Moles and Stoichiometry
Ch. 12
1
The Mole
 OBJECTIVES:
– Describe the mole concept
– Calculate the mass of a mole of
any substance.
– Convert between moles, grams,
liters, and particle number
2
What is a Mole 摩尔?
 We
measure mass in grams. g
 We measure volume in liters. L
 We
3
count pieces in MOLES.
mol
The Mole Concept
A Carbon-12 atom
has a mass of 12amu
C C
C C C
C
C CC CC
C
C C
CC C
C C C C
C C
C
C CC C C C
4
C
= 12amu
A mole of Carbon atoms
has a mass of 12 grams
Gram Atomic Mass
 Equals
the mass of 1 mole of an element
 Written on the Periodic Table
 12.01 grams of C has the same number of
pieces as 1.008 grams of H and 55.85
grams of iron.
 We can write this as
12.01 g C = 1 mole C
 We can count things by weighing them.
5
The Mole Concept
Atomic Mass
Mole of atoms
in grams
A mole of aluminum = 27.0g
A mole of gold
= 197g
A mole of silver
= 108g
A mole of boron
= 10.8g
6
The Mole Concept
So,
1. How many atoms are in a mole?
2. Does a mole of H have the same
number of atoms as a mole of He?
7
If a Helium atom is 4 times
heavier than a Hydrogen atom,
He
H
then a dozen Helium atoms is 4 times
heavier than a dozen Hydrogen atoms.
8
And a mole of Helium atoms is 4 times
heavier than a mole of Hydrogen atoms.
9
So, the number of atoms in a mole
is always the same
The number of atoms in a mole is called,
Avogadro's Number
Avogadro's Number = 6.022 x
10
23
10
Mole – Particle Conversions
The mole concept allows us to count atoms,
molecules, formula units and ions.
1 mole of Al = 6.022 x 1023 atoms of Al
So how many atoms in 3 mol of Al?
11
Mole – Particle Conversions
1 mole of Al = 6.022 x 1023 atoms of Al
So how many atoms in 3 mol of Al?
3 mol Al
6.022 x 1023 atoms of Al
1 mol Al
= 1.8066 x 1023 atoms of Al
12
Conversions
1 mole of Al = 6.022 x 1023 atoms of Al
So how many moles are 3.0 x 1023 atoms of Al?
3.0 x 1023 atoms Al
1 mol Al
6.022 x 1023 atoms of Al
= .5 mol of Al
13
Converting into Moles
How many moles of Na atoms are there
in 46.0g of Na?
46.0g Na
1 mol Na
= 2 mol Na
23.0g Na
2: Write
3:
Look up
Place
the
the
units
atomic
you
need
mass
cancel
on
Step 1:
a grid
and
place
thetogiven
number
on
the
the
bottom,
periodic
andleft
table
multiply
andpart
by
findthe
topgrid
and divide
into
the
upper
hand
ofout
the
howthemany
by
bottom
grams = 1mole
14
Mole-Mass Conversations
• Dimensional Analysis- converting between
physical properties using ratios
• How many grams are in 3.5 mol CO2?
3.5 mol CO2 44.0 g CO2
=154g CO2
1 mol CO2
• How many moles are in 150g in NaBr?
150 g NaBr 1 mol NaBr
= 1.5 mol NaBr
15
102.9 g NaBr
Examples
 How
much would 2.34 moles of carbon
weigh?
28.1 g C
 How
many moles of magnesium is
24.31 g of Mg?
1 mol Mg
 How
Li?
16
many atoms of lithium is 1.00 g of
8.60 x
22
10
atoms Li
What about compounds?
In 1 mole of H2O molecules there are 2
moles of H atoms and 1 mole of O atoms
 To find the mass of one mole of a
compound
– determine the moles of the elements they
have
– Find out how much they would weigh
– add them up

17
Types of Molar Mass
• Gram Molecular Mass (GMM)- the total
number of atoms in a molecular compound
• Cl2 = 2 x 35.4 g = 70.8 g
• CH4= (1 x 12.0g) + (4 x 1.0g) = 16g
• Gram Formula Mass (GFM)- the total
number of atoms in an ionic compound
• NaCl = (1 x 23.0g) + (1 x 35.4g) = 58.4g
• Ca(OH)2= (1 x 40.0g) + (2 x (16.0g+1.0g))= 74g
18
Molar Mass
 Molar
mass is the generic term for the
mass of one mole of any substance (in
grams)
 The same as:
– 1) gram molecular mass
– 2) gram formula mass
– 3) gram atomic mass
19
Examples

