Honors Chemistry
Chapter 3: Mass relationships in
Chemical Reactions
3.1 Atomic Mass
• Too tiny to mass individual atoms
• Unit needed to compare masses of atoms
• Atomic Mass Unit
• Early amu had several standards
• Now defined as 1/12 the mass of carbon-12
• Labeled u to signify unified amu
• Average Atomic Mass
• Weighted average of all isotopes of element
• Used for most calculations
3.2 Avogadro’s Number
• Amedeo Avogadro
• Looking for a convenient relationship between
grams and number of particles
• Varies by atomic mass
• 1 atomic mass in grams = 6.022 x 1023 atoms
• Avogadro’s number: NA = 6.022 x 1023
• The Mole
• 1 mol = NA particles
• Unit of counting particles, similar to the
unit “dozen”
3.3 Molecular Mass
• Sum of all atoms in a molecule
• E.g., H2O = 2 (1.008) + 1 (16.00) = 18.02 u
• Try this: Find M of NaHCO3
• Mole Conversions
• 1 mol = NA particles
• 1 mol = (molecular mass) g
• mass  moles  particles
M
NA
• Solve using dimensional analysis
3.3 Molecular Mass
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How many grams are in 0.75 mol NH3?
M = 1 (14.01) + 3 (1.008) = 17.03 g/mol
Therefore, 1 mol = 17.03 g
0.75 mol
17.03 g
------------ x ----------- = 13 g
1
1 mol
• Try this…
• How many moles are in 15.0 g N2O?
3.3 Molecular Mass
• How many molecules are in 125 g
CH3COOH?
• M = 2 (12.01) + 4 (1.008) + 2(16.00)
• M = 60.05 u
• 125 g
1 mol
6.022 x 1023 molecule
-------- x ---------- x ------------------------1
60.05 g
1 mol
• = 1.25 x 1024 molecules
3.4 Mass Spectrometer
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Device that separates particles by mass
helped verify the existence of isotopes
Used to identify unknown substances
Not terribly important to us!
3.5 Percent Composition
• Percent by mass of each element in the
compound
• Find % composition of CaCl2
• 1 Ca = 40.08
2 Cl = 70.90
•
110.98 u
• % Ca = 40.08 / 110.98 = 36.11 %
• % Cl = 70.90 / 110.98 = 63.89 %
3.6 Empirical Formula
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Based on laboratory data
Simplest whole number ratio of elements
E.g., glucose is C6H12O6
Empirical formula is CH2O
Subscripts represent ratio by mole, not
mass
3.6 Empirical Formula
• A compound contains 0.900 g Ca and 1.60 g Cl.
Find the empirical formula.
• Ca: 0.900 g
1 mol
---------- x ---------- = 0.0225 mol
1
40.08 g
• Cl: 1.60 g
1 mol
-------- x ---------- = 0.0451 mol
1
35.45 g
• Divide by lowest value to get whole numbers
• 0.0451 / 0.0225 = 2, so formula is CaCl2
3.6 Empirical Formula
• Try this…
• Find the empirical formula for a compound that is
composed of 53.73% Fe and 46.27% S.
• Remember that it is percent by mass, so those
percents can be treated as grams!
• There is a trick at the end of this one.
3.6 Molecular Formula
• Need to know the molecular mass
• Divide molecular mass by empirical mass
to find how much “bigger” the molecular
formula is
• Multiply empirical formula by that number
• Try this…
• A compound is 40.0% C, 6.67% H, 53.3% O.
• Molecular mass = 180.1 u.
• Find molecular formula.
3.7 Chemical Equations
• Chemical reaction
• Process in which a substance is changed into one
or more new substances
• Chemical equation
• Arrangement of chemical symbols to represent
what happens during a chemical reaction
• Example reaction
• Carbon reacts with oxygen to yield carbon dioxide
• C + O2  CO2
3.7 Chemical Equations
• States of substances
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(s) = solid
(l) = liquid
(g) = gas
(aq) = aqueous solution
• HgO (s)  Hg (l) + O2 (g)
• Note that the number of oxygens is not
conserved. We have to fix that!
3.7 Balancing Equations
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Consider: H2 + O2  H2O
We’ve lost an oxygen!
Can we change it to H2 + O2  H2O2 ?
Can we add an O: H2 + O2  H2O + O ?
A better solution: H2 + O2  2 H2O
Now fix the hydrogens: 2 H2 + O2  2 H2O
Two H2 molecules react with one O2
molecule to form two water molecules
3.7 Balancing Equations
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Try these…
NaClO3  NaCl + O2
2 NaClO3  2 NaCl + 3 O2
CO2 + H2O  C6H12O6 + O2
6 CO2 + 6 H2O  C6H12O6 + 6 O2
CaCl2 + Na3PO4  Ca3(PO4)2 + NaCl
3 CaCl2 + 2 Na3PO4 Ca3(PO4)2 + 6 NaCl
3.8 Stoichiometry
• Quantitative study of chemical reactions
• Realize that reaction coefficients are really
moles!
• 15.0 g oxygen react with excess hydrogen. How
much water is produced?
• 2 H2 + O2  2 H2O
• 15.0g O2 1 mol O2 2 mol H2O 18.0 g H2O
------------ x ------------- x -------------- x ------------1
32.0 g O2 1 mol O2
1 mol H2O
• = 16.9 g H2O produced
3.9 Limiting Reagents
• When reactants are not present in the
exact stoichiometric ratio, one reactant will
be used up before the others
• Limiting reagent – reactant used up first
• Excess reagent – reactant that is present
in larger quantity than needed
• Convert both to moles to find out which is
the limiting reagent
3.9 Limiting Reagents
• Try this…
• 5.0 g hydrogen react with 20.0 g nitrogen.
How much ammonia is produced?
• N2 + 3 H2  2 NH3
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Find how many mol NH3 can be made from the H2
Find how many mol NH3 can be made from the N2
Use the lesser amount (limiting reagent)
The other substance is present in excess
3.10 Reaction Yield
• Theoretical yield
• Amount you should have gotten from reaction
• Result of mol calculation
• Actual Yield
• What you actually got from the reaction
• Percent Yield
• Actual / theoretical x 100%