Antoine Laurent Lavoisier
(1743 – 1794)
A Father of Modern Chemistry
Lavoisier’s Lab
Lavoisier:
Discovered oxygen
Did combustion experiments
Showed matter is conserved
The law of conservation of matter
Matter cannot be created or destroyed under normal
circumstances.
The numbers and kinds of reactant ATOMS must always
equal the numbers and kinds of product ATOMS.
Atoms can be rearranged into new substances, but they
never change their INDIVIDUAL identity.
Balancing chemical equations conserves matter.
Balanced Chemical Equations
Chemical symbols are chemists’ letters.
Symbols make formulas, which are words,
Formulas make equations, which are sentences.
Balanced chemical equations are complete paragraphs.
Solving stoichiometric problems are chemical stories.
The Parts of a Chemical Equation
Reactants
Product(s)
2 C2H6 + 7 O2 → 6 H2O + 4 CO2
Subscript
Coefficient
Separator
Balancing Chemical Reactions
Two basic methods of balancing chemical reactions:
Inspection/trial and error method
Algebraic method
I will show both methods; you may use either.
Steps to Balancing Chemical Equations
A. Identify the reactants and products and, using chemical symbols, write the
basic chemical equation.
B. Do an atom inventory. (This means both reactants AND products.)
C. Check to see if all atoms are in balance: reactant to product.
D. If out of balance, pick one atom at a time to balance. Change the number
of atoms by changing only the coefficient.
E. Every time you change a coefficient, re-do the atom inventory.
F. Repeat steps B – F until all the atoms have the same numbers on each
side of the equation.
Steps to Balancing Chemical Equations
A. Identify the reactants and products and, using chemical
symbols, write the basic chemical equation.
Cesium, when added to nitrogen gas, produces cesium nitride
Cs + N2 → Cs3N
Steps to Balancing Chemical Equations
B. Do an atom inventory.
Cs
+
N2
→
Cs3 N
Cs
1
Cs
3
N
2
N
1
Steps to Balancing Chemical Equations
C. Check to see if all atoms are in balance: reactant to
product.
Cs
+
N2
→
Cs3 N
Cs
1
Cs 3
Out of Balance
N
2
N
Out of Balance
1
Steps to Balancing Chemical Equations
D. If out of balance, pick one atom at a time to balance.
Change the number of atoms by changing only the
coefficient. Let’s balance Cs !
3 Cs
+
N2
→
Cs3 N
Cs
1
Cs 3
N
2
N
1
Steps to Balancing Chemical Equations
E. Every time you change a coefficient, re-do the atom
inventory.
3 Cs
+
N2
→
Cs3 N
Cs
1 3
Cs 3
N
2
N
1
Steps to Balancing Chemical Equations
F. Repeat steps B – F until all the atoms have the same
numbers on each side of the equation.
Check for balance!
3 Cs
+
N2
Nitrogen is out of balance.
→ 2 Cs3 N
Cs
1 3
Cs 3
N
2
N
1
Change the
coefficient to
balance the
nitrogen.
Steps to Balancing Chemical Equations
Change a coefficient – so do a new inventory.
Check for balance.
3 Cs
+
N2
N is good – Cs is now out of balance.
→ 2 Cs3 N
Cs
1 3
Cs 3
6
N
2
N
2
1
Steps to Balancing Chemical Equations
Continue the process. The number of Cs atoms changed,
so the reactant side needs to be re-adjusted.
6
N2
→ 2 Cs3 N
Cs
1 3 6
Cs 3
6
N
2
N
2
3 Cs
+
1
Steps to Balancing Chemical Equations
Change a coefficient, do a new inventory and check for
balance.
6
N2
→ 2 Cs3 N
Cs
1 3 6
Cs 3
6
N
2
N
2
3 Cs
+
1
Steps to Balancing Chemical Equations
It’s done! The numbers and kinds of atoms in the reactants
equals the numbers and kinds of atoms in the products.
NOTE: When balanced, all coefficients must be in LOWEST
whole number ratio!
