Transcript 2016Ch_3_Notesx

AP 9/21 TEST TODAY
• Turn in your review
• Pick up a periodic table
• As soon as we do good things, you will receive
your test. You will have ONE hour to complete
the test. Work must be shown in the space
provide on your answer document. NO
WORK=NO CREDIT.
• After the test, we will start equation writing.
• HW: lab report due Friday. NO EXCEPTIONS.
review memorization stuff
BONUS #1: 2ZnS(s) + 3O2(g)  2ZnO(s) + 2SO2(g)
If the typical yield is 86.78%, how much SO2 should be
expected if 4897 g of ZnS are used?
BONUS #2
AP: 9/6
• Today we will grade your quizzes in class (3rd period
will need to finish due to the assembly on Friday)
• You will need a calculator, a marker, your chapter 3
notes, and a blank piece of paper
• After we grade the quiz, we will cover section 3.1 and
3.2
• Tomorrow we will cover sections 3.3, 3.4, and 3.7.
Please bring your notes tomorrow.
• NO HW TONIGHT (YAY!). The remainder of the
stoichiometry packet is due by Friday, I will give it
back to you tomorrow.
Quick Assignment…
• Draw a mass spectrometer and label the parts
• What is this piece of equipment used for?
• Describe the five stages of this device.
You will have no more than 15 minutes to finish
this.
AP: 9/7
• Today you will need the a blank piece of paper
form the side table, a calculator, your chapter 3
notes
• Turn in your mass spectrometer HW
• We will cover sections 3.1, 3.2, 3.3, and 3.4 today
• HW: finish the NMSI packet 7-12 (due Friday)
• Lab on Friday (don’t get to excited, it’s probably
the most boring lab you will perform in this
class. It is meant to help you learn how to write a
lab report)
Naming Practice
• On a piece of paper, answer the following:
▫ 1. Write the formula for the following compounds:




A. lead (II) sulfite
B. barium hydroxide
C. calcium chlorate
D. hydronitric acid
▫ 1. Name the following compounds:




A. HNO2
B. SnO2
C. Ag2CO3
D. N2O4
CHAPTER 3:
STOICHIOMETRY
AP
CHEMISTRY
Chapter 3 Objective
• In this chapter, we will look at the quantities of
materials consumed and produced in chemical
reaction. The study of these quantities is called
chemical stoichiometry.
3.1 Counting By Weighing
• If we were to take the individual masses of five
nails, each nail would probably have a small
deviation in mass, but would be relatively the
same, let’s say an average mass of 2.5g per nail.
If someone needed 1,000 nails for a building
project, it is easier to take the mass of 1,000
nails, or 2,500g, than to count out all those nails.
Objects do not have to have identical masses to
be counted by weighing; they behave as if they
are all identical. This is very lucky for chemists.
Average Atomic Mass
% Isotope A (mass of A) + % Isotope B (mass of B) + …=
avg. atomic mass
• Just like our nails deviated slightly, atoms of the same
element may have different masses. We call these
isotopes.
• Average Atomic Mass- an average of each isotope of an
element, based on their percent abundance
▫ Ex. Find the average atomic mass of element “X” if 1.40%
of the sample is and isotope with a mass of 203.973 amu,
24.10% is an isotope with a mass of 205.9745 amu, 22.10%
is an isotope with a mass of 206.9759 amu and 52.40% is
an isotope with a mass of 207.9766 amu.
0.0140(203.973) + 0.2410(205.9745) + 0.2210(206.9759) +
0.5240(207.9766) = 207.21 amu (probably lead)
3.2 Atomic Masses
• So if scientists can’t see atoms with their eyes, how
do they know all these isotopes exist? They use a
mass spectrometer.
• The mass spectrometer heats a sample of atoms
into a gas, passes them through a beam of highspeed electrons, which knock off some of the atom’s
electrons giving them positive charges, and then
these positively charged atoms pass through an
electric field. The atoms are deflected based on
their size, and then detected on a computerized
plate.
Figure 3.2
Neon Gas
Note that these spikes on the readout reflect how
much of each isotope is present in a sample.
