Examples
1. What is the mass in grams of 0.250 mol of
MgCl2?
2. How many moles of NH4Cl are in 76.5 g?
3. How many moles of nitrate ions are
present in 2.5 mol of Al(NO3)3?
4. How many grams of nitrate ions are
present in 10.0 g of Al(NO3)3?
1
Converting from Moles to Numbers of Formula
Units, Molecules, or Atoms
Moles can be converted into numbers of formula units
(for ionic compounds) OR numbers of molecules OR
numbers of atoms.
How many formula units of Mg(OH)2 are in 1.5 moles of the
compound?
1.5 mol Mg(OH)2 x 6.022 x 1023 formula units = 9.0 x 1023 formula units
1 mol Mg(OH)2
How many atoms of O are in 1.5 moles of Mg(OH)2?
1.5 mol Mg(OH)2 x 2 mol O atoms x 6.022 x 1023 atoms = 1.8 x 1024 atoms
1 mol Mg(OH)2
1 mol O atoms
2
Examples
1. How many molecules of HCOOH are
present in 0.0772 mol of the compound?
2. Calculate the mass of 6.626 x 1026 formula
units of NH4Cl.
3
The Map
Number
of
things
Moles
conversion factor:
1 mol = 6.022 x 1023
molecules or
atoms
(Avogadro’s number)
Mass
conversion factor:
Mass in grams of 1 mol
(molar mass)
4
Finding Empirical Formulas
5
Empirical Formulas
• What is the empirical formula of B2H6?
• Answer: BH3
• Relative number of atoms of each
element.
• Can restate that as:
3 moles of H per/to 1 mole of B
6
Calculating Empirical Formulas
One can calculate the empirical formula from the
percent composition.
7
Empirical Formula
• Chemical contains
87.5% N and 12.5% H by mass
• Same as: in 100 g of that compound
87.5 g N and 12.5 g H
Next:
Convert to moles using molar mass:
8
Grams to mole
87.5 g N
1 mol N
=
6.25 mol N
=
12.5 mol H
14.00 g N
1 mol H
12.5 g H
1.00 g H
9
Next:
mol : mol ratio of elements in compound
12.5 mol H
moles of H
=
moles of N
=
2
6.25 mol N
What does that “2” mean?
2 moles hydrogens per 1 mol nitrogen in this compound
NH2
Empirical formula of compound
10
Calculating Empirical Formulas
The compound para-aminobenzoic acid (you may have
seen it listed as PABA on your bottle of sunscreen) is
composed of
carbon (61.31%)
hydrogen (5.14%)
nitrogen (10.21%)
and oxygen (23.33%).
Find the empirical formula of PABA.
11
Calculating Empirical Formulas
Assuming 100.00 g of para-aminobenzoic acid,
C:
H:
N:
O:
1 mol
12.01 g
1 mol
5.14 g x
1.01 g
1 mol
10.21 g x
14.01 g
1 mol
23.33 g x
16.00 g
61.31 g x
= 5.105 mol C
= 5.09 mol H
= 0.7288 mol N
= 1.456 mol O
12
Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number
of moles:
C:
5.105 mol
0.7288 mol
= 7.005  7
H:
5.09 mol
0.7288 mol
= 6.984  7
N:
0.7288 mol
0.7288 mol
= 1.000
O:
1.458 mol
0.7288 mol
= 2.001  2
13
Calculating Empirical Formulas
These are the subscripts for the empirical formula:
C7H7NO2
14
Sample Exercise
Calculating Empirical Formula
Ascorbic acid (vitamin C) contains
40.92% C, 4.58% H, and 54.50% O by mass.
What is the empirical formula of ascorbic acid?
Solve We first assume, for simplicity, that we have exactly
100 g of material (although any mass can be used). In
100 g of ascorbic acid, therefore, we have
40.92 g C, 4.58 g H, and 54.50 g O
15
Second, we calculate the number of moles of each
element:
16
Third, we determine the simplest whole-number ratio of
moles by dividing each number of moles by the smallest
number of moles, 3.406:
The ratio for H is too far from 1 to attribute the difference
to experimental error; in fact, it is quite close to 1 1/3.
This suggests that if we multiply the ratio by 3, we will obtain whole
numbers:
C : H : O = (1 : 1.33 : 1) x3 = 3 : 4 : 3
17
The whole-number mole ratio gives us the
subscripts for the empirical formula of
ascorbic acid:
C3H4O3
18
Ascorbic Acid
• Empirical formula: C3H4O3
• Molecular formula?
