Download System International Base Units

Spring 2012
Pre-AP Chemistry Study Guide
SYSTEM INTERNATIONAL (SI) BASE UNITS
Property
Length
Mass
Temperature
Amount of substance
Time
Unit
Meter
Kilogram
Kelvin
Mole
Second
Abbreviation
m
Kg
K
mol
s
DERIVED UNITS
Property
Derived From
Common Unit Abbreviations
Volume
Length x Length x Length
m3, cm3, mL, L*
Density
Mass/Volume
Kg/m3, g/cm3, g/L
2
2
Energy
Mass x Length / Time
J*, cal*, Cal*
Solution Concentrations
Amount / Total amount
Mass %*, M*
(Mass x Length) / (Time2 x Length2)
Pressure
Pa*, KPa, mmHg, Torr, psi*
Bold are SI units for volume, density, energy, and pressure
* L-Liters; J-Joules; cal-calorie; Cal-kilocalorie; Mass %-Mass percent, M-molarity; Pa-Pascal, psipounds/square inch
IMPORTANT TEMPERATURES
Freezing Point of water 0°C, 32°F, or 273 K; Boiling Point of Water 100°C, 212°F, or 373 K
Normal Body Temperature, 37°C, 98.6°F, or 310 K; Room Temperature 25°C, 77°F, 298 K
CONVERSIONS
Metric Prefixes
 Mega (M)
=
1,000,000 times larger than its nonprefixed unit
 Kilo
(K)
=
1000 times larger than its base unit
 Some nonprefixed units are: gram (g), meter (m), second (s), mole (mol), Kelvin (K), Joule (J),
Liter L, Molarity (M), Pascal (Pa)…
 Deci
(d)
=
10 times smaller than the nonprefixed unit
 Centi (c)
=
100 times smaller than the nonprefixed unit
 Milli
(m)
=
1,000 times smaller than the nonprefixed unit
 Micro (μ)
=
1,000,000 times smaller than nonprefixed base unit
 Nano (n)
=
1,000,000,000 times smaller than the nonprefixed base unit
 Pico
(p)
=
1,000,000,000,000 times smaller than the nonprefixed base unit
Other Conversion Relationships
 1 mL = 1 cm3
 1 atm = 101.3 kPa = 760 mmHg = 760 torr = 14.69 psi
 ºC + 273 = K
K - 273 = ºC
 °F = 9/5 °C + 32 °C = 5/9 (°F-32)
 1 cal = 4.184 J
 1000 cal = 1 Cal
 6.022x1023 representative particles = 1 mole
o Representative particles can be atoms, ions, formula units, or molecules
 1 mole = molar mass of an element or compound
 1 mole of any gas = 22.4 L of gas at STP
o STP for gases is 101.3 kPa and 0°C
1
Spring 2012
CONVERSION PROBLEM SOLVING Methods below will work with any conversion problems
Sample Question 1 (Top Unit Changes)
 Coca-Cola Classic™ contains 1718 J/m L of energy. How many cal/mL does the soft drink
contain?
o Determine your given quantity and its unit
 The given quantity is 1718 and the unit is Joules (J)
o Determine the relationship between the given unit (J) and the wanted unit calorie
(cal)

o
If the unit you are changing is on top in the problem, divide the given unit on both
sides of the relationship

o
𝟏=
𝟒.𝟏𝟖𝟒 𝑱
𝟒.𝟏𝟖𝟒 𝑱
=
𝟏.𝟎𝟎𝟎 𝒄𝒂𝒍
𝟒.𝟏𝟖𝟒 𝑱
Multiply the given quantity and its unit by the conversion factor (bold fraction
above)

o
𝟒. 𝟏𝟖𝟒 𝑱 = 𝟏 𝒄𝒂𝒍
𝟏𝟕𝟏𝟖 𝑱
𝟏 𝒎𝑳
×
𝟏 𝒄𝒂𝒍
𝟒.𝟏𝟖𝟒 𝑱
=
𝟏𝟕𝟏𝟖 𝑱 ×𝟏 𝒄𝒂𝒍
𝟏 𝒎𝑳 ×𝟒.𝟏𝟖𝟒 𝑱
=
𝟏𝟕𝟏𝟖 ×𝟏 𝒄𝒂𝒍
𝟏 𝒎𝑳 ×𝟒.𝟏𝟖𝟒
= 𝟒𝟏𝟏. 𝟎 𝒄𝒂𝒍⁄𝒎𝑳
Notice that given unit ends up on opposite sides of the division line and cancels out
leaving only the wanted unit on top. THE GIVEN UNIT SHOULD ALWAYS
CANCLE OUT IF YOU SET UP YOUR CONVERSION CORRECTLY!
Sample Question 2 (Bottom Unit Changes)
 Coca-Cola Classic™ contains 1718 J/m L of energy. How many J/L does the soft drink
contain?
o Determine your given quantity and the unit you are changing
 The given quantity is 1718 and the unit is milliliters ( mL)
o Determine the relationship between the given unit milliliters (mL) and the wanted
unit liters (L)

