Download Formulas, Reactions, Equations, and Moles

Chemical Names, Formulas,
Reactions, Equations, and
Stoichiometry
Chemical Names and Formulas
Do Now
• Take out Reference Tables and open to Table E
– QUIZ ON POLYATOMIC IONS ON WEDNESDAY
• Turn to Table S – Where is the oxidation
number noted?
• How many aluminum atoms combine with
how many sulfur atoms?
How do we figure this out?
OXIDATION NUMBER
• Oxidation number = the charge an atom
would acquire if all its bonds were treated as
ionic bonds.
• To determine how many atoms combine with
one another in a compound we must
determine each element’s OXIDATION
NUMBER.
Predicting Ionic Charges
Group 1: Lose 1 electron to form 1+ ions
H+
Li+
Na+
K+
Predicting Ionic Charges
Group 2: Loses 2 electrons to form
2+ ions
Be2+
Mg2+
Ca2+
Sr2+
Ba2+
Predicting Ionic Charges
B3+
Al3+
Ga3+
Group 13: Loses 3
electrons to form
3+ ions
Predicting Ionic Charges
Group 14:
Lose of 4
electrons or gain
of 4 electrons?
Neither! Group 14
elements rarely form
ions.
Predicting Ionic Charges
N3- Nitride
P3- Phosphide
As3- Arsenide
Group 15: Gains 3
electrons to form
3- ions
Predicting Ionic Charges
O2- Oxide
S2- Sulfide
Se2- Selenide
Group 16:
Gains 2
electrons to form
2- ions
Predicting Ionic Charges
F1- Fluoride
Br1- Bromide
Cl1-Chloride
I1- Iodide
Group 17:
Gains 1
electron to form
1- ions
Predicting Ionic Charges
Group 18:
Stable Noble gases
do not form ions!
Predicting Ionic Charges
Groups 3 - 11: Many transition elements
have multiple oxidation states.
Iron(II) = Fe2+
Iron(III) = Fe3+
Predicting Ionic Charges
Groups 3 - 11: Some metals
have only one possible oxidation state.
Silver = Ag+
Zinc = Zn2+
Rules for assigning Oxidation Numbers
1. The atoms in a pure element have an oxidation number of zero.
2. Alkali metals always have an oxidation number of +1; alkaline earth
metals always have an oxidation number of +2.
3. Fluorine always has an oxidation number of -1.
4. Oxygen has an oxidation number of -2 in almost all compounds.
Exceptions are in compounds with a halogen, when it has an
oxidation number of +2, and in peroxides (H2O2), when it has an
oxidation number of -1.
5. Hydrogen has an oxidation number of +1 in almost all compounds
except when combined with a metal when it has an oxidation number
of -1.
6. The sum of all the oxidation numbers in a neutral compound is zero.
7. The sum of all the oxidation numbers in a polyatomic ion is equal to
the charge of the ion.
ASSIGNING OXIDATION NUMBERS
• A compound has a total charge of ZERO so set your
equation equal to ZERO.
• Assign the variable X to your unknown oxidation number.
• Given a compound, find the oxidation number of every
element you know for certain. Then solve for others
using algebra.
• KMnO4
• CaCO3
Do you notice a pattern with the
elements in these compounds?
H2+1 O1-2
Mg1+2F2-1
Al2+3S3-2
Cu3+1P1-3
Ca1+2Cl2-1
Fe2+3O3-2
CRISS-CROSS METHOD
to determine the chemical formula
1. Write the symbols for the elements side by side.
2. Write the oxidation states of each element to the top
right of the symbol. When the nonmetal is combines
with a metal, the oxidation state will always be the first
number (the negative one) in the list of oxidation states.
3. Criss cross the charges DOWN and use the absolute
values (-2 becomes 2).
4. Check to make sure the subscripts are the lowest ratio.
(Empirical formula)
Practice Criss-Cross Method
1. Na and S
2. K and P
3. Al and S
4. Mg and Br
5. Al and O
What about Polyatomic Ions?
• Write down the only cations on Table E:
(positively charges ions)
• Write down the only polyatomic ions that end
in –ide.
Polyatomic Ions Chart (Table E)
Find the charge of the polyatomic ion using Table E
Put parenthesis around the polyatomic ion.
1.
PO4
2.
CO3
3.
SO3
4.
NH4
5.
ClO
6.
ClO2
7. HSO4
8. ClO4
9. CN
10. OH
11. S2O3
12. SCN
Polyatomic Ions
• If you see a group of atoms together with a charge it is a
polyatomic ion from Table E.
• Put the polyatomic ion within Parenthesis.
• Find the charge of the polyatomic ion.
• Use the criss-cross method to determine subscripts.
