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EIT ReviewF2007
Dr. J.A. Mack
www.csus.edu/indiv/m/mackj/
Part 1
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Matter & Measurement
Matter is the physical material of the universe.
Matter is made up of relatively few elements.
Matter consists of atoms and molecules.
Each element is a unique atom.
Atoms combine to form molecules
Matter is anything that occupies space (volume) and has mass.
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Classifying Matter:Classifying Matter:
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2
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STATES OF MATTERSTATES OF MATTERSTATES OF MATTER
•• SOLIDSSOLIDS —— have rigid shape, fixed volume. have rigid shape, fixed volume. External shape can reflect the atomic and External shape can reflect the atomic and molecular arrangement.molecular arrangement.
•• LIQUIDSLIQUIDS —— have no fixed shape and may have no fixed shape and may not fill a container completely. not fill a container completely.
•• GASESGASES —— expand to fill their container. expand to fill their container.
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•• Physical PropertiesPhysical Properties can be determined without changing the chemical make–up of the sample.
•• Examples of physical properties are:Examples of physical properties are:
–Melting Point,Boiling Point, Density, Mass, Touch, Taste, Temperature, Size, Color, Hardness, Conductivity.
•• Examples of physical changes are:Examples of physical changes are:
–Melting, Freezing, Boiling, Condensation, Evaporation, Dissolving, Stretching, Bending, Breaking
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•• Physical PropertiesPhysical Properties can be determined without changing the chemical make–up of the sample.
•• Examples of physical properties are:Examples of physical properties are:
–Melting Point,Boiling Point, Density, Mass, Touch, Taste, Temperature, Size, Color, Hardness, Conductivity.
•• Examples of physical changes are:Examples of physical changes are:
–Melting, Freezing, Boiling, Condensation, Evaporation, Dissolving, Stretching, Bending, Breaking
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Chemical PropertiesChemical Properties are those that do change the chemical make–up of the sample.
Examples of chemical properties are:Examples of chemical properties are:
Burning, Cooking, Rusting, Color change, Souring of
milk, Ripening of fruit, Browning of Apples,
photography, Digesting food
Note: Chemical properties are actually chemical changes.
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Temperature:Temperature:
(phase change)(phase change)
(all motion stops)
Temperature is a measure of Temperature is a measure of particle motionparticle motion
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Units of Measure:Units of Measure:English Customary Weights and Measures
The International System of Units (SI)
distance
inch
foot
yard
mile
meter
micrometer
centimeter
kilometer
= 12 inches
= 3 feet
= 5280 feet
= 10−6 meters
= 10−2 meters
= 103 meters
Units with in a system can be represented by units within the system.Units within the metric system are related by powers of 10
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You need to memorize these values and have the ability to convert between.
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What are Significant Figures?
Significant figures communicate the uncertainty in a measurement.
45. 8724g ± 0.001g
Given the mass: With the uncertainty:
Which numbers are certain ? (significant)
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Counting Significant Figures
1. All non zero numbers are significant
2. All zeros between non zero numbers are significant
3. Leading zeros are NEVER significant. (Leading zeros are the zeros to the left of your first non zero number)
4. Trailing zeros are significant ONLY if a decimal point is part of the number. (Trailing zeros are the zeros to the right of your last non zero number).
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These digits are certain
45. 8724g ± 0.001
The error (±) tells us which digit is uncertain:
The uncertainty occurs at the thousands place:
Therefore, 45. 8724
This digit is uncertain
The blue digits are significant, or we say there are 5significant figures
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not trapped by a not trapped by a decimal place.decimal place.
zeros written zeros written behind the decimal behind the decimal are significant…are significant…
Determine the number of Sig. Figs. in the following numbers
4 sf
7 sf
3 sf
5 sf
3 sf
4 sf
4 sf
1256
1056007
0.000345
0.00046909
1780
770.0
0.0804021
Now count each step as a power of ten to find the exponent.
563,490.
Five steps: the exponent is 5
5.63490 × 105
Scientific notation:
5
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Sig. Figures in CalculationsMultiplication/Division
The number of significant figures in the answer is limited by the factor with the smallest number of significant figures.
Addition/Subtraction
The number of significant figures in the answer is limited by the least precise number (the number with its last digit at the highest place value).
NOTE: Defined numbers (numbers from tables and references) never limit calculations.
