Transcript Chapter 3 PowerPoint

Chemical Mathematics
Counting by Weighing
• Small items can be tedious to count.
• Items like nuts and bolts are priced by the
kilogram after someone initially counts how
many are in 1 kg.
• This only is done once and keeps the customer
from having to do it.
• There is an example on the next slide.
Counting by Weighing
• Let’s say that a bolt is priced at $ 5.00 per kg.
• The customer grabs a handful of the amount
that is needed and it gets weighed. Let’s say
the customer needs 1.6 kg of bolts.
• We would set up the following proportion.
1 kg = 1.6 kg
x = $ 8.00
$ 5.00
x
Atomic Mass
• Atoms are so small, it is difficult to discuss
how much they weigh in grams.
• Use atomic mass units.
• An atomic mass unit (amu) is one twelfth the
mass of a carbon-12 atom.
• This gives us a basis for comparison.
• The decimal numbers on the table are atomic
masses in amu.
They are not whole numbers
• Because they are based on averages of atoms
and of isotopes.
• Can figure out the average atomic mass from
the mass of the isotopes and their relative
abundance.
• Add up the percent as decimals times the
masses of the isotopes.
Examples
• There are two isotopes of carbon 12C with
a mass of 12.00000 amu (98.892%), and
13C with a mass of 13.00335 amu (1.108%).
What is the atomic wt. of Carbon ?
• There are two isotopes of nitrogen , one
with an atomic mass of 14.0031 amu and
one with a mass of 15.0001 amu. What is
the percent abundance of each ?
• All of these are solved with algebra !
Solved Example # 1
• There are two isotopes of carbon 12C with a
mass of 12.00000 amu (98.892%), and 13C
with a mass of 13.00335 amu (1.108%).
• Here is how we would set up the equation.
• (.98892 x 12.0000) + ( .01108 x 13.00335) =
• 12.01 grams
Solved Example # 2
• There are two isotopes of nitrogen , one with
an atomic mass of 14.0031 amu and one with
a mass of 15.0001 amu. What is the percent
abundance of each ?
• This one uses the same ideas but we have to
look at it differently.
• We know from the periodic chart that the
atomic wt. of Nitrogen is 14.01 grams.
Solved Example # 2
• Here is the set-up
• 14.01 = (14.0031 ● (1-x) ) + ( 15.0001 x )
• Look at how the two different percentages
were handled. Since they have to total up to
100 % (or 1 as a fraction). We know that if
one of the values is = to x, then the other
must be 1 – x.
Solved Example # 2
•
•
•
•
•
•
Now we can solve.
14.01 = (14.0031 ● (1-x) ) + ( 15.0001 x )
Simplify to
14.01 = 14.0031 – 14.0031x + 15.0001x
.0069 = .9970 x
x = .00692 or .692 % 15N Therefore,
• The % of 14N is = to 99.308 %
• Does this make sense ? Why ?
Formula Weight
• Alternate names: Molar Mass, Atomic Weight,
Molecular Mass, Formula Weight.
• Use units of grams or grams / mole
• Using the Periodic Chart, determine the
molecular weights of various examples, going
from the simplest to the more complex.
Formula Weight
• You can use the terms weight and mass
interchangeably here.
• The most general term is FORMULA WT.
• Molar mass is another common term.
• Atomic weight is the weight of an atom.
• Molecular weight is the weight of a molecule.
• Ionic weight is the weight of an ion
• Etc., etc., etc.
Molar mass
• Mass of 1 unit of a substance, called the mole.
More on that in a bit
• Often called molecular weight.
• To determine the molar mass of an element,
look on the table.
• To determine the molar mass of a compound,
add up the molar masses of the elements that
make it up.
Determining Formula Weight
•
•
•
•
•
•
Sn
NaBr
Cl2
K2O
Al2O3
4 Al2O3
118.71 g
23 + 79.9 = 102.9 g
35.45 x 2 = 70.9 g
(39.1 x 2) + 16 = 94.2 g
(27 x 2) + (16 x 3) = 102 g
Same as above – coefficients don’t
change the formula weight
• Ca(NO3)2 40.08 + 2 (14 + (3 x 16)) = 164.08 g
• (NH4)2S 2 (14 + ( 4 x 1)) + 32.06 = 68.06 g
Formula Weight of a Hydrate
• MgSO4 ● 7 H2O (hydrates).