Calculate the molar mass of the
following and tell what type it is:
 Na2S
= 78.1 g
 N2O4
= 92.0 g
 Ca(NO3)2 = 164.1 g
 C6H12O6
= 180.0 g
 (NH4)3PO4 = 149.0 g
20
Examples
 How
many moles is 4.56 g of CO2?
0.104 mol CO2
 How
many grams is 9.87 moles of
H2O?
178 g H2O
 How
21
many molecules is 6.8 g of CH4?
0.425 mol CH4
Gases
 Many
of the chemicals we deal with are
gases.
–They are difficult to weigh.
 Need to know how many moles of gas we
have.
 Two things effect the volume of a gas
–Temperature and pressure
 We need to compare them at the same
temperature and pressure.
22
Standard Temperature and
Pressure
and 1 atm pressure （1个大气压
，约为101.325kPa）
 0ºC
0oC = 32oF= 273 K
1 atm= 101.3 kPa = 760 mmHg (torr)
STP 标准状况
 At STP 1 mole of gas occupies 22.4 L
 Called the molar volume 摩尔体积
23  1 mole = 22.4 L of any gas at STP
 abbreviated
Examples
 What
is the volume of 4.59 mole of
CO2 gas at STP?
4.59 mol CO2
22.4 L CO2
1 mol CO2
=103 L CO2
 How
many moles is 5.67 L of O2 at
STP?
5.67 L O2
24
1 mol O2
22.4 L O2
=0.253 mol O2
Ideal Gas Law





Connection between the number of particles (n)
and the pressure (P), volume (V), and
temperature (T) of a gas
PV=nRT
R= 8.31 kPa∙L/ mol∙K
T= ALWAYS in Kelvin, so convert oC to K
-50oC + 273 = 223K
How many moles of helium are contained in a
5.00-L canister at 101 kPa and 30oC?
n= PV/RT
n=
101kPa x 5.00-L
8.31 kPa∙L/mol∙K x 303K
25 n= 0.2 mol
Stoichiometry化学计量法
“Stoichiometry” is Greek for “Measuring
Elements”
It starts with a balanced equation
2H2 + O2 2H2O
 2 moles of hydrogen reacts with
 1 mole of oxygen
 Forming 2 moles of water.
26
Stoichiometry
The coefficients tell us how many moles of
each substance. Not Grams!
For example:
2H2 + O2 2H2O
2 g of H2 + 1 g of O2 = 2 g of H2O
3 g of reactants can’t make only 2 g of products
27
Mass of a Product
The Law of Conservation of Mass applies
Convert the moles to grams and the equation does
work.
2 moles H2
2H2 + O2  2H2O
2.02 g H2
= 4.04 g H2
1 mole H2
1 mole O2 32.00 g O2
= 32.00 g O2
1 mole O2
36.04 g H2+O2
28
Mass of a Product
2H2 + O2  2H2O
2 moles H2O
18.02 g H2O
= 36.04 g H2O
1 mole H2O
2H2 + O2  2H2O
36.04 g H2 + O2 = 36.04 g H2O
29
2Na + Cl2  2NaCl
How many grams of sodium are needed
to react with .071g of chlorine gas?
2Na (s) + Cl2(g) 2NaCl (s)
Mole Ratio
Mass Ratio
2
23 x 2=
46
1
2
35.5 x 2 x 1= 58.5 x 2=
71
117
0.071 g Cl2 1 mol Cl2 2 mol Na 46 g Na
71 g Cl2 1 mol Cl2 2 mol Na
30
= 0.046 g Na
Chemical Yield
The amount of product made in a
chemical reaction.
There are three types:
1.Theoretical yield 理论产量 - the amount
of Product that should be made (from
calculations)
2. Actual yield 实际产量- the amount of
Product formed in the laboratory (always
given)

31
Chemical Yield
3. Percent yield – a percentage/ratio
between the actual yield and the
theoretical yield.
Actual Yield
% Yield = Theoretical Yield
% yield tells us how “efficient” a reaction
is.
 % yield can not be bigger than 100 %.

