Transcript ppt - Wits Structural Chemistry

Chapter 3:
Stoichiometry
Lavoisier: Law of
conservation of mass!
Chemical Reaction
Mass before = Mass after.
Scientific contributions cut short – Antoine Lavoisier was guillotined in 1794
at the age of 60. Today, considered to be father of modern chemistry. 1
Chemical Equations
Chemical equations are descriptions of chemical
reactions.
There are 2 parts to an equation: reactants and products:
H2 + O2  H2O
To balance:
change only the stoichiometric coefficients
cannot change the chemical species
Stoichiometric coefficients: numbers in front
of the chemical formulas; give ratio of
reactants and products.
2
Consider the unbalanced chemical equation:
H2 + O2  H2O
The equation H2 + O2  H2O2 is a balanced equation but for
a different reaction.
DO NOT change subscripts!
Balancing options
H2 + 1/2 O2  H2O,
or
2H2 + O2  2H2O, or
4H2 + 2O2  4H2O,
etc.
Preferably, no fractions and smallest possible common
coefficients used.
3
Let’s consider combustion reactions which involve the
burning/oxidation of hydrocarbons to produce CO2 and
H2O always.
Hydrocarbons are organic compounds made up only
of carbon and hydrogen.
For burning/oxidation to occur O2 has to be present!
All combustion equations will include O2(g) as a reactant!
Example 1
Write a balanced equation for the combustion of octane, C8H18.
We know that our reactants include C8H18 and O2(g), and the
products will be CO2 and H2O.
But in what ratios will they react and produce?
Let’s balance!
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C8H18 + O2(g)  CO2 + H2O
Element
Reactants
Products
C
8
1
H
18
2
O
2
3
Start with C
& H. Do O
last because
it appears in
3 species.
We need to increase C on the product side to 8, using a
stoichiometric coefficient.
C8H18 + O2(g)  8CO2 + H2O
Element
Reactants
Products
C
8
8
H
18
2
O
2
17
We need to increase H on the product side to 18, using a
stoichiometric coefficient.
5
C8H18 + O2(g)  8CO2 + 9H2O
Element
Reactants
Products
C
8
8
H
18
9x2 = 18
O
2
16+9 =25
We now need to increase O2 on the reactant side. It needs to
change from 2 to 25. This can only be done using a fraction!
C8H18 + 12.5 O2(g)  8CO2 + 9H2O
Element
Reactants
C
8
H
18
O
12.5x2=25
Products
=
=
=
8
18
25
Now to remove the fraction, and keep coefficients as small as
possible we multiply ALL coefficients by 2!
2C8H18 + 25O2(g)  16CO2 + 18H2O
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Here is another set up of the
same example:
C8H18 + O2  CO2 + H2O
HINT: Balance O last (occurs in 3 species)
but C and H only appear in two!
 Balance C
C8H18 + O2  8CO2 + H2O
 Balance H
C8H18 + O2  8CO2 + 9H2O
 Balance O
C8H18 +
25/ O
2 2
 8CO2 + 9H2O
2C8H18 + 25O2  16CO2 + 18H2O
 We usually add the physical states of the reactants and products
2C8H18(l) + 25O2(g)  16CO2(g) + 18H2O(l)
7
In combustion
reactions:
Rapid reactions that
produce a flame.
Usually very “clean” and
products are
predictable, but
incomplete combustion
(leftover ash) can be very
harmful.
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Example 2
Write a balanced equation for the combustion of purine, C5H4N4.
C  CO2
H  H2O
N  N2
9
Patterns of chemical reactivity
Elements in the same group of the periodic table react in a
similar manner.
Lithium
Sodium
Potassium
Alkali metal + water 
2Na + 2H2O  2NaOH + H2
2K + 2H2O  2KOH + H2
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Alkali earth metal + water 
Mg + 2H2O  Mg(OH)2 + H2
Ca + 2H2O  Ca(OH)2 + H2
Note: These are the EXACT same products as for
alkali metals, except for the stoichiometry!!!