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Multiplication and Division:
Determine the correct number of sig. figs. in the following calculation, express the answer in scientific notation.
23.50 ÷ 0.2001 × 17
The sf in the result is limited to the number with the least amount of sf.
4 sf 4 sf 2 sf
The answer must be rounded to 2 sf.
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from the calculator:
23.50 ÷ 0.2001 × 17
1996.501749 10 sf
in sci. not.: 1.996501749 x 103
Rounding to 2 sf: 2.0 x 103
Your calculator knows nothing of sig. figs. !!!Your calculator knows nothing of sig. figs. !!!
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Sig. Figs. Addition and Subtraction
How many sig. figs. are allowed in the following calculation?
391 - 12.6 +156.1456To determine the correct decimal to round to, align the numbers at the decimal place:
391-12.6
+156.1456One must round the calculation to the least significant decimal.
no digits hereno digits here
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391-12.6
+156.1456
one must round to here
534.5456 (answer from calculator)
round to here (units place)
Answer: 535
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Combined Operations:
When there are both addition / subtraction and multiplication / division operations, the correct number of sf must be determined by examination of each step.
Example: Complete the following math mathematical operation and report the value with the correct # of sig. figs.
(26.05 + 32.1) ÷ (0.0032 + 7.7) = ???
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(26.05 + 32.1) ÷ (0.0032 + 7.7) = ???
(26.05 + 32.1) (0.0032 + 7.7)
=
1st determine the correct # of sf in the numerator (top)
2nd determine the correct # of sf in the denominator (bottom)
The result will be limited by the least # of sf (division rule)
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(26.05 + 32.1) (0.0032 + 7.7)
=
26.05+ 32.1
0.0032+ 7.7
3 sf
2 sfThe result may only have 2 sf
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58.150
7.70032
(26.05 + 32.1) (0.0032 + 7.7)
=
= 7.5516 = 7.62 sf
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The Representation of Matter:
In chemistry we use chemical formulas and symbols to represent matter.
Why?Why?
We are “macroscopic”: large in size on the order of 100’s of cm
Atoms and molecules are “microscopic”:
on the order of 10-12 cm
32DmitriDmitri MendeleevMendeleev (1834 (1834 -- 1907)1907)
The Periodic TableThe Periodic TableWhere do we begin…
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The modern periodic table is defined by:
Periods (rows across)
Groups (families)(columns down)
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UhUh--Oh! Oh! this is confusing…this is confusing…
Molecules: H2 =
O2 =
H2O =
CO2 =
Elements: H =
O =
C =
Chemical Symbols and Formula:
hydrogen
oxygen
carbon
hydrogen
oxygen
water
carbon dioxide
Yes it is…Yes it is…Get over it Get over it
and get used to it!and get used to it!
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Which elements do we need to know?
As many as possible!As many as possible!
Start here as a minimum…
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The Periods are labeled:11
2233
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The Groups are labeled:1A
2A 3A 4A 5A 6A7A 8A
The “A” refers to The “A” refers to the “main group the “main group elements”elements”
3B 4B 5B 6B 7B 8B 1B 2B
The “B” refers to theThe “B” refers to thetransition metal elements.transition metal elements.
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Periodic Table: Metallic arrangement
1IA
18VIIIA
12
IIA13
IIIA14
IVA15VA
16VIA
17VIIA
2
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IIIB4
IVB5
VB6
VIB7
VIIB8 9
VIIIB10 11
IB12IIB
4
5
6
7
MetalsMetals Nonmeta
ls
Nonmeta
ls
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Periodic Table: Metallic arrangement
1IA
18VIIIA
12
IIA13
IIIA14
IVA15VA
16VIA
17VIIA
2
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IIIB4
IVB5
VB6
VIB7
VIIB8 9
VIIIB10 11
IB12IIB
4
5
6
7
Metalloids somewhere in between metals and non-metals
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Atomic Number, ZAtomic Number, ZAn element’s identity is defined by An element’s identity is defined by
the number of protons in the the number of protons in the nucleus:nucleus:
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AlAl
26.98126.981
Atomic numberAtomic number
Atom symbolAtom symbol
Atomic weightAtomic weight
ZZ
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NOMENCLATUREFormat for naming chemical compounds using prefixes, suffixes, and other modifications of the names of elements which constitute compounds.
