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3.1 Counting by Weighing
• If the average mass of an amount of particles
is taken particles behave as though they were
all identical for the purposes of weighing
(refer to the jelly bean example on p. 77-78).
• Since atoms are small it makes more sense to
count them by mass than by getting out our
micro-tweezers.
3.2 Atomic Masses
• All masses are relative to 12C. This isotope of
carbon was assigned a mass of 12 atomic mass
units.
• The most accurate method of weighing atoms is
with a mass spectrometer.
– Uses magnets to deflect ions.
– Heavier ions are deflected less.
• Recall that the atomic mass on the periodic table
is a weighted average of all known isotopes.
3.2 Atomic Masses
• To find the average atomic mass for a given
element the mass of each isotope is multiplied
by its relative abundance. The products of
each known isotope are then added together
to give the average atomic mass.
• Here’s an example….
3.2 Atomic Masses
There are 2 naturally occurring isotopes of Cu
Cu-63 62.93 amu 69.09 % abundance
Cu-65 64.93 amu 30.91 % abundance
(.6909 x 62.93) + (.3091 x 64.93) = 63.55 amu
3.3 The Mole
• The mole is the number of unbound carbon
atoms found in 12 grams of 12C.
– This value is 6.022 x 1023 units or entities.
• The mole is defined such that a sample of a
natural element with a mass equal to its
atomic mass will contain 1 mole of atoms.
3.4 Molar Mass
• Molar mass is the mass in grams of one mole
of a compound.
• This is found by added up the atomic masses
of all atoms present in a molecule.
• CH4 = 16 g / mol
3.5 Solving Problems
• Identify ending point
• Identify starting point
• Choose an efficient route to get there (riding
the mole train) --- preferably using
dimensional analysis
• Check --- does your answer make sense?
• Check --- are units and sig figs correct?
(see back of mole train guide for more detail)
3.6 % Composition of Cmpds.
• We can either describe a compound by listing
the number of the atoms that make it up (the
formula) or by a list of atomic percentages
(usually from lab data).
• To find the percentage one take the total mass
of an element from a compound and divides it
by the total mass of the compound.
• Here’s an example….
3.6 % Composition of Cmpds
Find the % composition (by mass) of H2O
H: 2 mol x 1 g/mol = 2 g
O: 1 mol x 16 g/mol = 16 g
Molar Mass of H2O = 18 g
H: 2/18 x 100 = 11.1%
O: 16/18 x 100 = 89.9%
3.7 Formula Determination
• Rarely, if ever, do lab results yield the exact
formula for an unknown compounds.
• Instead mass data or ratio data is used to find
the formula for the unknown compound.
• You may be given the masses as percentages
(15% of O, 85% C) or you may be given masses
of the products.
• Here’s an example…about empirical and
molecular formulas.
3.7 Formula Determination
A compound contains 42 g N and 6.0 g H
Molar Mass = 32 g/mol
Find empirical and molecular formula
N: 42 g x 1 mol / 14 g = 3.0 mol
H: 6 g x 1 mol / 1.0 g = 6.0 mol
Empirical formula is NH2
3.0/3.0 = 1
6.0/3.0 = 2
3.7 Formula Determination
Molecular form =
emp form x (actual molar mass / molar mass of emp form)
NH2 x 32/16 = N2H4
3.8 Chemical Equations
• All atoms must be accounted for on both sides
of the reaction arrow (in other words,
BALANCE THE EQUATION).
• This is mainly guess and check, but here are
some tips:
– Balance the biggest compounds first.
– Balance odd coefficients.
– Leave the smallest compounds (O2, H2O) for last.
3.8 Chemical Equations
Predicting products (some hints):
• Metal hydroxide  metal oxide + water
Zn(OH)2  ZnO + H20
• Metal chlorate  metal chloride + oxygen
2 KClO3  2 KCl + 3 O2
• Metal carbonate  metal oxide + carbon
dioxide
K2CO3  K2O + CO2
3.8 Chemical Equations
Predicting products (some hints continued):
• Decomposition of acids with S,N
H2SO4  H2O + SO3
• Alkali metal + water  metal hydroxide +
hydrogen
2 Na + 2 H2O  2 NaOH + H2
• Hydrocarbon + oxygen  carbon dioxide + water
CH4 + 2 O2  CO2 + 2 H20
3.10 Stoichiometric Calculations
• Once a chemical equation is balanced we can
use the masses of the reactants to determine
the masses of the products.
• The number of molecules in a given mass of a
compound is equal to the number of
molecules of a different compound with an
equal mass.
• So we must convert the masses given to
moles.
3.10 Stoich. Calculations
• Using mole ratios we can relate the given
moles of products and reactants.
What mass of water vapor forms when 32 grams of
methane undergoes complete combustion?
CH4 + 2 O2  CO2 + 2 H20
32 g CH4 x (1 mol CH4 / 16 g CH4) x (2 mol H20 / 1 mol CH4) x
(18 g H2O / 1 mol H2O) = 72 g H2O
3.10 Stoich. Calculations
• Steps to finding masses in a chemical reaction:
– Balance the reaction.
– Convert the known mass to moles.
– Determine the mole ratio between what is given
and what is asked for.
– Convert back to grams if necessary.
3.11 Limiting Reagent
• Considering the following Recipe:
6 strips of bacon + 6 leaves of lettuce + 4 tomato slices + 2
slices of toast
3.11 Limiting Reagent
• So given the following ingredients how many
of the GREATEST B.L.T. EVER can be made??
– 20 slices of bacon
– 43 leaves of lettuce
– 45 tomato slices
– 160 slices of toast.
3.11 Limiting Reagent
• Convert the masses of all reactants to moles.
• Compare moles to determine the LR and XS
reactants
• LR gets used up so …
– Use LR to determine amount of products formed
– Use LR to determine amount of XS reactants left
More details on stoichiometry guide
3.11 Limiting Reagent
100 grams of hydrogen gas 720 grams of oxygen gas
react as much as possible to form water vapor.
Write a balanced equation for the reaction.
What is the limiting reactant?
What reactant is in XS? How much of it is left over?
What mass of water vapor is produced