che121-lec8-sp120
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Chemistry 121
Lecture 8
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Outline
I. Solutions
- Molarity
- Stoichiometry
- Dilutions
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Aluminum Lab
Several concepts will be defined in the next
week or two.
Oxidation state = charge
In the ionic compound,NaCl
Na oxidation state = +1
Cl oxidation state = -1
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Molarity
Molarity (M)
A Conversion Factor between the amount of solute
and the volume of solution!!!
M = Molarity = moles of solute
L of solution
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Molarity Problems
How many liters of 0.300M KBr solution can be
prepared from 100.g of KBr?
A. 5.60 B. 1.40 C. 2.80 D. 0.600
0.300 mols KBr
1 L soln
100.g KBr = L soln
119.0g KBr
1 mol KBr 1 L soln
0.300 mols KBr
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Molarity in Chemical Reactions
In a chemical reaction,
The volume and molarity of a solution are used to
determine the moles of a reactant or product.
mol solute x volume (L) = moles solute
L
M
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Molarity in Chemical Reactions
In a chemical reaction,
If molarity, M (mol/L), and moles are given, the volume (L) can be determined.
moles solute x 1 L = volume (L) solution
moles solute
1
M
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Summary of Chemical Reactions
Copyright 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings
Basic gram to gram stoichiometry
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Summary of Chemical Reactions
Copyright 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings
Stoichiometry using Molarity
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Stoichiometric Equivalence - Titration
Reactant 1
Reactant 2 plus indicator (changes color when exactly
combined)
Acid/Base Reaction Lab
REDOX Reaction Lab
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Using Molarity of Reactants
How many mL of 3.00 M HCl solution are needed to completely react with 4.85 g CaCO3?
2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l)
B: 3.00 mols HCl
1L soln 4.85 g
Volume(mL)=?
Mols HCl
A:
0.0 g 0.0 g 0.0 g
0.0 g 0.0 g X Y Z
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Using Molarity of Reactants
How many mL of 3.00 M HCl solution are needed to completely react with 4.85 g CaCO3?
2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l)
3.00 mols HCl
1 L soln
1mol CaCO3
= mL soln 100.09g CaCO3
4.85g CaCO3
1mol CaCO3
2mol HCl
3.00mol HCl
1 L
1 L
1000mL
= 32.3 mL soln
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Learning Check
How many mL of a 0.150 M Na2S solution are needed to
completely react 18.5 mL of 0.225 M NiCl2 solution?
NiCl2(aq) + Na2S(aq) NiS(s) + 2NaCl(aq)
BEFORE:
18.5 mL -> 0.0185L
0.225mols
1L
Mols NiCl2
0.150mols
1L
Volume(mL) = ?
? Mols Na2S
AFTER: 0.0 0.0
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Learning Check
How many mL of a 0.150 M Na2S solution are needed to
completely react 18.5 mL of 0.225 M NiCl2 solution?
NiCl2(aq) + Na2S(aq) NiS(s) + 2NaCl(aq)
18.5 mL
1000 mL
1 L
0.0185L
= mL
NiCl2 solution Na2S solution
1 L
0.225 mols NiCl2 1molNa2S
1molNiCl2
1 L
.150 molsNa2S 1 L
1000mL
= 27.8mL Na2S
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More on Molarity
Solution preparation from a concentrated
stock solution - Solution Dilution
A concentrated solution (high Molarity) is
converted to a dilute solution (lower Molarity)
by adding solvent to it.
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More on Molarity - Dilution
Concentration = amount of solute
amount of solution
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More on Molarity - Dilution
Concentration = amount of solute
amount of solution
Solute Amount Remains the Same!
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More on Molarity
A concentrated solution (high Molarity) is
converted to a dilute solution (lower Molarity)
by adding solvent to it.
dilute concentrated
Mdil x Vdil = Mcon x V con
Moles of Solute = Moles of Solute
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More on Molarity
Muriatic acid, an industrial grade of concentrated HCl, is
used to clean masonry and cement. Its concentration is
11.7M. How would you go about making 11.4L of 3.5M
acid for routine use?