o
If the unit you are changing is on bottom in the problem, divide the wanted unit (L)
on both sides of the relationship

o
𝟏𝟎𝟎𝟎 𝒎𝑳
𝟏𝑳
=
𝟏𝑳
𝟏𝑳
=𝟏
Multiply the given quantity and it unit by the conversion factor (bold fraction above)

o
𝟏𝟎𝟎𝟎 𝒎𝑳 = 𝟏 𝑳
𝟏𝟕𝟏𝟖 𝑱
𝟏 𝒎𝑳
×
𝟏𝟎𝟎𝟎 𝒎𝑳
𝟏𝑳
=
𝟏𝟕𝟏𝟖 𝑱 ×𝟏𝟎𝟎𝟎 𝒎𝑳
𝟏 𝒎𝑳 ×𝟏 𝑳
=
𝟏𝟕𝟏𝟖 𝑱 ×𝟏𝟎𝟎𝟎
𝟏 ×𝟏𝑳
𝑱
= 𝟏𝟕𝟏𝟖𝟏𝟎𝟎𝟎 ⁄𝑳
Notice that the given unit ends up on opposite sides of the division line and cancels
out leaving only the wanted unit on bottom. THE GIVEN UNIT SHOULD
ALWAYS CANCLE OUT IF YOU SET UP YOUR CONVERSION
CORRECTLY!
2
Spring 2012
Sample Question 3 (Top Unit Change W/O Bottom Unit)
 A 12 ounce bottle of Dr. Pepper contains 40.0 g of table sugar (C12H22O11, molar mass
342.3g/mol ). How many moles of table sugar does the bottle contain?
o Determine your given quantity and the unit you are changing
 The given quantity is 40.0 and the unit is grams (g)
o Determine the relationship between the given unit grams (g) and the unit wanted unit
moles (mol)

o
If the unit you are changing is on Top in the problem, divide the given unit (grams)
on both sides of the relationship

o
𝟏 𝒎𝒐𝒍 𝑪𝟏𝟐 𝑯𝟐𝟐 𝑶𝟏𝟏
𝟑𝟒𝟐.𝟑 𝒈 𝑪𝟏𝟐 𝑯𝟐𝟐 𝑶𝟏𝟏
=
𝟑𝟒𝟐.𝟑 𝒈 𝑪𝟏𝟐 𝑯𝟐𝟐 𝑶𝟏𝟏
𝟑𝟒𝟐.𝟑 𝒈 𝑪𝟏𝟐 𝑯𝟐𝟐 𝑶𝟏𝟏
=𝟏
Multiply the given quantity and it unit by the conversion factor (bold fraction above)
𝟒𝟎 𝒈 𝑪𝟏𝟐𝑯𝟐𝟐𝑶𝟏𝟏
𝟏
o
𝟏 𝒎𝒐𝒍 𝑪𝟏𝟐 𝑯𝟐𝟐 𝑶𝟏𝟏 = 𝟑𝟒𝟐. 𝟑 𝒈 𝑪𝟏𝟐 𝑯𝟐𝟐 𝑶𝟏𝟏
×
𝟏 𝒎𝒐𝒍 𝑪𝟏𝟐𝑯𝟐𝟐𝑶𝟏𝟏
𝟑𝟒𝟐.𝟑 𝒈 𝑪𝟏𝟐 𝑯𝟐𝟐𝑶𝟏𝟏
=
𝟒𝟎 𝒈 𝑪𝟏𝟐𝑯𝟐𝟐𝑶𝟏𝟏 × 𝟏 𝒎𝒐𝒍 𝑪𝟏𝟐 𝑯𝟐𝟐𝑶𝟏𝟏
𝟏 × 𝟑𝟒𝟐.𝟑 𝒈 𝑪𝟏𝟐𝑯𝟐𝟐𝑶𝟏𝟏
=
𝟒𝟎 × 𝟏 𝒎𝒐𝒍 𝑪𝟏𝟐 𝑯𝟐𝟐𝑶𝟏𝟏
𝟏 × 𝟑𝟒𝟐.𝟑
= 𝟎. 𝟏𝟏𝟕 𝒎𝒐𝒍𝒆𝒔 𝑪𝟏𝟐 𝑯𝟐𝟐 𝑶𝟏𝟏
Notice that the given unit ends up on opposite sides of the division line and cancels
out leaving only the wanted unit on top. THE GIVEN UNIT SHOULD ALWAYS
CANCLE OUT IF YOU SET UP YOUR CONVERSIONS CORRECTLY!
ATOMIC STRUCTURE
Particle
Proton
Neutron
Electron



