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# Stoichiometry and Mole Calculation Notes

advertisement ```Stoichiometry Objectives
Objectives: students will be able to…
Stoichiometry
Complete Notes
Stoichiometry Vocabulary
(10 Words)
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actual yield
molar mass
percent yield
Mole
Stoichiometry
• excess reagent
1.
2.
3.
4.
5.
6.
7.
8.
9.
calculate the molar mass of a compound.
calculate the percent composition by mass of an element in a compound.
calculate the mass of an element present in a sample of a compound.
distinguish between an element's atomic mass and molar mass.
calculate the mass of a substance given the number of moles present.
calculate the number of moles present given the mass of a sample.
determine the mole ratio of compounds in a balanced chemical equation.
calculate stoichiometric quantities when given a balanced chemical equation.
identify the limiting reagent and calculate the quantity of products produced in
such a reaction.
10. calculate theoretical yield, actual yield, and percent yield.
Stoichiometry Notes
• Stoichiometry is a process used by chemists to make
predictions about the quantities of substances that
react and are given off as products in chemical reactions
and relating those quantities to one another.
• Stoichiometry relates reactants to products in amounts
not masses.
– Therefore the problems are worked in terms of moles not
grams.
– However, the information both provided and requested is
often in terms of grams.
– Therefore, the ability to convert between these two
quantities is essential.
Steps to Solve Stoichiometry Problems
1.
2.
Identify what you know and what you do not know and write it down.
Convert any masses given to moles.
•
3.
Mass of substance (g)
= Number of Moles
Molar Mass of substance (g/mol)
Determine the Mole Ratio needed.
• Coefficient of unknown substance in BALANCED chemical equation
Coefficient of known substance in BALANCED chemical equation
• mol unknown molecule
# mol known molecule
• # mol comes from BALANCED chemical equation NOT the problem itself.
4.
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Basic Stoichiometry Practice
1. How many moles of sodium will react with
water to produce 4.0 mol of hydrogen in the
following reaction? (8.0 mol Na)
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Use the mole ratio to determine the amount of product the known
reactant will produce.
• Known Moles X Mole Ratio = Moles of Product
5.
Convert moles of product to grams.
• Number of Moles X Total Molar Mass = Mass of substance (g)
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Basic Stoichiometry Practice …continued
2. Aluminum will react with sulfuric acid in the
following reaction. How many moles of sulfuric
acid will react with 18 moles of Al? (27 mol H2SO4)
2Al(s) + 3H2SO4(l) → Al2(SO4)3(aq) + 3H2(g)
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Basic Stoichiometry Practice …continued
3. Aluminum will react with sulfuric acid in the
following reaction. Based on your previous
answer, how many moles of each product will be
produced? (27 mol H2 and 9 mol Al2(SO4)3)
2Al(s) + 3H2SO4(l) → Al2(SO4)3(aq) + 3H2(g)
You
Basic Stoichiometry Practice …continued
4. How many moles of lithium chloride will be
formed in the reaction of chlorine with 0.046 mol
of lithium bromide in the following reaction?
(0.046mol LiCl)
2LiBr(aq) + Cl2(g) → 2LiCl(aq) + Br2(l)
Homework
1. Propane burns in excess oxygen according to the
following reaction:
C3H8 +5O2 → 3CO2 + 4H2O
a.
b.
How many moles of CO2 and H2O are formed from 3.85
moles of propane?
If 0.647 moles of oxygen is used in the burning of propane,
how many moles of CO2 and H2O are produced? How many
moles of propane are consumed?
2. How many moles of ammonium sulfate can be made
from the reaction of 30.0 mol of NH3 with H2SO4
according to the following equation?
2NH3 + H2SO4 → (NH4)2SO4
Homework Answers
End Day 1
1. Propane burns in excess oxygen according to the
following reaction:
C3H8 +5O2 → 3CO2 + 4H2O
a.
How many moles of CO2 and H2O are formed from 3.85
moles of propane? CO2 = 11.55 mol and H20 = 15.4 mol
b. If 0.647 moles of oxygen is used in the burning of propane,
how many moles of CO2 and H2O are produced? How many
moles of propane are consumed?
CO2 = 0.3882 mol and H20 = 0.5176 mol and C3H8 = 0.1294 mol
2. How many moles of ammonium sulfate can be made
from the reaction of 30.0 mol of NH3 with H2SO4
according to the following equation?
