Transcript Chapter 3 Sections 3.9-3.10 Powerpoint

Chocolate Chip Cookies
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2.25 cups flour
8 Tbsp butter
0.5 cups shortening
0.75 cups sugar
0.75 cups brown sugar
1 tsp salt
1 tsp baking soda
1 tsp vanilla
0.5 cups Egg Beaters
2 cups chocolate chips
Chocolate Chip Cookies
• 2.25 cups flour
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How much?
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8
0.5
0.75
0.75
1
1
1
0.5
2
Tbsp butter
cups shortening
cups sugar
cups brown sugar
tsp
salt
tsp
baking soda
tsp
vanilla
cups Egg Beaters
cups Chocolate chips
What units?
Of what?
Chocolate Chip Cookies
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How much?
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2.25
8
0.5
0.75
0.75
1
1
1
0.5
2
flour
butter
shortening
sugar
brown sugar
salt
baking soda
vanilla
Egg Beaters
Chocolate chips
Of what?
Chocolate Chip Cookies
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How much?
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2.25
8
0.5
0.75
0.75
1
1
1
0.5
2
cups flour
Tbsp butter
cups shortening
cups sugar
cups brown sugar
tsp
salt
tsp
baking soda
tsp
vanilla
cups Egg Beaters
cups Chocolate chips
What units?
Of what?
Get on with it!
What does this have to do with
CHEMISTRY?
2.25 cups flour + 8 Tbsp butter + 0.5 cups shortening +
0.75 cups sugar + 0.75 cups brown sugar + 1 tsp salt +
1 tsp baking soda + 1 tsp vanilla + 0.5 cups Egg Beaters
+ 2 cups Chocolate chips
unit
substance
coefficient
(177ºC)
1 batch of chocolate chip cookies!
(a synthesis reaction)
Welcome to
STOICHIOMETRY
•The coefficients in chemical equations represent numbers
of molecules or formula units, not masses of molecules or
formula units.
•When a reaction is to be run in a laboratory or chemical
plant, the amount of substances needed cannot be
determined by counting molecules directly.
•Counting is always done by weighing.
•We will see how chemical equations can be used to
determine the masses of reacting chemicals.
3NaOH (aq) + H3PO4 (aq)
Na3PO4 (aq) + 3 H2O
•In the above reaction, the coefficients give us the ratios by
moles of the reactants and products.
•The coefficients tell us that to make 1 mol of Na3PO4 from
1 mol of H3PO4, we must also use 3 mol of NaOH.
•We don’t however, have to carry out the reaction with
these actual numbers of moles.
•Whatever quantities we choose must be in the proportions
set by the coefficients.
•Regardless of the scale of the reaction, the coefficients of a
chemical equation give the ratio in which the moles of one
substance react with or produce moles of another.
3NaOH (aq) + H3PO4 (aq)
Na3PO4 (aq) + 3 H2O
•We can look at this equation as a calculating tool, because
its coefficients give us stoichiometric equivalencies
between the substances involved.
•For example, from the above equation we can generate
the following equivalencies:
3 mol NaOH ↔ 1 mol H3PO4
3 mol NaOH ↔ 1 mol Na3PO4
3 mol NaOH ↔ 3 mol H2O
1 mol H3PO4↔ 1 mol Na3PO4
1 mol H3PO4↔ 3 mol H2O
Any of these can be used to construct conversion factors
called mole ratios for stoichiometric calculations.
•Example: How many moles of sodium phosphate, Na3PO4,
can be made from 0.240 mol of NaOH by the following
reaction?
3NaOH (aq) + H3PO4 (aq)
Na3PO4 (aq) + 3H2O
•In practical work, a chemist is often confronted by a
question such as the following.
•“If I start with so many grams of reactant A, how many
grams of reactant B ought I use, and how many grams of a
particular product should be produced?”
•Notice that the question concerns grams, not moles, for
the practical reason that masses in grams are delivered by
laboratory balances.
•The coefficients of the desired reaction, however, know
nothing about grams, only about relative numbers of moles.
•If we know two facts, namely, the balanced equation and
the mass of any substance in it, we can calculate the
required or expected mass of any other substance in the
equation.
•Example: Portland cement is a mixture of the oxides of
calcium, aluminum, and silicon. The raw material for its
calcium oxide is calcium carbonate, which occurs as the
chief component of a natural rock, limestone. When
calcium carbonate is strongly heated it decomposes by the
following reaction. One product CO2, is driven off to leave
the desired CaO as the only other product.
CaCO3 (s)
heat
CaO (s) + CO2 (g)
A chemistry student is to prepare 1.50 x 102 g of CaO in
order to test a particular “recipe” for portland cement. How
many grams of CaCO3 should be used, assuming that all will
be converted?
Calculating Masses of Reactants and Products in Chemical
Reactions
1. Balance the equation for the reaction.
2. Convert the known mass of the reactant or product to
moles of that substance.
3. Use the balanced equation to set up the appropriate
mole ratios.
4. Use the appropriate mole ratios to calculate the number
of moles of the desired reactant or product.
5. Convert from moles back to grams if required by the
problem.
•Example: One of the most spectacular reactions of
aluminum, the thermite reaction, is with iron oxide, Fe2O3 , by
which metallic iron is made. So much heat is generated that
the iron forms in the liquid state. The equation is
Al (s) + Fe2O3 (s)
Al2O3 (s) + Fe (l)
A certain welding operation, used over and over, requires that
each time at least 86.0 g of Fe be produced. What is the
minimum mass in grams of Fe2O3 that must be used for each
operation? Calculate also how many grams of aluminum are
needed.