Always look for patterns in chemistry!
11
Combination Reactions
They are characterized by having fewer products
than reactants.
2Mg(s) + O2(g)  2MgO(s)
Mg has combined with O2 to form MgO.
Two reactants combine to form a single product.
Metal + oxygen 
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Non-metal + hydrogen 
C(s) + 2H2(g)  CH4(g)
Metal + halogen 
Ca(s) + Cl2(g)  CaCl2(s)
2Li(s) + F2(g)  2LiF(s)
13
Decomposition Reactions
They are characterized by having fewer reactants
than products.
2NaN3(s)  2Na(s) + 3N2(g)
Metal azide 
The NaN3 is ignited and rapidly decomposes into Na and N2 gas
Metal carbonate 
D
CaCO3(s)  CaO(s) + CO2(g)
limestone,
seashells
lime
14
15
Quantitative vs. Qualitative
Chemical formulae and equations both have qualitative and
quantitative significance.
Chemical formula says what compounds you are working with and the
equation gives a qualitative idea of how the reactants react, and what
products they produce.
N2 + 3H2  2NH3 (balanced)
Quality: We have nitrogen and hydrogen as reactants. The equation
tells us that they undergo a combination reaction to form ammonia.
Subscripts in formulae and coefficients in balanced equations
represent precise quantities.
Quantity: The product, NH3, is composed of 3H for every 1N atom.
The balanced equation shows that 2 molecules of NH3 are produced
for every N2 and 3H2 that are used up.
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Average Atomic Mass
REVISION FROM ISOTOPES!
Most elements occur as mixtures of isotopes.
To determine the average atomic mass of an element we must
consider each isotopes mass as well as their respective
abundances.
Before we consider real isotopes, lets create an analogy
using popcorn 
I have a bowl of popcorn where 93% of the pieces weigh 1.14
mg, 2% weigh 1.09 mg and 5% weigh 1.03 mg.
Which popcorn piece mass is in the greatest and lowest abundance?
Which mass will dominate in calculating the average mass of the
popcorn pieces?
So what do we expect the average mass to be?
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If the bowl contains 1000 popcorn pieces in it. Then 930 of
them would weigh 1.14 mg, 20 would weigh 1.09 mg and 50
would weigh 1.03 mg.
Let’s add those abundances together = 930 + 20 + 50
= 1000.
This shows that each isotope’s (popcorn-piece mass’)
abundance must make up 100% , which is 1000 in this case.
In isotopes be sure to check that your abundances ALWAYS
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add up to 100%
Now let us calculate the
average weight of the popcorn
pieces. (We are expecting it to
be just a little less than 1.14
mg.)
 abundance% 
  abundance% 

  m ass1   
  m ass2   ......
Averageweight  
100%
100%

 

1
2

 93 % 
  2 % 
  5 % 
 1.14 m g  
 1.09 m g  
 1.03m g
Averageweight  
 100% 1
  100%  2
  100% 3

Averageweight  1.06 mg  0.0218mg  0.0515mg
Averageweight  1.13mg
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Example 3
Three isotopes of silicon occur in nature, calculate the average
atomic weight of silicon.
28Si
(92.21%) mass = 27.97693 amu
29Si
(4.70%) mass = 28.97659 amu
30Si
(3.09%) mass = 29.97376 amu
 abundance% 
  abundance% 

  m ass1   
  m ass2   ......
Averageweight  
100%
100%

 

1
2
First make a logical prediction!
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Molecular and Formula Weights
Formula weight (Fr or FW) is the mass of a collection of
atoms represented by a chemical formula.
So it is the sum of atomic masses (Ar) of the atoms in the
chemical formula.
For example, K2Cr2O7
Fr (K2Cr2O7) = 2Ar(K) + 2Ar(Cr) + 7Ar(O)
= 2(39.10) + 2(52.00) + 7(16.00)
= 294.20 amu
Significant
figures!!