2NH3 + H2SO4 → (NH4)2SO4
(NH4)2SO4 = 15.0 mol
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Stoichiometry with Conversions Practice
1. Potassium chlorate is sometimes decomposed
in the laboratory to generate oxygen. The
reaction is 2KClO3(s) → 2KCl(s) + 3O2(g). What
mass of KClO3 do you need to produce 0.50
mol O2? 40.85 g KClO3
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Stoichiometry with Conversions Practice
…continued
3. Methane burns in air by the following
reaction: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
What mass of water is produced by burning
500.00 g of methane?
1122.7414 g H2O
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Stoichiometry with Conversions Practice
…continued
2. How many moles of aluminum will be
produced from 30.0 kg Al2O3 in the following
reaction: 2Al2O3→4Al + 3O2?
1176.9321 mol Al
You
Stoichiometry with Conversions Practice
…continued
4. Using the reaction 2Ag(s) + Cl2(g) → 2AgCl(s)
Calculate the mass of silver needed to react
with chlorine to produce 84.0 grams of silver
chloride.
63.2227g Ag
Limiting Reactants
End Day 2
• Substances can only react in a chemical reaction until
the reactants are used up.
• Reactions only occur when reactants are in a specific
ratio to one another and in multiples thereof.
o If 1 pencil reacts with 4 pieces of paper, then 2 pencils will
react with 8 pieces of paper. 3 pencils will react with 12
pieces of paper etc. at a continuous ratio of 1:4.
o If I have 32 sheets of paper, how many pencils do I need to
use them all up?
o If I have 1236 pencils and 5000 sheets of paper, which one
will be used up first?
• The one used up first is the limiting reactant.
Steps to Solve Limiting Reactants Problems
1.
2.
Identify what you know and what you do not know and write it down.
Convert any masses given to moles.
•
3.
4.
Steps to Solve Limiting Reactant Problems
…continued
6.
Mass of substance (g)
= Number of Moles
Molar Mass of substance (g/mol)
Determine the Mole Ratio needed.
• If there is more “Unknown” calculated than given, then it is the limiting reactant.
• If there is more “Unknown” given than calculated, then it is the excess reactant and the “Known” is
the limiting reactant.
Extension: If the problem continues to ask specifically how much one of the products in the
equation will be formed by the reaction…Steps 1-6 Same as Above and then:
• Coefficient of unknown substance in BALANCED chemical equation
Coefficient of known substance in BALANCED chemical equation
• mol unknown molecule
# mol known molecule
• # mol comes from BALANCED chemical equation NOT the problem itself.
7.
Use the mole ratio to determine the amount of product the known
reactant will produce.
8.
Convert moles of product to grams.
• Number of Moles X Total Molar Mass = Mass of substance (g)
Determine your new Mole Ratio using the Limiting Reactant identified above.
• Coefficient of unknown substance in BALANCED chemical equation
Coefficient of Limiting Reactant in BALANCED chemical equation
• mol unknown molecule
# mol Limiting Reactant
• # mol comes from BALANCED chemical equation NOT the problem itself.
Use the mole ratio to determine the amount of substance the Limiting Reactant will produce.
• Known Moles of Limiting Reactant X Mole Ratio = Moles of Unknown
• Known Moles X Mole Ratio = Moles of Product
5.
Compare the amount of “Unknown” calculated and compare it to the amount of “Unknown”
given in the problem.
9.
Convert moles of product to grams.
• Number of Moles X Total Molar Mass = Mass of substance (g)
Limiting Reactants Practice
I
1. Calcium hydroxide, used to neutralize acid spills,
reacts with hydrochloric acid according to the
following equation: Ca(OH)2 + 2HCl → CaCl2 + 2H2O.
If you have spilled 6.3 mol of HCl and put 2.8 mol of
Ca(OH)2 on it, which substance is the limiting
reactant?
We
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Limiting Reactants Practice …continued
2. Aluminum oxidizes according to the following
equation: 4Al + 3O2 → 2Al2O3.
0.048 mol of powdered Al is placed into a
container containing 0.030 mol O2. What is
the limiting reactant?
You
Limiting Reactants Practice …continued
Limiting Reactants Practice …continued
3. Chlorine can replace bromine in bromide
compounds forming a chloride compound and
elemental bromine. 2KBr + Cl2 → 2KCl + Br2 is
an example of this reaction.
When 0.855 g of Cl2 and 3.205 g of KBr are
mixed in solution, which is the limiting
reactant? How many grams of Br2 are formed?
4. A process by which zirconium metal can be
produced from the mineral zirconium(IV)
orthosilicate, ZrSiO4 , starts by reacting it with
chlorine gas to form zirconium(IV) chloride.
ZrSiO4 + 2Cl2 → ZrCl4 + SiO2 + O2
What mass of ZrCl4 can be produced if 862 g
of ZrSiO4 and 950.0 g of Cl2 are available? (You
must first determine the limiting reactant.)
