[MUSIC]
Lighthouse Scientific
Education presents a lecture
in the Mole Series.
The topic; Molecular
Weight and Conversions.
Material in this lecture
relies on an understanding
of the previous lectures;
The Mole as a Quantity
Atomic (Molar Mass)
Mass and Molecular Weight
and Dimensional Analysis.
The lecture begins with a
brief review of the definitions
and principles of molecular
weight, formula weight
and molar mass.
Then it is on to conversions
between moles and mass and
basic rules for
these conversions.
These rules are based on
generating unit factors from
the definition of molecular
and formula weights.
Several examples are shown.
The last topic is a
more complicated one.
It is the conversion
between number of particles
and the mass of
those particles.
These are multiple step
conversions that can be done
as a single equation or
using dimensional analysis.
4 examples are given.
To definitions. Molar Mass
is the mass of one mole of
an element or compound.
The units are grams per mole.
In the previous lecture
it was shown that the
mole is just a number
(a counting unit):
It is a very large number
because atomic scale
particles are very small.
It is called Avogadro's number
and is usually written
in scientific notation.
Often just the first 3
numbers of the coefficient,
6.02, are used.
Atomic mass is the
molar mass of an element.
It is the weight of a
mole of an element and it
encompasses the different
isotopes found in that element.
Molecular weight, abbreviated
Mr or MW is the sum of the
atomic masses of a molecule
or a polyatomic ion.
It is very similar to
molar mass and the two terms
are often interchanged.
Molecular weight is the
more commonly used term.
Formula weight is the sum
of the atomic masses from
a formula (molecular
weight is a formula weight).
Formula weight is a
bit broader because
it includes ionic compounds.
In dealing with molecules
any of the three terms,
molar mass, molecular
weight or formula weight
can be considered correct.
Use the one that your
instructor favors.
We'll be favoring
molecular weight.
The first type of conversion
covered in this lecture
is the conversion between 'the
mass' and 'the moles'.
In that molecular weight
is a conversion factor.
That also goes for molar
mass and formula weight.
Consider the molecular weight
of H2O: 18.02 g per mole.
More explicitly, it can
be written 18.02 g of H2O
per 1 mole of H20.
That is to say that 18.02
g of H2O is 1 mole H2O.
They are equal and as such
when they are written in a ratio
like molecular weight
they are a unit factor.
Unit factors are covered in
depth in the 'Dimensional
Analysis' lecture and
they are reviewed in the
'Mole as a
Quantity' lecture.
The term unit means
1 and the ratio
of 18.02 g of H2O
over 1 mole of H20
is equal to 1. Conversely,
1 mole of H20 over
18.02 g of H2O is also 1.
The two ratios are equal
and can be used to convert
between moles and
grams much in the way
as the unit factor
12 inches over 1 foot
can be used to convert a value
in feet into an
equivalent value in inches.
Unit factors work
because the starting units
cancel out and leave a new unit.
The math gives the
numerical relationship between
the values in the two units.
Since 22.2 feet is equal to
266 inches the unit factor
must be equal to 1.
There are a couple additional
points that should be covered
concerning molecular weight
as a conversion factor.
Most importantly is the
use of the molecular weight
in the correct orientation.
This can be better understood
by stepping back and
evaluating a conversion from
the perspective of units.
A conversion will
have a starting
value with units.
We will call those units the
'current unit' since
the conversion
has yet happened.
What those units
will be converted into
is called the
'desired units'.
The current and desired
unit arrangement can also be
termed the 'before units'
and the 'after units':
before and after
the conversion,
or the 'have units'
and 'want units'.
Each pair means the same thing.
We will use the current
and desired units.
To get a correct conversion
the unit factor will need to
have its current units
orientated in such a manner
that it cancels out
with the units of the
given or current value.
In this set up that means
the unit factor has the
current units in
the denominator.
Canceling out units is
the single best way to
prevent error in conversions.
If units do not cancel
then try flipping the
unit factor over and
see if that works.
