Download Chapter 11 Chemical Calculations

Chapter 11 Chemical Calculations
For the past several weeks we have been working on our qualitative understanding of
first atoms, then molecules, and finally chemical reactions. In this chapter we enter a
new phase: we begin to put numbers to everything and begin building up our
quantitative tools so we can determine actual amounts of chemicals to be used in
reactions.
11-1 The Mole
To start building up on quantitative understanding of chemistry we have to go
back to Chapter 2 where we discussed the mass of the atom.
If you remember, we left off this part of the story with defining the mass of one
atom of 126C as being 12 atomic mass units or 12 u. The numbers associated
with all the other atoms was then a relative number that told how much larger or
smaller the mass of another atoms was. And then we added the complication
that most elements in nature consist of a number of isotopes, and that the
numbers we write on the periodic table represent a weighted average of all the
isotopes.
Now that we know how molecules are built, let’s extend our concept of atomic
masses to molecular mass
If the mass of one atoms of 12C is 12 units, then the molecule CH4 which has one
C atoms and 4 H atoms is goin to be 12+(4x1) = 16 u, etc.
Key Concept:
A molecular mass is the sum of the atomic masses of all the atoms in a
molecule
Let’s generalize this one more step
A formula mass is the sum of the atomic masses of all the atoms in an atom,
molecule or ion.
Again we are talking relative numbers one molecule of CH4 is 16u and is 16u/12u
more massive than one atom of C
But dealing with ratios is tricky and abstract. I want something concrete and
absolute to work with. So I can weigh a chemical out on a balance and know
how much material I have.
The key to this absolute chemical scale is the mole
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Key concept:
The quantity of a substance whose mass in grams is numerically equal to the
formula mass of a substance is called a mole (abbreviated mol)
-AndThe molar mass of a compound is the number of grams need to make up one
mole of a compound.
Example:
CH4 Formula mass = 16.04 units (12.01 + 4(1.008))
1st definition above:
16.04 grams CH4 = 1 mole CH4
nd
2 definition above
Molar mass of CH4 = 16.04 grams of CH4
But I probably don’t have to tell you this; you have probably had this a zillion
times in high school, just maybe not presented in this manner.
The above equations, linking the molar mass of a substance in grams with the
mole is a key conversion factor that we will use over and over in chemical
calculation and fits right in with our Dimensional Analysis that you learned in
Chapter 1
Example 1:
I have 20 grams of CH4 How many moles is that
Given
want
grams
moles
Unit conversion
Grams X Moles = moles
Grams
16.04 grams CH4 = 1 mole CH4
Plug & chug
20 grams CH4
X
1 mole CH4 =1.25 moles CH4
16.04 g CH4
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Example 2:
I need .450 moles of NaCl for an experiment. How many grams do I need to
weigh out?
1 mole NaCl = 58.44 g NaCl
Given
moles NaCl
Conversion
moles NaCl x
Need
Grams NaCl
Grams NaCl = grams NaCl
Moles NaCl
Plug & Chug
.450 moles NaCl x 58.44 grams NaCl = 26.3 g NaCl
1 mol NaCl
Clicker question:
gram to mole conversion Ba(OH)2
show 4 equations each with a different answer have them pick out correct
equation
11-2 Avogadro’s number
So our definition of a mole was that 12g of 12C is 1 mole of C
But what is it really Now that we know that the C is composed of atoms of C,
how many atoms of C are in that 1 mole or 12 g of 12C?
Key Concept:
1 mole of a thing is 6.022x1023 things.
So 12.04 g of CH4 contains 6.022x1023 molecules of CH4
and 58.44 g of NaCl contains 6.022x1023 g of NaCl(s) or 6.022x1023 Na+ ions
and 6.022x1023 Cl- ions
And now we have another conversion factor
1 mole of stuff = 6.022x1023 items of stuff
A mole is a lot of stuff but I think you can handle it.
Example calculations:
I have a 10 pound lead weight that I use for an anchor on my boat. How many
moles are in that anchor? How many atoms?
1 lbs= 453.59 g
1 mole Pb = 207.2 g
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Given
lbs
Set up
lbs
x grams
lbs
Want
Moles
x moles
grams
=Mole
Plug & Chug
10.0 lbs Pb x 453.59 g Pb x 1 mole Pb = 21.9 moles Pb
1 lbs Pb
207.2 g Pb
If I wanted to go all the way to atoms
Given
lbs
Set up
lbs
x grams
lbs
x moles
gram
x atoms
Mole
Want
Atoms
= atoms
Plug & Chug
10.0 lbs Pb x 453.59 g Pb x 1 mole Pb x 6.022x1023 atoms = 1.32x1025 atoms Pb
1 lbs Pb
207.2 g Pb
1 mole Pb
Clicker question
Start with Giga molecules of Calcium Chloride (written not formula)
Give 3 wrong and 1 correct equation to convert to grams
11-3 Simplest (Empirical) Formulas
Back in chapter 2 we talked about analyzing a compound using mass percent
where mass % = (mass of element A)/(total mas of all elements in a compound) X100%
This is one of the simplest and oldest ways that chemists use to analyze
compounds.
