Chemistry Review for
Pennsylvania STEM
Multi-Region Grant
Charles Mahler
March 2015
Predicting Chemical Properties from the
Periodic Table
https://commons.wikimedia.org/wiki/File:
Periodic_table_%28polyatomic%29.svg
Ionic Bonding
• Ionic bonding involves positive cations
(that have lost electrons) and negative
anions (that have gained electrons)
• Usually this involves a compound formed by a metal (on the left side
of the periodic table, red, orange, grey, pink, and purple) and a nonmetal (on the right side of the periodic table, gold, green, and yellow)
• Examples: sodium bromide (NaBr) has positive Na +1 cations and
negative Br -1 anions; zinc sulfide (ZnS) with Zn +2 and S -2; and
aluminum chloride (AlCl3) with Al +3 and three Cl -1.
• Convention is to list the positive cation first, then the negative anion
Covalent Bonding
• Covalent bonds involve sharing (pairs of)
electrons, usually between two non-metals
(gold, green, and yellow elements)
• Sharing can be equal, especially if the two atoms are the same (as in
H2 or O2) or unequal if one atom has a greater portion of the electron
density in the bond (like HF). Unequal sharing leads to polar bonds.
• Electronegativity explains unequal sharing. Electronegativity (the
tendency of an atom to attract electrons to itself within a molecule)
increases going across (right) the periodic table and decreases going
down. Fluorine (F) is the most electronegative element, Cs the least.
Octet rule
• Elements that have the same number of
valence electrons as a noble gas (shown
in light blue) are especially stable
• This is often 8 valence electrons, so it is known as the “octet rule”
• The octet rule helps explain both ionic and covalent bonding
• For example, oxygen with 6 valence electrons, will gain 2 electrons (to
get to 8) and form a -2 anion (ionic bonding).
• Oxygen can also share 2 additional electrons to get to 8 (in water,
H2O, oxygen shares two electrons from the two H atoms to get to 8)
Octet Rule
Practice Problems I
• Predict the ion each element will form
•S
K
Mg
Al
•N
Br
Cs
C
• Te
P
H
Kr
Octet Rule
Practice Problems I Answers
• Predict the ion each element will form
• S -2
K +1
Mg +2
Al +3
• N -3
Br -1
Cs +1
C +4 OR -4
• Te -2
P -3
H +1
Kr 0 (it already is a noble gas)
Octet Rule
Practice Problems II
• Count the number of valence electrons in
•C
H
O
F
•N
B
S
P
K
Mg
• Predict compounds between:
C&H
O &H
•N&H
Mg & F
B&F
K&S
Octet Rule
Practice Problems II Answers
• Count the number of valence electrons in
• C4
H1
O6
F7
•N5
B3
S6
P5
K1
Mg 2
• Predict compounds between:
C & H CH4
O &H H2O
• N & H NH3
Mg & F MgF2
B & F BF3
K & S K2S
Octet Rule
Practice Problems II Answers
• Predict compounds between: B & F BF3
• Note that Boron only has 3 valence
electrons and forms an electron deficient
compound (instead of an octet, B only 6 electrons around it, though
each F has a full 8).
Moles, Mass, Stoichiometry,
Percent Yield, and Limiting Reactants
https://commons.wikimedia.org/wiki/File:Combustion_reaction_of_methane.jpg
Mole concept I
• There are a huge number of atoms or molecules in even a small amount of
a substance.
• A mole is Avagadro’s number (6.022 x 1023) of anything, and is a counting
number (similar to a dozen (12) or a score (20))
• The mole is chosen so that a mole of a chemical substance has the same
mass (weight) in atomic mass units (amu) and grams per mole
• So one atom of 12C weighs 12.00 amu and one mole of of 12C weighs 12.00
grams
• Element masses are given on the Periodic Table and are weighted averages
(since different isotopes of the same element have different masses)
Mole Problems I
• How many atoms are in 5.36 moles of argon (Ar)?
• There are 5.62 x 1028 atoms of nickel (Ni) in a mine, how many moles?
• How many grams of Ni are in the mine?
Mole Problems I
• How many atoms are in 5.36 moles of argon (Ar)?
• (5.36 mol)(6.022 x 1023 atoms/mol) = 3.23 x 1024 atoms
• There are 5.62 x 1028 atoms of nickel (Ni) in a mine, how many moles?
• (5.62 x 1028 atoms)/(6.022 x 1023 atoms/mol) = 9.33 x 104 moles
• How many grams of Ni are in the mine?