• These are easy if you understand what the
formula represents. There are 7 water
molecules attached to the single MgSO4
molecule.
• The raised dot DOES NOT mean multiplication.
• (24.3 + 32.06 + (4 x 16)) + 7 ( 18) = 246.36 g
Find the formula weight of
• CH4
• Mg3P2
• Ca(NO3)2
• Al2(Cr2O7)3
• CaSO4 · 2H2O
Answers
• CH4
16 g
• Mg3P2
134.9 g
• Ca(NO3)2
• Al2(Cr2O7)3
• CaSO4 · 2H2O
164.08 g
702 g
172.14 g
The Mole
• The mole is a number.
• A very large number, but still, just a
number.
• 6.02 x 1023 of anything is a mole
• Similar in idea to a dozen.
– A dozen is always 12 items
– A mole is always 6.02 x 1023 items
• A mole is in exactly 12 grams of carbon-12.
The Mole and the Magic Formula
• From a practical standpoint, we want to be
able to convert grams to moles and moles
to grams.
• THIS IS OF UTMOST IMPORTANCE
IN CHEMISTRY !!
• We need to introduce the most used formula in
Chemistry.
n=g/M
Moles = grams / formula wt.
Solving Mole problems
• Using simple algebra and n = g / M solve the
following:
• How many moles are in 154 g of CO2 ?
• How many grams are in 12.6 moles of NaCl ?
• How many moles are in .255 grams of H2O ?
• How many grams are in .6 moles of H2SO4 ?
• What is the formula wt. of a substance if .85
moles of has a weight of 42 grams ?
Solving Mole problems - answers
• 154 g of CO2 ? n = 154 / 44 n = 3.5 moles
• 12.6 moles of NaCl ?
12.6 = g / 58.45 g = 736.47 g
• .255 grams of H2O ? n = .255 / 18 n = .0142 n
• .6 moles of H2SO4 ? .6 = g / 98.06 g = 58.84 g
• What is the formula wt. of a substance if .85
moles of has a weight of 42 grams ?
• .85 = 42 / M
M = 49.41 g
• There are problems in real life like this !
Percent Composition
• Percent of each element a compound is
composed of.
• Find the mass of each element, divide by
the total mass (formula wt.), multiply by
100.
• Find the percent composition of
• CH4
Al2(Cr2O7)3
CaSO4
Percent Composition - answers
• CH4 – Formula Wt. of 16 g
C
H
12 / 16 = .75 .75 x 100 % = 75 % Carbon
4 / 16 = .25 .25 x 100 % = 25 % Hydrogen
• Al2(Cr2O7)3 – Formula Wt. of 702 grams
Al
Cr
O
54 / 702 = .0769 x 100 % = 7.69 % Al
312 / 702 = .4444 x 100 % = 44.44 % Cr
336 / 702 = .4786 x 100 % = 47.86 % O
Percent Composition - answers
• CaSO4 – Formula Wt. of 136.14
Ca
S
O
40.08 / 136.14 = .2944 x 100 % = 29.44 %
32.06 / 136.14 = .2355 x 100 % = 23.55 %
64 / 136.14 = .4701 x 100 % = 47.01 %
Notice that the sums of the percents always
will equal 100 %
So if a compound such as NaCl is known to be
39.35 % Na, we can simply subtract that from
100 % to determine that it is 60.65 % Cl.
Empirical Formula
• From percent composition, you can determine
the empirical formula.
• Empirical Formula is based on mole ratios and
is the lowest ratio of atoms in a molecule.
• It is important to know that Empirical Formulas
are not necessarily real chemical compounds.
• Why then do we bother ? Because it is a great
analytical tool to analyze an unknown
compound.
How to determine Empirical Formulas
• No matter what unit is given, you need it to be
in moles.