Atom Structure
Charge
1+
0
1-
Location
Nucleus
Nucleus
Electron could
Atomic number = number of protons
Mass number = number of protons + neutrons
Electrons = number of protons when atom is neutral
Elements
 All atoms can be divided into metals, nonmetals, and metalloids; study the periodic table to
make sure you can categorize them.
 All atoms of an element have the same number of protons
 Atoms of an element that have different numbers of neutrons are isotopes
 Atomic mass of an element = ((% occurrence of isotope A ÷ 100)*Mass Number of Isotope A) +
((% occurrence of isotope B ÷ 100)*Mass Number of Isotope B)…
Chemical Symbolism
3
Spring 2012
Valence electrons = group A group number. Example: Li in group 1A has 1 valence electron. Fluorine in
group 7A has 7 valence electrons.
Lewis dot structure for period 2 elements
 Notice dots equal their number of valence electrons and do not pair up until after all four
quadrants have at least one dot
Li
C
B
Be
N
O
F
Ne
Lewis structures for compounds
 Draw element with most unpaired electrons in center
 Draw other elements around center element (s) such that their unpaired electrons face each other,
then connect the unpaired electrons
 All atoms should have eight electrons around them (remember dashes represent 2 electrons)
Electron configurations
Find the electron configuration of
Ruthenium (Ru)
1s



2
2s2
Find Ruthenium’s atomic number
Ruthenium has an atomic number of 44
Follow the arrows adding the subscripts
until you reach 44
2p6
Answer 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d6

3s2
3p6
3d10
4s2
4p6
4d10
4f14
Notice Only 6 electrons were needed for the 4d
sublevel
5s2
5p6
5d10
5f14
Rearrange electron configuration in order of
increasing energy level
6s2
6p6
6d10
6f14
Notice 2 + 2 + 6 + 2 + 6 + 2 + 10 + 6 + 2 + 6 = 44
Answer 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d6 5s2
When doing electron configurations for ions, subtract the charge of the ion to its atomic number
to get the number of electrons needed for the electron configuration. You should notice the
electron configuration of group A ions are always the same as their nearest noble gas
o Example: Write the electron configuration for the 𝑶𝟐− ion.
 Oxygen has an atomic number of 8. So 𝟖 − (−𝟐) = 𝟏𝟎 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏𝒔
 𝑶𝟐− electron configuration is then 1s2 2s2 2p6, the same as Ne
4
Spring 2012
Ions – are formed from atoms that have lost electrons and become positively charged or atoms that have
gained electrons becoming negatively charged
 Metals form cations (positively charged ions due to loss of their valence electrons) in reaction
with nonmetals
o For group A metals, their ion charge is equal to their group number.
 Nonmetals form anions (negatively charged ions due to gaining enough electrons to have 8
valence electrons) in reaction with metals
o For group A nonmetals, their charge is equal to their group number minus eight
 Metals and nonmetals react to form ionic compounds called formula units with ionic bonding
 Nonmetals and Nonmetals react to form molecular compounds called molecules with covalent
bonds.
NOMENCLATURE
Ions

Type I Cations (Group A Metals)
o Type I Cations: name of element followed by word “cation”
 Mg2+ is the magnesium cation
 Al3+ is the aluminum cation

Type II Cation (Transition Metals)
o Type II Cations, Stock System: name of element is followed by Roman numeral
indicating its charge.
 Fe2+ is the Iron (II) cation
 Fe3+ is the Iron (III) cation

Type II Cations, Classical System: element’s suffix is changed to “ous” if ion is the lesser of the
two possible charges or “ic” if ion is greater of the two possible charges.
o Fe2+ is the Ferrous cation
o Fe3+ is the Ferric cation
Common Type II Cations to Memorize
Fe2+
Cu+
Co2+
Sn2+
Pb2+
Hg22+
Ag+
Zn2+
Cd2+

Fe3+
Iron (III) Cation
Cu2+
Copper (II) Cation
Co3+
Cobalt (III) Cation
Sn4+
Tin (IV) Cation
Pb4+
Lead (IV) Cation
Hg2+
Mercury (II) Cation
Type II cation with only one charge
Type II cation with only one charge
Type II cation with only one charge
Iron (II) Cation
Copper (I) Cation
Cobalt (II) Cation
Tin (II) Cation
Lead (II) Cation
Mercury (I) Cation
Silver Ion
Zinc Ion
Cadmium Ion
Anions (Nonmetals)
o Change the suffix of the element name to “ide” followed by word “anion”
 F- is the fluoride anion
 O2- is the Oxide anion
 N3- is the Nitride anion
5
Spring 2012