Atom:Atom: The smallest divisible unit of an elementCompound:Compound: A substance made of two or more atomsIon:Ion: A charged atom or moleculeCation:Cation: Positive ionAnion:Anion: Negative ion
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Metals form CationsCations non-metals form anionsanions MetalloidsMetalloids can do both
inert or “noble” gasses:
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Ion Charges:
+1+2
+3 -3 -2 -1
Variable
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Compounds fall into one of two classes:
Inorganic Salts Molecules
metal cation
+
non-metal orpolyatomic anion
non-metal
+
non-metal(no cations or anions)
The two use different formalisms for naming…
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Binary Compounds: Metal & non-Metal
Metal of fixed oxidation (charge) state combined with a non-metal.
Examples:
NameFormulaAnionCation
K+ Cl− KCl Potassium chloride
Ca2+ O2- CaO Calcium Oxide
Na+ S2- Na2S Sodium sulfide
Al3+ S2- Al2S3 Aluminum sulfide
non-metal takes on “ide” suffix
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Metals of variable charge (transition) with a non-metal
Examples:
NameFormulaAnionCation
Pb2+ Cl− PbCl2 lead (II) chloride
Pb4+ Cl− PbCl4 lead (IV) chloride
pronounced:pronounced: lead lead -- two two -- chloridechloride
Fe3+ O2− Fe2O3 Iron (III) oxide
modify transition metal name with roman numeral
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NH4+ ammonium
H3O+ hydroniumCO3
2– carbonateHCO3
2– hydrogencarbonate or bicarbonateNO2
– nitriteNO3
– nitrateSO4
2– sulfateSO3
2– sulfitePO4
3– phosphateC2H3O2
– acetate
Some common polyatomic ions:
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Ternary Compounds: Those with three different elements
Type A: metal of fixed charge with a complex ion
NameFormulaAnionCation
K+ OH− KOH Potassium hydroxide
Ca2+ OH− Ca(OH)2 Calcium hydroxide
Na+ 24SO − Na2SO4 Sodium sulfate
Al3+ 24SO − Al2(SO4)2 Aluminum sulfate
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Metal of variable charge transition) with a complex ion
NameFormulaAnionCation
Fe3+ Fe(NO3)3 Iron (III) nitrate3NO−
Fe2+ Fe(NO2)2 Iron (II) nitrite2NO−
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NonNon--metal with a nonmetal with a non--metalmetal
When non-metals combine, they form molecules.They may do so in multiple forms:
CO CO2
Because of this we need to specify the number of each atom by way of a prefix.
1 = mono 2 = di 3 = tri 4 = tetra
5 = penta 6 = hexa 7 = hepta
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Examples:
BCl3 boron trichloride
Formula Name:
SO3sulfur trioxide
NO nitrogen monoxide
we don’t write: nitrogen monooxideor mononitrogen monoxide
N2O4 dinitrogen tetraoxide
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D) Writing formulas for acids and Bases
•An acidacid is a substance that produces H+ when dissolved in water.•Certain gaseous molecules become acids when dissolved in water.
•A basebase produces OH− when dissolved in water.•Bases often are Group I and Group II hydroxide salts.
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Type I Acids: Acids derived from ––ideide anions.
The names for these acids follows the formula:
“hydro” + the root of the ide anion + ic “acid”
HCl
Anion: Acid: Name:
chloride hydrochloric acid
HFfluoride hydrofluoric acid
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H+ and S2-
H2S
it takes 2 H+ to cancel one S2-
hydro sulfuric acid
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Anion: Acid: Name:
Examples:
3NO−(nitrate) HNO3 nitric acid
(sulfate) 24SO − H2SO4 sulfuric acid
(acetate) 2 3 2C H O− HC2H3O2 acetic acidvinegar
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Practice:
N2SO4 sodium sulfate
barium carbonate BaCO3
FeO Iron (II) oxide
zinc phosphide Zn3P2
NiBr2 nickel (II) bromide
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CO2 carbon dioxide
carbon monoxide CO
P2O5 diphosphorous pentaoxide
nitrogen trihydride NH3
(ammonia)
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Common Names:
H2O water
ammonia NH3
CH4 methane
NO nitric oxide
N2O nitrous oxide
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please go to my chem. 1A web site and download a nomenclature practice worksheet if you need more review.
http://www.csus.edu/indiv/m/mackj/chem1a/chem1A_lab.html
The link to the worksheet is at the top of the page.