21
If the chemical formula of a substance is the same as its
molecular formula, then the formula weight is also called the
molecular weight.
Molecular weight (Mr or MW) is the mass of a collection of
atoms represented by a chemical formula for a molecule.
What is the molecular formula for glucose?
MW(C6H12O6) =
Formula weight and molecular weight are virtually
identical and are used interchangeably.
When dealing with an ionic compound (3D lattice)
we do not use molecular formulas to name them,
and so we could not calculate a molecular weight.
In this case we use the formula weight.
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Water of Crystallization
Some compounds have water of crystallization. These are
water molecules associated with the solid as it crystallizes
from solution.
CuSO4•5H2O
5 water molecules per CuSO4 unit
(Add them into molecular weight!)
Can often be driven off:
CuSO4•5H2O → CuSO4 + 5H2O
blue
grey
anhydrous copper sulphate
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The Mole
Mole: a convenient measure of enormous numbers.
Defined as the amount of matter that contains as many objects
(atoms/molecules etc.) as the number of atoms in 12g of 12C
12 g = 6.02214 x 1023 atoms
6.022 x 1023 is known as Avogadro’s Number, NA
NEW PUBLIC HOLIDAY!
Mole Day: October 23 from 6:02am to 6:02pm, invented May 15, 1991.
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The Mole
1 mole of anything = 6.022  1023 of that thing.
1 mole 12C atoms = 6.022  1023 12C atoms.
1 mole H2O molecules = 6.022  1023 H2O molecules.
1 mole NO3- ions = 6.022  1023 NO3- ions.
Example 4
Calculate the number of carbon atoms in 0.350 mol of C6H12O6.
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Example 4
Calculate the number of carbon atoms in 0.350 mol of C6H12O6.
Let’s look at this more mathematically now!
Avo’s number provides a conversion factor between the number of moles and
molecules of a given compound.
1 mole sugar = 6.022 x 1023 molecules of sugar.
So we can use the “GIVEN and DESIRED” formula! (Given: mol, Desired molecules!)
Desired unit
 Given unit  Desired unit
Given unit
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Molar Mass
The mass in grams of 1 mol of substance (units g/mol).
The molar mass (in g) of any substance is always numerically equal
to its formula weight (in amu).
One HCl molecule weighs 36.46 amu  One mol of HCl weighs
36.46 g.
Can you see where this comes from?
Remember we said that amu and g.mol-1 are equivalent and can be
used interchangeably?
So…..
HCl weighs 36.46 amu = 36.46 g.mol-1.
36.46 g.mol-1 x 1 mol = 36.46 g.
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28 g
1 mol of N2
18 g
58.45 g
1 mol of H2O
1 mol of NaCl
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MAKE SURE YOU UNDERSTAND
THESE RELATIONSHIPS!!!
29
Molar Mass
Inter-converting masses, moles and number of
particles:
Converting from mass to moles and moles to mass is straight
forward:
1. Calculate the molar mass of the substance
2. Use the mole concept
number of moles = mass / molar mass
n = m / Mr
Once the number of moles of the substance is calculated, use
Avogadro’s number to attain number of molecules:
number of molecules = n x AVO
Now we use the number of molecules to solve for number of atoms,
or ions within the substance.
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Molar Mass
TAKE NOTE OF THESE PARTICLES AND HOW
THEY RELATE TO EACH OTHER!
Sugar  C6H12O6
This is 1 MOLECULE of sugar.
This molecule contains 24 ATOMS; 6 carbon, 12 hydrogen and 6
oxygen.
Sulphuric Acid  H2SO4
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Example 5
Calculate the number of H atoms in 20.00 g of C6H12O6.
Strategy
Mass
Mols
Molecules
n = moles
n = moles
m = mass (grams)
NA = Avogadro’s Number
MW = molecular weight
or molar mass (g/mol)
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MUST know two things to figure this out…
(1) mass (g) and
(2) molecular weight (molar mass) (g.mol-1)
…of the compound in question.