Mdil x Vdil = Mcon x V con
Concentrated
Dilute
(11.7mols HCl)
1L
(11.4L)(3.5 mols HCl)
1L
= V con
V con = 3.4L
To Make the Solution, Take 3.4L of concentrated acid and
add enough water (11.4L - 3.4L = 8.0L) to make 11.4L
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Dilution Problem
How many mL of 18.0 M H2SO4 are required to prepare 650. mL of
6.00 M H2SO4 by dilution?
mmol(before dilution) = mmol(after dilution)
mL 18.0 M = 650. mL 6.00 M
mL = (650. mL 6.00 M) / 18.0 M
= 217 mL
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Dilution Problem
How many mL of H2O must be added to 375 mL of a 3.50 M NaCl
doln to prepare a 1.25 M NaCl soln?
mmol(before dilution) = mmol(after dilution)
375 mL 3.50 M = mL 1.25 M
mL = (375 mL 3.50 M) / 1.25 M
= 1050 mL
H2O(added) = Vfinal Vinitial = 1050 ml - 375 mL
= 675 mL
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Additional Molarity Problem
Assuming additive volumes, what is the molarity of a KBr soln
prepared by mixing 200. mL of a 0.053 M KBr soln with 500. mL of a
0.072 M KBr soln?
mol KBr soln 1 = 0.200 L 0.53 M = 0.011 mol KBr
mol KBr soln 2 = 0.500 L 0.072 M = 0.036 mol KBr
_______ _____________
0.700 L 0.047 mol KBr
Mfinal soln = 0.047 mol KBr / 0.700 L = 0.067 M
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Safety Tip When working with Acids
Generally speaking - care must be taken
when adding water to an acid!!!
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CHEM 121 - EXAM I REVIEW
Chapters 1, 2 & 3
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EXAM I - REVIEW
Bring pencil & calculator (no cell phones)
Periodic Table & Scantron form will be provided.
Scantron (problem 5 - start with #1 on form)
Multiple Choice 1-19
You may leave (quietly) when finished.
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Chapter 1 - Matter & Measurement
Matter -
Physical/Chemical
Elements/Compounds (Ch2)
Mixtures (Ch2)
Math in Chemistry
Measurement & Sig Figs
Dimensional Analysis
Density
Percent
Not On Exam: Energy
Scientific Method
Temperature
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Speed as a Conversion Factor
The distance from the Earth to the moon is approximately
240,000 miles. The peregrine falcon has been measured as
traveling up to 350 km/hr in a dive. If this falcon could fly to the
moon at this speed, how many seconds would it take? (Useful conversion: 1 mile = 1.609km)
240,000 miles = seconds
1 mile
1.609km
350 km
1 hour 60 minutes
1 hour
60 secs
1 minute
A. 4.0 x 106 B. 3.97 x 106 C. 1.45 x 104 D. None of these
Distance Time Speed
2 sig figs 4 sig figs 2 sig figs exact exact
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Density Learning Check
A 32.65 g sample of a solid is placed in a flask. Toluene (a liquid),
in which the solid is insoluble, is added to the flask so that the total
volume of solid and liquid together is 50.00mL. The solid and
toluene together weigh 58.58 g. The density of toluene at the
temperature of the experiment is 0.864g/mL. What is the density of
the solid?
Density = Mass Volume
A. 1.17g/mL B. 0.653g/mL C. 1.63g/mL D. None of these
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Density Learning Check
A 32.65 g sample of a solid is placed in a flask. Toluene (a liquid),
in which the solid is insoluble, is added to the flask so that the total
volume of solid and liquid together is 50.00mL. The solid and
toluene together weigh 58.58 g. The density of toluene at the
temperature of the experiment is 0.864g/mL. What is the density of
the solid?
Find mass of Toluene, so that you can find the volume of Toluene:
58.58 g (toluene + solid) - 32.65 g (solid)
25.93g (toluene) = Volume Toluene
0.864g
1 mL = 30.0mL
A. 1.17g/mL B. 0.653g/mL C. 1.63g/mL D. None of these
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Density Learning Check
A 32.65 g sample of a solid is placed in a flask. Toluene (a liquid),
in which the solid is insoluble, is added to the flask so that the total
volume of solid and liquid together is 50.00mL. The solid and
toluene together weigh 58.58 g. The density of toluene at the
temperature of the experiment is 0.864g/mL. What is the density of
the solid?