Figure 3.3 (NMSI Exercise 1)
• When a sample of natural copper is vaporized and injected into a
mass spectrometer, the results are shown. Use the data to calculate
average atomic mass of natural copper. The mass values are Cu-63
(62.93) and Cu-65 (64.94).
Figure
3.3
Mass
Spectrum
of
Natural
Copper
Calculating the Mass of an Element
• Example. Use the mass spectrum shown on the
left to calculate the average atomic mass of this
element, then identify the element.
Mass
90
91
92
94
96
Percent Abundance
50
11
18
19
2
0.50(90) + 0.11(91) + 0.18(92) + 0.19(94) + 0.02(96) = 91.35 or 90
Its zirconium!
3.3 The Mole
• The mole is the chemist’s way of counting
atoms. Remember that the mole (or
Avogadro’s number) is equal to 6.022 x 1023
atoms of any element, which is then equal
to the average atomic mass on the periodic
table.
• The mole is standardized as the number of carbon
atoms in 12.0g of pure carbon-12.
• 6.022 x 1023 amu = 1 g
• The mass of one mole of an element is equal to its
atomic mass in grams.
Taken by Paul Jung (AP ’09) at the Golden Gate Bridge in San Francisco
Take out the blank piece of paper
• We are going to review AMG by drawing a mole
map and the AMG chart
Ex. Cody found a gold nugget that had a
mass of 1.250 oz. How many moles was
this? How many atoms?
1.250 oz Au 1 lb 453.59g 1 mol Au
16 oz 1 lb
196.97g Au
= 0.1799 moles
0.1799 moles Au 6.022 x 1023 atoms
1 mol Au
= 1.083 x 1023 atoms
3.4 Molar Mass
• mass in grams of one mole of a substance
• may also be called molecular weight
Ex. Calculate the molecular mass of
cisplantin, Pt(NH3)2Cl2.
Pt
N
H
Cl
1 x 195.08 = 195.08
2 x 14.01 = 28.02
6 x 1.01 = 6.06
2 x 35.45 = 70.90
300.06 g or amu
The compound cis-PtCl2(NH3)2
cisplatinum is a platinum-based
chemotherapy drug used to treat various
types of cancers, including sarcomas,
some carcinomas (e.g. small cell lung
cancer, and ovarian cancer), lymphomas
and germ cell tumors.
Ex. How many grams are 3.25 moles
of cisplantin?
3.25 mol cisplantin
= 975g
300.06g
1 mol cisplantin
AP 9/12
• Pick up the paper from the side table
• Take out your lab, a blank piece of paper, and
pick up a calculator
• Today we will perform part 1 of your lab,
practice writing a hypothesis, perform more
AMG problems, and possibly practice percent
composition problems
• HW tonight (MAYBE): molar conversions and
percent composition (paper you picked up from
side table)
AP 9/13
• Take out your lab, your calculation HW, the
molar conversion problems, and a calculator
• We will finish the lab today and then practice
problems over mole conversions
• We will also go over percent composition and
empirical and molecular
• HW tonight: Molar conversions WS
Calculate the following:
• 1. The number of moles AND the mass in grams of 3.25
X 1026 atoms of sodium phosphate.
• 2. The mass in grams of 5.85 X 1032 units of iron (II)
phosphide.
• 3. The number of moles of 6.210 grams of dinitrogen
pentoxide.
• 4. The mass in grams of 4.86 moles of calcium
carbonate.
• 5. For question #4, calculate the mass of the carbonate
ions present.
NMSI #7 (new type of question)
• Isopentyl acetate (C7H14O2), the compound
responsible for the scent of bananas, can be
produced commercially. Interestingly, bees
release about 1 X 10-6 g of this compound when
they sting. The resulting scent attracts other bees
to join the attack.
• (a) calculate the number of molecules of
isopentyl acetate released in a typical bee sting.
• (b) Calculate the number of carbon atoms
present
AP 9/14
• You will need a calculator, the paper from the
side table, a white board, a marker, your chpt 3,
and your chapter 3 notes.
• Today we will cover % composition,
empirical/molecular formula, and
stoichiometric calculations
• Before we get started, we will pass back papers
• HW: molar conversion paper AND whatever
problems we don’t finish in class from the paper
you picked up today
Percent Composition
▫ “mass percent” or “percent by mass”
Ex. Find the percent composition of all
elements in cisplantin, Pt(NH3)2Cl2.
