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Lecture 3
Stoichiometry: Calculations with Chemical Formulas and Equations
Text: Sections E, F, G, and H
3.1 Atomic and Molecular Masses (sometimes called “Weights”)
The Atomic Mass Scale
1 amu = 1.66054x10-24 g
and
1
1.0078 amu
H
1.6735x10-24 g
1 g = 6.02214x1023 amu
Average Atomic Masses
Determine average atomic mass by using masses of various isotopes and their relative
abundances.
carbon is 98.892% 12C and 1.108% 13C
12
C is 12 amu (exactly) and 13C is 13.00335 amu
(0.98892)(12 amu) + (0.01108)(13.00335 amu) = 12.011 amu
The average atomic mass of each element (expressed in amu) is also known as its atomic
weight.
Formula and Molecular Weights
The formula weight of a substance is merely the sum of the atomic weights of each atom in its
chemical formula.
ex: H2SO4 has a formula weight of 98.1 amu
FW
= 2(AW of H) + (AW of S) + 4(AW of O)
= 2(1.0 amu) + 32.1 amu + 4(16.0 amu)
= 98.l amu
If the chemical formula is the molecular formula, then the formula weight is also called the
molecular weight.
ex: glucose, C6H12O6, has a molecular weight of 180.0 amu
MW
= 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu)
= 180.0 amu
Chapter 3 Stoichiometry
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With ionic substances such as NaCl, it is inappropriate to speak of molecules. We will use the
formula weight
FW
= 23.0 amu + 35.5 amu
= 58.5 amu
Percentage Composition from Formulas
Percent composition- the percentage by mass contributed by each element in the substance.
C6H12O6
%C 
612.0 amu 
 100  40.0%
180.0 amu
%H 
121.0 amu
 100  6.7%
180.0 amu
%O 
616.0 amu
 100  53.3%
180.0 amu
3.2 The Mole
A mole is defined as the amount of matter that contains as many objects (atom, molecules, or
whatever objects we are considering) as the number of atoms in exactly 12 g of 12C.
That number is 6.0221421 x 1023.
This number is given a special name: Avagadro's number.
We will use 6.02 x 1023.
Molar Mass
The mass of single atom of an element (in amu) is numerically equal to the mass (in grams) of 1
mol of atoms of that element.
One 12C atom weights 12 amu  1 mol of 12C weighs 12 g.
One 24Mg atom weights 24 amu  1 mol of 24Mg weighs 24 g.
One 197Au atom weights 197 amu 1 mol of 197Au weighs 197 g.
Chapter 3 Stoichiometry
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The mass in grams of 1 mol of a substance is called its molar mass.
The molar mass (in grams) of any substance is always numerically equal to its formula weight
(in amu):
One H2O molecule weighs 18.0 amu1 mol of H2O weighs 18.0g.
One NO3– ion weighs 62.0 amu  1 mol of NO3– weighs 62.0 g
One NaCl unit weighs 58.5 amu  1 mol of NaCl weighs 58.5 g.
Interconverting Masses, Moles, and Numbers of Particles
Conversion of mass to moles and moles to mass are made easy by use of dimensional analysis.
Example: How many moles of trinitrotoluene, C7H5N3O6 (TNT) are there in 214.0 g of TNT?
Solution: The molar mass can be used as a conversion factor for converting grams to moles. The
molar mass of TNT is 227.0 g.
1 mol TNT = 227.0 g TNT.
 1 mol TNT 
 0.9427 mol TNT
Moles TNT  214.0 g TNT  
227.0 g TNT 
Example: What is the mass of 3.125 mol of CO2?
Solution: The molar mass of CO2 is 44.0 g. 1 mol of CO2
44.0 g CO2 
 137.5 g CO2
grams CO2  3.125 mol CO2  
1 mol CO2 
Conversion factors can also be used to find the number of atoms, molecules, or ions.
Example: Find the number of copper atoms in a copper penny, which weighs 3 g.
1 mol Cu6.02  10 23 Cu atoms

