Download ATOMS, MOLES AND STOICHIOMETRY

1
Book 2
- Amount of Substance
...................The MOLE!!!
2
Valency
This is a number assigned to each element that indicates the number of
electrons that it has lost/gained in order to become stable. Valencies are
used to write chemical formulae and to determine the ions formed
Sodium tends to lose
electron where possible. We know this because
sodium is in group
so has
electron in its outer shell.
We would say that sodium has a VALENCY of 1
Valencies of elements in groups
Group
1
2
3
4
5
6
7
Number of
outer shell
electrons
1
2
3
4
5
6
7
Valency
Ion
formed
Name of ion
1
2
3
4
3
2
1
+1
+2
+3
-3
-2
-1
Sodium
Magnesium
Aluminium
Nitride
Sulphide
chloride
Group 8 elements are called the noble gases.
They do not react
with other elements as they have
a full octet
Valencies of elements not in groups (transition metals)
Metal
Copper
Zinc
Iron
Manganese
Chromium
Silver
Valency
1 or 2
2
2 or 3
2, 4 or 7
3 or 6
1
Ion formed/ name
Cu+, copper (I), Cu2+, copper (II)
Zn2+, Zinc
Fe2+, Iron (II), Fe3+, Iron (III)
Mn2+, Manganese (II)
Cr3+, Chromium (III)
Ag+, Silver
Learn these!!
Complex ions contain more than one element, you will probably be familiar
with some of these. You must learn the formula, charge & name of the ion.
3
Name
formula
Valency
Carbonate
Hydrogencarbonate
Sulphate
Sulphite
Hydrogensulphate
Nitrate
Hydroxide
Dichromate
Manganate
Ammonium
Phosphate
CO32HCO3SO42SO32HSO4NO3OHCr2O72MnO4NH4+
PO43-
2
1
2
2
1
1
1
2
1
1
3
complete the following table
Name
iodide
Suphate
sulphide
aluminium
dichromate
Formula
of ion
ISO42S2Al3+
Cr2O72-
Valency
1
2
2
3
2
Name
Formula
of ion
magnesium
Mg2+
ammonium
NH4+
nitride
N3hydroxide
OHIron (II)
Fe3+
Valency
2
1
3
1
3
When we know what ion an element forms we can write equations to show
the process. These are called half ionic equations
e.g. magnesium
Mg 
Mg2+ + 2ewrite ½ eqns to show the following forming ions
a) Lithium
Li

b) Chlorine
Cl2
+
c) Sulphur
S
+
Li+
2e2e-
e-
+
2Cl-


S2-
Reading and writing chemical formulae
4
Naming compounds
Rules
1. metal name goes first and doesn’t change
2. non-metal elements become ‘ides’ to show that they are negative
ions e.g. chloride
3. ate means oxygen is present e.g. sulphate
4. ite also means oxygen present but LESS than ate
5. when non-metals form positive ions they are called ‘ium’ e.g.
ammonium
6. if a metal has more than one valency it has a Roman numeral in the
name of the compound to show which ion is present.
e.g. Cu2O = copper (I) oxide
Reading
Formula
K2O
CaSO4
Mg(NO3)2
CuSO4 .5H2O
Na2CO3.10H2O
Name of compound
Potassium oxide
Calcium sulphate
Magnesium nitrate
Copper (II) sulphate
Sodium carbonate
Composition
2K and 1 O
1 Ca, 1 S and 4 O
1 Mg, 2N and 6 O
1 Cu, 1S 9O and 10 H
2 Na, 1C, 13 O and 20 H
Writing
There are two methods, one involves using valencies, the other - ions.
Example: magnesium chloride
Valency method
ion method
Mg
Cl
2
1
swap = MgCl2
Mg2+
Cl-
to be neutral need 2 Cltherefore
= MgCl2
5
Complete the table
Name
Strontium fluoride
formula
Sr F
2
1
Na SO4
1
2
Si Cl
4
1
Al CO3
3
2
NH4 PO4
1
3
Cu
O
2
2
Sodium sulphate
Silicon chloride
Aluminium carbonate
Ammonium phosphate
Copper (II) oxide
Formula
CsF
CaCO3
FeI3
CuSO4
NaHCO3
FeO
= SrF2
= Na2SO4
= SiCl4
= Al2(CO3)3
= (NH4)3PO4
= CuO
name
Caesium fluoride
Calcium carbonate
Iron (III) iodide
Copper (II) sulphate
Sodium hydrogen carbonate
Iron (II) oxide
WRITING EQUATIONS
Chemical equations must be BALANCED and often include state symbols
(g)
(l)
(s)
(aq)
Gas e.g. H2, N2, O2, CO2, Cl2, F2 and group 8
Liquid H2O, Hg, Br2
Solid all metals, C , S, P, I2, Si
Aqueous (dissolved in water), all acids and alkalis, most salts,
Write and balance the following equations and add state symbols:Magnesium + hydrochloric acid  magnesium chloride + hydrogen
Mg(s)
+
nitrogen
N2(g)
+
2HCl(aq)
+
H2(g)

