[9/8/08]
Units of measurement
When a number represents a measured quantity the units must be specified
SI: System international
Basic Units:
Quantity
Unit
Mass
Kilogram (Kg)
Temperature
Kelvin (K)
Length
Meter (m)
Time
Second (s)
Amount of substance
Mole (mol)
Electric current
Ampere (Amp)
Luminous intensity
Candela (cd)
Prefix:
Mega – 106
Kilo – 1000
Deci – .1
Centi – .01
Milli – .001
Nano – 10-6
Micro – 10-9
Ways to convert
K h D b d c m (kittens hate dogs because dogs cant meow)
Kilo
Heca
Deci
Basic
Deci
Centi
Milli
How to use: put decimal after the letter that you are using and move over to
whatever you desire and then move your answer that many decimal places
Quantity
Mass- measure of the amount of material in an object
SI unit- kilogram
Temperature- measure of hotness or coldness of an object
Heat flows spontaneously, from higher temperature to lower
temperature
Celsius: 0° C – freezing pt. of water, 100° C- melting pt. of water
Kelvin: 0 K – lowest attainable temp
K = °C + 273
°C = 9/5(°F-32)
°F = 9/5(°C) + 32
[9/10/08]
Density Lab
[9/11/08]
went over scientific notation and how to do a lab report
[9/12/08]
Significant figures
Significant figures
 All digits that are known + a last digit which could be estimated
o Ex. 2.15
 Certain about the 2.1 but not the .05
 Still the 2.15 is a significant figure
 Three significant figures
Depends on instrument of measure
Point is to be able to round off
All non zero numbers are significant
 Zeros between other nonzero digits are significant
 Zeros in front on nonzero digits are not significant
o Instead of .008  8 x 10-3
 Zeros at the end of the number and also to the right of the decimal
are significant
o Shows how exact something is

Zeros at the end of the number but to the left of a decimal place
are not significant (unless there is a decimal which then they are
significant ex. 10.)
Rounding rules
 Addition and subtraction
o Look at number of decimal places that each number that
someone is adding or subtracting and use the smallest
number of decimal places used
 Multiplication or division
o The final answer has the same number of significant figures
as the measurement having the smallest number of
significant figures
[9/15/08]
For inexact measurements, there is always some uncertainty are either
inaccurate or not precise
o Accuracy- measure of how close a measurement is to the
actual value
o Precision- measure of how close a measurement is relative to
other measurements
Unit conversions
[9/16/08]
went over more unit conversions
Chapter 1
Yossi Quint
[9/18/08]
Chemistry- the study of the composition of matter and the changes matter
undergoes (both chemically and physical)
 Matter- something that has mass and occupies space
Scientific method: logical, systematic approach to find a solution to a
scientific problem
 1. Make observation
 2. Ask a question
 3. Make a hypothesis = proposed exclamation for an observation
 4. Experiment
o Independent variable- change during experiment
o Dependent variable- observing
 5. Interpret results
 6. Conclusion
Theory = well tested explanation for a broad set of observations
Three states
 Solid
o Definite shape and volume



Shape doesn’t depend on shape of container
Liquid
o Indefinite shape but definite volume
 Shape is dependent on shape of container
Gas
o Indefinite shape and volume
 Contains both the shape and volume of container
o * Vapor
 gaseous state of a substance that is generally liquid or
solid at room temperature
[9/19/08]
Mixture: physical blend of two or more compounds
(Ex. chicken noodle soup, air)
 Two types:
o Heterogeneous
 Composition is not uniform throughout
o Homogeneous
Composition is uniform
 Same thing as a solution
Separating mixtures:
 Differences in physical properties can be used to separate mixture
o 1. Filtration: process that separates a solid from a liquid in a
heterogeneous mixture
o 2. Distillation: a liquid is boiled to produce a vapor that is
then condensed into a liquid
 solid substances that were dissolved in water remain in
the flask because their boiling points are higher

[9/24/08]
element- the simplest form of matter that has a unique set of properties
  cannot be broken down into simpler substances
compound- substance made of 2 or more elements chemically combined in a
fixed proportion
  can be broken down into simpler substances by chemical means
chemical change- change that produces matter with a different composition
than the original
 ex. heating, electricity
[9/25/08]
Chemical property- one that a substance displays only by changing its
composition via a chemical change = Chemical reaction (rxn)
Physical property- one that substance displays without changing composition
  during a physical change composition of matter never changes
chemical Rxn (reaction)
 reactants  products
o  = yield
law of conservation of mass
 mass cannot be created or destroyed, only changed
  during any chemical reaction, the mass of products is equal to
mass of reactants
ex.
a. Salt and sugar ground to fine power
 mixture, heterogeneous
b. glucose
 pure substance, compound
c. stainless steel
 mixture, homogeneous
d. air
 mixture, homogeneous
ex. physical or chemical
a. distillation of slat water
 physical
b. rusting of iron nail
 chemical
c. burning of wood
 chemical
d. salt precipitate
 physical
Chapter 2
5/27/2009 7:44:00 PM
[9/25/08]
2 laws that led to atomic theory
 conservation of mass
 Law of definitive proportions
o  in samples of any chemical compounds the masses of the
elements are always in the same proportions
 Law of multiple proportions
o Whenever the same two elements form more then one
compound the different masses of one element that combine
wit the same mass with the other element are in the ratio of
small whole numbers
[9/26/08]
atom- smallest particle of an element that retains its identity in a chemical
reaction
 Discovered by John Dalton (1766-1844)
o Daltons Atomic Theory
1. All elements are composed of tiny particles called atoms
2. Atoms of the same element are identical
3. Atoms of different elements can physically, chemically
combine in simple whole # ratios to form compounds
4. Chemical reactions occur when atoms are separated,
joined or rearranged
 Radius of most atoms is between 50 and 200 Pico meters (pm)
Structure of an atom
 Sub atomic particles
o Protons
o Neutrons
o Electrons

