Catalyst
1. To fill up a cup with water, let’s
say you need 5 billion molecules
of water. How many atoms of
hydrogen is that? How many
atoms of oxygen?
2. When you heat up food in the
microwave, how do you know
how long to set it for?
What is balanced in a chemical
equation?
Three things
are balanced
in a chemical
equation:
1. Atoms
2. Mass
3. Charge
Balancing Chemical
Equations
Atoms must be conserved!
 For
now, you will only be concerned
with balancing atoms. Remember, the
number of atoms of each element on
the reactants side must equal the
number of atoms of each element on
the products side!
Counting Atoms
• How many atoms of each element are in the following
molecules?
1.
2.
3.
4.
5.
H2SO4
CaOH2
NaCl
(NH3)3PO4
3H2O
Answers
 1.
2 hydrogen, 1 sulfur, 4 oxygen
 2. 1 calcium, 1 oxygen, 2 hydrogen
 3. 1 sodium, 1 chlorine
 4. 3 nitrogen, 9 hydrogen, 1 phosphorus,
4 oxygen
 5. 6 hydrogen, 3 oxygen
Is this equation balanced?
• NaCl + H2O  NaOH + Cl2
• left side has 1Na, 1Cl, 2H, and 1O
• right side has 1Na, 1O, 1H, and 2Cl
• NO! It is not balanced!
Is this equation balanced?
• HCl + NaOH NaCl+ H2O
• left side has 2H, 1Cl, 1Na, and 1O
• right side has 1Na, 1Cl, 2H, and 1O
• Yes! It is balanced!
Is this equation balanced?
• Ca + H2O Ca(OH)2 + H2
• left side has 1Ca, 2H, 1O
• right side has 1Ca, 2O, 4H
• No! It is not balanced!
• How can we make it balance?
Ca + H2O  Ca(OH)2 + H2

We need more H’s, but we can’t add just the H
atom, we have to add the entire molecule the H is
attached to:
Ca + H2O Ca(OH)2 + H2
H2O
Count the boxes. These are the coefficients:
1 Ca + 2 H2O  1 Ca(OH)2 + 1 H2
1 Ca + 2 H2O  1 Ca(OH)2 + 1 H2

Just like in math, this equation may be written in
a simpler way:

Ca + 2 H2O Ca(OH)2 + H2
The End

Let’s practice!
1. __H3PO4 + __KOH  __K3PO4 + _H2O
2. __K + __B2O3  __K2O + __B
3. __Na + __NaNO3  __Na2O + __N2
4. __Na + __O2  __Na2O
5. __H3PO4 + __Mg(OH)2  __Mg3(PO4)2 + __H2O
6. __NaOH + __H2CO3  __Na2CO3 + __H2O
7. __Al(OH)3 + __H2CO3  __Al2(CO3)3 + __H2O
8. __Rb + __NO3  __Rb2O + __N2
Percent Composition
• If the formula of a compound is known, it is a
fairly straightforward task to determine the
percent composition of each element in the
compound. For example, suppose you want to
calculate the percentage hydrogen and oxygen in
water, H2O. First calculate the molecular mass
of water:
• 1 mol H2O = 2 mol H + 1 mol O
• Substituting the masses involved:
• 1 mol H2O = 2 (1.0079 g/mol) + 16.00 g/mol
= 18.0158 g/mol
• percentage hydrogen = [mass H/mass H2O] ×
100 = [2(1.0079 g/mol)/18.0158 g/mol] × 100
= 11.19% H
• percentage oxygen = [mass O/mass H2O] ×
100= [16.00 g/mol/18.0158 g/mol] × 100
= 88.81% O
• As a good check, add the percentages together.
They should equal to 100% or be very close.
• Determine the mass percent of
each of the elements in C6H12O6
• ANSWER
– Formula mass= 180.158 amu
– %C =(6C atoms)(12.011
amu/atom)/(180.158 amu) x 100% =
40.002%
– %H =(12H atoms)(1.008
amu/atom)/(180.158 amu) x 100% = 6.714%
– %O =(6O atoms)(15.9994
amu/atom)/(180.158 amu) x 100% =
53.2846%
– Total =100 001 . %
Empirical Formula
• The empirical formula tells us what elements
are present in the compound and the simplest
whole-number ratio of elements. The data may
be in terms of percentage, or mass, or even
moles. But the procedure is still the same:
convert each to moles, divide each by the
smallest number, then use an appropriate
multiplier if needed.
Molecular (Actual) Formula
• If the actual molecular mass is known, dividing
the molecular mass by the empirical formula
mass gives an integer (rounded if needed) that is
used to multiply each of the subscripts in the
empirical formula. This gives the molecular
(actual) formula, which tells which elements
are in the compound and the actual number of
each.
Example
• For example, a sample of a gas was analyzed and
found to contain 2.34 g of nitrogen and 5.34 g
of oxygen. The molar mass of the gas was
determined to be about 90 g/mol.
Answer
• ( 2.34g N)x(1mol N/14.0 g N) = 0 .167 N
• (1.67/1.67) = 1 N
• (5.34 g O) (1 mol O/ 16.0 g O) = 0.334 mol O
• (0.334 / 0.167) = 2 O
• Therefore Empirical Formula = NO2
Answer
• The molecular formula may be determined by
dividing the actual molar mass of the compound
by the empirical molar mass. In this case the
empirical molar mass is 46 g/mol.
• Thus (90g/mol / 46g/mol) = 1.96 which, to
one significant figure, is 2. Therefore, the
molecular formula is twice the empirical
formula—N2O4.
PRACTICE PROBLEMS
1.) What is the percent composition of N2O4?
2.) What is the empirical formula of methyl acetate
97.28 g carbon, 16.22 g hydrogen, and 86.40 g
oxygen.
3.) A compound containing barium, carbon, and
oxygen has the following percent composition:
69.58g Ba, 6.09g C, 24.32g O. What is the
empirical formula for this compound?
4.) What is the empirical and molecular formula of
Vitamin D3 if it contains 84.31% C, 11.53% H,
and 4.16% O, with a molar mass of 384 g/mol?
ANSWERS
1.) 30.4% N and 69.6% O
2.) C3H6O2
3.)BaCO4
4.)C27H44O (both)
NOMENCLATURE
• The flowchart gives explicit
instructions on how to name
compounds
• Work backwards on the flowchart to
determine the formula
• Change ternarny to tertiary
For each of the following
questions, determine whether the
compound is ionic or covalent
and name it appropriately.
•
•
•
•
•
1) HClO4
2) P2O5
3) NH3
4) FeSO4
5) SiO2
•
•
•
•
•
6) GaCl3
7) CoBr2
8) B2H4
9) CO
10) HClO
For each of the following questions,
determine whether the compound is
ionic, covalent or acidic and write the
appropriate formula for it.
• 11) dinitrogen
trioxide
• 12) nitrous acid
• 13) hydroiodic acid
• 14) lithium acetate
• 15) phosphorus
trifluoride
• 16) vanadium (V)
oxide
• 17) aluminum
hydroxide
• 18) Arsenous acid
• 19) silicon
tetrafluoride
• 20) silver phosphate
For each of the following
questions, determine whether the
compound is ionic or covalent
and name it appropriately.
•
•
•
•
•
1) BBr3
2) CaSO4
3) HBr
4) Cr(CO3)3
5) Ag3P
•
•
•
•
•
6) H2SO3
7) VO2
8) PbS
9) CH4
10) N2O3
Write the formulas of the
following chemical compounds:
• 11) tetraphosphorus
triselenide
• 12) potassium acetate
• 13) iron (II)
phosphide
• 14) disilicon
hexabromide
• 15) titanium (IV)
nitrate
• 16)
hydrophosphoric
acid
• 17) copper (I)
phosphate
• 18) gallium sulfide
• 19) tetrasulfur
dinitride
• 20) bromic acid