hydrogen
MgCl2(aq)

 NH3(g)
+
H2(g)
ammonia (NH3)
sodium hydroxide + sulphuric acid  sodium sulphate
NaOH(aq)
+
H2SO4(aq)

Na2SO4(aq)
+ water
H2O(l)
6
THE MOLE!
An amount of substance is
the quantity whose unit is the mole
the amount of any substance containing as many
particles as there are carbon atoms in exactly 12g of
carbon-12
A mole is
Symbol = n
A mole is an actual amount, it is 6.02 x 1023. (Compare: a dozen = 12, a
score = 20, a pair = 2) This is called the Avogadro constant.
the number of atoms per
The Avagadro constant, NA, is
mole of the carbon-12 isotope.
So one mole of magnesium contains 6.02 x10 23 magnesium atoms.
Two moles of magnesium would contain
1.2. x 1024
Mg atoms
BEWARE! Because oxygen exists as O2 if we have one mole of O2 we
would have 2 x 6.02 x 1023 = 1.2 x 1024 atoms
Questions:
No. of atoms =
How many atoms are there in
a) 3 moles of Na
moles of
x 6.02 x 1023
3 x 6.02 x 1023 = 1.81 x 1024
atoms
b) 0.25 moles of Al
0.25 x 6.02 x 1023 = 1.51 x 1023
c) 2 moles of H2
2 x 2 x 6.02 x 1023 = 2.41 x 1024
Relative molecular mass, Mr is
the average mass of a molecule
compared to carbon-12
This is calculated using relative atomic masses. It is used for simple
molecular (covalent) molecules
e.g. CO2
Relative formula mass is
also calculated using Ar
7
This is used for giant molecular (covalent) molecules.
e.g. diamond
Calculating moles:
The relationship between moles, mass and Mr is used continuously in
chemical calculations
in grams
Moles =
(n)
mass
Mr
Calculated from the periodic table
Calculate the Mr of
a)
H2SO4
(2 x 1) + 32.1 + (4 x 16) = 98.1
b)
Mg(OH)2
24.3 + {(16 + 1) x 2] = 58.3
Calculate the mass of
a) 6 moles of rubidium
6 x 85.5 = 513g
mass =
moles x molar mass
b) 0.25 moles of CO2
0.25 x 44 = 11g
c) 0.01 moles of H2SO4
0.01 x 98.1 = 0.981g
how many moles are present in
a) 1.2g of Mg
n = 1.2/24.3 = 0.0494
b) 30g of C
n = 30/12 = 2.5
c) 6.7g of PCl5
Mr PCl5 = 208.5
n = 6.7/208.5 = 0.0321
Molar mass
Molar mass is the mass of one mole of a substance, it is the same as Mr
except that Mr does NOT have units, molar mass has units of g mol-1
(grams per mole)
Molar mass is the mass per mole of a substance
Units are g mol-1
(g/mol)
8
To calculate the molar mass we use the relative atomic masses of the
atoms from the periodic table.
Using moles in calculations
Steps:
1. calculate moles of the reactant
2. use the balanced eqn to deduce moles of product that should
be made (theory)
3. calculate the mass of product that should be made (theory)
Zinc reacts with hydrochloric acid producing zinc chloride and hydrogen
gas.
Write a balanced symbol equation, taking care with the formula of zinc
chloride.
Zn
+
2HCl 
ZnCl2
+
H2
The equation tells us that:
 one mole of zinc reacts with 2 moles of HCl
 mole ratio is 1 : 2 : 1 : 1
so 1 mole of zinc makes 1 mole of ZnCl2
If 6.5g of zinc are used, how many moles of HCl are needed to react with
n of Zn =
it?
As ratio is 1 : 2
Zn : HCl
6.5
65.4
= 0.0994 ( to 3 s.f.)
n of HCl = 2 x 0.0994 = 0.199
How many moles of zinc chloride should be made?
As ratio
Zn : ZnCl2
1 : 1
n of ZnCl2 = 0.0994
What mass of zinc chloride should we make (in theory) in this reaction?