The Wisdom of Gallagher
Why are there Interstate
Highways in Hawaii?
Why are there floatation
devices under plane seats
instead of parachutes?
Why do we drive on parkways
and park on driveways?
Why do hot dogs come ten to a package
and hot dog buns only eight?
Hot Dogs in the News
Takeru Kobayashi of
Japan downed 44½
hot dogs in 12
minutes.
One hot dog = one hot dog
+ one bun.
WHAT IF…
Mr. Kobayashi didn’t do his math correctly. He bought 5 packs of hot dogs (10
per package) and 5 packs of hot dog buns (8 per package). How many hot dogs
(according to the official formula) could he have eaten?
Source: CNN.com
Hot Dogs in the News
One hot dog = one hot dog + one bun.
WHAT IF…
Mr. Kobayashi didn’t do his math correctly. He bought 5
packs of hot dogs (10 per package) and 5 packs of hot dog
buns (8 per package). How many hot dogs (according to the
official formula) could he have eaten?
5 hot dog packs
x
10 hot dogs
=
50 hot dogs
1 hot dog pack
5 bun packs
x
8 buns
1 bun pack
=
40 buns
40 possible
hot dogs
Source: CNN.com
Let’s Revisit the Cookies (again)…
For 1 batch:
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2.25 cups flour
8 Tbsp butter
0.5 cups shortening
0.75 cups sugar
0.75 cups brown sugar
1 tsp salt
1 tsp baking soda
1 tsp vanilla
0.5 cups Egg Beaters
12 oz. Chocolate chips
In my pantry, I have:
• 5.5 cups of flour
• 16 Tbsp of butter
• lots of everything else
How many batches of cookies can I make?
Let’s Revisit the Cookies (again)…
For 1 batch:
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2.25 cups flour
8 Tbsp butter
0.5 cups shortening
0.75 cups sugar
0.75 cups brown sugar
1 tsp salt
1 tsp baking soda
1 tsp vanilla
0.5 cups Egg Beaters
12 oz. Chocolate chips
How many batches of cookies can I make?
EXCESS
5.5 c flour
x
1 batch cookies
=
2.25 c flour
2.4 batches
LIMITING
16 Tbsp butter
x
1 batch cookies
8 Tbsp butter
2.0 batches
=
Now I Want to Bake a Cake!
But do I have all the ingredients I
need?
How much flour do I have left after baking all those cookies?
5.5 c flour
1 batch cookies
x
=
2.4 batches of
cookies
2.25 c flour
GONE!
16 Tbsp butter
x
1 batch
=
2.0 batches of
cookies
8 Tbsp butter
2.0 batches
x
2.25 cups flour
SOME
FLOUR LEFT
OVER…
=
4.5 cups flour used
1 batch cookies
5.5 cups – 4.5 cups = 1.0 cups left
•Many times a chemist will mix reactants in a mole ratio that
does not agree with the coefficients of the equation.
•Some reactions proceed better when one reactant is in
stoichiometric excess, for example.
•One such reaction is the preparation of ammonia, NH3,
from its elements.
N2 (g) + 3H2 (g)
2NH3 (g)
•Suppose a chemist mixed 1.00 mol of N2 with 5.00 mol of
H2. What is the maximum number of moles of product that
could form?
N2 (g) + 3H2 (g)
2NH3 (g)
•Note the coefficients tell us that 1 mol of N2 consumes 3
mol of H2.
1 mol N2 ↔ 3 mol H2
•But 5 mol of H2 was used, not 3, so there will be 2 mol of H2
left over.
•Once the 1 mol of N2 taken is consumed, no additional NH3
can form.
•Therefore, the reactant that is completely consumed limits
the amount of product that forms, so it is called the limiting
reactant.
•In this reaction, N2 is the limiting reactant, it limited the
amount of NH3 that was formed.
•Example: In an industrial process for making nitric acid,
the first step is the reaction of ammonia with oxygen at high
temperature in the presence of a platinum gauze. Nitrogen
monoxide forms as follows:
4NH3 + 5O2
4NO + 6H2O
How many grams of nitrogen monoxide can form if a
mixture of 30.00 g of NH3 and 40.00 g of O2 is taken initially?
How many grams of the excess reactant is left over?
•In most experiments designed for chemical synthesis, the
amount of a product obtained falls short of the calculated
maximum amount.
•Losses occur for several reasons.
•Some are mechanical, such as materials sticking to
glassware.
•But one of the most common causes of obtaining less than
the stoichiometric amount of a product is the occurrence of
a competing reaction.
•It produces a by product, a substance made by a reaction
that competes with the main reaction.
•The actual yield of desired product is simply how much is
obtained or isolated, stated in mass units.
•The theoretical yield is the amount of product formed
when the limiting reactant is completely consumed and no
losses occur.
•When less than the theoretical yield of product is obtained,
chemists generally calculate the percentage yield or percent
yield to describe how well the reaction went.
•The percent yield is the actual yield calculated as a
percentage of the theoretical yield.
actual yield
Percent yield =
x 100%
theoretica l yield
•Example: A chemist set up a synthesis of phosphorus
trichloride by mixing 12.0 g P with 35.0 g Cl2 and obtained
42.4 g of PCl3. Calculate the percent yield of this compound.
The equation for the main reaction is:
2P (s) + 3Cl2 (g)
2PCl3 (l)