J.J. Thomson discovered Electrons
 Using cathode ray tube
 Two electrodes
o Cathode
 Negative charge
o Anode
 Positive charge


At the end he saw negatively charged
electrons
o He knew this since it went towards
the anode (and opposites attract)
He found the ratio of the charge/mass
[10/3/08]
 Millikan oil drop experiment
o He found the mass of an electron
 M = 9.11 x 10-28 grams or 10-31 kilograms
o  charge on electron = q = -1.6 x 10-19 C
 -1 elementary charges

symbol Remote
charge
Relative
mass
Actual
mass (g)
Mass Amu
Electron E-
1-
1/1840
9.11 x 10-28 1
Proton
1+
1
1.67 x 10-24 1
0
1
1.67 x 10-24 0
P+
Neutron N0

Thomson’s atomic model (plum pudding)
o Sphere with a positive charge, and little electrons spread
around it

Rutherford gold foil experiment
o  Atom is mostly empty space
o Atom Model
 1. Protons = electrons in number
 2. Most volume = empty space
 3. Most mass in nucleus
 Positive charge in center (protons, neutrons)
 Electrons are distributed around the nucleus, and
occupy most of the area
Atomic number (symbolized as Z)
 Number of protons in the nucleus of an atom of that element
o Used to identify, element
[10/6/08]
Mass number (symbolized as A)
 It is the protons and neutrons
Mass number − atomic number = number of neutrons
Ex. Oxygen
8
atomic number
chemical mass
Oxygen
15.944
name
Mass number
16
8


number on top mass number
number on bottom – atomic number
o so theirs 8 neutrons, 8 protons, and 8 electrons
Isotopes
 An atom that has the same number of protons but different number
of neutrons
o Three known types for hydrogen
 11H
 hydrogen, 0 neutrons
2
 1H
 deuterium, 1 neutron
3
 1H

tritium, 2 neutrons
AMU



All masses are related to 126C
o Its mass is exactly 12 amu
If we were to use grams instead of this we would have very long
numbers
1 amu = 1/12 mass of carbon atom
o ex.


Helium: 4 amu = 1/3 mass of carbon
Nickel: 60 amu = 5x mass of carbon
Atomic Mass
 Weighted average mass of the atoms in a naturally occurring
sample of the element
Ex.

Carbon 12, Carbon 13
o Abundance: 98.89 %, 1.11 %
o Mass: 12 amu, 13.003 amu

Atomic mass: (12*.9889) + (13.003*.0111)
o (atomic mass * relative abundance)
 in % so divide by 100
[10/7/08]
Periodic table: Arrangement of element in which the elements are separated
into groups and periods based on a set of repeating properties
 Elements listed in order of increasing atomic number, listed from
left to right and top to bottom 



Horizontal row – period
Vertical column – group/family
o Elements have similar chemical and physical properties
Elements are grouped in three categories: metals, nonmetals,
metalloids
o Metals
 good conductors of heat and electricity
 malleable (can be hammered into thin sheets) and
ductile (can be drawn into wires)
80 % of elements
solids at room temperature (except mercury, which is
liquid)
o Non-metals (upper right corner, except H)
 Greater variation in physical properties
 Most are gases at room temperature
 Solid: Sulfur + phosphorus
 Liquid: bromine


In general poor conductors of heat (exception in
carbon)
o Metalloids
 Elements that staircase the line, but not all
 B, Si, Ge, As, Sb, Te, At
 Properties similar to metal + nonmetals








Group 1A – Alkali metals
Group 2A – Alkaline earth matters
Group 7A (17) – halogens
Group 8A (18) – noble gases
3-12B – transition elements
o not arranged according to valence electrons
Ions
o Formation of cat-ions
 Positive ion
 Formed by the loss of an electron
o Formation