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Definitions
AMOUNT  number of moles (must always compare moles!)
n = mass / molar mass, or n = concentration/volume.
PERCENT COMPOSITION
Calculate by dividing the atomic weight (Ar) for each
element by the formula weight (Fr) of the compound, and
express as a percentage:
% Element 
 No. of atoms of Element    A r 
Fr of Compound
100
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Example 6
What is the % by mass (% composition) of O in K2Cr2O7?
First determine atomic and formula weights (or MWs).
% Element 
 No. of atoms of Element    A r 
Fr of Compound
100
Then apply equation (works the same in amu or g/mol).
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Percent Composition
If you are battling to understand what is meant by
percentage (by mass) contributed by each element in the
compound then just think about a box of colourful smarties!
What percentage of the
box of smarties is yellow?
# yellowsm arties
% Yellow 
100
Total # sm arties
So it’s the same as ELEMENTAL composition, except that
we need to consider mass when we use elements!!!
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Empirical Formulas from Analyses
Strategy
Mass %
of elements
Empirical formula
Assume
have 100 g
sample
Moles of each
element
Mole ratio
Grams of
each element
in sample
m
n= M
r
Mr of that specific
element!
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Example 7 (p. 85 8th Ed; p. 93 9th Ed; p. 98 10th Ed)
Ethylene glycol, the substance used in automobile antifreeze,
is composed of 38.7% C, 9.7% H, and 51.6% O by mass. Its
molar mass is 62.1g/mol.
Information given in problem
Ethylene glycol:
38.7% C
9.7% H
51.6% O by mass
Molar mass (from mass spectrometry): 62.1 g mol-1
Calculate:
(a)
(b)
empirical formula
molecular formula
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C
38.7%
38.7g
38.7g
12.01 g mol-1
H
9.7%
assume 100 g
9.7g
number of moles
9.7g
1.008 g mol-1
O
51.6%
51.6g
51.6g
16.00 g mol-1
3.22 mol
9.62 mol
3.23 mol
mole ratio (divide by smallest no. mols)
3.22/3.22
9.62/3.22
3.23/3.22
1
2.99
1.003
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Experimental error is to be expected, so
C
1
H
2.99
O
1.003
H
3
O
1
is likely to be
C
1
The empirical formula is therefore CH3O
which has the formula weight 1 x 12.01 + 3 x 1.008 + 1 x 16.00
= 31.03 g/mol
But molecular weight = 62.1 g/mol, which is twice this value
The molecular formula must be C2H6O2.
40
Combustion Analysis to Determine
Experimental Data
CxHyOz
Catalysis: complete
oxidation of C to CO2
Hydrogen in CxHyOz is
oxidized to H2O and
absorbed in the H2O
absorber
C in CxHyOz is
absorbed here
O in CxHyOz is
determined by mass
difference
41
Example 8
8th ed., p. 99, 3.54(a); 9th ed., p. 107, 3.48(a); 10th ed., p. 114, 3.52(a)
Combustion of 2.78 mg of ethyl butyrate produced 6.32 mg
CO2 and 2.58 mg of H2O. What is the empirical formula of this
compound?
Consider the given information separately. Let’s look at CO2 first. Recall
that we can only determine C and H content from information given!
Ask yourself: How much of the 6.32 mg of CO2 is due to C?
42
Example 8
Combustion of 2.78 mg of ethyl butyrate produced 6.32 mg
CO2 and 2.58 mg of H2O. What is the empirical formula of this
compound?
Now let’s deal with H2O, noting we can only determine H content.
How much of the 2.58 mg of H2O is due to H?
43
Example 8
Combustion of 2.78 mg of ethyl butyrate produced 6.32 mg
CO2 and 2.58 mg of H2O. What is the empirical formula of this
compound?
Now, determine how much O is present from the difference in mass.