Polyatomic ions
o Say the name of the ion followed by the word cation if it has a positive charge or anion
if it has a negative charge.
o NO3- (below) is the nitrate anion; NH4+ (below) is the ammonium cation
o Memorize the common polyatomic ions
Table 9.3
Naming Binary Ionic Compounds (made from cations and anions)
 Combine the name of the cation and anion while dropping the words “cation” and “anion.”
o MgF2 Magnesium cation and fluoride anion make magnesium fluoride
o MgSO4 Magnesium cation and sulfate anion make magnesium sulfate
o Fe(ClO3)3 Iron (III) cation and chlorate anion make iron (III) chlorate
Naming Acids (it is all about the anion as the cation is always H+).
 If the anion does not contain oxygen in its chemical formula:
o Add the prefix “hydro” to the anion name.
o Change the suffix in the anion name from “ide” to “ic”
o Then say the work “acid”.
 Example: HCl contains the chloride (Cl-) ion. As an acid HCl is called
hydrochloric acid
 If the anion contains oxygen and ends with the suffix “ite”:
o Change the “ite” to “ous” and say the word acid. (DO NOT SAY HYDRO)
 Example: HClO2 contains the chlorite ion: As an acid, HClO2 is called
Chlorous acid
 If the anion contains oxygen and ends with the suffix “ate”:
o Change the “ate” to ‘ic” and say the word acid. (DO NOT SAY HYDRO)
 Example: HClO3 contains the chlorate ion: As an acid, HCLO3 is called
Chloric acid
Naming Bases
 Bases are named as normal binary ionic compounds
o Example: NaOH is sodium hydroxide
Naming Binary Molecular Compounds (Made from Nonmetals)
 Say the appropriate prefix indicating the number of the first element (except mono) followed by
the elements name
 Say the appropriate prefix indicating the number of the second element followed by the modified
name of the second element where its suffix has been changed to “ide”
 Numerical prefixes are: 1-mono, 2-di, 3-tri, 4-tetra, 5-penta, 6-hexa, 7-hepta, 8-octa, 9-nona
o Examples: Cl3O7 is trichlorine heptaoxide; N2O4 is dinitrogen tetraoxide; CS2 is
carbon disulfide; CS is carbon monosulfide
6
Spring 2012
Chemical Formulas from Chemical Names
 Ionic Compounds
o Write the chemical symbols of the elements or polyatomic ions making up the compound
and indicate their appropriate charges
o Use the absolute value of opposite ion’s charge for the current ions number subscript.
Known as the Criss-Cross Method shown below.
Calcium
Phosphide
3Ca2+
Sodium Nitride
Ca3P2
Na3N
Na1+
P


Aluminum
Bromide
3+
1-
N3-
Al
Br
AlBr3
Aluminum
Nitrate
Al3+ NO31-
Al(NO3)3
The criss-Cross method forces you to join the ions together in the right ration of
cations to anions such that the net charge of the compound is equal to zero
 Example: If we look at the number of ions and their respective charges
𝟑−
in the formula unit calcium phosphate, 𝑪𝒂𝟐+
𝟑 𝑷𝟐 , we see that the sum
each ions number multiplied by its charge equals zero; see blow:
o (𝟑 𝑪𝒂 × (+𝟐 𝒄𝒉𝒂𝒓𝒈𝒆)) + (𝟐 𝑷 × (−𝟑 𝒄𝒉𝒂𝒓𝒈𝒆)) = 𝟎
Molecular Compounds
o Write the chemical symbol for the each element followed by a subscript to indicate the
number of that element.
 Example: dinitrogen tetraoxide is N2O4
CHEMICAL REACTIONS
Oxidation Reduction Reactions - electrons transferred from reducing agent to oxidizing agent
 Combination Reaction A + B  AB
o Example: 2Na + Cl2  2NaCl
 Decomposition Reaction AB  A + B
o Example: 2H2O2  H2O + O2
 Combustion Reaction Reactant + O2  Products, (one of the reactants is always O2)
o Example: CH4 + 2O2  CO2 + 2H2O
o COMPLETE REACTION OF ANY COMPOUND CONTAINING ONLY CARBON,
HYDROGEN, AND OXYGEN, FORMS CO2 and H2O AS PRODUCTS
 Single Replacement Reaction A + BC  AC + B
o Example (metal displacement): 3Mg(s) + 2FeCl3(aq)  3MgCl2(aq) + 2Fe(s)
 Mg displaces Fe for Cl since it is higher on the activity series of metals
 Elements lower on the activity series cannot replace element higher up
 Activity series of metals:
Higher Li>Rb>K>Ba>Ca>Na>Mg>Al>Mn>Zn>Cr>Fe>Co>Ni>Sn>Pb>H2>Cu>Ag>Hg>Pt>Au Least
o
Example (halogen displacement): Cl2(g) + 2NaBr(aq)  2NaCl(aq) + Br2(l)
 Cl displaces Br because it is higher in group 7A
Double Replacement Reaction – ions recombine in solution to form a solid, liquid, or gas; reactions will
not occur if solid, liquid, or gas is not formed
 Precipitation Reaction AB(aq) + ED(aq)  AD(s) + EB(aq)
o Example: AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
o Use the solubility table to predict whether the two potential products are soluble in
water.
7
Spring 2012