Want to lose the amount of valence electrons (closest to the noble
gas-?)
o Group 1A – loses 1 electron (e-) to gain a charge of 1+
o Group 2A – loses 2 electron (2e-) to gain a charge of 2+
o Group 3A – loses 3 electron (3e-) to gain a charge of 3+
 But the only one that works that was is Al  alu3+
[10/23/08]
wasn’t here, got Russel’s notes
1 mole = 6.02*10 to the 23rd power
mole- quantity of a substance whos mass in grams is the same as it's
formula weight
moles=(representative particles)X(1mol/6.02X10 to the 23rd power rep.
particles)
anion always end with –ide
[10/24/08]
How many moles 2.8 x 1024 atoms of silicon?
 2.8 x 1024 * (1 mol / 6.02 * 1023 atoms) = 4.65 mol Si
1 molecule CO2
 3 atoms
o 1 carbon
o 2 oxygen
 1 mole of CO2  6.02 x 1023
 1 mole CO2  (3 x Na) atoms
 1 mole C  Na atoms
o (Na = 6.02 x 1023)
How many molecules are in 2.12 mol of C3H8 ? How many atoms are in 2.12
mol of C3H8 ?
2.12 mol x (6.02 * 1023 molecules / 1 mol) = 1.28 x 1024 molecules
2.12 mol x (6.02 * 1023 molecules / 1 mol) x (11 atoms / 1 molecule) = 1.4
x 1025 atoms
[10/27/08]
How many atoms are in 1.14 moles of SO3?
1.14 mol x (6.02 * 1023 molecules / 1 mol) x (4 atoms / 1 molecules) =
2.75 x 1024 atoms
Going between mass and moles (mass  moles)
 Molar mass (element): atomic mass of an element expressed in
grams = mass of 1 mole of the element
 Ex. Carbon: 12 g/mol
o Molar mass contains 1 mol or 6.02 x 1023 atoms of that
element
 Ex. Molar mass of compound:





o SO3
o Molar mass = molar mass sulfur + 3 (molar mass oxygen)
= 32.12 mol + 3 (16) = 80.12 g/mol
Molar mass (compound): mass is grams of 1 mole of compound
o (If it says find the mass of one mole then the answer is only
in g NOT g/mol)
Ex. Need 3 mol NaCl
o Molar mass = 58.5 g/mol
o Mass (grams) - # moles x (mass (g) / 1 mol)
 3 moles x (58.5 g / mol) – 175.5
Ex. What is the mass of 9.45 mol Al2O3?
o 2(27) + 3(16) x 9.45 mol
Ex. How many moles of Fe2O3 are contained in 92.2 g of Fe2O3?
o 92.2 g / 160 (g/mol) = .576 mol
EX. Calculate number of moles in 75 g N2O3?
o 75 g / 76 (g/mol) = .987 mol N2O3
[10/28/08]
# Moles = mass / molar mass
# Moles = (# atoms / molecules) /NA
Grams  moles and then moles  molecules
How many molecules are there in 450 grams of Na2SO4?
450 g / 142 g/mol = 3.17 moles 
3.17 x (6.02 * 1023 molecules / 1 mol) = 1.94 x 1024 molecules
(if asked for atoms then multiply by seven)
How many grams are there in 7.4 x 1023 molecules of AgNO3?
7.4 x 1023 molecules x (1 mol/6.02 * 1023 molecules) = 1.23 mol
1.23 mol x 170 g/mol = 209.1 g
Calculate molar mass of the following chemicals:
A. FeCl3 =160
B. BF3 = 66.7
C. Mg (OH)2 = 56
How many moles are in 22 grams of argon?
22 g/ 40 g/mol= .55 moles
How many grams are in 88.1 moles of magnesium?
88.1 moles x 24 g = 2114.4 g
How many moles are in 98.3 grams of aluminum hydroxide, Al(OH)3 ?
98.3g/ 78 g = 1.26 moles
Molecules, compounds + chemical equations
Yossi Quint
[10/29/08]
Compound: substance that contains two or more elements chemically
combined in a fixed proportion
Mixtures: have variable composition
A compound has a much different composition then the atoms that make it
up
 Ex. Sugar (made of C,H,O): sweet tasting, white, solid
o C: no taste, black, solid
o H: colorless, gas
o O: colorless, gas
How are compounds held together?
 By chemical bonds
o Two types
 Ionic
 Covalent

Ionic
o
o
o
bondsBetween a metal + nonmetal
Involve a transfer of electrons
Between cation + anion
 Cation – metals lose e- to become positively charged
 Anion – nonmetals gain an e- to become negatively
charged
 e- transferred from metal  nonmetal (Cation  Anion)
[11/3/08]
ex.
 Na+ + Cl-  NaCl
Covalent bondo Atoms do not gain or lose an eo “Tug of war” for electrons between atoms  which bonds
them together
o sharing of an eo between nonmetals + nonmeatals


o form molecular compounds
Chemical formula
 Representation of composition of substances
 Shows the kinds + numbers of atoms

Empirical Formula – gives the lowest whole number ration of the
atoms of the elements in the compound
o Ex. HO or CH2O (instead of H2O2 and C6H12O6)

Molecular Formula – gives the actual number of atoms of elements
in a compound
o Ex. H2O2 and C6H12O6
 Some can be both like H2O

Structural formula – represents covalent bonds by dashes and
shows the arrangement of covalently bonded atoms
o Ex.
 F2  F–F
CO2  O = C
 double bond, stronger then a single bond, two
shared electrons
Molecular  empirical
o C2H6  CH3
o C2H6O  C2H6O
o Hg2Cl2  HgCl