• If moles, you are ready. If grams, use n = g/M
• If a %, assume 100 grams and use n = g/M
• After moles are obtained, you have a proper
ratio. Simply make sure that it is in whole
numbers !
Empirical Problem
• If a sample is found to be 30.43 % N, and
69.56 % O
• What is its empirical formula ?
1. Since we need to get moles, we need to
remember that the percent composition is an
intensive property and doesn’t change no
matter what the size of the sample is.
Empirical Formula
2. So let’s choose a 100 gram sample to simplify
the math.
So  30.43 % N of a 100 gram sample =
30.43 grams of Nitrogen
Doing the same thing to the other element gives
us 69.56 grams of Oxygen
Empirical Formula
3. So we have the following:
30.43 g N
69.56 g O
4. Convert to moles using n = g / M
2.17 n N
4.34 n O
This is the ratio of atoms in the compound, but
there is clearly a problem if we write it as the
following : N2.17O4.34
We can’t have a fraction of an atom !
Empirical Formula
5. So we need to keep the same ratio of atoms
but have them expressed as a whole number.
We can do this by dividing each by the lowest, so
it will look like this.
2.17
4.34
2.17
2.17
This gives us a ratio of 1:2 So the empirical
formula is NO2
Empirical Formula
• A 0.2000 gram sample of a compound
(vitamin C) composed of only C, H, and O is
burned completely with excess O2 . 0.2998 g
of CO2 and 0.0819 g of H2O are produced.
What is the empirical formula?
• Follow the steps and analyze the problem.
• A general rule in Chemistry is that if a reaction
is taking place it is a good idea to write it out.
It might not help – but it will never hurt !!
Empirical Formula – Solved Problem
• The equation is: CxHyOz + O2  CO2 + H2O
• When you analyze the problem you see that the
Carbon in the CO2 had to come from the
Carbon in the Hydrocarbon. The Hydrogen in
the water had to come from the Hydrogen in
the Hydrocarbon.
• That is our clue to proceed.
• This is a very representative type of AP
Chemistry Problem for this topic.
Empirical Formula – Solved Problem
• So if we have .2998 grams of CO2 we can use
our knowledge of percent composition to
determine how much Carbon is in this reaction.
• CO2 is 27.27 % C so that is .0818 grams of
Carbon in the sample. The reactant must
contain this amount of Carbon.
• Doing the same for the H in H2O gives us .0091
grams of H in the reactant.
Empirical Formula – Solved Problem
• .0818 g. of C + .0091 g. of H = .0909 grams
• Since the total mass of the reactant is .200 g.
the remaining mass must be the mass of the
oxygen.
• .2000 - .0909 = .1091 g. of Oxygen are in the
reactant.
• Since it is in grams, the 100 g. sample is not
needed. We can solve directly for moles.
Empirical Formula – Solved Problem
• So this is what we have. The reagent is made
up of the following:
.0818 g C
.0091 g H
.1091 g O
Let’s convert to moles:
.0068 n C
.0091 n H
.0068 n O
We can’t have a formula of C.0068H.0091O.0068
because we can’t have a fraction of an atom.
Empirical Formula – Solved Problem
• Here is our current formula C.0068H.0091O.0068
• We need to express the same relationship in
whole numbers.
• We do that by dividing each value by the
lowest. (that is .0068 in this problem)
.0068/.0068
.0091/.0068
.0068/.0068
so we get the following
1 : 1.33 : 1
Empirical Formula – Solved Problem
so we get the following
1 : 1.33 : 1
• What we have to do is to finish the work to get
all whole numbers in the formula.
• It helps to know your decimal fractional
equivalents. We can rewrite the ratio above as
1 : 1 & 1/3 : 1
Empirical Formula – Solved Problem
1 : 1 & 1/3 : 1
If we multiply each factor by 3 we would get
3 : 4 : 3 which can be written as
C3H4O3 and we have our empirical formula
Empirical Formula
• We need to remember that what we solved
for is the empirical formula, or the ratio of
atoms in the compound. C3H4O3 may or may
not be the actual formula for Vitamin C.