If one or more potential products are insoluble in water, a precipitate is formed
driving the reaction forward.
If neither of the two possible products formed are insoluble in water, no reaction
will occur
Ion
Solubility Rule
NO3All nitrates are soluble
C2H3O2All Acetates are soluble (AgC2H3O2) only moderately)
All chlorides, bromides, and iodides are soluble with some
except when joined with: Ag+, Pb2+ Pb4+, and Hg22+.
Cl-, Br-, I(PbCl2 is slightly soluble in cold water and moderately
soluble in hot water)
All sulfates are soluble except when bound to Ba2+, Pb2+,
SO42Ca2+, Sr2+
Carbonates and Phosphates are insoluble except those of
CO32- and PO43Na+, K+, and NH4+ and H+
Hydroxides are insoluble the exception of Na and K. Ba 2+
OHand Ca2+ are slightly soluble.
Sulfides are insoluble except when bound to Na+, K+,
S2NH4+, and Mg2+, Ca2+, Sr2+, Ba2+
All salts of sodium ion, potassium ion and ammonium are
Na+, K+, NH4+
soluble except for a few uncommon ones.
Neutralization (acid-base) reaction HA(aq) + ZOH(aq)  ZA(aq) + HOH(l)
o Example: H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2H2O(l)
 ACID-BASE REACTIONS ALWAYS PRODUCE SALT AND WATER
Predicting Chemical Reactions
 Use the types of chemical reactions above to help you predict products
 When predicting the products of a chemical reaction, ALWAYS make new products according the
criss-cross method (as shown previously) or Lewis Structures to generate ELECTRICALLY
NEUTRAL COMPOUNDS (as above)
 Then balance the equation as needed
Ionic Equations
 Double Replacement Reaction Example:
o Complete Ionic Equations
Pb(NO3)2(aq) + 2NaCl(aq)  PbCl2(s) + 2NaNO3(aq)

Separate soluble (aq) compounds in molecular equation above into their
constituent ions making the complete ionic equation below.
Pb2+(aq) + 2NO3-(aq) + 2Na+(aq) + 2Cl-(aq)  PbCl2(s) + 2Na+(aq) + 2NO3-(aq)

o
Notice that PbCl2 is not ionized since it is an insoluble solid in solution.
Net Ionic Equations
Pb2+(aq) + 2NO3-(aq) + 2Na+(aq) + 2Cl-(aq)  PbCl2(s) + 2Na+(aq) + 2NO3-(aq)

To the complete ionic equation above, identify spectator ions and cross them
out, and then write what you have left as your net ionic equation below
Pb2+(aq) + 2Cl-(aq)  PbCl2(s)
8
Spring 2012

Single Replacement Reaction Example:
o Complete Ionic Equations
3Mg(s) + 2FeCl3(aq)  3MgCl2(aq) + 2Fe(s)

Separate soluble (aq) compounds in molecular equation above into their
constituent ions making the complete ionic equation below
3Mg(s) + 2Fe3+(aq) + 6Cl-(aq)  3Mg2+ + 6Cl-(aq) + 2Fe(s)

o
Notice that compounds in their element form (not ions) do not ionize. Notice
that equation is balanced with respect to charge and mass
Net Ionic Equations
3Mg(s) + 2Fe3+(aq) + 6Cl-(aq)  3Mg2+ + 6Cl-(aq) + 2Fe(s)

To the complete ionic equation above, identify spectator ions and cancel cross
them out, and then write what you have left as your net ionic equation below
3Mg(s) + 2Fe3+(aq)  3Mg2+ + 2Fe(s)

Notice that equation is balanced with respect to charge and mass
CHEMICAL QUANTIES AND STOICHIOMETRY



Avogadro’s number = 6.022x1023 particles = 1 mole
Molar mass of an element = mass of 6.022x1023 particles = atomic mass of an element in grams
o Example: 1 mole Carbon = 6.022x1023 atoms = 12.01 grams
Molar mass of a compound = the combined molar mass of all the atoms of the compound
o Example: C6H12O6
C 6 x 12.01g/mol  72 g/mol
H 12 x 1.0 g/mol  12 g/mol
O 6 x 16.0 g/mol  96 g/mol
72  12  96  180 g/mol

Use “CONVERION PROBLEM SOLVING” to convert between representative units, moles, and
mass or memorize the information below
o Moles x 6.022x1023 = Representative Particles (ions, atoms, formula units, molecules)
o Representative Particles ÷ 6.022x1023 = Moles
o Moles x Molar Mass = Mass
o Mass ÷ Molar Mass = Moles
 REMEMBER YOU CANNOT JUMP RELATIONSHIPS SUCH AS
CHANGING GRAMS TO REPRESENTATIVE PARTICLES. TO DO SUCH,
YOU WOULD FIRST CONVERT GRAMS TO MOLES AND THEN MOLES
TO REPRESENTATIVE PARTICLES