Finally, same strategy as before…we have
all three masses and can determine the
empirical formula.
44
45
C
H
O
0.289 x 10-3 g
1.008 g mol-1
0.76 x 10-3 g
16.00 g mol-1
Determine moles of each element
1.73 x 10-3 g
12.01 g mol-1
Divide each by smallest # moles to determine ratios
1.44 x 10-4 mol
4.75 x10-5 mol
2.87 x 10-4 mol
4.75 x10-5 mol
4.75 x 10-5 mol
4.75 x10-5 mol
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Quantitative Information from
Balanced Equations
In a balanced equation the coefficients can be interpreted
as the relative numbers of molecules involved in the
reaction AND as the relative number of moles.
When given any chemical equation you always ensure
that you BALANCE it 1st!
Then you use the given information (usually a mass
in grams, or a number of moles in mol), together
with your calculated molar masses, to solve for
the unknown.
47
Quantitative Information from
Balanced Equations
Example 9
What mass of CO2 is produced when 1.00 g of propane,
C3H8, is burned?
First, write the balanced equation
Then, look at what is given…grams of reactant. In order
to determine how much CO2 is produced, you need to
consider reaction STOICHIOMETRY.
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STRATEGY
49
What mass of CO2 is produced when 1.00 g of propane,
C3H8, is burned?
C3H8 + 5O2 → 3CO2 + 4H2O
1.00 g
??? g
Calculate moles of propane:
Next, consider the MOLAR relationship between propane and carbon dioxide.
It is NOT 1:1, but rather 1:3 (THIS is stoichiometry!).
Stoichiometric coefficients are MOLE relationships!!
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Finally, calculate mass (g) of CO2 produced…need molar mass!
Notice how we link or compare the number of
moles of each substance!! Not their masses,
or anything else!!
Follow-up problem:
What mass of O2 was consumed in the process? Ans: 3.63 g
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Limiting Reactants
If the reactants are not present in stoichiometric amounts, then
at the end of a reaction some reactants are still present.
EXCESS!
Limiting Reactant
One reactant that is consumed
completely in the reaction.
The limiting reactant/reagent hampers us
from continuing with the reaction!
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Non-Chemistry Limiting Reagent
Cheese Sandwich
Ideally…
2B + 1C = B2C
More likely…
4B + 1C = B2C + 2B
You’ve run out of cheese,
time to go to the store.
The cheese limits us from making another sandwich!
Therefore we call it the limiting reagent!
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H2(g) + O2(g) → H2O(l)
Suppose we have 10 mol H2 and 7 mol O2
54
Solving Limiting Reagant Problems
Strategy
Mass
reactants
moles
reactants
Identify limiting
reagent
Moles of product based
on limiting reagent
Mass of product
Example10
Al reduces Fe2O3 to Fe. How much Fe is produced from the
reaction of 30.0 g of Al and 100 g Fe2O3?
55
Example 10 Al reduces Fe2O3 to Fe. How much Fe is
produced from the reaction of 30.0 g of Al and 100 g Fe2O3?
1
2
3
Mass of
reactants
Moles of
reactants
Identify limiting
reagent
4
Moles product
based on
limiting reagent
5
Mass of
56
product
From reaction stoichiometry
2 Al : 1 Fe2O3
1
Mass
reactants
2
mols
reactants
3
4
5
Identify limiting
reagent
Mols of product
based on
limiting reagent
Mass of
product
__________is in excess, and
__________ is the limiting reagent!
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1
Mass
reactants
2
mols
reactants
Now compare stoichiometry of Fe to Al.
3
4
5
Identify limiting
reagent
Mols of product
based on
limiting reagent
Mass of
product
Finally, calculate mass of Fe produced.
m = n x Mr
=
=
of Fe
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Percentage Yield
The percent yield relates the actual yield (amount of
material recovered in the laboratory) to the theoretical
yield:
Actual yield
% Yield 
 100
Theoretical yield
Theoretical yield
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Percentage Yield
So, we conduct an experiment in the lab:
Let us make a metal oxide, calcium oxide!
