Without passion, you don*t have energy

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CHEMISTRY UNIT 3!
CLASS 2
“WITHOUT PASSION, YOU DON’T HAVE
ENERGY; WITHOUT ENERGY, YOU HAVE
NOTHING. NOTHING GREAT IN THE WORLD
HAS BEEN ACCOMPLISHED WITHOUT
PASSION”
– DONALD TRUMP -
MORE CALCULATIONS:

The true meaning of the ‘mole’…
Ie/ 1 mole of peas = 6.022 x 1023 peas.
Anything with 6.022 x 1023 units = 1 mol
Avogadro’s
number
KEYWORDS:
‘How many…atoms/ molecules are there’
ALSO …


We also know that m = n x Mr
This lets us find out how many molecules/atoms
there are from the MASS!
Mosquitoes have 47 teeth

Eg/ Sam weighed out 5.6g of potassium
dichromate (K2Cr2O7 ).
a) How many molecules are there?
b) How many dichromate molecules?
c) How many potassium ions are there?
d) How many oxygen atoms are there?
e) How many atoms are there?
STOICHIOMETRY…
QUESTIONS:

Aluminum is an active metal that when placed in
hydrochloric acid produces hydrogen gas and
aluminum chloride. How many grams of
aluminum chloride can be produced when 3.45
grams of aluminum are reacted with an excess of
hydrochloric acid?
ANOTHER FAMOUS MOLE FORMULA
n = CV
n= mol.
C = concentration
(Molarity – mol/L Why?)
V = volume (L)
Eg/ Tracey had 22.5 ml of hydrochloric acid it was at a
concentration of 5M. How many mole of HCl is there?
Note: We will do this in more detail as we start
volumetric analysis.
LIMITING / EXCESS REACTIONS:

Sometimes the reactants are not mixed in EXACT
ratio.
This is when we have to do all the calculations with
the LEAST amount
If I have 2 mole of X and 3 mole of Y
I cannot use 3 mole in my calculations
Since X runs out it is ‘impossible’ to create Z and Y.
HOW TO PICTURE THIS…
1)
Each molecule will
collide… once it does it
creates the ‘pink pill’
But there are only 3 ‘boxes’ and
6 ‘circles’
+
=
5 STEPS TO FIND LIMITING REACTANT
A PROBLEM:
HOW TO FIND OUT HOW MUCH REACTANT
IS LEFT OVER
EXAMPLE

What is the mass of excess left?
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
The mass of superoxide is 45 g and water is 32 g.
THEORETIC YIELD VS. ACTUAL YIELD
- Theoretic yield is the amount of product formed in
theory.
- Actual yield is the amount of product ACTUALLY
made.
- It will be different due to loss in precipitate, or
other unavoidable error.
A PROBLEM…

Suppose the theoretical yield for an experiment
was calculated to be 19.5 grams, and the
experiment was performed, but only 12.3 grams
of product were recovered. Determine the % yield.
PERCENTAGE BY MASS
UNIT CONVERSION
µg
Gg
ng
WORKED EXAMPLE:
0.0008 mg into grams.
 7458 μg into grams
 0.04 mg into μg
 560493 ng into mg
 456 k-ants to M-ants

Hint:
In chem. you generally work will SI units (grams
for mass, L for volume) unless otherwise stated
in the exam
Students
with 85 95
Students
that get
99 +
Students
below 70
Students
70-85
GRAVIMETRIC ANALYSIS
PRECIPITATE = CLOUDY
THE STEPS OF GRAVIMETRIC ANALYSIS
WHY MUST THERE BE EXCESS ?