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•Atoms are made of protons, neutrons and electrons.
•The nucleus of the atom carries most of the mass.
• It consists of the protons and neutrons surrounded by a
cloud of electrons.
The charge on the electron is –1
The charge on the proton is +1
There is no charge on the neutron
The Atomic NumberAtomic Number or number of protons in the nucleus defines an element.
Modern Atomic Theory:
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The Composition of an Atom: The Composition of an Atom:
••protons and neutrons in the nucleus.protons and neutrons in the nucleus.
••the number of electrons is equal to the the number of electrons is equal to the number of protons.number of protons.
••electrons in space around the nucleus.electrons in space around the nucleus.
The atom is mostlyThe atom is mostlyempty spaceempty space
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Isotopes, Atomic Numbers, and Mass Numbers
•Atomic number (Z) = number of protons in the nucleus.•Mass number (A) = total number of nucleons in the nucleus (i.e., protons and neutrons).•One nucleon has a mass of 1 amu
(Atomic Mass UnitAtomic Mass Unit) a.k.a “DaltonDalton”•Isotopes have the same Z but different A.•The elements are arranged by Z on the periodic table.
ZA XBy convention, for element X, we write
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Se7534
75 protons + neutrons
34 protons
protons = 34 electrons = 34
neutrons = 75-34 = 41
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AvagadroAvagadro’’ss NumberNumber
Since one mole of 12C has a mass of 12g (exactly), 12g of 12C contains 6.022142 x 1023 12C-atoms.
But carbon exists as 3 isotopes: C-12, C-13 &C-14
The average atomic mass of carbon is 12.011 u.
From this we conclude that 12.011g of carbon contains 6.022142 x 1023 C-atoms
Is this a valid Is this a valid assumption?assumption?
Yes, since NYes, since NAA is so large, is so large, the statistics hold.the statistics hold.
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Molar MassesSince we can equate mass (how much matter) with moles (how many particles) we now have a conversion factorthat relates the two.
The Molar Mass of a substance is the amount of matter that contains one-mole or 6.022 × 1023 particles.
The atomic masses on the Periodic Table also represent the molar masses of each element in grams per mole (g/mol)
mols × molar mass (g/mol) = grams
aka: Avogadro's number Avogadro's number (N(NAA))
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So if you have 12.011g of carbon…you have 6.022×1023 carbon atoms!
So if you have 39.95g of argon…you have 6.022×1023 argon atoms!
if you have a mole of dollar bills… you are Bill Gates…
you have 6.022×1023 bucks!
and if you have 6.022×1023 avocados…
you have… a “guacamole”a “guacamole”68
The molar mass of a molecular compound is the sum of the molar masses of its atoms.
Example:
The molar mass of CO2 is:
1 x (12.01 g/mol) + 2 x (16.00 g/mol)
= 44.01 g/mol
Molar Masses (Molecular Weights) of Compounds:
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How many oxygen atoms are there in 25.1g of chromium (III) acetate?
step 1: write the correct chemical formula…
Cr3+ & C2H3O2− Cr(C2H3O2)3
step 2: calculate the molar mass… 229.13 g/mol
step 3: use dimensional analysis to solve the problem…
2 3 2 325.1g Cr(C H O ) 2 3 2 3
2 3 2 3
1mol Cr(C H O )229.13g Cr(C H O )
×2 3 2 3
6mol O1mol Cr(C H O )
×
236.022 10 O atoms1mol O
× −× = 3.96 × 1023 O-atoms
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Percent Composition:
The relative amounts of each atom in a molecule or compound can be represented fraction of the whole.
Question: What is the weight % of each element in C2H6?
g30.07ofmassahasHC of mol1 62
2 61 mol C H2 6
6 mol H 1 mol C H
× 1.0079 g H 1 mol H
×
(2×12.01 + 6 ×1.008) g/mol
Next Next determine the mass of hydrogen in 1 mol of the compound:
FirstFirst determine the molar mass of C2H6:
= 6.047 g H
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Now relate the mass of H in one mol of the compound to the molar mass of the compound
2 6
130.07g C H
×6.047 g H 100× = 20.11% H
Since there is only C as the remaining element:
% C = 100
The compound C2H6 is 20.11% H & 79.89% C
− % H = 79.89 % C
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Determining a Formula from Percent Composition:
Given the relative percentages of each element in a compound,
10 % X, 20 % Y, 30 % Z etc…
one can find the empirical formulaempirical formula of the compound.