From theory we now know that we need to do a combination
reaction using a metal and O2 as our reactants.
The equation: Ca + O2  CaO
Balance: 2Ca + O2  2CaO
We are given 2.053 g of calcium, and the oxygen used is
atmospheric oxygen, and therefore is in excess (unlimited
quantity).
From this information we can calculate how much product
the stoichiometry of our equation predicts we will produce 
60
theoretical yield.
Percentage Yield
Example 11
Preparation of calcium oxide is achieved by reacting 2.053 g of calcium
with an excess of oxygen. Calculate the theoretical yield of calcium
oxide, and solve for the percentage yield, if the real mass of the product
was measured to be 2.26 g.
2Ca
+
O2

2CaO
m = 2.053 g
m = excess
m = desired quantity!
Mr = 40.078g/mol
Mr = 32.00 g/mol
Mr = 56.08 g/mol
n= mass/molar mass
n= n(Ca) /2
n(CaO) = n(Ca)
n=
n = 0.05123 mol
2
= 0.02561 mol
2.053 g
40.078 g.mol-1
= 0.05123 mol
n(CaO) = n(Ca)
n = 0.05123 mol
61
Example 11
Calculate the theoretical yield of calcium oxide, and solve for the percentage
yield, if the real mass of the product was measured to be 2.26 g.
2Ca
+
O2

2CaO
m = 2.053 g
m = excess
m = desired quantity!
Mr = 40.078g/mol
Mr = 32.00 g/mol
Mr = 56.08 g/mol
n= 0.05123 mol
n= 0.02561 mol
n= 0.05123 mol
mass(CaO) = n x Mr
= 0.05123 mol x 56.08 g/mol
= 2.873 g
Actual yield
% Yield 
 100
Theoretical yield
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Expressing Concentration
There are many ways in which you can express the
concentration of a solution.
Qualitatively OR Quantitatively
DILUTE or
CONCENTRATED
•Dilute = Low
concentration of solute.
•Concentrated = High
concentration of solute.
MANY WAYS TO
DESCRIBE THIS!
•Mass percentage
•Mole fraction
•Molarity
•Molality
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Quantitative Concentration
Mass Percentage
m assof com ponentin solution
Mass % of com ponent
100
Total m asssolution
A 24.00 % NaCl solution by mass contains 24.00 g of
NaCl in a 100.0 g solution.
So if our solution has a total mass of 634.0 g, then what
mass of NaCl does it contain?
24.00 % of the total mass is due to the NaCl.
Ans: 152.2 g of NaCl
The remaining mass is due to the SOLVENT (water).
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Parts per Million (ppm)
very dilute solutions!
The concentration of a solution in grams of solute per
106 (million) grams of solution.
This is equivalent to milligrams (mg) of solute, per
litre of solution (for aqueous solutions).
m assof com ponentin solution
ppmof com ponent
106
Total m asssolution
Example: A 3 ppm solution
 contains 3 particles of solute, for every 1 million
(106) particles of solution.
 3 g of solute for every million grams of solution.
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Parts per Billion (ppb)
even more dilute solutions!
The concentration of a solution in grams of solute per
109 (billion) grams of solution.
This is equivalent to micrograms (µg) of solute, per
litre of solution (for aqueous solutions).
m assof com ponentin solution
ppbof com ponent
109
Total m asssolution
Example: 329 ppb corresponds to 329 µg of
solute per litre of solution.
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Example 12
Calculate the mass percentage of NaCl in a solution
containing 1.50 g of NaCl in 50.0 g of water.
m assof com ponentin solution
Mass % of com ponent
100
Total m asssolution
67
Example 13
A bleaching solution contains 3.62 mass percent sodium
hypochlorite. Calculate the mass of sodium hypochlorite in
a bottle containing 2500 g of bleaching solution.
ASIDE what is the chemical formula of sodium hypochlorite?
