium reacts with 35g chlorine, what is the
empirical formula?
(Ar of Al = 27; Ar of Cl = 35)
Divide each mass by its own RAM
Al: 9/27 = 0.333
Cl: 35/35 = 1
• This tells us that one 'mole' of Cl atoms combines with 1/3
mole of Al atoms.
• Al needs to go up to a whole number ie 1 –
• To do this we multiply the 0.333 by 3. Therefore Cl also
needs to be multiplied by 3.
• The ratio of Al:Cl = 1:3
• So the empirical formula is AlCl3
Calculating masses of reactants and products
(Higher tier)
What to do:
1. Write out a balanced equation using the formula
2. Work out relative formula mass (RFM) for each
compound in the equation
3. Below the RFM write out the mass given in the qs
4. Work out the scale factor by dividing/multiplying
mass given by RFM of that compound
5. Do the same thing to the value you are trying to
calculate a mass for – using the scale factor
Next slide shows an example
Calculate masses of reactants and
products using scale factors
Calcium reacts with oxygen like this:
2Ca + O2  2CaO (balanced equation)
Qs: What mass of oxygen will react exactly with 60g
of calcium
(Ar of Ca = 40;
Ar of O = 16)
Ca + O2
(80 + 32)
60 + ?
60/80 = 0.75 (scale factor)
32 x 0.75 = 24g of O
Calculate % yield (Higher tier)
The yield of a chemical reaction describes how much product is made.
% yield compares the amount of product that the reaction really
produces with the maximum amount that it could possibly produce.
% yield = amount of product produced
maximum amount of product possible x 100%.
Qs: Using known masses of A and B it was calculated that a
chemical reaction could produce 2.5g of product C.
When the reaction is carried out only 1.5g of C is produced.
What is the % yield of this reaction?
% yield = amount of product
max amount of product possible x 100%.
% yield = 1.5
2.5 x 100 = 60%
Very few chemical reactions have a
100% yield because:
- the reaction may be reversible - as products
form they react to form reactants again
- some reactants may react to give unexpected
products
- some of the products may be left behind in
the apparatus
- the reactants may not be completely pure
- some chemical reactions produce more than
one product - so it maybe difficult to
separate the product that we want from
the reaction mixture
Calculate atom economy (Higher tier)
Chemical companies use chemical reactions to make products which they sell.
Making as much product as possible means making less waste.
The amount of the starting materials that end up as useful products is
called the atom economy. The aim is to achieve maximum atom economy.
% atom economy = Mr of useful product
Mr of total products
x 100%
Ethanol (C2H5OH) can be converted into ethene (C2H4) which can be used to make
polyethene. Calculate the atom economy: (Ar: C=12;
H=1;
O=16)
C2H5OH
46
 C2H4
= 28
+
H2O
= 18
% atom economy = 28/46 x 100 = 61%
It is important to maximise atom economy to conserve
resources and reduce pollution
Balancing equations (Higher tier)
a.) Ca
+O
b.) N2
+
CaO

H2

NH3
c.) NaOH + Cl2  NaOCl + NaCl + H2O