Percent composition of a compound by mass
o Determine the total mass of compound by adding the experimental masses of each
element together.
o Divide the experimental mass of each element in the compound by the total mass of the
compound and multiply by 100.
o The percent composition of each element in the compound should add to equal 100%
9
Spring 2012
o
Example: A 46.0 gram sample of an unknown liquid breaks down to form 24.0 g of
carbon, 6.00 grams hydrogen, and 16.0 grams oxygen, what is the percent composition of
the unknown compound?
𝑪 = 𝟐𝟒. 𝟎 𝒈 ÷ 𝟒𝟔. 𝟎 𝒈 × 𝟏𝟎𝟎% = 𝟓𝟐. 𝟐% 𝑪
𝑯 = 𝟔. 𝟎𝟎 𝒈 ÷ 𝟒𝟔. 𝟎𝟎 𝒈 × 𝟏𝟎𝟎% = 𝟏𝟑. 𝟎% 𝑯
𝑶 = 𝟏𝟔. 𝟎 𝒈 ÷ 𝟒𝟔. 𝟎 𝒈 × 𝟏𝟎𝟎% = 𝟑𝟒. 𝟖% 𝑶

Percent composition of a compound from a chemical formula
o Multiply the molar mass of each element in the compound by its subscript
o Add up the total molar masses of each element to find the compounds molar mass
o Divide each elements total molar mass by the compounds molar mass and multiply the
result by 100%
o When completed, the percent composition of each element in the compound should add
to equal 100%
Example: Find the percent composition of glucose (C6H12O6)
𝒈
𝒈
𝒈
𝒈
𝑪 = 𝟔 × 𝟏𝟐. 𝟎𝟎
= 𝟕𝟐. 𝟎
𝟕𝟐. 𝟎
÷ 𝟏𝟖𝟎. 𝟎
× 𝟏𝟎𝟎%
𝒎𝒐𝒍
𝒎𝒐𝒍
𝒎𝒐𝒍
𝒎𝒐𝒍𝒆
= 𝟒𝟎. 𝟎% 𝑪
𝒈
𝒈
𝒈
𝒈
𝑯 = 𝟏𝟐 × 𝟏. 𝟎𝟎
= 𝟏𝟐. 𝟎
𝟏𝟐. 𝟎
÷ 𝟏𝟖𝟎. 𝟎
× 𝟏𝟎𝟎%
𝒎𝒐𝒍
𝒎𝒐𝒍
𝒎𝒐𝒍
𝒎𝒐𝒍𝒆
= 𝟔. 𝟕% 𝑯
𝒈
𝒈
𝒈
𝒈
𝑶 = 𝟔 × 𝟏𝟔. 𝟎
= 𝟗𝟔. 𝟎
𝟗𝟔. 𝟎
÷ 𝟏𝟖𝟎. 𝟎
× 𝟏𝟎𝟎%
𝒎𝒐𝒍
𝒎𝒐𝒍
𝒎𝒐𝒍
𝒎𝒐𝒍𝒆
= 𝟓𝟑. 𝟑% 𝑶
𝒈
𝒈
𝒈
𝒈
𝟕𝟐. 𝟎
+ 𝟏𝟐. 𝟎
+ 𝟗𝟔. 𝟎
= 𝟏𝟖𝟎. 𝟎
𝒇𝒐𝒓 𝑪𝟔 𝑯𝟏𝟐 𝑶𝟔
𝒎𝒐𝒍
𝒎𝒐𝒍
𝒎𝒐𝒍
𝒎𝒐𝒍

Empirical formulas
% 𝒐𝒇 𝑨 → 𝒈 𝒐𝒇 𝑨 ÷ 𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑨 ÷ 𝒍𝒆𝒂𝒔𝒕 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒊𝒏 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅
= 𝒔𝒖𝒃𝒃𝒔𝒄𝒓𝒊𝒑𝒕 𝑨
% 𝒐𝒇 𝑩 → 𝒈 𝒐𝒇 𝑩 ÷ 𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑩 ÷ 𝒍𝒆𝒂𝒔𝒕 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒊𝒏 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅
= 𝒔𝒖𝒃𝒃𝒔𝒄𝒓𝒊𝒑𝒕 𝑩
% 𝒐𝒇 𝑪 → 𝒈 𝒐𝒇 𝑪 ÷ 𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑪 ÷ 𝒍𝒆𝒂𝒔𝒕 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒊𝒏 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅
= 𝒔𝒖𝒃𝒃𝒔𝒄𝒓𝒊𝒑𝒕 𝑪
Continue with D, E ... as needed and round off final answers to nearest whole number
o Example: What is the empirical formula of a compound that is 52.2% C, 13.0% H, and
34.8% Oxygen?
𝒈
= 𝟒. 𝟑𝟓 𝒎𝒐𝒍𝒆 ÷ 𝟐. 𝟏𝟕𝟓 𝒎𝒐𝒍𝒆 = 𝟐
𝒎𝒐𝒍
𝒈
𝑯 𝟏𝟑. 𝟎𝒈 ÷ 𝟏. 𝟎
= 𝟔. 𝟕 𝒎𝒐𝒍𝒆 ÷ 𝟐. 𝟏𝟕𝟓 𝒎𝒐𝒍𝒆 = 𝟔
𝒎𝒐𝒍
𝒈
𝑶 𝟑𝟒. 𝟖 𝒈 ÷ 𝟏𝟔. 𝟎
= 𝟐. 𝟏𝟕𝟓 𝒎𝒐𝒍𝒆 ÷ 𝟐. 𝟏𝟕𝟓 𝒎𝒐𝒍𝒆 = 𝟏
𝒎𝒐𝒍
Empirical formula is C2H6O
𝑪 𝟓𝟐. 𝟐𝒈 ÷ 𝟏𝟐. 𝟎