The empirical formulaempirical formula of a compound or molecule represents the simplest ratio of each element in 1 mol of the compound or molecule.
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Example: A compound is found to be 64.82 % carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound?
Solution:
1. Since the percentages for each element sum to 100%, if one equates % to grams (g), the sum of the masses must be 100g.
(i.e. one can assume 100g of the compound)
64.82 g C 21.59 g O 13.59 g H
determine X, Y & Z in (CXHYOZ)
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2. Convert the grams of each element to moles. (g element X → mole X etc…)
C64.82gC g 12.011
Cmol1× Cmol5.397=
O21.59gO g 16.00
Omol1× Omol1.349=
H13.59g Hmol13.48=H g 1.0079Hmol1
×
Example: A compound is found to be 64.82 % carbon, 21.59 % oxygen and 13.59 % hydrogen. What is the empirical formula for this compound?
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3. Divide each of the individual moles by the smallest number ofmoles to gain the molar ratios for each element in the compound.
These are the formula subscripts. (X2Y3 etc…)
**If the ratios are fractional (0.5, 1.5 or 0.333) multiply each ratio by a
whole number to get even number formula subscripts.
4.001 1.3495.397 X ==
1.349Z 1.0001.349
= =13.48Y 9.9921.349
= =
Subscript for CSubscript for C Subscript for HSubscript for H Subscript for OSubscript for O
0.5 × 2 = 1 0.25 × 4 = 1 0.333 × 3 = 1 Examples:
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The empirical formula is: CC44HH1010OO
The results of this calculation tells us only about the empirical formula of the compound.
To determine the molecular formula, we need more information. This will be shown in a later example.
X = 4.001 = 4
Rounding to the nearest whole numbers:
Y = 9.992 =10 Z = 1.000 = 1
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Mass is conserved in a chemical reaction.
Chemical Equations:
Total mass of Total mass of reactantsreactants
Total mass of Total mass of productsproducts=
Chemical equations must therefore be balance for mass
The number and type of atoms on either side of the equation mustbe equal!
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C2H6 + O2 → CO2 + H2O2 C’s & 6 H’s 2 O’s 1 C & 2 O’s 2 H’s & 1 O
balance H firstbalance last
___C2H6 + O2 → CO2 + ___ H2O336622balance C next
2C2H6 + O2 → ___ CO2 + 6H2O44
balance O
2C2H6 + ____ O2 → 4CO2 + 6H2O77
4 C’s 12 H’s 14 O’s 4 C’s 12 H’s 14 O’s
Balancing Chemical Reactions: Example
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Reduction and Oxidation Reactions: RedOxRedOx
OxidationOxidation involves an atom or compound losing electrons
ReductionReduction involves an atom or substance gaining electrons
Niether process can occur alone… that is, there must be an exchange of electrons in the process.
The substance that is oxidizedoxidized is the reducing agentreducing agent
The substance that is reducedreduced is the oxidizing agentoxidizing agent
Chemists use oxidation numbersoxidation numbers to account for the transfer of electronsin a RedOx reaction.
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Oxidation Numbers
Metals take on their formal charge:
+1
+2
Variable
Non-metals do the same
+3 -3 -2 -1
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Electrochemistry: Oxidation numbers
In the compound potassium bromate (KBrO3), the oxidation number of bromine (Br) is?
KBrO3
The compound is neutral so the sum of the oxidation numbers should be zero.
+1 3×(-2) = -6??
1+ ?? + (-6) = 0 ?? = 5
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O2 + → 2O2-
Balancing REDOX reactions:
oxidation half reaction:
Fe + O2 → Fe2O3
0oxidation states: 0 +3 -2
+ 3e-
reduction half reaction:
Fe → Fe+3
4e-
Balance electrons transferred then sum the half RXN’s:
{ } x 3
{ } x 4
4Fe + 3O2 + 12e- → 2Fe2O3 + 12e-
4Fe + 3O2 → 2Fe2O3
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Consider the following reaction:
HCl (aq)
Balacing:
HCl (aq) + Ba(OH)2 (aq) H2O (l) + BaCl2(aq)
How many moles of HCl are consumed if 1.50 g of BaCl2 are produced assuming that Ba(OH)2 is in excess?