• To determine this we will need more
information, which brings us to the next topic.
Empirical To Molecular Formulas
• Empirical is lowest ratio of atoms in a formula.
• Molecular is the actual molecule.
• What we need is the Formula Weight. This will
be given to you.
• Ratio of empirical to molar mass will tell you
the molecular formula.
• Must be a whole number because molecules
are made up of whole number of atoms.
Empirical to Molecular Formulas
• Here is a generic example.
• Let’s say we determined an empirical formula
to be CH2 and we are given that the formula
weight of the actual compound is 154 g.
• First we should look at how many possible
molecular formulas that have an empirical
formula of CH2 there are.
Empirical to Molecular Formulas
• Let’s start by listing all the possible molecular
formulas.
CH2
C2H4
C3H6
C4H8
C5H10
C6H12
C7H14
C8H16
C9H18
C10H20
C11H22
C12H24
C13H26
C14H28
C15H30
We can now compare the given molecular weight with all
the possible weights of the formulas above.
This can get tedious very quickly !!
Empirical to Molecular Formulas
• Here is a much quicker way.
1. Determine the empirical formula by the
method shown earlier.
2. Set up the following:
Molecular Wt.
Empirical Wt.
3. This will always give us a whole number.
Empirical to Molecular Formulas
• In our sample problem here, the math would
work like this:
154 / 14 = 11
Where the molecular formula was given
as 154. The empirical formula of CH2 is
solved for an empirical wt. of 14
Notice the answer is a whole number.
Empirical to Molecular Formulas
• So what we want is the 11th version of the
empirical formula.
• The key is to know what it will look like,
sometimes students format it the wrong way.
• There are three possibilities but only one of
them are correct.
Empirical to Molecular Formulas
• Here are the three options:
11 CH2
(CH2)11
C11H22
The 1st one is incorrect. This is saying that there
are 11 units of a small CH2 molecule.
The second one is also incorrect. This is simply
bad form all around. Some organic groups can
be written this way sometimes, but not an
entire formula.
Empirical to Molecular Formulas
So that leaves us with the third option as our
choice.
C11H22 has a formula weight of 154 g
(12 x 11) + (22 x 1)
It can be ‘factored down’ to CH2 which is the
empirical formula in question.
Molecular Formula – Solved Example
• A compound is made of only nitrogen and
oxygen. It is 30.4 % N by mass. Its formula
weight is 92 g/mol. What is its formula ?
• We should notice that we don’t have both
percentages, but we can easily determine the
% of oxygen.
• 100 – 30.4 = 69.6 % Oxygen
Molecular Formula – Solved Example
• We have 30.43 % N and 69.6 % O
• Find the empirical formula – assume 100 g of
sample
30.4 g N & 69.6 g O
Moles
2.17 n
4.35 n
Determine the ratio: We divide each by the
lowest and get an empirical formula of
NO2
Molecular Formula – Solved Example
• We determine the empirical weight of NO2 to
be 46 g/n.
92 / 46 = 2
So the molecular formula is
N2O4
Chemical Equations
•
•
•
•
•
Are chemical sentences.
Describe what happens in a chemical reaction.
Reactants  Products
Equations should be balanced.
Have the same number of each kind of atoms
on both sides because ...
• To obey the Law of Conservation of Mass.
Chemical Equations
• Here are some guides to balancing equations.
– Make sure that all species are correctly written.
– Remember that you can only change the
coefficients, which is the number of the species.
• Changing the subscript changes the identity of the
species and therefore changes the chemical equation.
– Start with the largest compound and assume that it
has a coefficient of one.
– Only balance one atom at a time
– Save atoms that are found in multiple species to the
end.