Molecular formulas
o Calculate the empirical formula’s molar mass
o Divide the molecular formula’s molar mass (given to you in the problem) by the
empirical formula’s molar mass and round the answer off to nearest whole number. The
result is the multiplier
o Multiply the multiplier to the empirical formula subscripts. The result is the molecular
formula
o Example: What is the molecular formula of a compound whose empirical formula is
C2H6O and has a molecular molar mass of 138grams/mole?
10
Spring 2012
𝒈
𝑪𝒂𝒍𝒖𝒍𝒂𝒕𝒆𝒅 𝒆𝒎𝒑𝒊𝒓𝒊𝒄𝒂𝒍 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒊𝒔 𝟒𝟔. 𝟎
𝒎𝒐𝒍
𝒈
𝒈
𝟏𝟑𝟖
÷ 𝟒𝟔. 𝟎
=𝟑
𝒎𝒐𝒍
𝒎𝒐𝒍
𝟑 × (𝑪𝟐 𝑯𝟔 𝑶) = 𝑪𝟔 𝑯𝟏𝟖 𝑶𝟑 = 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒇𝒐𝒓𝒎𝒖𝒍𝒂
Stoichiometric Relationships
 Example Problem
o If you have 0.5 moles of Cl2 how many moles of NaCl can you make?
 For the chemical equation 2Na + 1Cl2  2NaCl and following question, use the
equation below.
𝒙 𝒎𝒐𝒍𝒆 𝑮 ×
𝒃 𝒎𝒐𝒍𝒆 𝑾 𝒙𝒃
= ⁄𝒂 𝒎𝒐𝒍 𝑾
𝒂 𝒎𝒐𝒍𝒆 𝑮
o
 𝒙 = 𝟎. 𝟓 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐂𝐥𝟐 (𝐱 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐆𝐢𝐯𝐞𝐧)
 𝒃 = 𝟐 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝑵𝒂𝑪𝒍 (𝒃 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝑾𝒂𝒏𝒕𝒆𝒅)
 𝒂 = 𝟏 𝒎𝒐𝒍𝒆 𝒐𝒇 𝑪𝒍𝟐 (𝒂 𝒎𝒐𝒍𝒆 𝒐𝒇 𝑮𝒊𝒗𝒆𝒏)
“a” and “b” are the coefficients from the chemical equation above
o
So 𝒙𝒃⁄𝒂 𝒎𝒐𝒍𝒆 𝑾 = 𝟎. 𝟓 𝒎𝒐𝒍 𝑪𝒍𝟐 ×
𝟐 𝒎𝒐𝒍 𝑵𝒂𝑪𝒍
𝟏 𝒎𝒐𝒍 𝑪𝒍𝟐
= 𝟏 𝒎𝒐𝒍𝒆 𝑵𝒂𝑪𝒍
SOLUTIONS

Mass Percent
𝑴𝒂𝒔𝒔 % =

𝑴𝒂𝒔𝒔 𝒐𝒇 𝑺𝒐𝒍𝒖𝒕𝒆
𝑴𝒂𝒔𝒔 𝒐𝒇 𝑺𝒐𝒍𝒖𝒕𝒆
× 𝟏𝟎𝟎% =
× 𝟏𝟎𝟎%
(𝑴𝒂𝒔𝒔 𝒐𝒇 𝑺𝒐𝒍𝒗𝒆𝒏𝒕 + 𝑴𝒂𝒔𝒔 𝒐𝒇 𝑺𝒐𝒍𝒖𝒕𝒆)
𝑴𝒂𝒔𝒔 𝒐𝒇 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏
Molarity
𝑴𝒐𝒍𝒂𝒓𝒊𝒕𝒚 = 𝑴 =

𝑴𝒐𝒍𝒆𝒔
𝒎𝒐𝒍
=
𝑳𝒊𝒕𝒆𝒓𝒔
𝑳
Solution Dilutions
𝑴 𝟏 × 𝑽𝟏 = 𝑴 𝟐 × 𝑽𝟐
ACIDS AND BASES




Hydronium ion is equivalent to Hydrogen ion:
𝑯𝟑 𝑶+ = 𝑯+
Spontaneous Ionization of Water:
𝑯𝟐 𝑶 + 𝑯𝟐 𝑶 ↔ 𝑯𝟑 𝑶+ + 𝑶𝑯−
In pure water the concentration, [ ] in M, of [𝑯+ ] = 𝟏 × 𝟏𝟎−𝟕 𝑴 and [𝑶𝑯− ] = 𝟏 × 𝟏𝟎−𝟕 𝑴
The ionization constant Kw of water is 𝟏 × 𝟏𝟎−𝟏𝟒 = [𝑯+ ] × [𝑶𝑯− ]