Solution:
g BaCl2 mol BaCl2 mol HCl
using the Molar mass of BaCl2 using the Molar ratio for equation
22 22
+ Ba(OH)2 (aq) H2O (l) + BaCl2(aq)
convertto then convert to
Quantitative calculations: Mass and moles
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2 BaClg 1.502
2 BaClg 208.24
BaClmol1×
2 BaClmol 1HClmol2
× HClmol0.0144=
22HCl (aq) + Ba(OH)2 (aq) 22H2O (l) + BaCl2(aq)
g BaCl2 mol BaCl2 mol HCl
Molar mass BaCl2 Molar ratio for equation
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Conversions between masses & moles in chemical reactions.Stoichiometry:
Limiting Reactant:
When one reactant is present in an amount such that it is completely consumed before all other reactants, we say that it limitslimits the reaction.
The other reactants are said to be in excessexcess.
The Theoretical YieldTheoretical Yield is determined by the stoichiometry of the limiting reactant.
The limiting reactant can only be determined through molar ratios. It cannot be identified by mass.
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Reaction Yields:
The theoretical yield is the maximum product yield that can be expected based on the masses of the reactants and the reaction stoichiometry.
The actual yield is the experimentally measured amount of products that results upon completion of the reaction.
The percent yield is a measure of the extent of the reaction in terms of the actual vs. the theoretical yield.
Yield% =)r molesin grams o( YieldActual
)r molesin grams o( YieldlTheoretica100×
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The Behavior of Solutes:Strong Electrolytes: complete dissociation into ions
1 M Na3PO4(aq) 3M Na+ (a) + 1M PO43–
4M overall in ions
Non–Electrolytes: no dissociation into ions
1M CH3OH (aq)methanol
1M overall in molecules
Weak Electrolytes: partial dissociation into ions
1M HC2H3O2(aq) H+ (aq) + C2H3O2–(aq)
between 1 & 2 M overall88
Example:Example: A solution of sodium phosphate is added to a solution of aqueous barium nitrate. A white ppt is observed.
Unbalanced Equation:
Na3PO4 (aq) + Ba(NO3)2 (aq) → Ba3(PO4)2 (s) + NaNO3 (aq)
Molecular:
+ 6NaNO3 (aq)2Na3PO4 (aq) + 3Ba(NO3)2 (aq) → Ba3(PO4)2 (s)Ionic:
6Na+ (aq) + 2PO43- (aq) + 3Ba2+ (aq) + 6NO3
- (aq)
→ Ba3(PO4)2 (s) + 6Na+ (aq) + 6NO3- (aq)
Net Ionic:
//
//
2PO43- (aq) + 3Ba2+ (aq) → Ba3(PO4)2 (s)
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Solutions and Concentration
per liter of solution.Molarity: Moles of solute
Molarity (M) = moles of solute
L of Solution{units: mol/L}
molarity is a conversion factor that transforms units of volume to mole and vise–versa
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How many grams of sodium phosphate are in 35.0 mL of a 1.51 M Na3PO4 solution?
mL solution L mols Na3PO4 g Na3PO4
35.0mL 3
L10 mL
× 3 41.51 mol Na POL
×3 4
163.94g1 mol Na PO
×
3 48.66g Na PO=
use M as a conversion factor
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New Molarity =old Molarity × old Volume
new Volume
M2 = M1 × V1
V2
M1 × V1 = M2 × V2Rearranging:
Dilutions:Dilutions:
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The pH Scale:The pH Scale:The pH Scale:1pH log
H+
= [H+] = Molarity of H+
(or anything for that matter…)
Since 1log log(x)x
= −
pH = -log[H+]
In a In a neutralneutral solution, solution, [H[H++] = [OH] = [OH--] = 1.00 x 10] = 1.00 x 10--77 M at 25 M at 25 oCoC
pH = pH = -- log [Hlog [H++] =] = = 7= 7--log (1.00 x 10log (1.00 x 10--77)) = = -- [0 + ([0 + (--7)]7)]
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What is the pH of the 0.0515 M HCl solution that Jane made?
pH = -log[H+]
HCl (aq) →
Strong electrolyte!Strong electrolyte!