We can use the unit factor
from the molecular weight of
water in an example of
converting between mass
and mole using the
correct orientation of
the molecular weight.
Say that the current
unit is grams of water and
the conversion is
into moles of water.
The correct orientation
of the unit factor
(the molecular weight)
would have mass or grams
in the denominator so that
grams of water unit cancel out
leaving the unit
moles of water.
What if instead of grams
for the current unit we had
moles of water and
wanted to convert that
into grams of water.
The correct orientation of
the unit factor would have
a mole of water in
the denominator so that
moles would cancel out
leaving grams of water.
Keep track of the units and
conversions can become an
important and useful tool
in the study of chemistry.
An outline of the basic process
of conversions, including those
using molecular weights, can
be helpful in the beginning.
The first thing that needs
to be done is to identify
the units involved
in the conversion.
For this discussion we
will leave the conversion to
those between mass and moles.
These steps work regardless
of which of the unit types is
given as the current units.
Then identify the
relationship between the units.
For the conversion between
mass and moles that is
the molecular or formula
weight of the compound.
Now, set-up an equation with
current units on the left hand
side of the equal sign and
desired units on the right.
That provides a basis for
making the correct choice for
the orientation
of the unit factor.
Use the molecular weight
to make a unit factor
such that current
unit will cancel out.
What's left is to verify
the correct orientation of the
unit factor by canceling
out units and then solve the
numbers part of the equation
with some calculator work.
This is a straight forward
process that will work
every time if
implemented correctly.
Let's put it to work
with a couple of examples.
Beginning with 'How
many moles of H2O are
there in 55 g H2O'?
The process begins by
identifying what units
are in the conversion.
A review of the problem shows
that we are given a mass of H2O
and asked to convert
that value into moles.
The conversion
is grams to moles.
What is the relationship
between mass of water
and moles of water?
It is the molecular weight.
Earlier we saw that it
was 18.02 grams per mole.
Begin the conversion with
an equation that has the
current units, grams of water,
on the left of the equal
sign and the desired units
moles of water on the
right of the equal sign.
Using the relationship of
molecular weight as a guide,
orient the unit factor
such that the current units,
grams, cancel out.
That will occur when
grams are in the denominator.
This is the flip of how
the molecular weight
is usually written.
Question; is it always the
case that current unit will be
in the bottom of
the unit factor?
No, not always. If the
current units are themselves
in the denominator, such
as time units in velocity
(think miles per hour), then
the correct orientation of
the unit factor would have
current units in the numerator
of the unit factor.
What drives the placement
is the canceling of
the current units.
Here the orientation is
correct because in the math step
we cancel out current
units of grams leaving the
desired units of moles.
Taking the numbers to a
calculator: 55 divided by
18.02 gives the conversion
of 3.052 moles of H2O.
There is one more
consideration though.
Our value in the
current unit only has 2 digits
or 2 significant figures.
Since this is a multiplication
problem the rule says that
our answer can
only have 2 digits
or 2 significant figures.
The second significant figure
in the answer is a zero.
That digit place is as far
as this answer can be given.
That doesn't mean we should
just lop of the 5 and the 2.
We need to consider whether
the 0 needs to be rounded up.
In this case the 0 does
need to be rounded up to a 1
and our final answer
is 3.1 moles of H2O.
For a second example we
are given 2.55 moles of CO3
with a minus 2 charge
and asked to find how
many grams of carbonate.
Identify the type
of conversion.
Moles of carbonate
have been given and
grams of carbonate
have been asked for.
This one is a conversions
of moles into grams.
The relationship between
moles and grams is
the molecular weight.
The molecular
weight of carbonate
is 60.01 grams per mole.
The process of getting
that molecular weight is
not shown here so that the
focus remains on
the current lesson.
The next step is the
set up an equation
that has the current
units or, given units,
on the left hand side
and the desired or,
wanted units, on
the right hand side.
Success in correctly laying
out the equation makes
choosing the proper
orientation of the unit factor
straight forward.
