Chang Unit 1 Chapters 1 & 3
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Transcript of Chang Unit 1 Chapters 1 & 3
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Unit I: Ch. 1 & 3
Review of IntroductoryChemistry
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ChemicalFoundations
CHAPTER 1
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1. Introduction
2. Classification of Matter
3. Prefixes
4. Density
5. Scientific Notations (quick review)
6. Significant Figures
7. Dimensional Analysis
KEY CONCEPTS
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Chemistryis the study of matter and the
changes it undergoes
The scientific methodis a systematic approach
to research and discovery
A hypothesisis a tentative explanation for a set
of observations
A hypothesisbecomes a theoryif there are
many experimental results that upheld it
What is Chemistry?
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Common Prefixes and Values
Memorize these symbols & their meaning!!
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Density has unit of mass per volume
In a system of 2 phases, the denser phase will
sink and the less dense phase will float
Density of oil vs. water? Ice vs. water?
Density of water is ~ 1 g/mL (d depends on T)
Memorize: 1 mL = 1 cm3
density =mass
volumed=
mV
Density
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Scientific Notation
Used to express values too small or too big
Ex: 602,200,000,000,000,000,000,000= 6.022 x 1023
N, base #, a numberbetween 1 and 10
n, exponent, is a positive
or negative integer
N x 10n
N, base #, determines the # of significant figures
for the whole scientific notation
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Scientific Notation Manipulation
Converting normal notation to scientific notation:
1. Move decimal point (dp) so that 1 < N < 10
2. If dp is moved left, exponent n will be positive
3. If dp is moved right, exponent n will be negative
4. The value of n = # of digits dp is moved
5. If there is not a dp in a number, imagine it beingat the end of the number
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Scientific Notation Manipulation
Ex 1: 568.762 = 5.68762 x 102
(dp is moved 2 digits to left n = +2)
Ex 2: 0.00000772 = 7.72 x 10-6
(dp is moved 6 digits to right n = -6)
Ex 3: 215300000 = ?
Ex 4: 2.37 x 10-3 = ?
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Scientific Notation Manipulation
Addition or Subtraction
1. Write each quantity with the same exponent n2. Combine N1 and N23. The exponent, n, remains the same
Ex: 4.31 x 104 + 3.9 x 103 =
4.31 x 104 + 0.39 x 104 = 4.70 x 104
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Scientific Notation Manipulation
Multiplication
1. Multiply N1 and N22. Add exponents n1 and n2
(4.0 x 10-5) x (7.0 x 103) =(4.0 x 7.0) x (10-5+3) =
28 x 10-2 =2.8 x 10-1
Division
1. Divide N1 and N22. Subtract exponents n1 and n2
8.5 x 104 5.0 x 109 =(8.5 5.0) x 104-9 =
1.7 x 10-5
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Significant Figures (sf)
1. Needed to indicate the certainty of a measurement
Ex: 2.00 kg is scientifically not the same as 2.0 kg
2. Any value obtained from measurement (using an
instrument) has a limited # of sf
Ex: T, V, m measurements
3. Values obtained NOT using an instrument
(counting, exact #, by definition) have an infinite
(unlimited) # of sf
Ex: 10 people in this room, 1 L = 1000 mL142/23/2010
Significant Figures (sf)
4. How many sf a measurement should have?
All measurements must have 1 (and only 1)
uncertain digit Read all the readable (certain)
digits up to including the first guessed digit
Ex: Read the T of this room in F (~ 75 C) using a
thermometer calibrated to tenths of a degree (10
divisions between one degree and next). How
many sig. fig. should the measurement have?
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Rules for Significant Figures1. Any digit that is not zero is sf
1.234 kg 4 significant figures
2. Captive zeroes (between nonzero digits) are sf
60006 m 5 significant figures
3. Leading zeroes are never sf
0.08 L 1 significant figure
4. Trailing zeroes are sf only if there is a decimal
point; if not they are not sf
0.00420 mg 3 significant figures
12000 g 2 significant figures162/23/2010
How many sf in each of the following measurements?
24 mL
3001 g
0.0320 m3
6.4 x 104 molecules (careful!)
560 kg
How to write 64,000 with 4 sf?