In this section you will learn how to go from a mass % to an empirical formula
But first, what is an empirical formula?
Key Concept:
An Empirical formula is the simplest possible formula for a compound. It
expresses the ratio of one atom to another, but it does not convey the total
amount of each element in a compound
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Example:
Benzene and acetylene have the same empirical formula CH. This says that
there is 1 C atom for every H atom in both of these compound
The actual molecular formula of Benzene is C6H6 and that of acetylene is C2H2
so you can see how the actual molecular formula is a multiple of the empirical
formula
The way McQuarrie explains how to go from a mass % to an empirical formula
makes sense, but he hasn’t really boiled it down to a simple step-by-step
procedure for you to follow. So let’s try doing it this way:
Key Procedure:
Calculating Empirical Formulas from mass %
1. Calculate the numbers of grams of each element that would occur if the
sample was 100 g. We do this to convert from % to g in one easy step.
2. Calculate the Number of moles for each element in a 100 gram sample
3. Find the element with the smallest number of moles, and then divide the
number of moles for each element by that smallest number.
- This should result in small whole numbers that will represent the
subscripts for the elements in the empirical formula
- If you do not have whole numbers, multiply all the numbers in the
equation by integers until you get whole numbers for all coefficient in the
equation
Example calculation:
Let’s try an example, hydrazine is 87.42% nitrogen and 12.58% hydrogen, what
is its empirical formula?
1. The first step is to convert % to grams assuming you have a 100 gram sample
100g X .8742 = 87.42 g N
100 g X 12.58 = 12.58 g H
2. Second step - convert from grams to moles
87.42 g N/14.01 g/mol = 6.24 moles N
12.58 g H / 1.008 g/mol = 12.48 moles H
3. Divide through by the smallest number of moles
12.48 moles H/ 6.24 moles of N
So 2 moles of H Ô 1 mole of N
Ô means is stoichiometically equivalent to
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Three things to notice.
A. Always put the largest number over the smallest. If you have more
than 2 elements choose the smallest overall and use that as the
denominator for all calculations
B. You will usually get nice whole numbers, but sometimes may end up
with ratios like 1.5/1. In this case you need to multiply both the numerator
and the denominator by a single number that will give you all nice whole
numbers. In this case 2(1.5)=3 and 2(1) =2, so the actual ratio is 3/2
C. The resulting formula is called the empirical formula. It is not the
molecular formula you have the ratio right but the overall numbers can be
factors of this NH2, (N2H4, N3H6 etc.)
A second example (if you want it)
An unknown liquid has 12.5% H, 27.5% C and 50% O, what is the empirical
formula of this compound?
H : 100g x 12.5% = 12.5g
C: 100g x 37.5% = 37.5g
O: 100g x 50% = 50.0g
H: 12.5g x 1 mole/1.008 g = 12.5 mole
C: 37.5g x 1 mole/12.01g = 3.125 mol
O: 50 g x 1 mole/16.00 g = 3.125
12.5/3.125 = 4 units of H
3.125/3.125 = 1 unit of C
3.125/3.125 = 1 unit of O
4 moles of H Ô 1 mole of C Ô1 mole of O
Empirical formula = CH4O
Clicker question? Long an complicated break into steps?
The book does some additional problems where you burn a compound in a gas
and look at the resulting oxide or nitride. I think I will skip this.
11-4 Using the empirical formula to determine the atomic mass of an element in a
compound
While this may be a fairly standard lab technique, its not one that we do so I will
skip this section
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11-5 Molecular Formulas
So if we can only get an empirical formula from a mass %, how do we get an
actual molar formula? We need one more piece of information, the molar mass.
You have seen that the empirical formula is the simplest possible molecular
formula. You can think of it as a ‘lowest common denominator’. The actual molecular
formula is some multiple of this:
Key Equation
molecular formula = empirical formula x N
molecular mass = empirical mass x N
Where N is some whole number
Example:
A compound is 85.7% C and 14.3% H and has a molar of 42. What is the
molecular formula of this compound?