Look at the equation
and then look at the molecular
weight of the carbonate.
That molecular weight,
as a conversion factor,
needs to be oriented so
that the given unit of
moles cancel out.
As currently written
the molecular weight has
moles in the denominator.
This allows the canceling
out of the mole unit
leaving the desired gram units
A bit of multiplication
and a valued of 153. 0 grams
of carbonate is had.
2.55 moles of carbonate is
153.0 g of carbonate.
Are we done?
Perhaps, but a check of
significant figures is in order.
Our starting value has 3
significant figures meaning
our final value cannot have
more than 3 significant figures
The 3 is the third significant
figure in our answer so
we need to consider whether
we need to round that digit.
Being followed by a 0
we should leave the 3 as
it is and our final answer
is 153 grams of carbonate.
A final example of a
conversion with molecular weight
starts with 97 g of the
sugar C6 H12 O6 and asks for
the number of
moles of C6 H12 O6.
The conversion is one
going from grams to moles.
The relationship, as in
the other examples, is the
molecular weight of the sugar.
If the student is still
developing their skill in
getting molecular weights
then perhaps they should pause
the lecture, get out their
Periodic Table, and
give this molecule a try.
(PAUSE)
Onto the equation
for the conversion.
It has the starting value
of 97 g of the sugar as the
current units on the left
hand side of the equation.
Moles of C6 H12 O6 are on
the right of the equal sign.
Using the relationship
of the molecular weight
as a guide insert a unit factor
such that the current units
of grams will cancel out.
This orientation is another
example of writing the
flip of the molecular
weight as the unit factor.
In the math step the student
should always verify that
the current unit cancels
out before getting out
the calculator and doing
the division problem.
97 divided by
180.16 gives a value
of 0.584 moles of the sugar.
A check of significant
figures and we are done.
The starting value of 97 grams
has 2 significant figures
so the converted value
cannot have more than that.
The 8 in the mole value
is the second sig. fig.
It is followed by a 4 so
we will not be rounding
the 8 to a 9. 97
grams of the sugar
C6 H12 O6 is 0.58
moles of the sugar.
As with many of the process
involved in the study chemistry
it is absolutely to the
students benefit to develop
a strategy and to rigorously
apply it in all circumstances.
That will be especially true of
the last topic in this lecture;
conversion between
particles and mass
because there is no
direct conversion between
number of particles
and grams of particles.
While that is not exactly true,
the process described here
is significantly easier.
In this type of conversion we
will see that the mole plays
the intermediate between
the very large numbers
associated with atomic scale
particles and the mass
of these particles.
To be more specific we are
considering the conversion
between a count, or
number, of particles
that goes through
two conversions to
get to mass, as in
grams, of those particles.
One of the conversion
was covered in the
'Mole as a Quantity'
lecture and the other was a topic
of this lecture.
That is, the first conversion
goes from number of particles
into moles of particles.
In the 'The Mole as a Quantity'
lecture, going from number
of particles to moles of
particles is done using
a unit factor from the
definition of moles;
1 mole of particle equal 6.022
times 10 to the 23rd particles.
That's
Avogadro's number.
The second conversion takes
the newly found units of moles
and, with the appropriate unit
factor, converts it into mass.
That unit factor is the one we
just covered in this lecture.
It is the molecular
weight of the particle.
So this overall conversion
uses moles twice.
Each of the unit factors
will have moles in them.
Moles are useful in handling
quantities in chemistry.
They may initially
seem burdensome.
An analogy would be having
to do all math with numbers
in terms of dozens.
That would mean that
each individual
number would have to be
initially be
converted into dozens,
Some math would be
done to the dozens
and then the answer would
be converted back out of
dozens into regular numbers.
Burdensome.
So while working in
moles has this same labor,
once mastered the
mole show itself as an
indispensable tool.
That might just have
to be taken on faith.
From here on in we are in
the world of moles and this
common refrain we be
used again and again;
Get into moles...get
out of moles.
To be complete we should ask
if two conversion are necessary
when starting with grams
of particles and asking for
count of particles?