Molecular Elements – Diatomic molecules exist in nature as
molecules ; 2 atoms of an element bonded together
o ex. F2, Cl2, O2, N2, Br2, I2, H2

Molecular Compound – made of 2 or more nonmetals that are
covantly bonded
Polyatomic ion – group of atoms that has a positive or negative
charge

[11/4/08]
Polyatomic ions
 Ex. classify as atomic element, molecular element, molecular
compound , ionic compound (metal, nonmetal)
o Neon

o CO

o F2

o C2H6

Atomic element
Molecular compound (Two nonmetals)
Molecular element
Molecular compound (Two nonmetals)
o KCl


Ionic compound (One is a metal the other is a
nonmetal)
Got handout with the common polyatomic ions
How to write formulas for ionic compounds
 1) write the cation (+) and anion (-)
o (symbol and charge: you get charge from the table if a
compound, if it’s a element look at element,
 group 1  1+. 2  2+. 3  Al3+, 7  1-, 6  2-, 5 


N3-)
2) adjust subscripts to balance overall charge
3) check that cation charge = anion charge  neutral molecule
o ex. Calcium and Chlorine
 1) Ca2+Cl 2) CaCl2
 3) 2+ = 2o ex. Magnesium and Oxygen
 Mg2+O2 MgO
 2+ = 2o ex. Aluminum and Oxygen
 Al3+O2 Al2O3
 6+ = 6Naming Ionic Compounds
 Name of metal + stem+ide
o Ex.




NaCl- sodium Chloride
AlN- Aluminum Nitride
NH4Cl- Ammonium chloride
MgSO4- Magnesium Sulfate (polyatomic ion- look at
chart for name)
o Ex. Calcium + nitrate
 Ca2+NO3-



Ca(NO3)2
2+ = 2Calcium Nitrate
[11/5/08]
 naming transition elements and compounds
o Fe2+  Iron (II)
o Fe3+  Iron (III)
o Cu+  Copper (I)
o Cu2+  Copper (II)

o Fe2+ + Cl-  FeCl2  Iron (II) Chloride
o Fe3+ + Cl-  FeCl3  Iron (III) Chloride
o Cu2O  Copper (I) Oxide
o Tin (IV) Oxide  SnO2
o CuS  Copper (II) Sulfide
o Lead (II) Bromide – PbBr2
Hydrated Ionic compounds
o Hydrate- contains a specific number of water molecules
 Ex. CuSO4 * 5H2O

 Copper (II) Sulfate pentahydrate
Name of ionic compounds + prefix-“hydrate”
 Prefixes
 1. Mono
 2. Di
 3. Tri.
 4. Tetra
 5. Penta
 6. Hexa
 7. Hepta
 8. Octa
 9. Nona
 10. Deca
 ex. FeSO4 * 7H2O
 Iron (II) Sulfate Heptahydrate
Naming Molecular compounds
 Prefix (number) + name of 1st element
Prefix + stem + ide
o Ex. CO2
 When first element is one don’t write mono
 Carbon Dioxide
o P2O3
 DiPhosphurus TriOxide
o SO2
 Sulfur Dioxide
o P2O5
 Diphosphurs Pentoxide (leave off a when a vowel is
next)
o Nitrogen Dioxide
 NO2
o Dichlorine Heptaoxide
 Cl2O5
[11/6/08]
Naming Acids
 Acid = compound that contains one or more hydrogen atoms and
produce hydrogen ions (H+) when dissolved in water


For compounds were hydrogen is the first element add hydro- to
the front and –ic to the end
o Ex. HBr
 Hydrobromic acid
o Ex. H2S
 Hydrosulfuric acid
o Ex. HF
 Hydrofluroic acid
 anion contains oxygen
o Ex. H2SO4
 Sulfuric acid
 Any polyatomic that ends in –ate  -ic
o Ex. H2SO3
 Sulfurous acid
 -ite  -ous
o Ex. HNO3
 Nitric acid
o Ex. HNO2
 Nitrous acid
[11/17/08]
Percent Composition – relative amounts of elements in a compound
expresses as a percent (% by mass of each element in a compound)
 % Mass of an element = (mass element/mass compound) x 100
o Ex. K2CrO4
 K – 40.3 % Cr – 26.8 % O – 32.9 %
 Percent must total 100%
 (2(39) / 184) x 100 = 40.3
 (molar mass of K x 2 / total compound) X 100
o Ex. When a 13.6 g sample of a compound containing only
magnesium and oxygen (Mg, O) is decomposed, 5.4 g of
oxygen is obtained. What is the percent composition of this
compound
 O – 39.7 % Mg – 60.3 %
o Ex. calculate the mass of carbon in 82 g of C3H8
 82 g of C3H8 X (81.8 C / 100 g 82 g of C3H8) = 67.1 g C

o Ex. 25.9 % nitrogen, 74.1 % oxygen. What is the empirical
formula?
 1) Assume 100g of a compound 
 25.9 g N, 74.1 g O
 2) Divide each one by molar mass 
 25.9 g N / 14 g/ mol = 1.85 mol N
 74.1 g O/ 16 g / mol = 4.63 Mol O
 3) Divide by lowest number so one will each one
 N1.85O4.63
 Divide by 1.85
 4) Multiply by smallest number to get whole numbers
 N1O2.5
 This case 2
 Answer: N2O5
o To find molecular formula: Divide molar mass by empirical
formula mass  this gives us # to multiply empirical formula
by to get molecular formula