Pt 195.08 x 100 = 65.01%
300.06
N 28.02 x 100 = 9.34%
300.06
H 6.06 x 100 = 2.02%
300.06
Cl 70.90 x 100 = 23.63%
300.06
Practice:
• Find the percent composition of the following:
▫
▫
▫
▫
A. nitric acid
B. glucose
C. sodium phosphite
D. percent of bromine in zinc (II) bromide
Practice: NMSI 8 and 9
3.7 Determining the Formula of a
Compound
Empirical Formula Poem
• Empirical formulasimplest whole number
ratio of the various types
of atoms in a compound
• Molecular formula- the
exact formula or
(empirical formula)x
Percent to mass,
Mass to moles,
Divide by smallest,
And round (or multiply)
‘till whole
Remember:
• The empirical formula is the true formula for
ionic compounds but may not be for molecular
compounds.
• The formula to calculate molecular formula is:
▫ MF formula=MF mass/EF mass
▫ The MF mass will be given
Conceptual Practice:
• Which of the following have NO molecular
formula:
▫
▫
▫
▫
A. calcium sulfate
B. methanol (CH3OH)
C. aluminum sulfide
D. dinitrogen pentoxide
Ex. A sample of a compound contains 11.66g of iron and 5.01g
of oxygen. What is the empirical formula of this compound?
(NOTE: you can check your answer BC we know it can only
have two forms)
11.66g Fe 1 mol Fe = 0.2088 mol Fe
55.85g Fe
5.01g O 1 mol O
= 0.3131 mol O
16.00g O
Fe:O 0.2088 : 0.3131
1:1.5 or 2:3
0.2088 0.2088
Fe2O3
Ex. What is the empirical formula of hydrazine, which
contains 87.5% N & 12.5%H? What is the molecular
formula if the gram formula mass is 32.07g?
87.5gN 1 mol N = 6.246 mol N
14.01g N
12.5g H 1 mol H = 12.38 mol H
1.01 g H
N:H = 6.246 : 12.38 1 : 1.98 = 1 : 2
6.246 6.246
NH2
NMSI # 10, 11, 12
Combustion Analysis
Suppose you isolate an acid from clover leaves and know
that it contains only the elements C, H, and O. Heating
0.513g of the acid in oxygen produces 0.501g of CO2 and
0.103g of H2O. What is the empirical formula of the
acid? Given that another experiment has shown that the
molar mass of the acid is 90.04g/mol, what is its
molecular formula?
C + O2  CO2
0.501g CO2 1 mol CO2 1 mol C
12.01gC
44.01g CO2 1 mol CO2 1 mol C
= 0.1367 g C in compound
2H2 + O2  2H2O
0.103g H2O 1 mol H2O 2 mol H 1.01g H
18.02g H2O 1 mol H2O 1 mol H
= 0.01155 g H
0.513g cmpd - (0.137 + 0.012)= 0.364g O
0.364g O 1 mol O = 0.02275 mol O
16.00g O
0.137g C 1 mol C = 0.01141 mol C
12.01g C
0.01155g H 1 mol H = 0.01144 mol H
1.01 g H
C:H:O
0.01141: 0.01144 : 0.02275
0.01141: 0.01144 : 0.02275
0.01141 0.01141 0.01141
1: 1: 1.98
1:1:2
CHO2
efm = 12.01 + 1.01 + 32.00 = 45.02
90.04/45.02 = 2
C2H2O4
3.8 Chemical Equations
• During a chemical reaction, atoms are
rearranged when existing bonds are broken and
reformed in a different arrangement.
• Since atoms are not lost or gained during a
reaction, a balanced equation must have the
same number of atoms of each element on each
side.
• Symbols representing physical states: (s) (l)
(g) (aq)
Observe the diagram to the right.
Balance the equation and then draw in
the correct number of expected
products.
3.9 Balancing Chemical Equations
• We balance equations by adding coefficients,
never by changing formulas.
• Most equations can be balanced by inspection.
Some redox reactions require a different
method.
Let’s Practice! Try these on your own!
Ex.