Yes they are and they will
also go through
moles of particles.
In this situation the first
conversion, mass to moles,
will use a molecular
weight unit factor and
the second conversion will
use the unit factor with
Avogadro's number.
These conversion are
really just the reverse of
the conversions that started
with number of particles.
The unit factors are the
flips or reciprocals.
That same refrain
of getting into moles
and then getting
out of moles is used.
The two conversion can be
done separately or combined
into a multiple step
dimensional analysis equation.
We will cover 4 examples and
do so using both ways with
the first 3. The last
example will be done with just
dimensional analysis.
Our first conversion asks us
to find how many H2O molecules
are there in 25 grams of H20.
The overall conversion is going
from mass, grams of particles,
to numbers of particles with
molecules being the particles.
Since we do not have a
direct conversion between
these units we need
more than 1 conversion.
We will need 2
conversions to be exact.
Fortunately, we have our
refrain to help us along.
The first conversion will be
to get into moles,
and the second
will be to get out of moles and
into number of molecules.
These conversions can be
done one after the other,
use the results of the first
conversion as the start of
the second or with 1
multiple step equation.
We will first do the problem
as individual conversions.
Starting with the first
conversion and using the
same steps there were used
earlier in the lecture;
identify the type
of conversion.
It is mass, 25 grams,
converted into
moles of particles, molecules.
The relationship between
mass and moles of particles
is the molecular weight of H2O.
1 mole of H2O is
18.02 grams of H2O.
Now set up an equation
that relates current
units to desired units.
The current unit is grams of
water. Given are 25 grams of
water which is equal to some
amount of moles of water.
With this set up determine
the correct unit factor.
Success in this step is
found by arranging the
molecular weight so that
the unit grams cancel out.
This will be the case if
grams are in the denominator.
We see that this unit factor
is the reciprocal or the flip
of the molecular weight
from the relationship step.
It is not as simple as
inputting the relationship.
The correct arrangement
of units in the denominator
and in the numerator
also has to be considered.
Moving on to the math step,
and canceling out grams of
water, leaving moles of water
and just a bit of arithmetic.
25 divided by 18.02, yields the
answer of 1.39 moles of water.
25 grams of water is
1.39 moles of water.
Since we are going to use
the 1.39 value in another
calculation we should hold off
checking significant figures.
That is done just once
and always at the end.
Before the start the
second calculation, a minor
modification to the
original question can be made
that prepares us for
the next calculation.
Without altering the value
in the problem, the 25 grams
can be replaced with an
equal value of 1.39 moles
Now the second calculation,
getting out of moles,
looks like a conversion
we covered before.
First identify of the
type of conversion at hand.
It is those 1.39
moles into particles.
Is there a known relationship
between moles and
number of particles?
From the introductory
lecture on moles we see
that it is the definition
of a mole. 1 mole equals
Avogadro's number and as
written here, is a unit factor.
Writing out the equation has
us adding the current units
on the left; 1.39 moles.
And the desired units?
Right, molecules of H2O
on the right hand side.
The correct orientation of the
unit factor will have the moles
in the denominator so
that they cancel out.
It is already in
that orientation.
Be sure to include the
water along with the numbers
and the units in
the unit factor.
The current units, by the way,
do cancel out in the math step.
The multiplication is 1.39
times 6.02 times 10 to the 23rd
will gives a value of
8.37 times 10 to the
23rd molecules of water.
That is how many water
molecules are in 1.39 moles
or 25 grams of water.
What's left is
a check of sig. figs.
From the original problem
it can be seen that the
given value of 25
only has 2 sig. figs.
That means that the answer
cannot have more than
2 sig. figs. That 3
is in the last position
of the final answer.
Does that 7 that follows the 3
mean that the 3 is moved up to
a 4 or should it stays as a 3?
Since the 7 is greater than
or equal to a 5 the 3 is
rounded to a 4 and
the final is answer
or the final final answer is
8.4 times 10 to the
23rd molecules of water.