NaOCl
68
Example 14
In an experiment 2.569 mg of CuSO4 was dissolved in
water to form a 2.000 kg solution. Express the
concentration of the solution in parts per million.
m assof com ponentin solution
ppmof com ponent
106
Total m asssolution
RULE: Both masses MUST be in the same unit of mass!
Convert 2 kg into mg:
1st convert kg  g and then g  mg
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Quantitative Concentrations in terms of moles
Mole Fraction
m olesof com ponent
Mole fractionof Com ponent
Total m olesof all com ponents
We often express mole fraction using an X.
Example: the mole fraction of HCl in a solution of
HCl  XHCl.
1.00 mol of HCl, dissolved in 8.00 mol of water,
gives a mole fraction of HCl of 0.111.
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Quantitative Concentrations in terms of moles
Mole Fraction
The sum of the mole fractions of all the
components in a solution must equal 1!
Let’s test this theory using our previous example!
1.00 mol of HCl dissolved in 8.00 mol of water gives a
mole fraction of HCl of 0.111.
So lets calculate the mole fraction of the water!
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Quantitative Concentrations in terms of moles
Molarity
Molesof solute
Molarity
Litres of solution
Units  mol.L-1(or more
commonly: M).
Example 15: Dissolving 2.36 g of CuSO4
in enough water to fill a 250 ml volumetric flask.
1.
2.
3.
Weigh out 2.36 g of
these crystals.
Place in volumetric
flask.
Fill to the line with
water.
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Quantitative Concentrations in terms of moles
Molarity
Example 15: Dissolving 2.36 g of CuSO4 in enough
water to fill a 250 ml volumetric flask.
1. Calculate the number of copper sulphate moles (n)
2. Remember to convert your volume of solution (ml L).
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Quantitative Concentrations in terms of moles
Molality
NB: Notice molaLity not molaRity!
Molesof solute
Molality
kg of solvent
There are two major differences between Molality and Molarity!
They both appear in the denominator of the equation:
MOLARITY
Consider the VOLUME of the entire solution
(always in L)!
MOLALITY
Consider the MASS of the solvent (always in kg)!
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Be sure that you are able to
calculate concentrations using each
of these methods.
Example 16:
A solution containing equal masses of glycerol (C3H8O3)
and water has a density of 1.10 g/ml. Calculate the
a) Molality of glycerol
b) Mole fraction of glycerol
c) Molarity of glycerol
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What information has been given?
•We have 1 solution. It contains 2 components.
•Each component has the same mass.
We don’t know what mass though!
•The solution’s density is 1.10 g/mol.
Mass
Density 
Volume
Because we know the masses of glycerol and water are
equal….we can make an assumption!
Assume that each component has a mass of 10.00 g, and
therefore the mass of the entire solution must be 20.00 g.
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Example 16:
A solution containing equal masses of glycerol (C3H8O3) and water has a density of
1.10 g/ml. Calculate the
a) Molality of glycerol
b) Mole fraction of glycerol
c) Molarity of glycerol
a) Molality involves the number of moles of the component
in question, and the mass in kg of the solvent.
From our assumption:
Mass of solvent = 10.00 g
= 20.00 x 10-3 kg
We can now calculate the number of moles of glycerol.
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Example 16:
A solution containing equal masses of glycerol (C3H8O3) and water has a density of
1.10 g/ml. Calculate the
a) Molality of glycerol
b) Mole fraction of glycerol
c) Molarity of glycerol
b) Calculating the mole fraction of glycerol involves
calculating the number of moles of glycerol, and the total
number of moles within the solution (nglycerol + nwater).
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Example 16:
A solution containing equal masses of glycerol (C3H8O3) and water has a density of
1.10 g/ml. Calculate the
a) Molality of glycerol
b) Mole fraction of glycerol
c) Molarity of glycerol
c) To calculate the molarity we need the number of moles of
glycerol, and the volume of the solution in litres!
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