Al(s) + Cl2 (g)  AlCl3(s)
2Al(s) + 3 Cl2 (g)  2AlCl3(s)
N2O5(s) + H2O(l)  HNO3(l)
N2O5(s) + H2O(l)  2 HNO3(l)
Pb(NO3)2(aq) + NaCl(aq)  PbCl2(s) + NaNO3(aq)
Pb(NO3)2(aq) + 2 NaCl(aq)PbCl2(s) +2 NaNO3(aq)
You should be able to write and balance an
equation when given the reactants and products in
words.
Ex. Phosphine, PH3(g) is combusted in air
to form gaseous water and solid diphosphorus
pentoxide. Try this one!
PH3(g) + O2(g)  H2O(g) + P2O5(s)
2PH3(g) + 4O2(g)  3H2O(g) + P2O5(s)
When ammonia gas is passed over hot liquid sodium
metal, hydrogen is released and sodium amide, NaNH2, is
formed as a solid product.
NH3(g) + Na(l)  H2(g) + NaNH2(s)
2NH3(g) + 2Na(l)  H2(g) + 2NaNH2(s)
AP 9/16
• Turn in the molar conversion homework BUT NOT the
other 3 stoichiometry problems.
• You will need your chapter 3 notes, a calculator, and
something to write with
• We are practicing stoichiometric calculations today.
Grab a marker and whiteboard if you prefer to do your
work on that.
• HW: Stoichiometry practice you picked up today. Only
do # 1,2,3,5,7,8,10,12,13
• NEXT WEEK: Monday we will go over limiting and
excess reactants. Tuesday we will go over combustion
analysis. Wednesday you will have your first test.
3.10 Stoichiometric Calculations
• Although coefficients in balanced chemical
equations tell us how many atoms will react to form
products, they do not tell use the actual masses of
reactants and products we will use or expect to
produce. For this, we need to relate the reactants
and products in terms of their mole ratios.
▫ The mole ratio = moles required/moles given.
Remember:
• THE MOLE RATIO IS USED IN
STOICHIOMETRY.
• There are four types of stoichiometric problems:
▫
▫
▫
▫
MOLE TO MOLE
MOLE TO MASS
MASS TO MOLE
MASS TO MASS
PRACTICE
1. For the following reaction write the mole ratio for lead
(II) nitrate and sodium nitrate
Pb(NO3)2(aq) + 2 NaCl(aq)PbCl2(s) +2 NaNO3(aq)
2. Write the mole ratio of aluminum chloride to chlorine
gas
2Al(s) + 3 Cl2 (g)  2AlCl3(s)
Mole to mole
Mass given  mole given  mole unknown  mass unknown
• Given quantity is in moles
• Unknown quantity is in moles
• Need 1 conversion factor to solve
▫ Mole ratio to convert between mole given & mole
unknown
# mol given mol unknown
x
 mol unknown
1
mol given
Practice:
• Ammonia is formed from the synthesis of
nitrogen gas and hydrogen gas in the Haber
process. How many moles of hydrogen are
needed to react with 2.00 mol of nitrogen?
Practice:
• Hydrogen gas reacts with oxygen to synthesize
water. If how many moles of water will be
produced from 5.52 moles of hydrogen?
Mass to mole
Mass given  mole given  mole unknown  mass unknown
• Given quantity is a mass (g, kg, etc.)
• Unknown quantity is in moles
• Need 2 conversion factors to solve:
▫ Molar mass to convert from mass given to mole given
▫ Mole ratio to convert from mole given to mole unknown
# g given 1 mol given mol unknown
x
x
 mol unknown
1
g given
mol given
Practice:
• Sodium hydroxide neutralized sulfuric acid to
form a salt and water. How many moles of
sodium hydroxide are needed to form 23.0 g of
the salt?
Mole to mass
Mass given  mole given  mole unknown  mass unknown
• Given quantity is in moles
• Unknown quantity is mass (g, kg, etc.)
• Need 2 conversion factors to solve:
▫ Mole ratio to convert from mole given to mole unknown
▫ Molar mass to convert from mole unknown to mass unknown
# mol given mol unknown mass unknown
x
x
 mass unknown
1
mol given
1 mol unknown
Practice:
• Hydrogen peroxide decomposes to form water
and oxygen gas. This reaction proceeds faster
when a catalyst is added. How many grams of
water will be produced from 2.25 moles of H2O2?