Balancing equations
 CO2 + H2O
CH4 + O2
Reactants
Products
1 C 1
4 H 2
2 O 3
Balancing equations
CH4 + O2  CO2 + 2 H2O
Reactants
Products
1 C 1
4 H 2 4
2 O 3
4
Balancing equations
CH4 + 2O2  CO2 + 2 H2O
Reactants
Products
1 C 1
4 H 2 4
4 2 O 3
4
Abbreviations & Symbols
•
•
•
•
•
•
(s) or
(g) or
(l)
(aq)
heat
catalyst
Solid
Gas
Liquid
Aqueous
Heat added as a catalyst
Any other catalyst
Practice
• Ca(OH)2 + H3PO4  H2O + Ca3(PO4)2
• Cr + S8  Cr2S3
• KClO3(s)  KCl (g) + O2(g)
• Solid iron(III) sulfide reacts with gaseous
hydrogen chloride to form solid iron(III)
chloride and hydrogen sulfide gas.
Bal. Equation Answers
• Ca(OH)2 + H3PO4  H2O + Ca3(PO4)2
-------------------------------------------------------------------------------------------------------------------------------------------------------
• Start with Ca3(PO4)2 because it is the largest.
• 3 Ca(OH)2 + H3PO4  H2O + Ca3(PO4)2
• 3 Ca(OH)2 + 2 H3PO4  H2O + Ca3(PO4)2
– Oxygen is not a good choice to balance next
because it is found in every species.
• 3 Ca(OH)2 + 2 H3PO4  6 H2O + Ca3(PO4)2
Bal. Equations Answers
• Cr + S8  Cr2S3
_____________________________________________________________________________________________________________________________________________________________________________
• We have 8 S  3 S so we can use lowest
common multiple
• Cr + 3 S8  8 Cr2S3
• This leaves Cr to balance.
• 16 Cr + 3 S8  8 Cr2S3
Bal. Equations Answers
• KClO3(s)  KCl (g) + O2(g)
• Note: The phases don’t affect the balancing
____________________________________________________________________________________________________________________________________________
• Start with KClO3 It is the largest.
• 2 KClO3(s)  KCl (g) + 3 O2(g)
• We notice that we have 2 K and 2 Cl on the
left. It fits nicely with the KCl product.
• 2 KClO3(s)  2 KCl (g) + 3 O2(g)
Bal. Equations Answers
• Solid iron(III) sulfide reacts with gaseous
hydrogen chloride to form solid iron(III)
chloride and hydrogen sulfide gas.
• The only difference here is that we have to use
our formula writing skills to correctly write out
a skeleton (unbalanced) equation.
• Fe2S3 + HCl  FeCl3 + H2S
Balancing Equations
• If we write a formula incorrectly one of two
things can happen (none of them good).
1. We will find it impossible to balance.
2. We will be able to balance it, but we won’t be
balancing the desired equation.
Balancing Equations
• A balanced equation can be used to describe a
reaction in molecules and atoms.
• NEVER GRAMS ! In other words:
• 2 H2 + O2  2 H2O means
• 2 molecules (or moles) of Hydrogen react with
1 molecule (or mole) of Oxygen to form 2
molecules (or moles) of Water
• It doesn’t mean 2 grams of H + 1 gram of O to
form 2 grams of water !!
Stoichiometry
• Given an amount of either starting material or
product, determining the other quantities.
• We must have a balanced equation.
• Use conversion factors from
– molar mass (g - mole)
– balanced equation (mole - mole)
• Keep track by setting up proportions in moles.
Stoichiometry Example # 1
• One way of producing O2(g) involves the
decomposition of potassium chlorate into
potassium chloride and oxygen gas. A 25.5 g
sample of Potassium chlorate is
decomposed. How many moles of O2(g) are
produced ?
• How many grams of potassium chloride ?
• How many grams of oxygen ?
Stoichiometry Example # 1
• First we need a balanced equation:
KClO3  KCl + O2 is the unbalanced equation.
2 KClO3  2 KCl + 3 O2 is balanced.
We need to organize the problem. Here is one
suggestion how to do it.
25.5 grams
?
?
2 KClO3  2 KCl + 3 O2
This clarifies what we have and what we need.
Stoichiometry Example # 1
25.5 grams
?
?