• (9.33 x 104 mol)(58.693 g/mol) = 5.48 x 106 g (or 5.48 x 103 kg)
Mole Problems II
• What are the molar masses (molecular weights) of water, H2O, and
methane, CH4, and salt, NaCl?
Mole Problems II Answers
• What are the molar masses (molecular weights) of water, H2O, and
methane, CH4, and salt, NaCl?
• Molar masses: H 1.0079, C 12.011, O 15.999, Na 22.990, Cl 35.453
• Water, H2O:
15.999 + (2 x 1.0079) = 18.015 g/mol
• Methane, CH4:
12.011 + (4 x 1.0079) = 16.043 g/mol
• Salt, NaCl:
22.990 + 35.453 = 58.443 g/mol
Stoichiometry I
• Moles are useful because they allow us to relate things we can
measure (like masses) to numbers of atoms and molecules
• A balanced chemical reaction equation gives the number of atoms
and molecules present as reactants and as products – this is called
the stoichiometry of the reaction
• In the reaction pictured in the top right corner, the stoichiometry is
1 CH4 + 2 O2  1 CO2 + 2 H2O
• In this reaction for every one mole of methane (CH4) that reacts, we
need two moles of oxygen (O2)
Stoichiometry Problems
• In the reaction CH4 + 2 O2  CO2 + 2 H2O, if we had 3.47 moles of
methane, how many moles of oxygen would we need?
• In the reaction CH4 + 2 O2  CO2 + 2 H2O, if we had 2.57 g methane
and enough oxygen, how many grams of water would be produced?
Stoichiometry Answers
• In the reaction CH4 + 2 O2  CO2 + 2 H2O, if we had 3.47 moles of
methane, how many moles of oxygen would we need?
• There are 2 moles of O2 that react for ever mole of CH4, so
(3.47 mol CH4)(2 mol O2 / 1 mol CH4) = 6.94 mol O2 needed
• In the reaction CH4 + 2 O2  CO2 + 2 H2O, if we had 25.7 g methane
and enough oxygen, how many moles of water would be produced?
• (25.7 g)(1 mol CH4/16.043 g)(2 mol H2O/1 mol CH4) = 3.20 mol H2O
• Can also find (3.20 mol H2O)(18.015 g/mol) = 57.7 g H2O
Stoichiometry and
Limiting Reactant
• In the real world, amounts of rectants are often not exactly what is needed,
and one reactant will run out first – we call this the “limiting reactant” (or
“limiting reagent”), often abbreviated LR.
• LR because it limits / controls the amount of products made – once the LR
is gone, the reaction stops and no more product is made.
• An everyday example – if you were making cheese sandwiches and had 3
cheese slices and 10 bread slices, the stoichiometry would be 1 cheese slice
per 2 bread slices. You could make 3 sandwiches based on cheese, and 5
based on bread. Cheese is the LR, once it is gone (3 sandwiches), no more
sandwiches can be made (4 bread left over).
• LR is always based on the smallest amount of product that can be made
Finding the Limiting Reactant
• LR is always based on the smallest amount of product that can be
made
• To find LR for a chemical reaction, you need a balanced equation
• First calculate moles of each reactant
• Use the stoichiometry to convert moles of each reactant to moles of
one product
• The reactant that makes the smallest number of moles of product is
the LR (limiting reactant)
• Can also calculate mass of a product for each, least mass is still LR
Sample Limiting Reactant Calc
• In the reaction CH4 + 2 O2  CO2 + 2 H2O, there are 20 g each of methane
and oxygen, what is the limiting reagent? How much water is made?
• (20 g)(1 mol CH4/16.043 g)(2 mol H2O/1 mol CH4) = 2.493 mol H2O
(from CH4)
• (20 g)(1 mol O2/31.998 g)(2 mol H2O/2 mol O2) = 0.625 mol H2O
(from O2, LR)
• Oxygen yields less product (water), so it is the Limiting Reactant, and only
0.625 moles of water is made
Limiting Reactant Problem
• In the reaction SO3 + H2O  H2SO4, there are 20.0 g SO3 and 10.0 g
H2O. What is the LR and how many grams of H2SO4 are produced?
Limiting Reactant Solution
• In the reaction SO3 + H2O  H2SO4, there are 20.0 g SO3 and 10.0 g H2O.
What is the LR and how many grams of H2SO4 are produced?