We need excess AgNO3 to make sure that all the salt
within the kebab undergoes the reaction
The more AgNO3 we add the greater the
probability that there will be a collision with the
salt.
This will increase the
chance that 100% of the salt
turns to AgCl (s)
This way error is decreased.
THE KEBAB STORY….PART 2:



The NaCl from the dissolved kebab will react with the
excess AgNO3
This will create a precipitate – this precipitate must
be STABLE as if it reacts it will look like there is
less of NaCl
This will create a precipitate
– this precipitate must be
STABLE as if it reacts it will
look like there is less of NaCl
We then filter the solution…
KEBAB’S ADVENTURE PART 3…

All chemists care about is the solid precipitate
because that’s the only PRACTICLE thing they can
measure – i.e. weighing

But before weighing –
1) The precipitate must be washed: To remove
contaminants such as free ions that can effect
weight
2) The precipitate must be dried: Water on
the AgCl will increase mass – we only care
about AgCl … all the water must be removed.
A FEW NOTES ON THE PRECIPITATE:

The precipitate must be inert (i.e. non reactive!!!)
otherwise you cannot accurately measure it.
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The precipitate must have a large molar mass –
The larger it is the less error there will be in the
calculations.
The precipitate must be FULLY INSOLUBLE –
some are partial DO NOT use them as you will not
get an accurate answer.
The precipitate must be stable when
heated and must have a KNOWN molar
mass.
HOW DO WE GET RID OF THE WATER?

1) Weigh the precipitate + water (initial)

2) The precipitate must be heated in an oven
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
This way the water will evaporate
Nothing should happen to the precipitate (no
shrinking, charring etc)
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3) Take it out – weigh + record
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4) Put it back into oven – weigh + record

5) Repeat cycle 3) & 4) until constant
mass.
A PROBLEM…

Jack wanted to calculate the salt content of a
packet of fries from Mc Donald’s:
a) Write a flow chart of what he should do.
b) Jack decides to heat the solution when the
precipitate is forming. Why?
WHAT DO WE USED THE WEIGHED
PRECIPITATE FOR?
A PROBLEM:

Jenny wants to calculate the amount of salt in
pizza. The pizza weighs 0.78 kg. Her doctor
advised her that she shouldn’t eat any foods that
have 3.5% of salt by mass (her blood pressure is
too high). If the mass of the precipitate weighed
1097 mg … can Jenny eat the pizza?
OTHER TYPES OF QUESTIONS:

Calculate the grams analyte of the precipitate for
the following:
i) P (at wt =30.97) in Ag3PO4 (FW = 711.22)
ii) Bi2S3 (FW 514.15) in BaSO4 (FW = 233.40)

Trick: Look for the common element. Make sure
that the elements are equivalent in number!!!
ANOTHER EXAMPLE:

Phosphate in a 0.2711 g sample was precipitated
giving 1.1682 g of (NH4)2PO4.12 MoO3
(FW = 1876.5).
i) Find percentage P (at wt = 30.97)
ii) Find the percentage P2O5 (FW = 141.95) in the
sample.
YET ANOTHER:

Manganese in a 1.52 g sample was precipitated
as Mn3O4 (FW = 228.8) weighing 0.126 g.
i) Find percentage Mn2O3 (FW = 157.9)
ii) Find percentage of Mn (at wt = 54.94) in the
sample
AND ANOTHER……

A mixture containing only FeCl3 (FW = 162.2)
and AlCl3 (FW = 133.34) weighs 5.95 g. The
chlorides are converted to hydroxides and ignited
to Fe2O3 (FW = 159.7) and Al2O3 (FW = 101.96).
The oxide mixture weighs 2.26 g.
Calculate:
i) the percentage Fe (at wt = 55.85)
ii) the % of Al (at wt = 26.98) in the sample.

A 0.4960 g sample of a CaCO is dissolved in an
acidic solution. The calcium is precipitated as
CaC2O4 .H2O and the dry precipitate is found to
weigh 0.6186 g.
What is the percentage of CaO in the sample?
WHAT I WANT YOU TO DO:
The questions Set 2
 Go over the lectures (both)
 Attempt the topic test for lecture 1
 Study for at least 30 mins. a day for chemistry.


If you get any SAC’s send a PDF file to my email
and I will go over the important things you must
include in email.
NEXT WEEK WE GO ON WITH:

Gravimetric analysis – we finish it

Empirical formulae

Gas equations

Begin volumetric analysis

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