2 KClO3  2 KCl + 3 O2
•
•
•
•
•
Convert to moles  .2081 n of KClO3
We need a proportion
2 KClO3 / 3 O2 = .2081 n KClO3 / x
x = .3121 moles of O2
Using n = g / m 9.988 grams of O2
Stoichiometry Example # 1
•
•
•
•
•
•
•
•
To get the KCl, we follow a similar procedure.
2 KClO3 / 2 KCl = .2081 n KClO3 / x
x = .2081 moles of KCl
Using n = g / m gives us 15.514 g of KCl
Let’s verify by the Law of Cons. Of Mass
Total mass of reactants = 25.5 g
Total mass of products = 15.51 + 9.99 = 25.50 g
Our answers look good !
Stoichiometry Example # 2
• A piece of aluminum 5.11 cm. by 3.23 cm.
x 0.0381 cm. is dissolved in excess HCl(aq).
How many grams of H2(g) are produced ?
• We have some preliminary work to do.
• The volume of the Aluminum piece is
.6289 mL
OK - That’s nice, I suppose !
• What is now the next step ?
Stoichiometry Example # 2
•
•
•
•
•
•
We can use the Density of Al to get the mass.
D = m / V 2.7 = g / .629 mL mass = 1.70 g.
Here is our balanced equation:
2 Al + 6 HCl  2 AlCl3 + 3 H2
1.7 grams of Al = .0629 moles
Our proportion is going to contain Al because
we have info on it and H because that is what
we are looking for.
Stoichiometry Example # 2
• Here is the proportion:
• 2 Al / 3 H2 = .0629 n / x
x = .0944 n of H2
• Convert to grams using n = g / m
• We obtain .1887 grams of H2 as our answer.
Stoichiometry Example # 3
• How many grams of each reactant are needed
to produce 15 grams of iron from the following
reaction ?
Fe2O3(s) + Al(s)  Fe(s) + Al2O3(s)
• First we balance:
Fe2O3(s) + 2 Al(s)  2 Fe(s) + Al2O3(s)
We also need to know moles of Fe
n = g / m  n = 15 / 55.85 n = .267 n Fe
Stoichiometry Example # 3
• Now that we have moles we can move forward
with the problem. We can do the Fe2O3 first
Our proportion is: 1 Fe2O3 / 2 Fe = x / .267 n Fe
Solving gives us .1335 moles of Fe2O3
Converting to grams gives us 21.32 g of Fe2O3
Doing similar work for the Al gives us
2 n Al / 2 n Fe = x / .267 n Fe x = .267 n of Al
.267 n Al  7.21 grams of Al
Stoichiometry Example # 4
• K2PtCl4 + NH3  Pt(NH3)2Cl2 + KCl
• What mass of Pt(NH3)2Cl2 can be produced from 65 g
of K2PtCl4 ?
• How much KCl will be produced ?
• First we must balance
• K2PtCl4 + 2 NH3  Pt(NH3)2Cl2 + 2 KCl
• Following the steps gives as in the previous problem:
• 65 g of K2PtCl4  .1566 moles
Stoichiometry Example # 4
•
•
•
•
•
•
•
1 n K2PtCl4 / 1 n Pt(NH3)2Cl2 = .1566 / x
x = .1566 moles of Pt(NH3)2Cl2
Convert to grams (FW = 299.98 g)  46.977 g
FOR KCl
1 n K2PtCl4 / 2 n KCl = .1566 / x
x = .3132 moles of KCl
.3132 n of KCl  23.35 g of KCl
Limiting Reagent
• The concept of limiting reagent is that if you
have two or more reactants in a chemical
equation. When they start reacting, one of
them are often going to run out first.
• This occurs in non – chemical situations as well.
• If we are making bicycles and have 40 seats and
50 handle bars, we can only make 40 bicycles
because once the seats run out, we can’t make
any more, no matter how many handle bars you
have.
Limiting Reagent
• The term limiting reagent is used for the
material that runs out first.
• Since the reaction stops when a reactant is
missing, the limiting reagent governs the
amount of product produced.
• The other material that has some left over is
said to be in excess.
• Sometimes abbreviated as inxs.
Limiting Reagent - Summary
• Reactant that determines the amount of
product formed. It governs the reaction.