Mass = n x Mr =
0.0994 x [65.4 + (2 x 35.5)] = 13.6g
If only 7.3g of zinc chloride is ACTUALLY made, what is the percentage
yield?
% yield = 7.3 x 100 = 53.7%
13.6
9
Percentage yield calculations
given in question
% Yield = actual mass
theoretical mass
x 100
the one you calculate
Questions
1. when 20g of calcium carbonate are thermally decomposed, only 4g
of calcium oxide are actually made. Calculate % yield.
CaCO3  CaO + CO2
Step 1
n of CaCO3 =
step 2
ratio is 1 : 1 so moles of CaO
step 3
theoretical mass CaO
step 4
20
= 0.200 (3s.f.)
100.1
=
=
= 0.200
0.2 x 56.1
11.2g
% yield = 4
x 100
11.2
= 35.7%
2. when 30g of chlorine react directly with excess iron, 25g of the
product iron (III) chloride is actually made. Write a balanced
symbol eqn and calc % yield.
2Fe + 3Cl2 
2FeCl3
n of Cl2 = 30 = 0.423 mol
70
n of FeCl3 = 0.423 x 2/3
theoretical mass =
% yield =
25
45.8
0.282
x 100
x
=
=
0.282 mol
162.3
54.9%
=
45.8g
10
MOLES AND GASES
The ideal gas equation
Volume
Pressure
Moles
Temp
PV = nRT
1000
cm3
R = Universal gas constant 8.31jk-1mol-1
dm3
11
When using the ideal gas equation we must use certain units
of volume, pressure and temperature we call these SI units
Temperature: Kelvin
C
o
(+273)
k
Pressure: Pa (Pascals) 100kpa = 1000,00pa
Kpa
x103
pa
Volume: m3
cm3
x10-3
dm3
x10-3
m3
1. Rearrange the ideal gas equation to find Volume
V = nRT
P
2. Rearrange the ideal gas equation to find Pressure
P = nRT
V
12
3. Rearrange the ideal gas equation to find Moles
n = PV
RT
4. Rearrange the ideal gas equation to find Temperature
T = PV
nR
13
Question:
Potassium Nitrate decomposes on strong heating forming
Oxygen and Potassium Nitrite as the only products
a) A 1.00g sample of KNO3 (Mr 101.1) was heated
strongly until fully decomposed. Calculate the number
of moles of KNO3 in the sample
b) At 25OC and 100kpa the oxygen gas produced in the
decomposition occupied a volume of 12.6cm3. State
the ideal gas equation and use it to calculate the
number of moles of O2 produced.
(R = 8.31J k-1mol-1)
1. n KNO3 = 1.00 / 101.1 = 9.89X10-3
2. Convert units (k, Pa and m3)
25OC + 273 = 298k
12.6cm3 x10-6 = 1.26x10-5m3
100kpa x103 = 100,000pa
4.Apply the ideal gas equation (rearranged)
n = PV
RT
100,000pa X 1.26x10-5m3 = 5.088x10-4
8.31JK-1mol-1 X 298k
C) Calculate the % yield of Oxygen in the reaction
KNO3KNO2 + ½ O2
% yield = Actual moles / Theoretical moles x 100
5.088x10-4 /
100 = 10.3%
4.945x10-3 (9.89x10-3 / 2 due to 1:0.5 ratio) x
14
MOLES AND CONCENTRATION
The concentration of a solution is the amount of solute, in mol,
dissolved in 1 dm3 (1000cm3) of solution.
The units are mol dm-3
( mol/dm3)
We can calculate the concentration of any solution if we know
 number of moles (n)
 volume in dm3
concentration
moles
C =
n
v
volume in dm3
[ n = c x v ]
Acids are said to be concentrated or dilute.
one in which a large amount of acid is
in each dm3 of the solution.
A concentrated acid is
one in which a small amount of acid is in
each dm of the solution.
A dilute acid is
3
Calculate the concentration of the following:
a) 9.4g of NaCl dissolved into 500cm3 of water.
9.4 = 0.161 mol
58.5
b) 10g of KOH dissolved into 2dm3 water
n = 10 = 0.178
56.1