Cu atoms  3 g Cu 

63.5 g Cu
1 mol Cu

 3  10 22 Cu atoms
Chapter 3 Stoichiometry
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3.3 Empirical Formulas from Analyses
Analysis gives the amount of each element as a percentage. If we assume the sample to be 100 g,
we can divide these masses (the percentages in grams) by the appropriate atomic weight to
obtain the number of moles of each element in 100 g.
We then divide the larger mole numbers by the smallest mole number to obtain the empirical
formula. The ratios may not be exact due to experimental errors.
Analysis of an aluminum compound gives 15.77% Al, 28.11 %
S, and 56.12% O. Calculate its empirical formula.
Chapter 3 Stoichiometry
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Molecular Formulas from Empirical Formulas
The subscripts in the molecular formula of a substance are always a whole-number multiple of
the corresponding subscripts in the empirical formula.
Example: The empirical formula of ascorbic acid is C3H4O3, giving a empirical formula weight
of 88.0 amu. The experimentally determined molecular weight is 176. amu. The molecule has
twice the mass of the empirical formula. The molecule must contain twice as many atoms as the
empirical formula.
Molecular Formula = n ( Empirical Formula)
? = n (C3H4O3)
Molecular Formula Weight = n ( Empirical Formula Weight)
176. amu = n (88.0 amu)
n = 176. amu / 88.0 amu = 2
? = n (C3H4O3) = 2 (C3H4O3) = C6H8O6
Calculate the molecular formula of quinine, whose molecular weight is 324.41
g/mol. Analysis of quinine gives 74.04 % C, 7.46% H, 9.86 % O, and 8.63% N.
Chapter 3 Stoichiometry
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(G)
3.4 Solutions
Solutions in which water is the dissolving medium are called aqueous solutions.
General Properties of Aqueous Solutions
solution – a homogeneous mixture of two or more substances.
solvent – usually the component that is present in greater quantity
solutes – the other substances in the solution
Concentrations of Solutions
Molarity
Molarity 
moles solute
volume of soln in liters
Expressing the Concentration of an electrolyte
The concentration of an electrolyte in solution can be specified either in terms of the compound
used to make the solution or in terms of the ions that that the solution contains.
Ex: 1.0 M Na2SO4 or 2.0 M Na+ or 1.0 M SO42–.
Interconverting Molarity, Moles, and Volume
Use dimensional analysis
moles = liters  molarity
moles = liters 
moles
liters
Dilution
Solutions of lower concentration can be obtained by adding water to concentrated solutions, a
process called dilution.
Toolbox G.1
Moles of solute before dilutoin = moles of solute after dilution.
Chapter 3 Stoichiometry
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Moles  molarity  volume
Mi  Vi = Mf  Vf
Vi =
M f  Vf
Mi
Chapter 3 Stoichiometry
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(H)
3. Chemical Equations
2 H2
 O2
 2 H2O
reactants
Symbols:
products
+ means "reacts with"
 means "produces"
coefficients - indicate amount of substance
Equations must be balanced. Equal amounts of each element on each side of the equation.
2 H2
 O2
4H
2O
 2 H2O
4H, 2O
When balancing an equation, subscripts should never change, as this changes the chemical
identity.
Changing coefficients only changes amount of a substance, not the identity.
H2O
2 H2 O
O
H
H
O
H
O
H
H
H
H
H2O2
O
O
H
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CH4  O2  CO2  H2O unbalanced
Balance those species that occur in the fewest formulas on each side.
C C
H H
First try:
1
1
4
Double H in H2O
1
1
Double H in H2O
1
1
2
O O
2
3
4 2(2)
2
4 (! 2 from both)
4 2(2)
2(2)
4
CH4  O 2  CO2  H 2O
very simple example!!! Do Self test H.1 !!!

Answers must have integer coefficients.
Chapter 3 Stoichiometry
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