This problem was solved in
two separate conversions.
It can also be solved by
combining those two separate
conversions into a single
equation in a process
called dimensional analysis.
This process is covered in
depth in its own lecture in
the Measurement
and Numbers series.
It is essentially the same
set of steps used in the
two conversions, which
are gathered in the lower
rectangle on the screen.
There is one additional
understanding though
and we will get to that.
The dimensional analysis
equations begins by identifying
the overall conversion.
The given value, or the current
value, is the 'grams of H20'.
That will be converted
into moles of H2O
which will then
be converted into
desired units of
molecules of H2O.
Dimension analysis has us
solve individual conversions,
in order, one at a time,
moving left to right.
The first conversion is the
'grams to moles' conversion
and the logic and the look is
exactly like the first of the
single conversions
already solved below.
In that conversion the unit
factor of molecular weight
is added to the equation.
A slight change in syntax is
also shown. Instead of
a multiplication sign
followed by the unit factor
a set of parenthesis with the
unit factor inside is used.
It is the same thing since
a value directly in front
of parentheses is multiplied
by what is in the
parenthesis. As such,
the unit grams of water
cancel out just like
they did in the first of the
two individual conversations.
Doing the division in the
conversion would of course
give the same 1.39
moles of water.
However, in a dimensional
analysis equation,
doing the math is held
off until the entire set of
conversions is complete.
So far these are
identical equations.
The additional understanding
in the dimensional analysis
is that the equation we
have so far is also equal to
the starting value of
the second conversion.
They are all 1.39 moles
of water. This means we can
leave the dimensional
equation as is but treat it
as if it was the start
of the second equation.
Which means it is time to
get out of moles and that is
done by adding in the same
unit factor used in the
second conversion.
It is inserted inline,
next to the first unit factor.
Yes, parentheses are added.
Note that the moles of water
cancel out, got out of moles,
leaving the units of
molecules of water.
The equation is complete with
the exception of the math.
And the math is more complex
with dimensional analysis.
A calculator hint with this
math would be to do all the
multiplication first (25 times
6.02 times 10 to the 23rd)
and then do all the
division (divide by 18.02).
Multiplying or dividing by 1
does not change the value so
the 1's can be
omitted in the calculation.
The results should be
8.4 times 10 to
the 23rd molecules.
Both methods produce
the same result
and either way can be used.
Be forewarned though that
when the difficult subject of
stoichiometry comes up having
practiced dimensional analysis
with the getting
into and out of moles
will greatly reduce the
initial confusion of that topic.
One last point concerning
dimensional analysis.
Here we used parenthesizes
to hold the unit factors.
Some teachers do not
use parenthesis to teach
dimensional analysis.
Instead they use a grid where
all the numerators and
non-fractions are in the top row
and, all the denominators,
in the bottom row.
It makes the crossing
out of units easier but,
in every other
way, it is identical
to the process covered here.
That was a lot to take in
so we should try a few more
examples to solidify
our understanding.
Our second example asks
'How many grams of CO3-2
(carbonate) are there in
1.00 x10 24th particles
(ions) of carbonate?'
Identification of such problems
as 'conversion problems'
should become easier with
each new problem. The
given or current value
of the overall conversion
is the number of particles.
What it asks for is mass.
What might take
a while to master
with conversions is
gauging which type
of conversion can be
done with one unit factor
and which require more than 1.
The number-to-mass conversion
is of the latter type.
This conversion will
require the mole.
That is get into moles
from particles and
then get out of
moles into mass.
It is a 2 conversion
problem and we will do them
individually before
combining them into a single
dimensional analysis equation.
Starting with the
first conversion.
Identify the type
of conversion.
It is number of particles
going into moles.
What is the relationship
between numbers of particles
and moles of particles?
Yes, the definition of the mole.
1 mole is the number
6.02 times 10 to the 23rd.
Return to the question
and write down the
outline of the equation of
the specific conversion.
The number of particles is
on the left hand side
and the number of moles
is on the right.