Mass to mass
Mass given  mole given  mole unknown  mass unknown
 Given quantity is a mass (g, kg, etc.)
 Unknown quantity is a mass
 Need 3 conversion factors to solve:
 Molar mass to convert from mass given to mole given
 Mole ratio to convert from mole given to mole unknown
 Molar mass to covert from mole unknown to mass unknown
# g given 1 mol given mol unknown mass unknown
x
x
x
 mass unknown
1
g given
mol given
1 mol unknown
Practice:
What mass of NH3 is formed when 5.38g of Li3N
reacts with water according to the equation:
Li3N(s) + 3H2O  3LiOH(s) + NH3(g)
Ex. What mass of NH3 is formed when
5.38g of Li3N reacts with water according
to the equation:
Li3N(s) + 3H2O  3LiOH(s) + NH3(g)?
5.38g Li3N 1 mol Li3N 1 mol NH3 17.04g NH3
34.83g Li3N 1 mol Li3N 1 mol NH3
= 2.63g NH3
AP 9/19
• Pick up the packet from the side table, turn in
your stoichiometry HW.
• You will need a calculator, something to write
with, and possibly your chapter 3 notes.
• Today we will review a few stoichiometry
problems and then go over limiting/excess and
theoretical yield.
On a piece of paper, work the
following:
1.If excess sulfuric acid reacts with
30.0 g of sodium chloride in a double
replacement reaction, how many grams
of hydrochloric acid are produced?
2. How many moles are required to
replace 4.00 g of silver (I) nitrate which
is dissolved in water.
3. C6H6(l) + Cl2(g)  C6H5Cl(s) + HCl(g)
When 36.8g C6H6 react with an excess of Cl2, the
actual yield of C6H5Cl is 38.8g. What is the
percent yield?
3.11 The Concept of Limiting
Reagent
• Limiting reagent- reagent that restricts or
determines the amount of product that can be
formed
• If in a stoichiometry problem you are given
amounts of two or more reactants and asked to
determine how much product forms, you must
determine which reactant is limiting and use it to
work the problem.
Which is limiting? Excess?
• To decide which reactant is limiting in a rxn, use
one of your givens to solve for the other.
• Then compare how much you have (given) to
how much you need under ideal conditions
(solved for).
▫ If you have more than you need, the 1st
substance is limiting.
▫ If you have less than you need, the 2nd
substance is limiting.
Limiting reactant example
• CO(g) + 2H2(g)  CH3OH
If 500. mol of CO and 750. mol of H2 are present,
which is the limiting reactant?
▫ Solve to determine how much H2 would be needed to
completely react 500. mol CO.
500. mol CO 2 mol H 2
x
 10 0 0 mol H 2
1
1 mol CO
▫ Do you have 1000 mol H2? No, you only have 750. mol.
▫ H2 is the limiting reactant.
Limiting reactant example #1
3ZnCO3(s) + 2C6H8O7(aq)  Zn3(C6H5O7)2(aq) + 3H2O(l) + 3CO2(g)
• If there is 1 mol of ZnCO3 & 1 mol of C6H8O7, which is the
limiting reactant?
• Answer:
▫ 1 mol ZnCO3 could react with 0.67 mol C6H8O7, which is less
than is available.
▫ ZnCO3 is limiting.
Pre-AP 9/20
•
•
•
•
Turn in your HW
Pick up the test review
Take out a piece of paper
Today you will review for the test by performing a
gallery walk on the back tables. You will answer
those questions on the blank piece of paper. Further
instructions will be given once class starts.
• HW: answer the questions on the test review. You
will take the atomic structure test on Thursday.
9/20 AP
• Pick up the review from the side table and take
out the limiting and excess problem you were
asked to calculate for HW.
• Today we will go over that problems and then
you will have time in class to review.
• HW: The stoichiometry “story packet” is due
Monday. You only need to do three problems of
your choice. However, I would read the
information in the packet.
• LAB REPORT IS DUE FRIDAY. NO
EXCPETIONS
Limiting reactant example #2
• Aspirin, C9H8O4, is synthesized by the rxn of salicylic
acid, C7H6O3, with acetic anhydride, C4H6O3.