32
x 100
Example

According to your calculations the
theoretical yield for the production of
NaCl is 13.6 grams. In the laboratory
your actual yield is 11.8 grams of NaCl.
What is the percent yield?
11.8 g NaCl
x 100
13.6 g NaCl
33
Answer = 86.7 %
Calculating Percent Composition质
量分数of a Compound
 Like
all percent problems:
Part
x 100 %
whole
 Find the mass of each component,
 then divide by the total mass.
34
Calculating Percent Composition质
量分数of a Compound
35

Find the mass percent for the elements in
Sodium hydrogen carbonate:

NaHCO3
-molar mass is 84 g/mol
-mass of: Na= 23g
H= 1g
C= 12g
O= 3 x 16= 48g
%C = (12g/84g) x 100= 14.3%
%Na = (23g/84g) x 100= 27.4%
%H = (1g/84g) x 100= 1.2%
%O = (48g/84g) x 100= 57.1%
The Empirical Formula 经验式
 The
36
lowest whole number ratio of
elements in a compound.
 The molecular formula = the actual
ratio of elements in a compound.
 The two can be the same.
 CH2 is an empirical formula
 C2H4 is a molecular formula
 C3H6 is a molecular formula
 H2O is both empirical & molecular
Calculating Empirical
 Just
find the lowest whole number ratio
 C6H12O6
 CH4N
 It is not just the ratio of atoms, it is also
the ratio of moles of atoms.
 In 1 mole of CO2 there is 1 mole of carbon
and 2 moles of oxygen.
 In one molecule of CO2 there is 1 atom of
C and 2 atoms of O.
37
Calculating Empirical
 We
can get a ratio from the percent
composition.
 Assume you have a 100 g.
 The percentages become grams.
 Convert grams to moles.
 Find lowest whole number ratio by
dividing by the smallest mole value
38
Example
 Calculate
39
the empirical formula of a
compound composed of 38.67 % C, 16.22 %
H, and 45.11 %N.
 Assume 100 g so
 38.67 g C x 1mol C
= 3.220 mole C
12.01 g C
 16.22 g H x 1mol H
= 16.09 mole H
1.01 g H
 45.11 g N x 1mol N = 3.219 mole N
14.01 g N
Example
 The
ratio is 3.220 mol C = 1 mol C
3.219 mol N
1 mol N
 The ratio is 16.09 mol H = 5 mol H
3.219 mol N
1 mol N
=
C1H5N1
 Caffeine
40
= CH5N
is 49.48% C, 5.15% H, 28.87%
N and 16.49% O. What is its empirical
formula?
Empirical to molecular
Since the empirical formula is the lowest ratio,
the actual molecule would weigh more.
 By a whole number multiple.
 Divide the actual molar mass by the empirical
formula mass.
 Caffeine has a molar mass of 194 g. what is its
molecular formula?
C4H5ON2 = 97.11g
Ratio= 194g/ 97.11g = 2
2(C4H5ON2) = C8H10O2N4; molecular formula

41
Example
A
compound is known to be
composed of 71.65 % Cl, 24.27% C
and 4.07% H. Its molar mass is
known (from gas density) to be 98.96
g. What is its molecular formula?
42
Final Exam
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43
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What is matter?
Compounds; elements; substances vs. mixtures (Homogeneous
and Heterogeneous)
Atomic symbols; Naming compounds
Periodic table trends: Electronegativity, Atomic Radius, Ion
Size, etc…
Reaction energy; exothermic vs. endothermic; activation energy
Solvents, solutes, and solutions
Balancing Equations
Valence electrons
Structure of an Atom; Molecular structures
Alpha, Beta, and Gamma Decay
Group names: Halogens, transition metals, alkali metals, etc…
Properties of metals, non-metals, metalloids, and semiconductors
Final Exam
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44
Lewis Dot Structures
Types of reactions
Orbitals and sub-orbitals
Ionic vs. Covalent bonding
Temperature and Kinetic motion
Equilibrium and shifting the direction of reactions
Calculate:
– Density
– Wavelength and Frequency
– Atomic Mass
– Half life
– Radioactive decay
– Moles; Mass percent, Percent yield, Mass-Mole
conversions