calc moles

Calculate conc.
C = n/v
c = n = 0.161 = 0.322 mol n = mass
v
0.500
Mr
c = 0.178 = 0.0890 mol dm
2
What mass of solid must be added to make the following?
a) 200cm3 of a 0.1 mol dm-3 solution of NaOH
n = 0.1 x 200/1000 = 0.0200 mol
mass = 0.0200 x 40 = 0.8g
b) 750cm3 of a 2 mol dm-3 solution of Na2CO3.10H2O
-3

n = c

Mass
Calc moles
x v/1000
Calc mass
= n x Mr
15
n = 2 x 750 / 1000 = 1.5 mol
[Mr for Na2 CO3.10H2O
= (23 x 2) + 12 + (16 x 3) + (10 x 18) = 286 ]
mass = 1.5 x 286 = 429g
TITRATIONS
There are several types of titration; we will only focus on ACID-BASE
titration’s.
Titration’s are used to find the concentration of an acid or alkali (base)
An indicator is used to show the ‘end-point’ which is when the acid has
reacted with all of the alkali
Acid name
Formula
Hydrochloric acid
HCl
Sulphuric acid
H2SO4
Nitric acid
HNO3
Note: all acids and alkalis are given the state symbol (aq). The salts made
when these react are also usually (aq).
Example
25cm3 of 0.05mol dm-3 NaOH was added to a conical flask. This required
exactly 17.5cm3 of HCl to react with it. Calculate the concentration of
the HCl used.
Steps
1. write a balanced symbol eqn
2. calculate moles of the KNOWN (i.e. the thing we know conc.
and vol for)
3. use the stoichiometry of the eqn to DEDUCE the number of
moles of UNKNOWN used
4. calc the conc. of the unknown
calculation:
1.
NaOH + HCl

NaCl + H2O
2.
n of NaOH = 0.05 x 25/1000 = 1.25 x 10-3 ( 0.00125)
3.
as ratio 1 : 1 n of HCl = 1.25 x 10-3
4.
conc of HCl = 1.25 x 10-3/ 0.0175 = 0.0714 mol dm-3
16
Questions:
1.
25cm3 of 0.2moldm-3 NaOH reacts with 11.2cm3 of H2SO4.
Calculate the conc. of the sulphuric acid. (work out the formula of the
salt carefully)
2.
A student has 50cm3of 0.01 moldm-3 KOH, she wishes to exactly
neutralise this with 0.1 moldm-3 HNO3. What volume of the acid will she
need?
Answer on the back of a page.
1.
2NaOH
+
25 cm3
0.2 mol dm-3
H2SO4

11.2 cm3
?
n of NaOH = 0.2 x
n of H2SO4
25 =
1000
KOH
+
50 cm3
0.01 mol dm-3
+
2H2O
5.0 x 10-3 ( 0.00500)
ratio 2 : 1 = 0.00500 =
2
conc H2SO4 = 0.00250
0.0112
2.
Na2SO4
0.00250
= 0.223 mol dm-3
HNO3