It may be tempting to write
grams of particles as
the desired unit but this
is a multiple part conversion
problem and getting grams
is not the current focus.
The single most important
factor in getting the
problem right is the
correct assignment of units.
Choosing a unit factor has
us using the relationship with
an orientation that
cancel outs the given units
of particles of ions.
That is 1 mole
of carbonate ions
over is 6.02 times 10 to
the 23rd carbonates ions.
In the math step the
correct orientation of the
unit factor is verified with
the canceling out of units.
Dividing takes a knowledge
of one's calculator,
especially in
handling the exponent.
The answer should be a value
of 1.66. Clearly an easier
number to deal with than 1 x10
24th which is exactly why
we use the mole.
Again, we wait until
all the math is done
before checking sig. figs.
While it is not necessary, we
can return to the problem and
update it with the
equivalent amount of moles.
Onto the second conversion,
getting out of moles and
we will be using
this 1.66 moles.
Let's specifically identify
the conversion of interest.
We have moles and we
want mass in grams.
That relationship, between
moles of a substance and
the mass of that substance,
is the molecular weight.
60.01 grams of carbonate
is 1 mole of carbonate.
An equation of the conversion
will start with the solution
to the previous problem;
1.66 moles of carbonate.
It is our current unit
in the second conversion.
The desired units are
what the problem asks for;
grams of carbonate.
Almost there.
We need the correct unit
factor as outlined in the
relationship step. Moles of
carbonate need to cancel out
so the molecular weight
as written will be correct.
Do the units cancel out?
Yes they do. And a more
reasonable math problem
awaits with a final value
of 99.7 g of carbonate.
A final check of sig.
figs. has us return to
the original given value with
its 3 sig. figs.
Remember, exponent plays
no role in significant figures.
3 sig. figs. is also what
the final answer has so
no change is needed.
The two conversions
can be combined into a single
dimensional analysis equation.
In considering the
conversion as a whole
the given or currents units is
the number of particles, ions.
The wanted or desired units is
the mass in the form of grams.
Bringing up the individual
conversions will show
that dimensional analysis is
really just the combination
of the two equations.
The logic of the
dimensional analysis
equation is the same logic.
We do not have a direct
conversion from count to mass
so we will need to
go through moles.
The first conversion is the
template for the start of
the dimensional analysis
equation. They both have
count of carbonate
as a current unit and
they both use the relationship
of 1 mole of carbonate
to Avogadro's
number of carbonate ions.
Units cancel out
and so far these
two equations are identical.
That is, with the exception
of the parenthesis.
As such, they yield the
same 1.66 moles of carbonate.
That is also the starting
value of the second conversion.
The dimensional analysis
equation, as currently written,
is essentially the same thing.
That means we
can leave it as is
and proceed with
getting out of moles.
The intellectual leap in moving
from individual conversions
to dimensional analysis
is recognizing what the
'current units' and what
the 'desired units' are for
the particular conversion.
The current units at
this particular juncture in
the equation are the moles
since the count of
carbonate has canceled out.
The correct unit factor between
moles and the desired units
of grams is the molecular
weight of carbonate
just like it is in
the second conversion.
Units of moles of carbonate
cancel out leaving grams
of carbonate and multiplying
all numerators together
and dividing by all
the denominators gives
99.7 grams of carbonate.
Both the dimensional
analysis and the two
separate conversions
give the same results.
Onto example 3.
15 grams of carbon is
how many carbon atoms?
Identification of the
overall conversion is one of
grams of particles (atoms) into
number of count of particles.
Since we do not have a
relationship between these
units it falls on us to
build a bridge between them
and that bridge is the mole.
getting into moles ...
getting out of moles.
The overall logic of
the problem is set.
Solving the problem in
separate equations, begins by
identifying the type
of the first conversion.
It is mass of carbon
into moles of carbon.
What relationship is
there between these units?
Well carbon is an element
so atomic molar mass
is the relationship.
A quick survey of
the Periodic Table
shows the atomic molar
mass of carbon to be
12.01 grams per mole.