• The one you run out of first.
• Makes the least product.
• If it is not easily apparent which reactant is the
limiting reagent, you can always set up 2
proportions and see which one fits. One will
fit and the other one won’t – ALWAYS.
Limiting reagent
• There are two possible ways to work a Limiting
Reagent problem
1. To determine the limiting reagent requires
that you do two stoichiometry problems.
– Figure out how much product each reactant
makes.
– The one that makes the least is the limiting
reagent.
Limiting Reagent
• The second way is to set up two proportions
and analyze them to determine which is the
limiting reagent.
• This is the preferred way and the method that
we will explore in class.
• However, either way is acceptable.
Limiting Reagent – Example
• Ammonia is produced by the following
reaction
• N2 + H2  NH3
• What mass of ammonia can be produced from
a mixture of 500 g N2 and 100 g H2 ?
• How much unreacted material remains?
Limiting Reagent – Example
• The first thing we need is a balanced equation.
• N2 + 3 H2  2 NH3
• 500 g of N2 = 17.857 moles
• 100 g of H2 = 50 moles
• The key is setting up the proportions, and
equally important is how to analyze them to
obtain the desired information.
Limiting Reagent
• N2 + 3 H2  2 NH3
• 500 g of N2 = 17.857 moles
• 100 g of H2 = 50 moles
• At first glance it is tempting to say that the N2
must be the limiting reagent simply because
there is much less of it.
• However, that is not often the case.
Limiting Reagent
• What we need to do is to set up proportions.
• The balanced equation is: N2 + 3 H2  2 NH3
1 n N2 / 3 n H2 = 17.857 n of N2 / x n of H2
You create the proportion by using the
coefficients from the balanced equation on the
left hand side, and the information in the
problem on the right hand side.
Limiting Reagent
• 1 n N2 / 3 n H2 = 17.857 n of N2 / x n of H2
• It is very important to realize what this
proportion is asking for.
• It is solving for how much H2 would be needed
IF all the N2 reacts.
• We need to solve for x and then analyze the
answer.
• Doing the math gives us 53.57 moles of H2
Limiting Reagent
• If we need 53.57 moles of H2 but we only
have 50 moles of H2 what does that mean ?
• We need more H2 than we have, so that is the
material in short supply.
• That makes the H2 the limiting reagent and
the material that controls the reaction.
Limiting Reagent
• Just to verify that it wasn’t just luck in
choosing the correct material, let’s set up the
proportion using the other reagent.
• 1 n N2 / 3 n H2 = x n of N2 / 50 n of H2
• Solving for x gives us 16.67 moles of N2
needed. However since we have 17.857 n of
N2 given to us initially, we have more N2 than
we need and so N2 is in excess (INXS).
• So that makes the H2 the Limiting Reagent.
Limiting Reagent
• So now we can see that the key is not choosing
the ‘magic’ material to put in the proportion –
it doesn’t matter which one we choose.
• The key is analyzing what we get when we
solve the proportion and use it accordingly.
• We need to use the limiting reagent in the
proportion because it governs the reaction.
• When it runs out – the reaction stops.
Limiting Reagent
• So let’s go back and finish the problem.
• Here is our proportion:
1 n N2 / 3 n H2 = x n of N2 / 50 n of H2
• Solving for x gives us 16.67 n of N2 reacting.
• If we are asked to determine the amount of
material in excess, we can simply subtract the
amount given minus the amount reacted.
• 17.86 – 16.67 = 1.19 moles of N2
• We can convert to grams if needed. (33.32 g)
Limiting Reagent
• Here is what we would do to find how much of
the product is formed.
• N2 + 3 H2  2 NH3
• Use the limiting reagent in the proportion. For
this problem, we established that it was H2
• 2 n NH3 / 3 n H2 = x n of NH3 / 50 n of H2
• Solving for x gives us 33.33 n of NH3
• Convert to grams as needed: 566.61 g NH3
More about Stoichiometry
• Sometimes they might simplify a problem by
saying one of the materials in excess.