Percent Yield
End Day 3
• In theory, everything is perfect. People do what is right, even
if it hurts them, simply because that is what is right. If a
person works hard toward a goal, one day he will achieve it.
If I buy expensive, energy efficient light bulbs that are
supposed to last for 5 years, I will only have to buy light bulbs
once every 5 years.
• Unfortunately, we do not live in a theoretical world. In our
world, things go wrong. And as we’ve seen with many other
things, as goes the world, goes chemistry.
• Therefore, our perfectly balanced chemical equations that
say if we put in so many grams/moles of X we will get back so
many grams/moles of Y are victims of our imperfect world
also. In the real world, we will only get a percentage of the Y
out that the chemical equation predicted for us.
Percent Yield…continued
• Chemical Equation produces → Theoretical (perfect) Yield
Steps to Solve Percent Yield Problems
1.
2.
• Real World produces → Actual (imperfect) Yield
• Identify what you know and what you do not know and write it down.
• Convert any masses given to moles.
• The actual yield can never be greater than the theoretical yield.
• The actual Yield can never be more than 100% of the theoretical yield.
•
• Coefficient of unknown substance in BALANCED chemical equation
Coefficient of Limiting Reactant in BALANCED chemical equation
• # mol unknown molecule
# mol Limiting Reactant
• # mol comes from BALANCED chemical equation NOT the problem itself.
Reactants or products leak out, especially when they are gases.
The reactants are not 100% pure.
Some product is lost when it is purified.
There are also many chemical reasons, including:
o The products decompose back into reactants (as with the ammonia process).
o The products react to form different substances.
o Some of the reactants react in ways other than the one shown in the equation.
– These are called side reactions.
• The reaction occurs very slowly. This is especially true of reactions
involving organic substances.
Mass of substance (g)
= Number of Moles
Molar Mass of substance (g/mol)
• Determine the Mole Ratio needed.
• The discrepancy is not due to an incorrect equation. It is due to
imperfections in the process such as:
•
•
•
•
Identify the Limiting Reactant in the reaction.
Calculate the Theoretical Yield of the substance in question.
• Use the mole ratio to determine the amount of product the known reactant will produce.
• Known Moles of Limiting Reactant X Mole Ratio = Moles of Unknown
• Convert moles of product to grams.
• # of Moles Calculated X Total Molar Mass (g/mol) = Theoretical Mass of substance (g)
3.
Determine % Yield
•
Mass of Actual Yield (g) X 100 = % Yield
Mass of Theoretical Yield (g)
Percent Yield Practice
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1. Dichlorine monoxide, Cl2O, is sometimes used
as a powerful chlorinating agent in research. It
can be produced by passing chlorine gas over
heated mercury (II) oxide according to the
following equation: HgO + Cl2 → HgCl2 + Cl2O
What is the % yield, if the quantity of reactants
is sufficient to produce 0.86g of Cl2O but only
0.71g is obtained?
Percent Error Notes
Percent error is a calculation of how efficient your
lab technique is. Just like % Yield, it compares your
Theoretical yield to your Actual yield. However, the
focus is on what was not produced instead of what
was produced.
Steps to Solve % error Problems:
1. Identify your Actual and Theoretical Yields
2.
Mass of Theoretical Yield(g) - Mass of Actual Yield (g) X 100 =
Mass of Theoretical Yield (g)
You
Percent Yield Practice …continued
2. Acetylene, C2H2, can be used as an industrial starting
material for the production of many organic
compounds. Sometimes, it is first brominated to form
1,1,2,2-tetrabromoethane, CHBr2CHBr2, which can
then be reacted in many different ways to make other
substances. The equation for the bromination of
acetylene follows: C2H2 + 2Br2 → CHBr2CHBr2
If 72.0g of C2H2 reacts with excess bromine and 729g
of the product is recovered, what is the % yield of the
reaction?
Percent Error Practice
We
1. Dichlorine monoxide, Cl2O, is sometimes used
as a powerful chlorinating agent in research. It
can be produced by passing chlorine gas over
heated mercury (II) oxide according to the
following equation: HgO + Cl2 → HgCl2 + Cl2O
What is the % yield, if the quantity of reactants
is sufficient to produce 0.86g of Cl2O but only
0.71g is obtained?
You
Percent Error Practice …continued
2. Acetylene, C2H2, can be used as an industrial starting
material for the production of many organic
compounds. Sometimes, it is first brominated to form
1,1,2,2-tetrabromoethane, CHBr2CHBr2, which can
then be reacted in many different ways to make other
substances. The equation for the bromination of
acetylene follows: C2H2 + 2Br2 → CHBr2CHBr2
End Day 4
If 72.0g of C2H2 reacts with excess bromine and 729g
of the product is recovered, what is the % yield of the
reaction?