6
N2
→ 2 Cs3 N
Cs
1 3 6
Cs 3
6
N
2
N
2
3 Cs
+
1
Balance the following equations:
1. Na + FeCl2 → NaCl + Fe
2 Na + FeCl2 → 2 NaCl + Fe
2.
PbS + O2 → SO2 +
PbO
2 PbS + 3 O2 → 2 SO2 + 2 PbO
1. Balance compounds first; leave “single” elements
for last.
2. Leave elements in “multiple” substances for last.
3. Treat polyatomic ions as “single” units.
1. Balance compounds first; leave “single” elements
for last.
Al + Fe3N2 → AlN + Fe
2. Leave elements in “multiple” substances for last.
H2O + CO2 → C7H8 + O2
3. Treat polyatomic ions as “single” units.
(NH4)3 PO4 +
Pb(NO3)4 →
Pb3 (PO4)4 +
NH4 NO3
Consider the reaction:
C4H10 + O2 → CO2 + H2O
This equation could be balanced by inspection / trial and error:
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
To check for balance:
C - Reactant:
C - Product:
Coefficient (2) x subscript (4) = 8 C atoms
Coefficient (8) x subscript (1) = 8 C atoms
H- Reactant:
H- Product:
Coefficient (2) x subscript (10) = 20 H atoms
Coefficient (10) x subscript (2) = 20 H atoms
O- Reactant:
O- Product:
Coefficient (13) x subscript (2)
= 26 O atoms
(Coef: (8) x sub (2) + Coef: (10) x sub (1)) = 26 O atoms
Let’s apply this balancing check to find another way to determine
coefficients - the algebraic approach.
To balance a reaction algebraically:
1. Start by putting unknown coefficient
variables in front of each molecular
species in the equation:
x C4H10 + y O2 → z CO2 +w H2O
x C4H10 + y O2 → z CO2 + w H2O
2. Write down the balance conditions for each element
in terms of the unknowns. (Coefficients x subscripts)
In this case, there are four unknowns: x, y, z and w.
3. For carbon to balance, we have the condition:
x4 = z1
For hydrogen to balance condition is:
x 10 = w 2
For oxygen to balance condition gives:
y2 =z2 + w1
(We only need three conditions for three elements.)
4x = z
10 x = 2 w
2y=2z + w
• Have created a set of only three equations for four unknowns.
▪ The fourth condition (unknown) recognizes that chemical equations specify
relative amounts of reactants and products.
▪ One coefficient is arbitrary: all the others can be expressed relative to it.
▪ Process: Take one of the coefficients to be equal to 1, arbitrarily.
▪ Determine values of other three unknowns from the three conditions.
x C4H10 + y O2 → z CO2 + w H2O
x C4H10 + y O2 → z CO2 + w H2O
If we let x = 1, then it is easy to see that
4x = z
4x1=z,
ergo,
z = 4
10 x = 2 w
10 x 1 = 2 w,
ergo,
w = 5
2y=2z + w
y = 6.5
Which gives us the balanced equation:
C4H10 + 6.5 O2 → 4 CO2 + 5 H2O
C4H10 + 6.5 O2 → 4 CO2 + 5 H2O
Of course, we want coefficients to be in the lowest whole number
ratio, so our balanced chemical equation becomes:
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
Your turn!
Balance algebraically:
w FeS + x O2 → y Fe2 O3 + z S O2
Fe:
S:
w = 2y
w=z
O:
2x=3y+2z
Your turn!
Balance algebraically:
w FeS + x O2 → y Fe2 O3 + z S O2
w=2y
w=z
2x=3y+2z
If we let y = 1
Then w = 2
z=2
and x = 3.5
W=2
X = 3.5
Y=1
Z=2
Your turn!
And:
w FeS +
x O2 → y Fe2 O3 + z S O2
2 FeS + 3.5 O2 →
Fe2 O3 + 2 S O2
But we need the lowest WHOLE number ratio:
4 FeS +
7 O2 → 2 Fe2 O3 + 4 S O2
Is it now balanced?