There are a couple of ways to do this. First the long way
gC in 100g sample = 85.7g
gH in 100g sample = 14.3 g
mole C = 85.7/12.01 = 7.14 moles C
mole H = 14.3/1.008 = 14.19 moles H
Dividing by 7.14
1 mole C Ô2 moles of H
Empirical formula = CH2 ; Empirical mass = 14
Molecular mass = empirical mass x N
42 = 14x N
42/14=N
N=3
Molecular formula = empirical formula x N
= (CH2) x 3
=C3H6
I like the long way because you see that there are three empirical units in
the molecular unit.
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Now the short way
Multiply the molar mass by the %
42 g/mole x 85.7%/100% = 36.0 g/mol C
42 g/mol x 14.3%/100% = 6.00 g/mol H
Divide the g/mol fo reach element by the atomic mass
36.00 g/mol C /12 g/atom = 3 atoms/mole = C3
6 g/molH / 1g/atom H = 6 atoms/mol = H6
Molecular formula = C3H6
Key calculation
Given % composition and molar mass be able to determine the molecular
formula. This can be done by two different procedures choose the one that you
like best.
11-6 Combustion Analysis
This is another fairly standard technique that is used in many labs, but I think we
will skip for now.
11-7 Coefficients in Chemical Equations
Back in chapter 3 you learned how to balance chemical equations. In this
process you came up with a set of balancing coefficients, or numbers in front
of every molecule that told you how many of each molecule you needed so that
all the atoms that appeared in the reactant molecules were also present in the
product molecules so that the mass of every atom was conserved
These coefficients are also the key to our stoichiometry problems. For example
let’s start with the balanced equation for the burning butane:
2C4H10(l) + 13O2(g) 6 8CO2(g) + 10H2O(g)
The equation says that
2 molecules of C4H10 Ô 13 molecules O2 Ô 8 molecules CO2 Ô 10 molecules H2O
Which the same as
2 moles of C4H10 Ô 13 moles O2 Ô 8 moles CO2 Ô 10 moles H2O
This gives you a wealth of conversion factors.
A cigarette lighter might have 5 grams of butane in it. How much O2 is required
to burn this amount of butane?
First, how many moles of butane do I have
What is the Molecular mass of butane? 4(12.01) + 10(1.01) = 58.24
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5 g x 1 mole/58.24 g = 0.08585 moles
Now from the above reaction we know that 2 moles Ô 13 moles of Oxygen (or we
also say is equivalent to). Now just as we did unit conversions, lets set this up as
a proportionality
2 moles butane = 13 moles oxygen
2 moles butane/13 moles oxygen = 1
-or13 moles oxygen/2 mole butane =1
Now with these conversions, how do we convert moles of butane to moles of
oxygen?
0.08585 moles butane X 13 moles oxygen= .5580 moles of oxygen
2 moles of butane
Since O2 has a molecular mass of 32.0, our weight of O2 is
.5580 moles O2 x 32g O2
1 mol O2
= 17.86 grams O2
Out of curiosity, lets continue this to calculate the amount of water made in this
reaction. From our chemical reaction we know that 2 moles of butane Ô 10
moles of water, so we have the proportion
2 mol butane/10 mol water = 1 -or- 10 mol water/2 mol Butane =1
If we have
.08585 mol butane x 10 mol water = .429 moles of water
2 mol butane
Molar mass of water= 18.02 so
.429 mole water x 18.02 g water = 7.735 grams of water
1 mole water
Key Procedure:
Stiochiometric calculations
1. Balance the equation
2. Convert the mass of given substance to moles
3. Use balanced equation to set up molar conversion factors
4. Use appropriate conversion to calculate the moles of substance you want
5. Use molecular mass of substance to convert moles into grams of substance.
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11-8 Stoichiometry
Figure 11.7 Puts the above procedure into a nice flow chart. But I can’t seem to
grab it from the computer
Let’s try this in another example; the thermite reaction. Used in rockets,
underwater welding and incendiary bombs
Fe2O3 (s)+Al(s) 6 Fe(l) + Al2O3(s) (Unbalanced)
(review names iron(III) oxide and Aluminum oxide (why no (III)?)
If I want to generate 15 grams of Fe, how much Al and how much Fe2O3 is
needed
Step 1. balance reaction
Fe2O3 (s)+2Al(s) 6 2 Fe(l) + Al2O3(s)
Fe
2
2
Al
2
2
O
3
3
Fe2O3 (s)+2Al(s) 6 2 Fe(l) + Al2O3(s)
Step 2. Convert 15 g Fe to moles
15 g Fe X 1 mole Fe = 0.269 moles Fe
55.85 g Fe
Step 3 Convert moles Fe to moles Al
0.269 mol Fe X 2 mol Al = 0.269 mole Al;
2 mol Fe
Step 4 Convert mole Al to grams Al
0.269 mole Al x 26.98 g Al = 7.12 g Al
1 mol Al
Step 3 for Fe2O3 : Converting moles Fe to moles Fe2O3
0.269 mol Fe X 1 mol Fe2O3 = 0.1345 mol Fe2O3
2 mol Fe
Step 4 Convert moles Fe2O3 to g Fe2O3
0.1345 mol Fe2O3 X 159.7 g Fe2O3 = 21.5 g Fe2O3
Mole Fe2O3
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11-9 Stoichiometry without chemical equations
Using the principal of conservation of mass, it is sometimes possible to do a
stoichiometric calculation without a chemical equation!