The relationship
is therefore that 12.01
grams of carbon being equal to
1 mole of carbon.
Setting up an equation
has the 15 g of carbon
as the current unit
on the left hand side
and moles of
carbon on the right.
Selecting the correct
orientation of the unit factor,
atomic molar mass,
is a matter of having the
unit grams in
position to cancel out.
That requires a flip of the
atomic molar mass so that
grams are in the denominator.
The math step verifies
our correct choice
by canceling out gram units.
15 divided by 12.01
gives a value of 1.25.
15 grams of carbon is
1.25 moles of carbon.
Holding off on the check
sig. figs. and moving into
the second conversion
(getting out of moles),
the 1.25 moles of carbon
stays and will be the
new current unit.
This second conversion is
identified as going
from moles to count.
The relationship between
those units is the definition
of moles with Avogadro's
number of particles.
The equation in this
conversion takes the solution
of the previous conversion
as the current value
of this conversion.
The number of carbon atoms
is the desired unit.
The orientation of the
relationship that makes an
appropriate unit factor will
have the unit of moles in the
denominator so that in the
math step the current unit
cancels out leaving
count of atoms.
Multiplication of 1.25
times Avogadro's number
divided by 1 produces a value
of 7.52 times 10 to the 23rd.
That is how many
carbon atoms there are
in 15 grams of carbon.
Speaking of 15 grams,
there are only 2 significant
figures in that value so
the final solution cannot
have more than 2
significant figures.
The 5 occupies the
second position.
Being followed by a 2 there
will not be rounding up
and the proper solution
to the problem is
7.5 times 10 to the
23rd carbon atoms.
To continue the practice
of dimensional analyses
we will repeat the conversion
of 15 grams of carbon
into a count of carbon atoms.
We do not have a direct
conversion from mass to count
and that means that there
will be more than 1 conversion
in the dimensional analysis.
The single conversions are
included to give perspective.
The getting into
moles conversion in the
dimensional analysis follows
the same steps that were
used in the first
individual conversion.
Units are identified, a
relationship is established
and the correct orientation
of the unit factor is inserted
Hopefully, it will
look like this.
That unit grams cancels out
is a good sign that
we got it right.
And once again we see that
the start of the dimensional
analysis looks just like
the first conversion.
They both equal
1.25 moles of carbon.
Since that is also the start
of the second conversion
the second part of the
dimensional analysis,
getting out of moles, will look
just like the second conversion
including the unit factor the
atomic molar mass of carbon.
Mole units cancel out
and 15 times 6.02 times 10
to the 23rd, divided by 12 .01
produces 7.5 times 10
to the 23rd carbon atoms.
As was seen in the
second conversion.
That is 3 examples of doing
the count to mass conversion
as single conversions and
using dimensional analysis.
The last example will be done
solely as a dimensional
analysis equation.
What is the mass of 2.3 x 10
the 25th calcium ions (Ca+2)?
Identification of
the overall process
has the given units as
number of particles, ions,
and the desired
unit is mass, grams.
Without a direct
conversion the process
will almost certainly
go through moles.
Get into moles
...getting out of moles.
The logic of this
conversion is the same as
the previous 3 examples.
A 2 part conversion using
dimensional analysis will
look very similar to
doing the conversions in
separate equations.
The main differences are
that the 'current unit' and
'desired units' change as the
equation proceeds and the
math is held off to the end.
Including the specifics in
the overall equation has the
count of ions as the
initial current unit and the
grams of calcium ion as
the final desired unit.
While our solution
will be in the form of a
dimensional analysis
equation, the first of the
multiple conversions
should be addressed as if it
were a single conversion.
Identify the units in
the conversion.
A big step in mastering
the dimensional analysis
technique is correctly
identifying what units are
in the conversion at hand.
In a multiple step conversion
they will never be both
the current units and
the desired units as
given in the problem.
In this conversion
it does use given
units in the problem
as the current unit but the
desired units are
not grams of calcium.