2C7H6O3 + C4H6O3  2C9H8O4 + H2O
• When 20.0g of C7H6O3 and 20.0g of C4H6O3 react, which is
the limiting reactant? How many moles of the excess
reactant are used when the rxn is complete? What mass in
grams of aspirin is formed?
 C7H6O3, 0.0724 mol, 26.1 g
Once we know how much of a product if
formed in a reaction, we can find the
percent yield of the reaction.
• Theoretical yield- amount of product that should
form according to stoichiometric calculations
• Percent yield = Actual yield
•
Theoretical yield
•
• Actual yield- experimental yield
x 100
Percent Yield example
• C6H6(l) + Cl2(g)  C6H5Cl(s) + HCl(g)
When 36.8g C6H6 react with an excess of Cl2, the actual yield
of C6H5Cl is 38.8g. What is the percent yield?
• Given:
36.8g C6H6, excess Cl2, actual yield=38.8g C6H5Cl
• Unknown: ? g C6H5Cl, percent yield
36.8 g C 6 H 6
1 mol N 2 O 1 mol C 6 H 5Cl 112.56 g C 6 H 5Cl
x
x
x
 53.0 g C 6 H 5Cl
1
78.12 g C 6 H 6 1 mol C 6 H 6
1 mol C 6 H 5Cl
% yield 
38.8 g C 6 H 5Cl
actual yield
x 100 
x 100  73.2%
theoretica l yield
53.0 g C 6 H 5Cl
Percent Yield example
• Methanol can be produced through the rxn of CO
and H2 in the presence of a catalyst.
CO(g)  2H2(g)  CH3OH(l)
catalyst
• If 75.0 g of CO reacts to produce 68.4 g CH3OH,
what is the percent yield?
• Answer: 79.7%
Percent Yield example
• 2ZnS(s) + 3O2(g)  2ZnO(s) + 2SO2(g)
If the typical yield is 86.78%, how much SO2 should
be expected if 4897 g of ZnS are used?
Theoretica l yield 
% yield 
4897 g ZnS 1 mol ZnS 2 mol SO 2 64.07 g SO 2
x
x
x
 3219.278 g SO 2
1
97.46 g ZnS 2 mol ZnS 1 mol SO 2
actual yield
(% yield)(the oretical yield)
x 100  actual yield 
theoretica l yield
100
(86.78)(32 19.278 g SO 2 )
actual yield 
 2794 g SO 2
100
Ex. How many moles of Fe(OH)3(s) can be produced
by allowing 1.0 mol Fe2S3, 2.0 mol H2O and 3.0 mol O2
to react?
2Fe2S3(s) + 6H2O(l) + 3O2(g)  4Fe(OH)3(s) +6S(s)
1.0 mol
2.0 mol
limiting
3.0 mol
1.0 mol Fe2S3 4 mol Fe(OH)3
2 mol Fe2S3
2.0 mol H2O 4 mol Fe(OH)3
6 mol H2O
3.0 mol O2
4 mol Fe(OH)3
3 mol O2
= 2.0 mol Fe(OH)3
= 1.3 mol Fe(OH)3
= 4.0 mol Fe(OH)3
In this case, we can see that H2O gave use the smallest amount of product, so it is the
limiting reagent. Thus, we use that answer as our overall answer.
Ex. If 17.0g of NH3(g) were reacted with 32.0g of
oxygen in the following reaction, how many grams of
NO(g) would be formed?
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l)
17.0g
32.0g
Xg
17.0g NH3 1 mol NH3 = 1 mol NH3
17.0g NH3
32.0 g O2 1 mol O2 = 1 mol O2
32.0g O2
We need 5 mol O2 to every 4 mol NH3, so O2 is
limiting.
1 mol O2 4 mol NO 30.01 g NO = 24.0g NO
5 mol O2 1 mol NO
Ex. In the reaction of 1.00 mol of CH4 with an
excess of Cl2, 83.5g of CCl4 is obtained. What is
the theoretical yield, actual yield and % yield?
Actual yield = 83.5g CCl4
CH4 + 2Cl2  CCl4 + 2H2
1.00 mol CH4 1 mol CCl4 153.81g CCl4
1 mol CH4 1 mol CCl4
= 154g CCl4 (theoretical yield)
83.5 x 100 = 54.2% yield
154