Stoichiometry
Review
Element
A substance which cannot be
split into anything simpler by
chemical means. A substance
in which all atoms have the
same proton number.
Atom
The smallest part of an element
that can take part in a chemical
reaction.
Molecule
The smallest particle of a compound
(combination of two or more elements).
Note: Ionic compounds (e.g. sodium
chloride) do not exist as molecules.
Ion
The name given to any atom or
molecule, that has become charged by
gaining or losing electrons.
• positively charged ions are cations
• negatively charged ions are anions
Review
Formula
A formula represents one molecule of a compound, or the simplest ratio of
the ions present.
The number of atoms or groups of atoms in a formula is given by putting a
small number just below and behind the symbol.
1s aren’t written (you put NaCl not Na1Cl1).
In some formulae brackets are used.
Aluminium sulphate has the formula Al2(SO4)3 to show that there are two
Al’s to every three SO4’s.
Review
Formulae
You need to be able to find the formula of many different compounds.
This involves thinking about the number of covalent bonds an atom makes,
or the charge of an ion.
eg. The formula for sodium oxide.
Sodium has one outer shell electron, which it gives away.
Oxygen has six outer shell electrons, and takes two from another
atom.
Two sodiums are needed for every one oxygen,
So the formula for sodium oxide is
Na2O
Review
Formulae
You need to be able to find the formula of many different compounds.
This involves thinking about the number of covalent bonds an atom makes,
or the charge of an ion.
eg. The formula for aluminium hydroxide.
Aluminium has three outer shell electrons, which it gives away,
giving it a 3+ charge. Al3+
Hydroxide is a compound ion, with a 1- charge. OH—
Three hydroxides are needed for every one aluminium,
So the formula for aluminium hydroxide is
Al(OH)3
Review
Formulae
You need to be able to find the formula of many different compounds.
This involves thinking about the number of covalent bonds an atom makes,
or the charge of an ion.
eg. The formula for when nitrogen and hydrogen bond.
Nitrogen has five outer shell electrons, and needs three more,
so it will form 3 covalent bonds.
Hydrogen has one outer shell electron, and needs one more,
so it will form 1 covalent bond.
Three hydrogens are needed for every one nitrogen,
So the formula is
NH3 (called ammonia)
Review
There are several acids, alkalis, and compound ions you need to know.
Hydrochloric acid: HCl
Hydrogencarbonate: HCO3—
Sulfuric acid: H2SO4
Sulfate: SO42—
Nitric acid: HNO3
Nitrate: NO3—
Sodium hydroxide: NaOH
Carbonate: CO32—
Calcium hydroxide: Ca(OH)2
Phosphate: PO43—
Ammonia: NH3
Hydroxide: OH—
Ammonium: NH4+
Balancing equations
Equations • show the formulae of the reactants and the products.
• show the relationship between the numbers of each substance;
this is known as the STOICHIOMETRY
• can show in which state the substances exist.
BALANCING EQUATIONS
1. Find the reactants and products. Word equations help.
2. Get the correct formula for each species. This can not change.
3. Check to see if it is balanced. An equation balances if the same number
of each type of atom appears on either side of the arrow.
4. Place large numbers in front of any formula to indicate if more than one
of it is required. This multiplies everything in the formula immediately
behind it.
Balancing equations
Step 1 hydrogen
+
oxygen →
water
Step 2 H2(g)
+
O2(g)
H2O(l)
→
Step 3 Count the atoms :
2 H, 2 O
→
2 H, 1 O
The equation doesn’t balance. An extra O is needed on the right hand side
(RHS). The formula must not change. You can only get extra O’s by having
two waters, so we multiply H2O by two.
Step 4 H2(g)
+
O2(g)
→
2 H2O(l)
Now we have too many H’s on the RHS; add another H2.
2 H2(g)
Now it is balanced!
+
O2(g)
→
2 H2O(l)
Balancing equations
Step 1 magnesium
+
oxygen
→
magnesium oxide
Step 2 Mg(s)
+
O2(g)
→
MgO(s)
Step 3 Count the atoms :
1 Mg, 2 O
O
→
1 Mg, 1
The equation doesn’t balance. An extra O is needed on the right hand side.
Step 4 Mg(s)
+
O2(g)
→
2 MgO(s)
Now the O balances, but we have too many Mg’s on the RHS. Add more on
the LHS.
2 Mg(s)
Now it is balanced!
+
O2(g)
→
2 MgO(s)
Balancing equations
Step 1 nitrogen
+
hydrogen
→
ammonia
Step 2 N2(g)
+
H2(g)
→
NH3(g)
Step 3 Count the atoms :
2 N, 2 H
→
1 N, 3 H
The equation doesn’t balance. An extra N is needed on the right hand side,
and extra hydrogen on the LHS. Pick one to start with. I picked N.
Step 4 N2(g)
+
H2(g)
→
2 NH3(g)
Now the N balances, but we have too many H’s on the RHS; add some H2.
N2(g)
Now it is balanced!
+
3 H2(g)
→
2 NH3(g)
Balancing equations
Step 1 sodium +
hydrogen
water
Step 2 Na(s)
+
H2O(l) →
+
H2(g)
Step 3 Count the atoms :
1 Na, 2 H, 1 O
→
→
sodium hydroxide
+
NaOH(aq)
1 Na, 3 H, 1 O
The equation doesn’t balance. An extra H is needed on the left hand side
(LHS). You can only get extra H’s by having two waters, so we multiply H2O
by two.
Step 4 Na(s)
+
+
H2(g)
2 H2O(l) →
NaOH(aq)
Now we have too many O’s and H’s on the LHS; add another NaOH.
Na(s)
+
H2(g)
2 H2O(l) →
2 NaOH(aq)
+
Relative atomic mass
The actual mass of an atom is very very small.
For example, one atom of hydrogen weighs 1.67⨉10-24 grams.
(0.00000000000000000000000167 grams)
It would be horrible to work with that kind of number, so we instead
talk about the relative mass of atoms.
Since carbon is a very important element, and carbon has a nucleon
number of 12, the mass of a carbon atom was decided to be 12
‘grams’.
Relative atomic mass
Relative atomic mass (Ar) is
the weighted average mass
of an atom of an element,
on a scale where an atom of
12C has a mass of 12.
It is the same as the mass
number.
Use your periodic table to find the
name of these elements, then
state their relative atomic masses.
12
6
1
1
C
Carbon. Ar = 12g
H
Hydrogen. Ar = 1g
5
11
53 B
127
I
Boron. Ar = 11g
Iodine. Ar = 127g
Relative molecular mass
Relative molecular mass (Mr)
is the weighted average mass
of a molecule on a scale
where an atom of 12C has a
mass of 12.
eg. The relative molecular
mass of water is found by:
It is the same as adding up
the mass numbers of the
atoms in the molecule.
2⨉1
1⨉16
H2O
2⨉Hydrogens
2
16
Mr = 18g
+ 1⨉Oxygen
+
+
Relative atomic and molecular masses
The atomic mass of some
elements are shown below.
H:1
C : 12
O : 16
Determine the molecular mass
of these molecules:
Methane: CH4
Carbon dioxide
Glucose: C6H12O6
Methane:
C + 4⨉H
12 + 4⨉1
= 16g
Carbon dioxide: CO2
C + 2⨉O
12 + 2⨉16
= 44g
Glucose:
6⨉C + 12⨉H + 6⨉O
6⨉12 + 12⨉1 + 6⨉16
= 180g