• (20.0 g)(1 mol SO3/80.06 g)(1 mol H2SO4/1 mol SO3) = 0.2498 mol H2SO4
(from SO3, LR)
• (10.0 g)(1 mol H2O/18.02 g)(1 mol H2SO4/1 mol H2O) = 0.5549 mol H2SO4
(from H2O)
• (0.2498 mol H2SO4)(98.08 g/mol) = 24.5 g H2SO4
Limiting Reactant and
Percent Yield
• The amount of product made in a reaction based on the limiting
reactant is called the theoretical yield
• Often the actual (experimental) yield in a chemical reaction is less
than this
• Percent yield = (actual yield/theoretical yield) x 100%
• So if a reaction was expected to produce 56.3 g of A, and actually only
made 42.5 g of A, the percent yield in this case would be:
Percent yield = (42.5 g/56.3 g) x 100% = 75.5%
Percent Yield Problem
• In the reaction N2 + 3 H2  2 NH3, 18.20 kg NH3 is produced when
6.00 kg H2 and excess N2 react. What is the reaction’s percent yield?
Percent Yield Problem
• In the reaction N2 + 3 H2  2 NH3, 18.20 kg NH3 is produced when
6.00 kg H2 and excess N2 react. What is the reaction’s percent yield?
• (6.00 kg H2)(1000 g/1 kg)(1 mol H2/2.016 g)(2 mol NH3/3 mol H2) =
1984 moles NH3
• (1984 mol NH3)(17.03 g/mol NH3)(1 kg/1000 g) = 33.79 kg NH3 (theory)
• Percent yield = (18.20 kg/33.79 kg) x 100% = 53.8% yield
Equilibrium and
Le Chatelier’s Principle
• Chemical equilibrium is a dynamic process
• The forward and reverse reactions continue at equilibrium, but the
concentrations do not change
• For a given reaction a A + b B  c C + d D, the equilibrium constant is
• K = [C]c[D]d
[A]a[B]b
• The reaction quotient Q is calculated the same way, but does not
have to be at equilibrium (so K is a special case of Q, where the
system is at equilibrium)
Le Chatelier’s Principle
• Le Chatelier’s principle applies to systems at equilibrium which are
disturbed in some way. If a system at equilibrium is disturbed by a
change (or stress), the system will respond so as to minimize the
change and return to equilibrium
• Silly example – most sitcoms have a well established plot scenario
(like equilibrium, i.e. Niles secretly loves Daphne on Frasier), the plot
usually involves some threat to the equilibrium (Niles says he adores
Daphne when they dance in Season 3), but the status quo is quickly
restored (Daphne thinks Niles is just putting on an act for the crowd
at the dance, and he is too shy to tell her the truth).
• P.S. They did not get together for good until Season 8!
Le Chatelier’s Principle
• Le Chatelier’s principle examples involving amounts of reagents
• If reactant is added, the system responds by making more products (it
shifts to the right)
• If product is added, the system responds by making more reactants (it
shifts to the left)
• If reactant is removed, the system responds by making more
reactants (it shifts to the left)
• If product is removed, the system responds by making more products
(it shifts to the right)
Le Chatelier’s Principle
• Ways to think about Le Chatelier’s principle
• In terms of the examples given on the previous slide (adding shifts to
the opposite side, removing material shifts back to the same side)
• Water in a U-shaped tube (if water is added to the left side, the right
side will soon go higher (add to the left, shifts right)
• Comparing K and Q (in that order) If K > Q, reaction shifts to the right
(where the greater than symbol is pointing), but if K < Q, the reaction
shifts to the left (where the less than symbol is pointing). If doing this,
it is helpful to pretend all equilibrium concentrations are 1, then a
change is easy to calculate (double A, Q = ½, K > Q, reaction )
Le Chatelier’s Principle
• For pressure changes, if gas phase species are present,
• an increase in overall pressure favors the side with fewer total gas
molecules
• a decrease in overall pressure favors the side with more total gas
molecules
• If both sides have the same number of gas molecules, a change in
pressure will have no effect (neither side is favored)
Le Chatelier’s Principle
• For exothermic and endothermic reactions, think of heat as a product
(if exothermic) or a reactant (if endothermic)
• Raising the temperature is like adding heat, so raising the
temperature will shift an exothermic reaction to the left (heat is a
product, add heat (product), make more reactants) and will shift an
endothermic reaction to the right (more reactant heat makes more
products)
• Lowering the temperature is like removing heat (exothermic ,
endothermic )
• This fits well with what we already know for adding / removing stuff
More Practice Problems
• Le Chatelier’s principle
http://www.sciencegeek.net/Chemistry/taters/LeChatelier.htm
• Limiting Reactant
• https://www.khanacademy.org/science/chemistry/chemicalreactions-stoichiome/limiting-reagentstoichiometry/e/limiting_reagent_stoichiometry