• Then we simply use the other material, which
is going to be the one with information
included with it.
• Only one material can be the Limiting Reagent.
• We can also verify our work by comparing the
amounts we obtained and verifying the Law of
Conservation of Mass.
More about Stoichiometry
• Here is a standard problem we worked earlier.
2 KClO3  2 KCl + 3 O2
We had 25.50 g of KClO3
We solved for 9.99 grams of O2 and 15.51
grams of KCl.
Total mass of products = 15.51 + 9.99 = 25.50 g
Mass of Reactants = Mass of Products
• Our answers look good !
More on Stoichiometry
• Let’s do the same thing for a limiting reagent
problem.
• N2 + 3 H2  2 NH3
• We had a 500 g. of N2 and 100 g. of H2 (600 g.
total).
• We solved the problem and got 566.61 g of
NH3.
• We appear to be violating the Law of
Conservation of matter.
More on Stoichiometry
N2 + 3 H2  2 NH3
•
500 g + 100 g  566.61 g
600 g  566.61 g
• But we have material in excess (N2).
• The excess material doesn’t react and
essentially just sits there in the vessel.
• We solved for the amount in excess earlier and
got 33.32 g.
• 600 – 33.32 = 566.66 grams, which is close
enough with rounding.
Review of Terms - Excess Reagent
• The excess reactant is the one you don’t run
out of.
• The amount of product you make is the yield.
• The theoretical yield is the amount you would
make if everything went perfect. This is
obtained by doing a stoichiometry problem
and sometimes called the stoichiometric yield.
• The actual yield is what you make in the lab.
Percent Yield
• % yield =
Actual
x 100%
Theoretical
Or stated another way
• % yield =
what you got
x 100%
what you should have got
Percent Yield – Example # 1
• Aluminum burns in bromine producing
aluminum bromide. During a laboratory 6.0 g
of aluminum reacts with excess bromine. The
student got 50.3 g of aluminum bromide as a
product. What is the percent yield ?
• We need a balanced chemical equation
• 2 Al + 3 Br2  2 AlBr3
Percent Yield – Example # 1
•
•
•
•
•
Here is our equation: 2 Al + 3 Br2  2 AlBr3
6 g of Al = .222 n of Al.
Set up a proportion: 2 Al / 2 AlBr3 = .222 / x
x = .222 moles of AlBr3 (59.21 g )
50.3 / 59.21 = 84.96 %
• Be aware that not all reactions go to
completion for a variety of reasons.
Percent Yield – Example # 2
• Years of experience have proven that the
percent yield for the following reaction is
74.3%
Hg + Br2  HgBr2
• If 10.0 g of Hg and 9.00 g of Br2 are reacted,
how much HgBr2 will be produced ?
• If the reaction did go to completion, how much
excess reagent would be left ?
• We need to do the Limiting Reagent problem
first to determine how much should be made.
Percent Yield – Example # 2
•
•
•
•
•
•
10 g of Hg = .0499 moles of Hg
9 g of Br2 = .0563 moles of Br2
Hg is the Limiting reagent (due to the 1:1 )
1 n Hg / 1 n HgBr2 = .0499 / x
x = .0499 moles of HgBr2 (17.98 g of HgBr2)
The yield is 74.3 % so we would expect to
obtain .743 x 17.98 g = 13.36 grams of HgBr2
Examples
• Commercial brass is an alloy of Cu and Zn. It
reacts with HCl by the following reaction
Zn(s) + 2 HCl(aq)  ZnCl2 (aq) + H2(g)
Cu does not react. When 0.5065 g of brass is
reacted with excess HCl, 0.0985 g of ZnCl2 are
eventually isolated. What is the composition
of the brass ?
• First think of what you are solving for.
• Plan the problem out.
Putting it Together
• ZnCl2 is 47.98 % Zn so .0985 g of ZnCl2
contains .0473 grams of Zn
• The total sample weighs .5065 grams so the
mass of the Cu in the sample must be .4592 g
• .4592 / .5065 = 90.66 %
• Note that there are many different types of
Brass and the percentage varies accordingly.