For example say I am going to react 10 g of sodium in water to make NaOH
If NaOH is the only product of this reaction that contains sodium, then the
principle of conservation of mass says that all the sodium I start off with has to
appear in the NaOH product. So
Na Ô NaOH
So, not even knowing the reaction I can do my stoichiometric calculation:
10g Na x 1 mole Na
23 g Na
= .435 mole Na
.435 mole Na x 1 mole NaOH = .435 mol NaOH
1 mole NaOH
.435 mol NaOH x 40 g NaOH = 17.4 g NaOH
1 mol NaOH
11-10 Limiting Reactant
When two or more substances react the mass of the product is determined by
the limiting reactant. Let’s go back to the thermite reaction. When we left that
problem we had determined that
7.12 g of Al and 21.5 g of Fe2O3 were required to make 15 of Fe using this
reaction
What would happen if we used 15 g of Al and 21.5 g of Fe2O3, would we make
any more product?
NO. Adding excess of one reactant doesn’t give you more product, because the
other reactant limits the amount of product that can be made.
Key Concept:
In a chemical reacion with more that one reactant, usually one reactant, called
the limiting reactant, will limit the amount of product made in the reaction.
Other reactants, that are present in excess are called excess reactants.
You will often be given problems like this, where you are give the gram amounts
of two reactants and you need to figure out which one is the limiting reactant and
which one is the excess reactant.
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So let me give you a procedure to figure out which is which
Key Procedure and definition:
Finding the limiting reactant
1. Pick a product and stick with it for the rest of the problem
2. Calculate how many moles of this product you can get from reactant A
alone
3. Calculate how many moles of this product you can get from reactant B
alone
4.
.....
C, D, E???
The reactant that give you the SMALLEST product is the limiting reactant. The
other reactants are all excess reactants.
The amount of product obtained with this limiting reactant is the
Theoretical Yield
Let’s try this with a variation of our butane lighter problem. If I start the butane
lighter with 0.5 grams of butane, and place it in a 5 gallon jug and seal the jug,
which will run out first, the air in the jug or the lighter fuel?
We already calculated that 0.5 grams of butane is 0.0086 moles of butane. And
we know from the balanced equation
2C4H10(l) + 13O2(g) 6 8CO2(g) + 10H2O(g)
That
2 moles butane Ô 13 moles O2 Ô 8 moles CO2 Ô 10 moles H2O
Now I will skip ahead a few steps on how to go from 5 gallons of air to moles of
O2 , and just give you the final answer that 5 gallons of air contains .187 moles of
O2
Step 1 pick a product
I will pick CO2
Step 2 calculate amount for product given by reactant A
.0086 m butane x 8 mol CO2 = .0344 moles CO2
2 mol butane
Step 3 calculate amount for product given by reactant B
.187 m O2 x 8 mol CO2 = .115 mole CO2
13 m O2
Butane is the limiting reagent because it makes the smallest amount of product.
O2 is the excess reagent because you have an excess amount of it.
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Which product do you choose? It doesn’t matter, and the result will be the
same. However, on a test, the next question is usually ‘how much is your
expected yield of X’ so if you chose the reagent that answers that question, you
are one step closer to answering that question.
11-11 Percentage Yield
In the above section I talked about the theoretical yield. That the amount of yield
you calculate from the limiting reactant. You actual yield almost never matches
this theoretical value.
Why?
Reaction fails
Reaction not complete
Side reaction that gives other products
You may lose some product by accident
Chemists like to keep track of how close their yield is to the theoretical with a
calculation called the % Yield
Key Calculation:
% yield = (actual yield/theoretical yield) x 100 %
Example:
Going back to the previous example. If our actual yield was .0301 moles of CO2
and our theoretical yield was .0344 mole of CO2, what is our % yield?
.0301/.0344 x 100 = 87.5%
Note 1:
You can calculate % yield based on grams of product or moles of product. It
doesn’t matter which one you use they should give you the same number of you
do the calculation correctly.
Note 2:
You might think that the % yield should always be less than 100% but guess
again. Sometimes it is >100%. How can that be? Sometimes when you purify
your product, you accidentally purify some ‘Kaka’ with it. Now the weight you
record is higher than it should be because of the added ‘Kaka’.