Empirical and Molecular Formula Notes
An Empirical Formula is a chemical formula for a
compound that reflects the atoms contained in the
compound in their smallest whole-number ratio.
•Examples: CH, BH, Ba(NO3)2
A Molecular Formula is the actual formula for a
molecular compound.
•Examples: C2H2, B12H12, Ba3(NO3)6
•An empirical formula may not be a correct molecular
formula.
Steps to Solve Empirical and Molecular
Formula Problems
Determining the correct Empirical Formula
1. Look at the formula subscripts. Are they in their
smallest whole-number ratio?
a)
b)
c)
If yes, then you are done.
If no, then divide each subscript by their largest common
denominator.
Example: C2H6F4 has en empirical formula of CH3F2 since
all of the original subscripts were divisible by 2.
I
Steps to Solve Empirical and Molecular
Empirical and Molecular Formula Practice
Formula Problems…continued
Determining the correct Molecular Formula
1. Determine the molar mass of the empirical formula.
2. Divide the molar mass of the sample by the molar
mass of the empirical formula.
3. Multiply the subscripts of the empirical formula by
the answer.
4. Example: the empirical formula of the sample is CH
and the sample molar mass is 104.16 g/mol.
a)
b)
c)
CH has a molar mass of 13.03 g/mol
(104.16 g/mol ) / (13.02 g/mol)= 8
The Molecular Formula is C8H8
Determine the empirical formula.
1. C6H6 C3H3
Determine the molecular formula.
2. A compound with an empirical formula of C2OH4
and a molar mass of 88 grams per mole. What is
the molecular formula of this compound? C4O2H8
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Empirical and Molecular Formula
Practice …continued
You
Empirical and Molecular Formula
Practice …continued
Determine the empirical formula.
Determine the empirical formula.
5. C14H42O14 CH3O C4H9
3. C6H18 C4H9
Determine the molecular formula.
4. A compound with an empirical formula of
C4H4O and a molar mass of 408 grams per mole.
What is the molecular formula of this compound?
C 8 H8 O 2
Determine the molecular formula.
6. A compound with an empirical formula of
CFBrO and a molar mass of 636.75 grams per
mole. What is the molecular formula of this
compound? C2F2Br2O2
End Day 5
Converting between Moles, Grams, and Molecules
Converting Moles to Grams:
1. Find the molar mass of the molecule.
 Find the molar mass of each element on the periodic table.
 Multiply the # of atoms of each element by the molar mass of each element.
 Add all of the Total Masses together.
2. Multiply the # of moles by the molecule’s molar mass.
→
Amount of Moles (mol)
X
Total Molar Mass (g/mol) = Mass of Molecule (g)
Converting Grams to Moles:
1. Find the molar mass of the molecule
 Same as above
2. Divide the mass in grams by the molecule’s molar mass.
→ Mass of Molecule (g)
= Number of Moles
Molar Mass (g/mol)
Converting Moles to Number of Molecules / Atoms:
1. Find the number of moles in the molecule.
 Could be given
 May have to convert from grams
2. Multiply by Avogadro’s Number
 6.022 x 1023
→ Number of Moles X 6.022 x 10
23
=
Number of Molecules / Atoms
Converting Number of Molecules/ Atoms to Moles:
1. Find the number of atoms/ molecules present in the element/ compound.
 Could be given
 May have to convert from grams
2. Divide by Avogadro’s Number
 6.022 x 1023
→ Number of atoms/ molecules = Number of Moles
6.022 x 1023
Converting Number of Molecules/ Atoms to Grams:
1. Limit this application to atoms only at this level.
 To apply to molecules:
1. Break molecule down into components.
2. Convert the moles of each element to number or atoms of each element
3. Determine the total molar mass of each element and convert each to grams
separately.
4. Add the separate masses together.
 Shorted to convert straight from moles to grams.
2. Find the total molar mass of the atoms.
 Same as above
3. Divide by Avogadro’s Number

6.022 x 1023
→ (Number of Atoms)(Molar Mass of Element)
6.022 x 1023
Converting Grams to Number of Molecules/ Atoms:
1. Limit this application to atoms only at this level.
2. Find the mass of the element in grams.
 Could be given
 May have to convert from moles
3. Find the molar mass of the element.
4. Multiply the mass of the element in grams by Avogadro’s Number.
5. Divide by the molar mass of the element.
→ (Mass of element in grams)(6.022 x 1023) = Number of Molecules/ Atoms
Molar Mass of 1 mol of the element
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