Ex. HO – 24 g / mol  34 /17 = 2 so 2-h 2-o
[11/18/08]
Calculate the molecular formula of a compound whose molar mass is 60
g/mol and empirical formula is CH4N?
 60 / 30 = 2  C2H8N2
Determine the molecular formula - 50. 7 % C, 4.2 % H , 45.1 % O, molar
mass 142 g
 50.7 g C , 4.2 g H, 45.1 G O
o 50.7/12 – 4.2
o 4.2/1 – 4.2





o 45.1/16 – 2.8
C4.2H4.2O2.8
C4.2H4.2O2.8 /2.8  C1.5H1.5O1
C3H3O2
142/71 = 2
Answer C6H6O4
Chemical equation
 Reactants  products
Balanced chemical equation:
 Each side has the same # of atoms of each element (mass is
conserved)
o  use coefficien ts (not subscribes)
[11/20/08]
put in phases into the equations when balancing
 (s) – solid
 (l) – liquid
 (g) – gas


(aq) – aqueous
o put them after the atoms or molecules
Remember: if (g) then it may be a diatomic molecules
Chemical Quantities and Aqueous RXNS
5/27/2009 7:44:00 PM
[11/24/08]
Stoichiometry
 Using a numerical relationship between chemical quantities in a
balanced chemical reaction to calculate other quantities
 we can predict amount of product formed based on amount of
reactant
 Moles to moles o 2Na + Cl2 2NaCl
 How many moles of NaCl will be produced from 3.4
moles of Cl2?
6.8 mol NaCl
 explained. 3.4 mol Cl2 (2 mol Nacl/1 mol
Cl2) = 6.8 mol NaCl
o H2 + Cl2  2HCl
 4.8 mol Cl2
 H2? 4.8 mol
 HCl? 9.6 mol
o 4Al + 3O2  2Al2O2
 3.7 2Al2O2 x (4 mol AL/2 mol 2Al2O2) = 7.4


mass- mass
o must convert to moles first
 ex. N2 + 3H2  2NH3
 5.4 g H2
 1) calculate mass NH3:
 5.4 g H2 (must convert to moles)
o 5.4 g H2 / 2 g/mol = 2.7 mol H2
 2) 2.7 mol H2 (2 mol NH3 / 3 mol H2) = 1.8 mol
 3)1.8 mol NH3 x 17 mol = 30.6 g NH3

2C8H18 + 25O2  16CO2 + 18H2O
o Calculate mass CO2 produced from 500g C8H18
 500 C8H18 /114 C8H18 = 4.4 mol C8H18
 4.4 mol C8H18 (16 CO2/2 C8H18) = 35.2 mol CO2
 35 mol CO2 x 44g/mol = 1540 g CO2
For the reaction shown, calculate how many grams of each product
form when 4.7 g Al completely reacts to form the following
products.

o 2 Al + Fe2O3  Al2O3 + 2Fe
 9.74 g
[12/1/08]
How many grams of sodium oxide can be synthesized from 17.4 g of
sodium?
 4Na +O2  2Na2O
o (17.4 g Na / 23 g/mol) = .76 mol Na
o .76 x (2 Na2O / 4 Na) = .38 mol Na2O
 .38 x 62 g / mol = g 23.45 Na2O
Limiting reactant problem (limiting reagent)
 The reactant that limits the amount of product we can get
[12/2/08]
 Limiting reactant –
o reactant that limits the amount of product in a reaction
 Theoretical yield –
o amount of product that can be made in a chemical reaction
based on the amount of limiting reactant
 Actual yieldo Amount of product that is actually produced in a reaction
 Percent yieldo (Actual yield/theoretical yield) x 100
[12/3/08]
 Consider the following reaction:
o 2Al+3Cl2  2AlCl3
 If we begin with .552 mol of Al and .887 mol of Cl, what is the
limiting reactant and theoretical yield of AlCl3 in moles
o .552 mol Al x (2 AlCl3 /2Al) = .552 AlCl3



o .887 mol Cl2 x (2 AlCl3 / 3O2) = .591 mol AlCl3
Consider the following reaction:
Cu2O + C  2Cu + CO
When 11.5 g of C are allowed to react with 114.5 g of Cu2O, 87.4 g
of Cu are obtained. Find the limiting reactant, theoretical yield, and
percent yield.
o 11.5 /12 = .96 mol
 .96 * (2/1) = 1.92