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Significant Figures in Calculations
Addition and Subtraction
The answer must have the same number of decimal
places (not sf) as the value that has the least # of
decimal places. From here, the # of sf in the final
answer can be determined
89.332 (3 dp)1.1 (1 dp)+
90.432 = ?
3.70-2.9133
0.7867 ?
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Significant Figures
Multiplication or Division
The number of significant figures in the result is set
by the number that has the smallestnumber of
significant figures
4.51 x 3.6666 = 16.536366 = ?
6.8 112.04 = 0.0606926 = ?
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Significant Figures
Many calculations involved
Apply the rule of sf one step at a time. The # of sf
in the final answer depends on the rule of the last
calculation
2.37 x 1.500 12.007 = ?= 16.536366 12.007 = ?
(0.0840 0.08206)
0.08206X 100% = ?
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Accuracy how close a measurement is to the truevalue
Precision how close a set of measurements are to each other
Accurate &precise
Precise butnot accurate
not accurate ¬ precise
In an unknown determination, accuracy is not known; can
only know precision consistent results are needed
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Dimensional Analysis As Method ofSolving Problems
1. To clearly show how result is obtained, and for the work to
be understandable by everyone (universal meaning)
2. Determine which conversion factor(s) are needed (can
convert one unit at a time, independently of other units)
3. A conversion factor is a ratio, = 1, has 2 different units
4. Carry units through calculation; cancel identical units
5. If all units cancel except for the desired unit(s), then the
problem was solved correctly.
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How many atoms of P are there in 2.50 g Ca3(PO4)2,MM =310.18 g/mol?
What is 2500 sf2 in cm2? 1 ft = 0.3 m (exactly)
Ex: The speed of sound in air is about 343 m/s. Whatis this speed in miles per hour? (1 mile = 1609 m)
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Stoichiometry
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1. Mole Concept
2. Molar Mass & Mass-Mole Conversion
3. Percent Composition
4. Empirical Formula & Molecular Formula
5. Stoichiometry
6. Limiting Reagent/Reactant
7. % Yield
KEY CONCEPTS
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Atomic mass: mass of an atom, depends on p+ & n
Ex: C-12 has atomic mass of 12.00 amu
C-13 has atomic mass higher than 12.00 amu
H-1 has atomic mass of 1.008 amu
Some elements have isotopes: C-12, C-13, C-14,
each have different atomic mass
Average atomic mass: mass of an atom takinginto account all the isotopes and their abundance
Atomic Mass
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Natural lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
7.42% x 6.015 + 92.58% x 7.016 = 6.941 amu
Average atomic massof lithium:
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QUESTION
The atomic mass of rhenium is 186.2. Given that
37.1% of natural rhenium is rhenium-185, what
is the other stable isotope?
1)183
75Re 4)
181
75Re
2)187
75Re 5)
190
75Re
3)189
75Re
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The mole (mol)is the amount of a substance that
contains as many elementary entities as there are
atoms in exactly 12.00 grams of 12C
1 mol = NA = 6.0221367 x 1023
(NA): Avogadros number
The Mole
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Molar mass: mass of 1 mole of an element or
compound in grams Molar mass for a given
compound is a constant
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li
Molar mass of a compound = sum of molar
masses of all atoms in the compound
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atomic mass (in amu)Molar mass (in g/mol)
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One Mole of:
C S
Cu Fe
Hg
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Which of the following is closest to the
average mass of one atom of copper?
a) 63.55 g
b) 52.00 g
c) 58.93 g
d) 65.38 g
e) 1.055 X 10-22 g
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Which of the following 100.0 g samplescontains the greatest number of atoms?
1. Magnesium
2. Zinc
3. Silver
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How many atoms are in 0.551 g of potassium (K) ?
How many H atoms are in 72.5 g of C3H8O ?
How many molecules are in 72.5 g of C3H8O ?
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M= molar mass in g/mol
NA = Avogadros number
variable const. variable const. variable
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Light
Light
Heavy
Heavy
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Percent compositionof an element in a cmpnd =
nx atomic mass of elementmolar mass of compound
x 100%
Atomic mass of O = 16.00 g/mol, not 32.00 g/mol
n: number of moles of an atom in 1 mole of the
compound (i.e., n is the subscript of the atom inthe compound)
total mass of elementtotal mass of compound
x 100% =
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C2H6O
%C =2 x (12.01 g)
46.07 gx 100% = 52.14%
%H =6 x (1.008 g)
46.07 gx 100% = 13.13%
%O =1 x (16.00 g)
46.07 g
x 100% = 34.73%
52.14% + 13.13% + 34.73% = 100.0%
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Evaluate Your Answer
Is it always true that a larger subscriptfor oxygen in a compound leads to a
greater percent oxygen by mass in acompound?