• Empirical formula weight=
3x C = 3x 12.01 amu = 36.03 amu
4x H = 4x 1.01 amu = 4.04 amu
3x O = 3x 16.00 amu = 48.00 amu
C3H4O3 = 88.07 amu
19
• Molecular weight of Ascorbic Acid
(by experiment) = 176 amu
Multiplier for subscript:
molecular weight
176
= 2
=
=
empirical formular weight
88
20
Molecular Formula of Ascorbic Acid
• Empirical formula subscripts x 2
• C3H4O3
 C 6H 8O 6
x2
Molecular formula of
ascorbic acid
21
Molecular Formula from
Empirical Formula
• Percent composition  empirical formula
• IF we also know molar mass or molecular weight
of compound  molecular formula
• Remember:
molecular formula B2H6
empirical formula BH3
NOTE: Subscripts in molecular formula are wholenumber multiples of corresponding subscripts in
empirical formula
22
Exercise
A 5.325-g sample of methyl benzoate, a
compound used in the manufacture of perfumes,
contains 3.758 g of carbon, 0.316 g of hydrogen,
and 1.251 g of oxygen.
What is the empirical formula of this substance?
Start with converting to mass percent:
5.325 g compound = 100% mass
So: (3.758 g C / 5.325 g compound)x 100% is
mass percent of C……
23
Solution
Carbon:
(3.758 g C/ 5.325 g compound)x100% = 70.57% C
If you have 100 g compound  70.57 g is C
grams of C  mol of C using molar mass
70.57 g C x (1mol C/12.01 g C) = 5.876 mol C
Hydrogen: 5.886 mol
Oxygen : 1.468 mol
Divide by smallest mol
C4H4O1
24
Limiting Reactants
25
How Many Cookies Can I Make?
• You can make cookies
until you run out of one
of the ingredients.
• Once this family runs
out of sugar, they will
stop making cookies
(at least any cookies
you would want to eat).
26
How Many Cookies Can I Make?
• In this example the
sugar would be the
limiting reactant,
because it will limit the
amount of cookies you
can make.
27
Limiting Reactants
Hydrogen and oxygen react to form water:
H2(g) + O2(g)
H2O (g)
Balanced equation:
2 H2(g) + O2(g)  2 H2O (g)
28
Complete Reaction
• For every 2 mol hydrogen we need 1 mol
oxygen, or for every 1 mol oxygen we
need 2 mol hydrogen.
• So: if we have 10 mol H2…..
1 mol O2
= 5 mol O2
• we will need 10 mol H2
2 mol H2
29
Limiting Reactants
• The limiting reactant is the reactant present in
the smallest stoichiometric amount.
– In other words, it’s the reactant you’ll run out of first (in
this case, the H2).
In the example above, the H2 would be the limiting
reagent, O2 would be the excess reagent.
30
Reactions
• Many reactions are carried out using
excess of one of reactants.
• Use balanced chemical equation to
determine which is limiting reaction.
• If amount of reactants are given,
determine amount of product(s).
31
The most important commercial process for converting N2
from the air into nitrogen-containing compounds is based
on the reaction of N2 and H2 to form ammonia (NH3):
N2(g) + 3 H2(g)→2 NH3(g)
How many moles of NH3 can be formed from
3.0 mol of N2 and 6.0 mol of H2?
32
Limiting Reactant?
Mol:Mol ratio…..3 mol N2 require ? mol of
H2
Only 6 mol H2 available, H2 is limiting
reactant
33
Maximum product from 6.0 mol H2
• Need mol-to-mol ratio for H2 and NH3
• Can go further:
How many grams of NH3 can be made?
Need molar mass of NH3.
4.0 mol NH3
x
17.04 g NH3
= 68 g NH3
1 mol NH3
34
Theoretical Yield
• The theoretical yield is the maximum
amount of product that can be made.
– In other words it’s the amount of product
possible as calculated through the
stoichiometry problem.
• This is different from the actual yield,
which is the amount one actually produces
and measures.
35
Percent Yield
One finds the percent yield by comparing
the amount actually obtained (actual yield)
to the amount it was possible to make
(theoretical yield).
Percent Yield =
Actual Yield
x 100%
Theoretical Yield
36
Percent yield
• From previous example;
6.0 mol H2 yield 4.0 mol = 68 g NH3
This is the theoretical (calculated) yield.
Actual yield may be less….
say for example: 63 g NH3
Percent yield =
63 g
x 100% = 93%
68 g
37