o 114.5 /142.9 = .80 mol
 .80 * (2/1) = 1.60
o 87.4 / 63.5 = 1.38 mol
o Limiting reactant – Cu2O
o Theoretical yield- 1.6 mol Cu 8 63.5 g/ mol = 101.7 g
o Actual yield – 87.4 g
o Percent yield - (87.4/101.7) * 100 = 85.9 %
Consider the following reaction:
o 2FeCl3 3H2S  6HCl + Fe2S3
90 g FeCl3
52 g H2S
a) limiting reactant – FeCl3
o 90/162 - .55 mol (6/2) = 1.65 mol
o 52/34 – 1.53 mol (6/3) = 3.06 mol
b) mass HCl produced – 60.6 g HCl
o 1.65 x 36.5 = 60.6 g
c) mass of excess reactant after RXN – 24 g
o we used .55 mol FeCl3 (3 mol H2S/2 mol FeCl3) =.825 mol
H2S is used
o 1.53 mol - .825 mol = .705 mol excess remaining
o .7 mol x (34 g /1 mol) = 24 g H2S remaining
[12/4/08]
2NH3 + CO2  CH4N2O + H2O
Find limiting reactant, theoretical yield, and actual yield and excess reactant:
136.4 kg NH3, 211.4 Kg CO2, actual – 168.4 CH4N2O
NH3 - 8023.5 mol * (½) = 4011.75 mol * 60 g/mol = 240705 g
CO2 - 4804.5 mol (1/1) = 4804.5 g
LR – NH3
TY – 240.705
Actual yield – 168.4 / 240.705 = .6996 * 100 = 69.96 %
Excess reactant – 8023.5 mol NH3 (1 CO2 / 2 NH3) = 4011.75 mol CO2 
4804.5 mol– 4011.75 = 792.75 mol CO2  792.75 mol CO2 * 44 g/mol
(CO2) = 34881 g CO2
[12/8/08]
Pg. 143 - 2C8H18 + 25O2  16CO2 + 18H2O
 48 Kg of C8H18
 Change to g  48,000 g
 48000 g / 114 g/mol = 421.05
 421.05 * 25/2 = 5263.16 mol O2
 5263.16 mol O2 * 32 g / mol = 168421 g O2
Solution Stochiometry
 Solute




o Substance that gets dissolved
 ex. coffee grinds in water
Solvent
o Substance that does dissolving
 (Present in excess)
 ex. Water
Solution
o Homogeneous mixture of a solvent and one or more solutes
Concentration
o Measure of the amount of solute present in a given quantity
(volume) of solution
 Increase amount of solute in a given amount of solution
 increases concentration
Molarity = concentration
o Denoted in brackets [x]
o Units: mol / L = (M)
 = number of mol in solute per 1 liter of solution
 Written as a capitol M
 Ex. 5M “molar”
o Molarity = moles of solute / liters of solution
 Ex. a solution has a volume of 250 mL and contains .7
mol NaCl. What is its molarity?
 .7 mol / .25 L = 2.8 M
 Ex. IV saline solutions :
 .9 g NaCl
 Cl 35.5 + Na 23 = 58.5


o .9/58.5 = .015
 100 mL solution
 = .1 L
 Molarity?
 .015 mol / .1 L = .15 M
ex. How many grams of sucrose C12H22O11 are
contained in 1.72 L of a .758 M sucrose solution
 C 12*12 = 144. H = 22 * 1 = 22, 16* 11 = 176
= 342 g / mol
 1.72 L x .758 M = 1.30 mol
 1.3 mol x 342 g/mol = 444.6 g
Diluting a solution
o We have “stock solution” and we want to change it to another
molarity
 M1V1 = M2V2 (molarity * volume… = mol)
 M1,V1 – molarity + volume of initial solute
 M2,V2 – molarity + volume of desired solution
o Ex. To what volume should you dilute a .1 L of 15M NaOH to
obtain a 1 M NaOH solution


.1*15 = x*1  1.5 =1x  1.5=x
Adding 1.4 L
[12/9/08]
 How many mL of aqueous 2 M MgSO4 solution must be diluted with
water to prepare 100 mL of aqueous.4 M MgSO4?
o 2 * x = .4 * .1
o x = .02 * 1000 = 20 mL
o  80 mL must be added (20  100)
Solution Stoichiometry (Continued)

Use mole ratio
o Ex. 2Na3PO4 + 3CuCl2  Cu3(PO4)2 + 6NaCl
 What volume of .175 M Na3PO4 solution is necessarily to
completely react with 95.4 mL of .102 M CuCl2?
 .0954 L x .102 M = .00973 mol CuCl2
 (since Mol / L = M  cross multiply)
  .00973 mol CuCl2 (2 mol Na3PO4 / 3 mol CuCl2) =
.006487 mol Na3PO4a
 1.75 M = (.006487 mol Na3PO4 / X L [x liters])
 .006487/1.75 = 3706 L  37.06 mL
o Ex. What amount in liters of 6 M H2SO4 is necessary to
produce 25 g H2 according to the following RXN?
 2 Al + 3 H2SO4  Al2(SO4)3 + 3H2
 25 g / 2 g/mol = 12.5 mol H2
 12.5 mol H2 x (3 mol H2SO4 / 3 mol H2)
 12.5 mol H2SO4 / x L = 6 M
 12.5/6 = 2.083 L
o Steps when doing Stoichiometry problem