Hydrogen ion concentration is always [𝑯+ ] =

Hydroxide ion concentration is always [𝑶𝑯− ] =







+]
𝟏×𝟏𝟎−𝟏𝟒
[𝑶𝑯− ]
𝟏×𝟏𝟎−𝟏𝟒
[𝑯∓ ]
The pH of a solution is 𝒑𝑯 = − 𝐥𝐨𝐠[𝑯
To find the hydrogen ion concentration from pH, [𝑯+ ] = 𝟏𝟎−𝒑𝑯
The pOH of a solution is 𝒑𝑶𝑯 = − 𝐥𝐨𝐠[𝑶𝑯− ]
To find the hydroxide ion concentration from pOH, [𝑶𝑯− ] = 𝟏𝟎−𝒑𝑶𝑯
pH plus pOH always equals 14, 𝒑𝑯 + 𝒑𝑶𝑯 = 𝟏𝟒
The pH of a solution is 𝒑𝑯 = 𝟏𝟒 − 𝒑𝑶𝑯
The pOH of a solution is 𝒑𝑶𝑯 = 𝟏𝟒 − 𝒑𝑯
11
Spring 2012
GASES


When solving gas law problems, the temperature must be in KELVIN!
V = volume, P = pressure, T = temperature, n = mole, R is the gas constant 𝟖. 𝟑𝟏
𝒂𝒕𝒎∙𝑳
𝒌𝑷𝒂∙𝑳
𝒎𝒐𝒍∙𝑲
or
𝟎. 𝟎𝟖𝟐𝟏
. When using the gas constant, make sure you covert your pressure to either kPa or
𝒎𝒐𝒍∙𝑲
atm depending on which R value you are using

Boyle’s Law – volume of a gas is inversely proportional to its pressure when temperature and
amount of gas remain constant
𝑽𝟏 × 𝑷𝟏 = 𝑽𝟐 × 𝑷𝟐

Charles’s Law – volume of gas is directly proportional to its temperature when pressure and
amount of gas remain constant
𝑽𝟏 𝑽𝟐
=
𝑻𝟏 𝑻𝟐

Gay-Lussac’s Law – The pressure of a gas is directly proportional to its temperature when
volume and amount of gas remain constant
𝑷𝟏 𝑷𝟐
=
𝑻𝟏 𝑻𝟐

Combined Gas Law – only the amount of gas is constant
𝑽𝟏 × 𝑷𝟏 𝑽𝟐 × 𝑷𝟐
=
𝑻𝟏
𝑻𝟐

Avogadro’s Law – The volume of gas is directly proportional to the amount of gas when the
pressure and temperature of the gas remain unchanged
𝑽𝟏 𝑽𝟐
=
𝒏𝟏 𝒏𝟐

Ideal Gas Law
𝑷×𝑽= 𝒏×𝑹×𝑻

Dalton’s Law of Partial Pressure
Ptotal = P1 + P2 + P3 +…

Gram’s Law of Effusion
12
Spring 2012
BIOCHEMISTY


Monomers are polymerized into polymers called macromolecules
Four major macromolecules in biological systems are nucleic acids, proteins, carbohydrates, and
lipids
o Examples are shown below:

The monomers making up the four major macromolecules are nucleotides that make nucleic acids,
monosaccharaides that make up complex carbohydrates, amino acids that make proteins, and fatty
acids that make up most lipids
o Examples shown below:
13
Spring 2012
DNA Replication
 DNA polymerase polymerizes nucleotides following rules of base paring (adenine pairs with
thymine and guanine pairs with cytosine) to form an new strand of DNA from a template strand of
DNA
o Results in new DNA molecules with one new stand and one old strand
Transmission of genetic information:
 Gene in DNA is transcribed into hundreds of RNA copies by RNA polymerase
 First AUG in mRNA encodes the amino acid methionine and the subsequent start of translation
where amino acids are hooked together based on the codon sequence of the RNA
 Translation occurs on a catalytic machine known as the ribosome which continues liking amino
acids into proteins until a stop codon is reached

Transmission of genetic information sample question
o Transcribe and translate the following DNA sequence:
DNA SEQUENCE
ATGGTACACCTGGTTTAATCGAATCACCGC
RNA TRANSCRIPT
UACCAUGUGGACCAAAUUAGCUUAGUGGCG
Codons are:
AUG UGG ACC AAA UUA GCU UAG
TRANSLATED: Methionine-Tryptophan-Threonine-Lysine-Leucine-Alanine (STOP)
o
Use the genetic code below to translate RNA. Start translation at first AUG in sequence
14