H+(aq) + Cl−(aq)
Therefore [H+] = [HCl]
pH log[H ] log(0.0515)+= − = −
= 1.29
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Reaction Quotient & Equilibrium Reaction Quotient & Equilibrium ConstantConstant
H2 + I2 → 2 HI 2 2H I 2HI+
Equilibrium EstablishedEquilibrium Established
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THE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANT
For any type of chemical equilibrium of the typeFor any type of chemical equilibrium of the typea A + b B a A + b B →→ c C + d Dc C + d D
the following is a CONSTANT (at a given T)the following is a CONSTANT (at a given T)
K =[C]c [D]d
[A]a [B]b
conc. of products
conc. of reactantsequilibrium constant
If K is known, then we can predict concentrations of products orIf K is known, then we can predict concentrations of products or reactants. reactants.
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The Reaction Quotient, QThe Reaction Quotient, QIn general, ALL reacting chemical systems are In general, ALL reacting chemical systems are
characterized by their characterized by their REACTION QUOTIENT, QREACTION QUOTIENT, Q..a A + b B a A + b B →→ c C + d Dc C + d D
If Q < K, then system will shift to the right, convert reactantsIf Q < K, then system will shift to the right, convert reactants to products.to products.If Q > K, then system will shift to the left, products convert tIf Q > K, then system will shift to the left, products convert to reactants.o reactants.
If Q = K, then system is at equilibrium.If Q = K, then system is at equilibrium.
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Forms of Chemical BondsForms of Chemical Bonds
•• There are 2 extreme forms of connecting There are 2 extreme forms of connecting or bonding atoms:or bonding atoms:
•• IonicIonic——complete transfer of 1 or more complete transfer of 1 or more electrons from one atom to anotherelectrons from one atom to another
•• CovalentCovalent——some valence electrons some valence electrons shared between atomsshared between atoms
•• Most bonds are somewhere in Most bonds are somewhere in between.between.
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When there exists a difference in the Electronegativity (EN) between the two bonding atoms, (∆EN) the bond is said to be polar.
Example: ∆EN Bond
Br2 0 non-polar
HCl 0.9 polar-covalent
NaF 3.1 ionic
24
99
Electronegativity:The measure of the tendency for a given atom to polarize the electrons in a covalent bond is called the Electronegativity (EN) of an atom. Electronegativity is related to the ionization energy and the electron affinity of an atom.
ENdecreases
EN increases
Periodic TablePeriodic Table
100
Rank the following bonds by increasing strength:
H−F, H−Cl and H−I
Each bond has hydrogen in common so we have a basis for comparison.From the periodic table, in terms of electronegativity:
F > Cl > I
inc. EN
One concludes that the H-F bond is the most polar, followed by H-Cl then H-I.
Bond StrengthBond StrengthH−I < H−Cl < H−F
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Multiple bonds in covalent molecules:
The oxygen and nitrogen that makes up the bulk of the atmosphere also exhibits covalent bonding in forming diatomic molecules.
O: + O: → O = O “double bond”
N. + N⋅ → :N≡N: “triple bond”
: :
: :: :
: :
: :
: :
Polyatomic Molecules (More than two atoms)
Carbon dioxide: CO2 O=C=O
: :
: :
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Bond Strength and Bond Properties:
Covalent bond strength increases with increasing ∆EN
example: HCl bond is stronger than the HBr bond
Covalent bond strength increases with increasing bond order
example: O=O bond in stronger than O–O bond
triple > double > single
Bond Order: 3 2 1
Bond length decreases with increasing bond order (Strength)
example: O=O bond is shorter than O–O bond
25
103
H–bonding in water brings about a network of interactions which explain phenomena such as:
capillary action surface tension why ice floats
104
Surroundings
Systemheat in
q > 0(+)
∆E > 0
heat out
q < 0(–)
∆E < 0
Heat Transfer:
105
E initial
E Final
ener
gy
q in
work in
Ef > Ei
∆Esystem > 0 (+)
q out
work out
Ef < Ei
∆Esystem < 0 (–)
E Final
E initial
Energy in Energy out
Thermochemistry:
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Heat and the Specific Heat Capacity:
TCmq ∆××=
The amount of heat (q) transfer is related to the mass and temperature by:
q = heat lost or gained (J) m = mass of substance (g)
C = the Specific Heat Capacity of a compound o
Jg C
⋅
When heat is absorbed or lost by a body, the temperature must change as long as the phase (s, g or l) remains constant.