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Emperical Formula vs. Molecular Formula
Emperical formula: a formula that gives the relative
ratio of atoms in a molecule
Should always be written as (CxHyOz)n where n =
1,2,3,4.
Molecular formula: a formula that clearly gives the #
of atoms in a molecule, hence also their ratios
To determine molecular formula, the emperical
formula and molar mass have to be known
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Emperical formula of a cmpnd can be determined if
the % composition is known (and vice versa)
STEPS:
1. Assume the % composition of each element in the
compound is the mass of that element
2. Determine the moles of each element
3. Divide moles of each element by the smallest
moles get ratio of moles of elements making up
the cmpnd empirical formula
4. Dont round up in step 3; instead multiply moleratios by an integer to make all mole ratios integers
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Ex : A compound is made up of 81.8% C and the restis H. What is the empirical formula of the compound?
Ex: A compound is known to make up of C, H and N.The %C = 65.47%, %H = 9.09%. If the molar massof the compound is 111 g/mol, what is its molecular
formula?
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3 ways of representing the reaction: pictures, words, symbols
Chemical Equations
A chemical equationuses chemical symbols to show the
changes in a chemical reaction (not how it happens)
reactants products
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How to Read Chemical Equations
2 Mg + O2 2 MgO
2 atoms Mg + 1 molecule O2 makes 2 molecules MgO
OR 2 moles Mg + 1 mole O2
makes 2 moles MgO
OR 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
BUT IS NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
IS
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Balancing Chemical Equations
1. Start with the most complicated cmpnd first
2. Change the numbers in front of the formulas
(coefficients) to make the number of atoms of
each element the same on both sides of the
equation. Do not change the subscripts.
3. Repeat step 2 for all cmpnds in the equation
4. Check to make sure that every atom is balanced.
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Ex 1: C3H6O2N + O2 CO2 + H2O + NO2
Ex 2: Al4C3 + H2O Al(OH)3 + CH4
Ex 3: PCl3 + H2O H3PO3 + HCl
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1. Write balanced chemical equation
2. Convert mass of known substances into moles
3. Use coefficients in balanced equation to convert to the
number of moles of compound sought
4. Convert moles of compounds sought into mass
Mass Changes in Chemical Reactions
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Ex 1: If 209 g of methanol are used up in the combustion
reaction below, what mass of water is produced?
2CH3OH + 3O2 2CO2 + 4H2O
Ex 2: When 20.0 g NH3 is reacted with excess oxygen, how
many grams of NO can be produced?
4NH3 + 5O2 4NO + 6H2O
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6 red left over
Limiting Reagents
Reactant that will be used up first; opposite of excess
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Mass Changes in a rxn with a Limiting Reactant
1. Using mass of each reactant, calculate the massof a common product formed (2 values obtained)
2. The smaller amount calculated (of the 2) is theamount of product that will be formed
3. The reactant giving this amount of product is thelimiting reactant
4. What is the meaning of the larger amount?
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Ex 1: 124 g of Al is reacted with 601 g of Fe 2O32Al + Fe2O3 Al2O3 + 2Fe
Calculate the mass of Al2O3 formed. Which is limiting?
Ex 2: 12.0 g NaCl and 21.0 g H2SO4 are reacted
2NaCl + H2SO4 Na2SO4 + 2HCl
Calculate the mass of acid formed. Which is limiting?
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Properties of a reaction (always
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Ex 1: A 5.95-g sample of AgNO3 is reacted with 6.52 g
BaCl2 according to the equation:
to give 3.14 g solid. What is the percent yield of the
reaction?
3 2 3 22 ( ) ( ) 2 ( ) ( ) ( ) AgNO aq BaCl aq AgCl s Ba NO aq+ +
Ex 2: When 150g ammonia is reacted with 150g oxygen,
87g of NO is obtained. Is the reaction efficient?
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g).