Find volume of the solid, and then compute the density:
- 30.0mL (toluene)
50.00mL (solid + toluene)
20.0 mL (solid)
Density (solid) = 32.65g
20.0mL
= 1.63g/mL
A. 1.17g/mL B. 0.653g/mL C. 1.63g/mL D. None of these
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Calculate the following:
[(5.031-4.96)(2.38)] 3.91 = _____
A. 0.04
B. 0.043
C. 0.0432
D. None of these
5.031 -4.96
0.07
1 sig fig
(0.07)x(2.38)
3.91
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Chapter 2 - Matter
Matter
Laws that lead to Atomic View of Matter
History of Atom
Atomic Symbols & Atomic Mass
Periodic Table
Ionic Compounds
Predicting formulas
Nomenclature
Polyatomic Ions
Hydrates
Covalent Compounds & Elements
Nomenclature
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Learning Check
Which one of the following is not a physical property of iron pentacarbonyl [Fe(CO)5]
a. color
b. density
c. melting point
d. flammability
e. boiling point
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Learning Check
Oxygen, O2, is an example of
a. an atom
b. a molecular compound
c. an ionic compound
d. a molecular element
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Learning Check
Which species has 54 electrons?
a. 132 Xe+
b. 123 Te2-
c. 118 Sn2+
d. 112 Cd2+
54
52
50
49
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36
Learning Check
Name each compound.
A. N2O4
B. MnCl2
C. (NH4)3PO4
D. Cu2CO3
Dinitrogen tetroxide (Covalent - Greek prefixes)
(Ionic - Variable charge on metal) Manganese (II) Chloride
(Ionic - Polyatomics) Ammonium Phosphate (Ionic - variable charge on metal & polyatomic)
Copper (I) Carbonate
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Chapter 3 - Stoichiometry
The Mole
Empirical/Molecular
Formulas (combustion)
Chemical Equations
Writing from scratch
balancing
Stoichiometry
Basic
Limiting Reactant & %
yield
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Learning Check
The combustion of 12.62 grams of a compound which contains only C,H and Br yields 12.09 grams of CO2 and
4.95 grams of H2O. What is the empirical formula of the
compound?
a. C4H12Br2 b. C5H10Br2 c. C5H10Br5 d. C7H10Br4
e. C5H8Br2
12.09g CO2
g C + g H + g Br
= ? g C
44.01g CO2
1 mol CO2
1 mol CO2
1 mol C
1 mol C
12.011g C
= 3.300g C
- 3.300g C 12.62 g
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Learning Check
The combustion of 12.62 grams of a compound which contains only C,H and Br yields 12.09 grams of CO2 and
4.95 grams of H2O. What is the empirical formula of the
compound?
a. C4H12Br2 b. C5H10Br2 c. C5H10Br5 d. C7H10Br4
e. C5H8Br2
4.95g H2O
- 3.300g C 12.62 g
g C + g H + g Br
= g H
18.02g H2O
1 mol H2O
1 mol H2O
2 mol H
1 mol H
1.0079 g H
= 0.554g H
- 0.554g H
8.77 g Br
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Learning Check
The combustion of 12.62 grams of a compound which contains only C,H and Br yields 12.09 grams of CO2 and
4.95 grams of H2O. What is the empirical formula of the
compound?
a. C4H12Br2 b. C5H10Br2 c. C5H10Br5 d. C7H10Br4
e. C5H8Br2
- 3.300g C 12.62 g
g C + g H + g Br
0.554g H
- 0.554g H
8.77 g Br
3.300g C
8.77 g Br
12.011g C
1 mol C
1.0079g H
1 mol H
79.90 g Br
1 mol Br
= 0.2747 mol C
= 0.550 mol H
= 0.110 mol Br
0.110
0.110
0.110
= 2.5 C
= 5 H
= 1 Br
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Learning Check
The combustion of 12.62 grams of a compound which contains only C,H and Br yields 12.09 grams of CO2 and
4.95 grams of H2O. What is the empirical formula of the
compound?
a. C4H12Br2 b. C5H10Br2 c. C5H10Br5 d. C7H10Br4
e. C5H8Br2
- 3.300g C 12.62 g
g C + g H + g Br
0.554g H
- 0.554g H
8.77 g Br
3.300g C
8.77 g Br
12.011g C
1 mol C
1.0079g H
1 mol H
79.90 g Br
1 mol Br
= 0.2747 mol C
= 0.550 mol H
= 0.110 mol Br
0.110
0.110
0.110
= 2.5 C
= 5 H
= 1 Br
X 2
X 2
X 2
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Learning Check
Analysis of a metal chloride, XCl3, shows that
it contains 67.2% Cl by mass. Calculate the
atomic mass of X .
100% - 67.2%(Cl) = 32.8%(X)
100 g sample
67.2g Cl
32.8g X
35.453g Cl
1 mol Cl = 1.90 mol Cl
1 mol X
Molar Mass X
= ? mol X
Smallest 1:3 ratio of atoms
? mol X
= 3 Cl
= 0.633 mol X
Molar Mass of X = 51.8g/mol
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Learning Check
Magnesium metal and oxygen react to form magnesium
oxide, which is a white solid. If 36.5g of Mg and 35.0g of
O2 actually produce in the lab 52.3 grams of magnesium
oxide
What is the percent yield of Magnesium oxide as
determined by the limiting reactant problem?