They are the objective of
the specific conversion.
They are, in this case,
moles of calcium ion.
Correctly identifying
the units, especially when
one unit type is currently
not written in the equation,
is a difficult step but
once it has to been made
the conversion moves on into
establishing a relationship
between the count of ions and
the number of moles of the ion.
And that relationship is
the definition of mole.
Avogadro's
number over 1 mole.
If an equation has not been
set up with current units
of count then do so. However,
the desired units of moles
are not written
in the equation.
This is one of the ways a
single conversion differs
from a multiple step
conversion. The desired units
will be brought in but
they will get in as part
of the correct unit factor. The
orientation of the unit factor,
as found in the relationship
step, needs to be one that has
count in the denominator
and moles in the numerator.
like this. Remember to
include the calcium
ion information.
Another difference in the
dimensional analysis approach
is the math step. Unless
this particular conversion
is the last conversion,
hold off on the math.
That doesn't mean
that units should not be
canceled out. That is how we
verify the correct
orientation of the unit factor.
Canceling also shows that we
end up with the units of moles.
The first conversion is
complete and it is now time
to get out of moles. The
second conversion repeats
the steps of the
single conversion.
Begin by identifying
the conversion.
It starts with the
units remaining from the
first conversion, moles, and
goes to the final desired units
of mass in the form of grams.
The relationship between
moles of ions and grams ions
is the atomic molar mass.
Ions, to a large degree,
have the same atomic molar
mass as their atoms do.
The Periodic Table shows the
element calcium to have an
atomic molar mass of
40.08 grams per mole.
40.08 grams of the calcium
ion is 1 mole of calcium ion.
And what should done
about the equation.
It already has current
and desired units so
nothing more is required.
All that remains
to complete the dimensional
analysis equation is
the unit factor that
converts moles into grams.
The atomic molar mass, as
written in the relationship,
will do since in the math
step moles cancel out.
The remaining unit is grams
which tells us that the
problem was set up correctly.
The calculator math can be
a bit daunting but,
as mentioned earlier,
a good practice in solving
these equations with
multiple multiplications
and divisions is to do all
the multiplication first and
then all of the divisions.
Additionally, the
multiplication and division,
of 1's can be skipped.
So, the calculation
would follow as...
2.3 times 10 to the 25th times
40.08 divided by
6.02 times 10 to the 23rd.
That gives 1531
grams of calcium ion.
All that is left is
a check of sig. figs.
The given value of
2.30 has 3 sig. figs.
The solution to
the problem has the
3rd sig. figs. being a 3.
Since it followed by a 1
there is no need to round up.
It stays a 3 and the correct
answer to the problem is 1530.
Yes there is a 4th digit here but
it does not carry significance.
It is needed because
removing it would leave
153 which is not
the right answer.
Develop dimensional analysis
as a tool and it will reward
you all way through
your studies.
To recap the lecture;
Molar Mass is the mass
of one mole of an element
or compound.The units
are grams per mole.
Molecular weight is
a conversion factor.
It is a unit factor since
numerator equals denominator.
There are convenient steps for
converting between mass and
moles. These steps can also
be used for other conversions.
Begin by identifying the units
involved in the conversion.
Get a relationship.
For the mole-mass conversion
that is the molecular weight
of the compound.
Next, set-up an equation:
current units on left;
desired units on the right.
Make a unit factor from
the relationship such that
current units will cancel out.
Cancel out the units and solve
the math. Not all conversions
can be done with a
single unit factor though.
There is no direct conversion
between number of particles
and grams of particles.
There is an intermediate
and it is the mole.
A common refrain in chemistry
when dealing with mass is
getting into mole and then
getting out of the mole.
This helps us remember
that the mole bridges
units of count and mass.
The 2 step conversion can
be from count of particles
through moles into mass or
the other way around starting
with mass of particle that
convert into moles allowing
a final conversion
into particles
And that concludes our lecture.
The best advice for this
material is to practice
practice practice and
watch and cancel your units!
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