KNO3
?
0.100 mol dm-3
+
H2O
n of KOH = 0.01 x 50
= 5.0 x 10 –4 (0.000500)
1000
n of HNO3 = 5.0 x 10-4 due to 1.1 ratio
vol = 0.00050 =
0.1
5 x 10-3 dm-3 (0.00500)
or 5cm3
17
Calculating formulae:
Copper has a valency of 1 or 2. Use the following data to calculate the
correct formula for an oxide made when copper is heated in air.
4.3g of copper was heated in air and 4.85g of solid oxide was formed.
Steps:
1.
2.
3.
subtract the mass of the metal from the mass of the solid
oxide formed to find mass of oxygen.
calculate the number of moles of each element.
divide the number of moles of each element by the smallest
value.
Mass of oxygen 4.85 – 4.3 = 0.55g
Cu
: O
4.3 : 0.55
63.5 : 16
0.0677 : 0.0344
1.97 : 1
( Ar)
( smallest)
formula = Cu2O
Iron has two possible valencies. If 7.21g of iron was heated in air the
solid made weighed 10.29g. Calculate the formula of the iron oxide.
Mass of oxygen = 10.29 – 7.21 = 3.08g
Fe
7.21
55.8
:
O
3.08
16
0.129 : 0.193
1 :
1.5
2
:
3
formula = Fe2O3
18
Calculating Stoichiometric relationships.
Sodium hydroxide, NaOH, reacts with an acid to form a salt and water. It
is known that the acid is either sulphuric acid or hydrochloric acid. Using
the information below, calculate the stoichiometric relationship between
the acid and NaOH and therefore identify the acid used.
Concentration/mol dm
Volume/cm3
Moles of NaOH = 0.01
Moles of acid =
-3
NaOH
0.01
25
Acid
0.005
25
x 25 / 1000
0.005 x 25 / 1000
=
=
2.50 x 10-4
1.25 x 10-4
Ratio of NaOH :
acid is
2.5 x 10-4 : 1.25 x 10-4
2
:
1
so reacting ration is 2 NaOH :1 acid
acid must be H2SO4
2NaOH + H2SO4
 Na2SO4
+ H2O
Potassium hydroxide reacts with either nitric acid or phosphoric acid,
H3PO4. Use the data in the table below, calculate the stoichiometric
relationship between the acid and KOH and therefore identify the acid
used.
KOH
Acid
-3
Concentration/mol dm
0.2
0.16
3
Volume/cm
20
25
Moles of KOH = 0.2
Moles of acid =
x 20 / 1000
=
4.0 x 10-3
0.16 x 25 / 1000
=
4.0 x 10-3
Ratio of KOH :
acid is
-3
4.0 x 10
: 4.0 x 10-3
1
:
1
so reacting ration is 1 KOH :1 acid must be HNO3
KOH + HNO3
 KNO3
+ H 2O
19
EMPIRICAL AND MOLECULAR FORMULAE
Empirical formula is the simplest whole number ratio of atoms
of each element in a compound.
Molecular formula is the actual number of atoms of each
element in a compound.
Determining the Empirical and Molecular formula.
A compound contains 6.75g of Al and 26.63g of Cl. What is its empirical
formula? The molar mass of the compound is 267gmol-1 , calculate the
molecular formula of this compound.
Steps
1. find moles of each element (from mass or from % of element)
2. divide by the smallest number
3. use the whole number ratio to deduce Empirical formula
4. use molar mass/Mr to calc molecular formula
Al
:
6.75 :
27 :
Cl
26.63
35.5
0.25 :
0.25 :
0.750
0.25
1
:
3
 smallest
Empirical formula = AlCl3 mass = 133.5
Mr = 267 so Ef x 2 = Mf
= Al2Cl6
Questions (answer over the page)
1. an organic molecule is found to contain 90.6% C and 9.4% H, calculate
the empirical formulae. The Mr of the molecule is 106, use this to deduce
the molecular formula.
2. A compound contains only carbon, hydrogen and oxygen. Its composition
by mass is 40%C and 6.7% H. calculate the Empirical formula. The molar
mass is 180 gmol-1. Use this data to calculate Molecular formula.
Answers to questions.
20
ANSWERS
1
C
90.6
12
:
:
:
H
9.4
1
7.55
1
:
:
4
:
9.4
1.25 (can’t round up, x 4 to
get whole number)
5
empirical formula = C4H5
Er mass = 53
Mr 106 106 / 53 = 2
2.
Mf = emp. Formula x 2
=C8H10
C
40
12
:
:
:
H
6.7
1
:
:
:
O
53.3
16
3.33
1
:
:
6.7
2
:
:
3.33
1
empirical formula = CH2O
Er mass = 30
Mr = 180
Mf = emp. Formula x 6
=
C6H12O6
21
Compounds containing water
Some compounds can exist in both the anhydrous and hydrated forms.
For example copper (II) sulphate is a white powder when anhydrous and
forms blue crystals when hydrated.
The difference between the two forms is due to the presence of water
of crystallisation.
Water of crystallisation: water molecules which make up
the part of the crystal structure of a compound, that
can be removed by heating.
CuSO4
+
5H2O