Get to moles
Use moles ratio
Convert that to whatever is needed (molarity, mass,
volume)
[12/10/08]
2Al + 3H2SO4  Al2(SO4)3 + 3H2
 How much of 2M H2SO4 to get 50 g H2, volume?
 50 g / 2 g/mol = 25 mol H2
 25 mol (3/3) = 25 mol H2SO4


Na2SO4





25 mol / x L = 2 M
25 mol / 2 M = 12.5 L
+ 2HCl  H2SO4 + 2NaCl
If we use 250 mL of a 6 M Na2SO4 Solution, how much of a 2M NaCl
solution will be produced (what volume)
X / .250 L = 6M
6 X .25 = 1.5 mol
1.5 mol (2 NaCl/1 Na2SO4) = 3 mol NaCl
3 mol / x L = 2M
 3/2 = x L  X = 1.5 L
If we use 50 mL fo 2M solution, how much water do I add to get a .5 M
solution?
 .05 * 2 = X * .5
 .1 = .5x
 x = .2 L
 200 mL
 add 150 mL
Solution process
 Ex. Crystal in water 
o Individual solute ions break away from the crystal
o  These cations and anions become surrounded by solvent
molecule (water surround each ion – water is polar)
o  ionic crystal dissolves
 electrolyte o compound that conducts an electric current when its in
aqueous solution
 ions are mobile

 all ionic compounds are electrolytes
nonelectrolyte –
o compound that doesn’t conduct electricity in aqueous
solutions
 ex. molecular compounds
Strong electrolyte
Weak electrolyte
 completely dissociates in
solution - (bulb glows brightly)
 conducts electricity poorly
(only fraction of solute exists as
ions) (bulb glows dimly)
Solubility
 Soluble – substance that dissolves in water
 Insoluble – doesn’t dissolve in water
[12/11/08]
Had a pop quiz
[12/12/08]
ex. soluable or insolubale
 AgBr - insoluble
 CaCl2 – soluble
 Pb(NO3)2 - soluble
 PbSO4 – insoluble
Precipitation Reactions
 Precipitateo When ionic compounds come out of solution when there
solubility is exceeded
o Solid
o  insoluble product of a reaction in a solution
 Precipitation reaction (rxn) –
o Reverse of dissolving
o  rxn that forms a precipitate upon mixing of two aqueous
solutions
[12/15/08]
Precipitation reactions
 Ex. AX(aq) + BZ (aq)  AZ + BX (Anion, Cation)
 Double replacement RXNu



One is a solid and the other one is aqueous
If both products are soluble  no precipitate RXN
Q: complete + balance RXN, determine precipitate forms
o Ex. Ca(OH)2 (aq) + FeCl3 (aq)  X
 3Ca(OH)2 (aq) + 2FeCl3 (aq)  3CaCl2 + 2Fe(OH)3
 2Fe(OH)3 is the precipitate and is a solid
 3CaCl2 is aqueous
o Ex. KI + NaCl 
 KI (aq) + NaCl (aq)  KCl + NaI
No RXN  since there is no precipitate (don’t need to
put products
o EX. Sodium Carbonate and Copper (II) Chloride solutions are
mixed
 Na2CO3 + CuCl2  2NaCl + CuCO3
 2NaCl is aqueous
 CuCl3 is a solid (or precipitate)
Molecular equation
o Shows the complete neutral formula for every compound in



RXN
o Ex. 3Ca(OH)2 (aq) + 2FeCl3 (aq)  3CaCl2 + 2Fe(OH)3
Complete Ionic equations
o Shows reactants + products as they are actually present in a
solution
o Ex. 3Ca2+ + 6OH- + 2Fe3+ + 6Cl-  2Fe(OH)3 (s) + 3Ca2+ +
6Cl 2Fe(OH)3 (s) is the precipitate

don’t combine things that aren’t the precipitate

Spectator ions
o Present but don’t participate in RXN to produce precipitate
 Net Ionic equation
o Omit spectator ions
o  only shows ions that react to produce precipitate
[12/18/08]
 Net Ionic Equation (continued…)
o Ex. 3Ca2+ + 6OH- + 2Fe3+ + 6Cl-  2Fe(OH)3 (s) + 3Ca2+ +
6ClNet ionic equation - 6OH- + 2Fe3+  2Fe(OH)3
Notes when balancing:
 subscripts are for charges, coefficients are for balancing
 Also always balance reactions
 When it says complete reactions  must write if it’s a solid, or if its
aqueous
 In complete reactions  get rid of spectator ions(?)
[until 1/6/09]
Went over sheets

5.1-5.5, 214, 215
Yossi Quint
[1/6/09]
Gases
Kinetic molecular theory – rules for an ideal gas
 1. Particles are so small when compared to the distances between
them that the volume of the individual particles can be assumed to
be negligible (pretend that the volume isn’t there)
o summation – volume of particles negligible compared to
volume of gas
[1/7/09] (cont…)
 2. Particles are in constant motion