∆T is the temperature change in degrees Celsius or Kelvin’s
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107
How many kJ of energy are released when 128.5 g of methane (CH4(g)) is combusted?
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ∆H = -802 kJ
4CH 128.5gg 16.04
CH mol 4× =×4CH 1mol
kJ 802 6.42×103 kJ
How many hours would this power a 100 W light bulb if one could use all of this energy?
100 W = 100 Js–1
6.42×103 kJkJ
J103× J100
s×
s60min
× =×min60hr 17.8 hrs
Enthalpy: ∆H
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P V constant for a set number of gas molesn T
×=
×
P V n R T × = × ×
L atmR "gas constant" 0.08206 mol K
⋅= =
⋅
PV = nRT
Ideal Gas Law
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Problems: How many grams of krypton (Kr) will it take to exert a pressure of 11.2 atm in an 18.5 L at 28.2 oC.
PV nRT=L atmR 0.082057 mol K
⋅=
⋅
PVn RT
= =11.2 atm 18.5 L×
L atm0.08206 mol K
⋅⋅
( )28.2 273.15 K× +8.3789 mols Kr=
83.80 g8.3789 mols Krmol
× = 702 g Kr (3sf)
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What wavelengths correspond to FM radio (93.5 MHz) signals?
λ υ c× =cλυ
=
λ =
8 m3.00 10 s×
93.5 MHz610 Hz
1 MHz×
11sHz
−
×
= 3.21 m
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What wavelengths correspond to FM radio (93.5 MHz) signals?
λ υ c× =cλυ
=
λ =
8 m3.00 10 s×
93.5 MHz610 Hz
1 MHz×
11sHz
−
×
= 3.21 m
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113
Each element’s outermost electrons (valence) are related to the elements position on the periodic table.
All of the subshells below the valence are full.
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We can use the periodic table to determine the electron configuration by counting:
N: 7 electrons
1 264 73 5
1: 1s1 2: 1s2 3: 2s1 4: 2s2 5: 2p1 6: 2p2 7: 2p3
1s2 2s2 2p3
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115
Electron Configurations cont…
Orbital box notation:
1s 2s 2p 3s 3p
Aluminum: Al (13 electrons)
1s
2s
2p
3s
3p
↑↓
↑↓
↑
↓↑
↓↑
↓
↑
↓
↑
↓
↑ ↑↓
↑
↓
↑↓
↑
↓ ↑↓
↑
1s2 2s2 2p6 3s2
Electron Configuration
3p1
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Periodic Trends
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The reaction rate law expression relates the rate of a reaction to the concentrations of the reactants.Each concentration is expressed with an order (exponent).The rate constant converts the concentration expression into thecorrect units of rate (Ms−1). (It also has deeper significance, which will be discussed later)For the general reaction:
aA + bB → cC + dD
x and y are the reactant orders determined from experiment.x and y are NOT the stoichiometric coefficients.
x yRate = k [A] [B]
Kinetics: Rate Law & Reaction Order
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A reaction order can be zero, or positive integer and fractional number.
Reaction Orders:
Order Name0 zeroth1 first2 second0.5 one-half1.5 three-half0.667 two-thirds
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119
Recognizing a first order process: A products
Whenever the conc. of a reactant falls off exponentially, the kinetics follow first order.
kto[A] [A] e−=
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A plot of ln[A] versus time (t) is a straight line with slope -kand intercept ln[A]o
( )ktoln [A] [A] e−=
Determining the Rate constant for a first order process
Taking the log of the integrated rate law for a first order process we find:
oln[A] ln[A] k t= − ×
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A certain reaction proceeds through t first order kinetics.The half-life of the reaction is 180 s.What percent of the initial concentration remains after 900s?
Step 1: Determine the magnitude of the rate constant, k.
0.693k
=
10.00385s−=
12
t ln 2k
=
12
ln 2kt
= ln 2180s
=
122
kt
o
[A] e[A]
−=10.00385 s 900 s
o
[A] e[A]
−− ×= = 0.0312
Using the integrated rate law, substituting in the value of k and 900s we find:
Since the ratio of [A] to [A]0 represents the fraction of [A] that remains, the % is given by:
100 × 0.0312 = 3.12%
A certain reaction proceeds through t first order kinetics.The half-life of the reaction is 180 s.What percent of the initial concentration remains after 900s?
k = 0.00385 s-1