CuSO4. 5 H2O
Anhydrous: the compound which remains when the water
is removed from a hydrated salt.
Means: being without water.
Hydrated: a complex molecule which has water molecules
incorporated
e.g. CuSO4 is anhydrous
CuSO4. 5H2O is hydrated
as it contains 5 waters of crystallisation.
22
Questions involving water of crystallisation
1. Calculate the percentage mass of water of crystallisation in copper
(II) sulphate-5-water [CuSO4. 5 H2O]
Mass of CuSO4 . 5H2O = 63.5 + 32.1 + (4 x 16) + 10 + ( 5 x 16)
= 249.5
mass of water in molecule = 5 x 18 = 90
% mass of H2O =
90
249.5
x 100 = 36%
2. Weighed samples of the following crystals were heated to drive
off the water of crystallisation. When they reached constant mass,
the following masses were recorded. Deduce the empirical formulae
of the hydrates
a) 0.942g of MgSO4. a H2O gave 0.461g of residue
mass of water removed= 0.942 – 0.461 =
0.481g
MgSO4
:
H 2O
0.461
:
0.481
120.4
18
0.00383
0.0267
1
7
formula = MgSO4 . 7 H2O
b) 1.124g of CaSO4. b H2O gave 0.889g of residue
mass of water removed = 1.124 – 0.889 = 0.235g
CaSO4
:
H 2O
0.889
:
0.235
136.1
18
0.00653
0.0130
1
2
formula = CaSO4 . 2H2O
c) 1.203g of Hg(NO3)2. c H2O gave 1.172g of residue
mass of water removed = 1.203 – 1.172 = 0.031g
Hg(NO3)2
:
H 2O
1.172
:
0.031
325
18
0.00360
0.00172
2
1
1
½
formula = Hg(NO3)2 . ½ H2O
23
ATOM ECONOMY
atom economy =
mass of desired product___ x 100
sum of masses of all products (or reactants)
Make sure equation is balanced or you will get the wrong answer!!!!
Mass of products = Mass of reactants as long as the equation balances.
e.g. ethene reacts with steam, the desired product is ethanol.
A single product is made in an addition reaction, giving 100% atom
economy.
e.g.
2-iodobutane undergoes a hydrolysis reaction with aqueous
sodium hydroxide; the desired product is butan-2-ol. Write a balanced
equation for this reaction and calculate the atom economy.
Step 1: calculate the Mass of the desired product
Step 2: divide this by the sum of the Mass of all the products
CH3CH2CHICH3 + NaOH  CH3CH2CHOHCH3 + NaI
Mr = 74
Mr = 149.9
Atom economy =
74
x 100
(74 + 149.9)
=
33.1%
24

The majority of the starting materials have turned in to waste.

Even if the reaction proceeded with a 100% yield, a large
proportion of the mass of atoms used would be wasted.
A high atom economy benefits society as it reduces the amount of
waste produced, and it reduces the cost to companies. Reactions are
more sustainable.
“Waste products” are the by-products formed during a reaction. Their
disposal can be costly. However, some by-products can be sold on or used
elsewhere in the chemical plant.
NB. Percentage yield and atom economy are not the same thing!!
Atom economy tells you the proportion of reactants that is
converted into useful products.
Percentage yield tells you the efficiency of this i.e of
converting reactants into products
Do not confuse them!!
25