CHEM 107-001
Summer 2017
Dr. Eric C. Booth
Moles, molar mass,
stoichiometry part 1
What is Known About Atoms?
 Atoms—and
only atoms—make up
all substances
 Atoms are extremely small
 Result: in any amount of material,
there will be an enormous number
of atoms
 Need a number that’s large enough
to describe how many atoms are in
typical amounts of matter
Avogadro’s Number
 …is
about 6.022 x 1023
 INCREDIBLY
huge quantity
 a gumball machine that contained
6.022 x 1023 ordinary gumballs would
be larger than the moon
 Defined
as number of carbon
atoms in 12 grams of 12C isotope
 …1 amu = 1 g / mol
Moles (not the facial or
burrowing kinds)
1
mole of something = Avogadro’s
number of that thing
 1 mole of atoms of an element =
about 6.022 x 1023 atoms of that
element
 1 mole of molecules of a
substance = about 6.022 x 1023
molecules of that substance
 “Mole” is abbreviated “mol”
Mass of Atoms Revisited
 The
atomic mass of each element
is directly related to its molar mass
 One mole of atoms of any element
will always have a mass, in grams,
that is equal to the number written
below that element on the periodic
table
Counting Atoms, Using Mass
 Since
we know the molar mass
of each element, finding the
mass for a sample of an
element will also give the
number of atoms in that sample
 Dividing by the molar mass
gives number of moles
 Multiplying by Avogadro’s
number gives number of atoms
Counting Atoms, Using Mass:
Example
 About
how many atoms are in
17.0 grams of:
 carbon?
 iodine?
 mercury?
Mass of a Molecule, Revisited
 If
you have a mole of molecules of
some specific chemical, what is
the mass of that sample?
 If we know the formula of the
compound, we know:
 what
elements the compound has
 how many atoms of each element
the compound has
Mass of a Molecule, Revisited
 We
also know what the molar
mass of each element is
 From all this data, we can
determine the molar mass of the
compound
Moles of Elements in a
Compound
 The
subscripts in a molecule’s
formula tell us how many
moles of each element will be
contained in one mole of that
molecule
Formula Mass: Example
 What
is the molar mass of:
 N2O2?
 CH4?
 BF3?
Using Molar Mass
 Having
molar mass of a substance
lets you make conversion ratios, to
go from mass quantities to number
quantities (and vice versa)
 Example: How many moles of H2O
are in 374.0 g of water?
Putting it all Together
 Always
“recipes” for making things
 Example: assembling an album
takes 3 things:
a
CD
 liner notes
 jewel case
 The
formula?
 1CD
+ 1 case +
1 set of liner notes
 1 album
Equations for Chemical
Reactions
 An
equation for a chemical
reaction, then, is just the recipe
for creating a new substance from
other, older ones
 The reactants are the chemical
ingredients that go in to process
 The products are the chemicals
we get out of the reaction
Parts of The Equation
States
Subscripts & Coefficients
 Two
numbers in an equation
determine # of any element’s atoms
 subscripts
 coefficients
 Subscripts
(the ‘little’ numbers)
 molar
proportions of elements inside
a molecule
 Coefficients
 number
reaction
(the ‘big’ numbers)
of molecules involved in
Moles & Masses in Reactions
 As
noted before, we use molar
masses to go from ‘regular’ mass
(measurable, but not useable) to
numbers of moles (useable, but
not measureable)
 Since mole proportions of
chemicals in a reaction are always
definite, you can find masses of
products from masses of
reactants (and vice versa)
1) Convert Masses to Moles
 In
the combustion reaction
CH4 (g) + 2O2 (g)CO2 (g) + 2H2O (l),
we have 32.086 g of CH4, and 128.0
g of O2. What mass of H2O is
made?


1 mol
  2.0000 mol CH4
32.086 g CH4 
 16.043 g CH4 
 1 mol 
  4.000 mol O2
128.0 g O2 
 32.00 g O2 
2) Use Mole-Mole Ratio to Get
Moles of New Substance
 Since
the reaction is
CH4 (g) + 2O2 (g)CO2 (g) + 2H2O (l),
this means there will be:
 two
moles of water created for every
one mole of methane consumed
 two moles of water created for every
two moles of oxygen consumed
2) Use Mole-Mole Ratio to Get
Moles New Substance (cont’d)
 If
the water/methane ratio is chosen
(for example), the result is
 2 mol H2O 

2.0000 mol CH4 
 1 mol CH4 
 4.0000 mol H2O
3) Convert Moles to Masses to
Get Mass of New Substance
 Using
the molar mass of water, we
calculate that
 18.016 g H2O 
4.0000 mol H2O 

1 mol


 72.064 g H2O