2Mg + O2 ----> 2MgO BEFORE: 36.5g 35.0g 0.0g
Actual Yield
AFTER: Theory Yield
% Yield = Actual x 100%
Theory
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Learning Check
2Mg + O2 ----> 2MgO BEFORE: 36.5g 35.0g 0.0g
Actual Yield = 52.3g MgO
AFTER: Theory Yield
36.5g Mg = g MgO
24.30g Mg
1 mol Mg
2 mol Mg
2 mol MgO
1 mol MgO
40.30g MgO
= 60.5g MgO
35.0g O2 = g MgO
32.00g O2
1 mol O2
1 mol O2
2 mol MgO
1 mol MgO
40.30g MgO
= 88.2g MgO
0.0g 60.5g
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Learning Check
2Mg + O2 ----> 2MgO BEFORE: 36.5g 35.0g 0.0g
Actual Yield = 52.3g MgO
AFTER:
36.5g Mg = g MgO
24.30g Mg
1 mol Mg
2 mol Mg
2 mol MgO
1 mol MgO
40.30g MgO
= 60.5g MgO
0.0g
52.3g
60.5g X 100% = 86.4% % Yield=
60.5g
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LIMITING REAGENT PROBLRM
A 30.8 g sample of PCl5 (MW = 208.22 g / mol) was reacted with
8.50 g H2O (MW = 18.02 g / mol) according to the following rxn:
PCl5 + H2O H3PO4 + HCl
What is the theoretical yield of HCl? Assuming a student isolated
13.0 g HCl, what is the %-yield?
PCl5 + 4 H2O H3PO4 + 5 HCl
mol PCl5 = 30.8 g 1 mol PCl5 / 208.22 g = 0.148 mol PCl5
0.148 mol / 1 = 0.148
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Mol H2O = 8.50 g 1 mol H2O / 18.02 g = 0.472 mol H2O
0.472 mol / 4 = 0.118 LR
Note: The smaller of the two values 0.148 and 0.118 is the LR.
g HCl = 8.50 g H2O 1 mol H2O / 18.02 g H2O
5 mol HCl / 4 mol H2O 36.45 g HCl / mol HCl
= 21.5 g
%-yield = (actual yield / theortical yield) 100
= 13.0 g HCl / 21.5 g HCl (100) = 60.5 %
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Mult-step Stoichiometry Problem
How many g of KClO3 would be needed to supply the proper amount
of oxygen to burn 33.2 g of methane (CH4) according to the following
reactions?
KClO3 KCl + O2
CH4 + O2 CO2 + H2O
First balance each reaction.
2 KClO3 2 KCl + 3 O2 CH4 + 2 O2 CO2 + 2 H2O
g KClO4 = 33.2 g CH4 1 mol CH4 / 1604 g CH4 2 mol O2 / mol CH4
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2 mol KClO3 / 3 mol O2 122.55 g KClO3 / mol KClO3
= 338 g KClO3
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Learning Check
If 22.8 mL of 0.100 M MgCl2 is needed to completely react with 15.0 mL
of AgNO3 solution, what is the molarity of the AgNO3 solution?
MgCl2(aq) + 2AgNO3(aq) 2AgCl(s) + Mg(NO3)2(aq)
A. 0.304M B. 1.20M C. 0.152M D. 0.405M
BEFORE: 22.8 mL
0.100 M
15.0 mL
Moles MgCl2 ? Moles AgNO3
? M
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Learning Check
If 22.8 mL of 0.100 M MgCl2 is needed to completely react with 15.0 mL
of AgNO3 solution, what is the molarity of the AgNO3 solution?
MgCl2(aq) + 2AgNO3(aq) 2AgCl(s) + Mg(NO3)2(aq)
A. 0.304M B. 1.20M C. 0.152M D. 0.405M
0.0228 L 0.100 mols MgCl2
MgCl2
= moles AgNO3
1 L
2 mol AgNO3
1 mol MgCl2
= 0.00456 moles AgNO3
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Learning Check
If 22.8 mL of 0.100 M MgCl2 is needed to completely react with 15.0 mL
of AgNO3 solution, what is the molarity of the AgNO3 solution?
MgCl2(aq) + 2AgNO3(aq) 2AgCl(s) + Mg(NO3)2(aq)
A. 0.304M B. 1.20M C. 0.152M D. 0.405M
Molarity = 0.00456 moles AgNO3
0.0150 L
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