Gases


o  collision of particles with wall of containers are the cause of
pressure
3. Particles are assumed to exert no forces on each other
o (no attraction or repulsion)
4. Average kinetic energy of gas particles is directly proportional to
the Kelvin temperature of gas
Uniformly fill a container (take shape of a container)
Compressible
 mix with other gases
 Lower density
 Exert a pressure on its surroundings
Atmospheric pressure
 results from collisions of molecules in air with objects
 decreases as climb up mountain because density of Earth’s
atmosphere decreases as the altitude increases
 depends on weather + altitude
 sea level + fair weather = 760 mm Hg (mercury)


o Barometer is used to measure pressure
Other units: 1 atm = 760 mm Hg = 101.3 kpa = 101,300 pa =
760 tor = 14.7 psi
o Problem:
 450 kpa = 4.44 atm = 3374.4 mm Hg
pressure = force/area
o P = F/A
= force per unit that results from collisions of a gas with
surrounding objects
absolute zero = 0 K = -273 C
o motion of particles stop


[1/8/09]
To describe gases:
 Temperature (T) – measured in K
 Volume (V) – measured in L
 Pressure (P) – measured in atm, kpa
 Amount in moles (n) – measured in moles
Amount of gas
 If we double amount of particle, pressure doubles
 Constant – T, V
Volume
 If we reduce volume, pressure increases
Temp
 Temp rises  pressure rises
Gas Laws – shows how a change in one property affects one or more of the
others
 Boyles Law – V  1 / P or PV = constant
o The volume of a gas and its pressure are inversely
proportional at constant temperature and amount
o P1V1 = P2V2
o Isotherm curve – curve on a graph that has the same
temperature throughout (blue line

o
Charles’s law – T & V
o The volume of a gas and its Kelvin temperature are direct
proportional
 V  T or V/T = constant
o Isobar – a line that pressure is held constant

o V1 / T1 = V2 / T2
Gay- Lussac’s laws – P & T
o Pressure of a gas is directly proportional to Temperature if
Volume is constant
 P  T or P / T = constant


P1 / T1 = P2 / T2
Isochore – a line on the graph where volume is held
constant
[1/9/09]
 Q: a sample of N2 gas has a pressure of 6.58 kpa at 539 K at 211
K, what is P?
o A: 6.58 / 539 = P2/211
 p = 2.58 kpa
Combined Gas Laws

PV / T = constant
o P1V1 / T1 = P2V2 / T2
 STP = Standard temperature + Pressure
o 273 K, 1 atm
o ex.
 153 kpa , 30 L, 313 K (and STP)
 1.51 * 30 /313 = 1x / 273
 x = 39.5 L
Avogadro’s Law


If we increase the amount of gas  volume increases
o If pressure and temperature is constant
V1 / n1 = V2 / n2
o N – amount of moles of gas
o Ex. 4.8 L, .22 mol He - How many additional moles of He
gas must be added to have volume 6.4?
 4.8 / .22 = .64 / n  n = .29
 .29 -.22 = .07 mol
Ideal Gas Law
 P1V1 / n1T1 = P2V2 / n2T2
o PV / nT = constant
 R : Ideal gas constant
o R = 8.31 (L * Kpa / k * mol)  what is in the parenthesis is
the unit
o R = .0821 (L * Kpa / atm * mol)
 PV = nRT
o N - moles
[1/12/09]
What is the pressure in atm of .18 mol of gas in a 1.2 l flask at 298 k?
 (.18 * .0821 * 298) / 1.2
o 3.7 atm
How many kg of CH4?
 Given: 2.24 x 106 L methane, 1.5 x 103 kpa, 315 k
o (1.5 x 103 kpa * 2.24 x 106 L) /(8.31 * 315) = x mols
o 3.36 x 109 / 2617.65 = 1.28 x 106
 1.28 x 106 x 16 g/mol = 2.05 x 107 = 2.05 x 104 kg
find molar mass?

Given: .126 g, .112 L, 298 K, 1.06 atm
o 1.06 * .112 = .0821 * 298 * x mol
 mol x = .00485
 .136/.00485 = 28 g/mole
Ideal gas
 Particles have no volume
 No attraction between particles
 * At high temperature and low pressure  real gases act like ideal
o we assume ideals
Material for final ends
5/27/2009 7:44:00 PM
Format


What is

40 multiple choice
short answer
on it
Sig figs, unit conversions, safety, density (m/v), prefixes, only need
C  K, scientific notation, accuracy vs. precision, mixtures, Dolton’s
atomic theory, JJ Thompson, discovery electrons, protons,
neutrons, weighted average, periodic tables, properties of parts of
it, metals - non metals – metalloids, Alkali, alkaline, halogens,
transitions, ions, mols, NA number, empirical, molecular, diatomic
atoms, naming for ionic compounds, naming acids, percent
composition (for each element). Balancing reactions, stoichiometry,
limiting reactant, theoretical yield, actual yield, molarity, dilute,
solubility, all gas laws
What we are given
 polyatomic, solubility, periodic elements