Na +1 CO3-2
How many atoms are present in the compound?
NH4 +1 S-2
How many atoms are present in the compound?
Rules for writing the formula of the
compound composed of ions
1. Place all Polyatomic Ions in Parenthesis (Table
E)
2. Determine all oxidation numbers of elements
and polyatomic ions
3. Use Criss-Cross Method
4. Reduce to Empirical Form (simplest ratio)
Practice
1. NH4 S
2. Na NO3
3. Cu Br
4. Al SO4
5. Fe CO3
6. Pb PO4
7. Ag ClO
8. Ca F
9. NH4 SO3
10. Cu OH
11. Ni I
12. Zn SO4
13. Pb ClO2
14. H I
15. Fe HSO4
16. Cu CO3
17. NH4 O
18. Ag S
19. Al ClO4
Do Now
• What is the name of H2O?
• What is the name of NaCl?
• What is the name of NH4?
Naming Compounds
• Many chemical compounds have common
names.
• H2O = water
• NaCl = table salt
• NH4 = ammonia
Review
• Chemical formula = the relative number of
atoms of each kind in a chemical compound
• Empirical formula = the elements appear in
the smallest whole number ratio.
• Molecular formula = the number and kind of
atoms in a molecule (not necessarily the
smallest whole number ratio)
• Structural formula = shows how the atoms in a
molecule are bonded to one another
NAMING IONIC COMPOUNDS
• An Ionic compound can quickly be determined
if a METAL (or cation) is bonded to a
NONMETAL (or anion).
• Naming Ionic compounds with metals that
have only ONE oxidation state is fairly simple.
Of the following metals listed below, check off
all the elements that have more than one
oxidation state:
a. Mn___
b. Zn ___ c. K ____
d. Pb___
e. Mg___
f. Au ___ g. Ag ___
h. Ga__
i. Sn __ j. Li ____ k. U ____
l. Cr ___
Naming Ionic Compounds that contain
metals with 1 oxidation state
• Binary Ionic Compounds = ionic compounds
with only 2 different elements.
• Name the metal and end the nonmetal in –
ide.
• For example: CaBr2 = _______________
Naming Binary Ionic Compounds
*using the IUPAC system
• Naming salts is very easy, because they are binary ionic compounds
(made up of two elements).
– The cation is named by borrowing the name of the element.
– The anion named by combining the name of the element with
an –ide ending.
• The name of compound is made up of both the cation and anion
name
– Ex: NaCl = sodium chloride
– Ex: ZnS = zinc sulfide
– Ex: K2O = potassium oxide
– Ex: Mg3N2 = magnesium nitride
– Ex: Al2S3 = aluminum sulfide
Name the following Binary
compounds:
1. MgO= ________________________
2. CaCl2 = _______________________
3. AlBr3 = _______________________
4. Ag3N = _______________________
5. Al2O3 = ______________________
6. LiI = _________________________
7. BaF2 = _______________________
8. Zn2C = _______________________
9. Ba3N2 = ______________________
10.CdO = _______________________
11. Ga2S3 = _______________________
12. K3N = ________________________
13. SrO = ________________________
What about Polyatomic ions? (Table E)
• These are ions consisting of more than one
atom. The names of polyatomic ions end in –
ate or –ite, except for two (ammonium and
cyanide)
Polyatomic ion chart (Table E)
Ternary Ionic Compounds
•
•
•
•
Ionic Compounds with 3 different elements.
They usually contain Polyatomic Ions (Table E)
Name the following Polyatomic Ions:
a. NO3- _________
d. SO42- ___________
• b. ClO2- _________
e. SO32- __________
• c. CO32- _________ f. SCN- ____________
Naming Ternary Compounds
• Name the metal and then name the
polyatomic ion (if it has a negative oxidation
number).
• For example: KNO3 = ________________
Naming Ternary Compounds
• Name the polyatomic ion (if it has a positive
oxidation number) and then name the
nonmetal.
• For example: NH4Cl = ________________
Naming Ternary Compounds
• If there are two polyatomic ions, name the
positive polyatomic ion first and then name the
negative polyatomic ion.
• For example: NH4NO3 = ______________
• DO NOT CHANGE THE ENDINGS OF POLYATOMIC
IONS!
Name the following Ternary compounds:
1. NaC2H3O2 _____________________
2. AgHCO3 _______________________
3. LiNO2 _________________________
4. Ga2(S2O3)3 ____________________
5. Ca3(PO4)2 _____________________
6. ZnSO3 __________________________
7. KClO3 ___________________________
8. Al(OH)3 _________________________
9. RbSCN __________________________
10. SrCO3 _________________________
Naming Ionic Compounds with Metals with Multiple
oxidation states (multiple charges):
Using the STOCK SYSTEM
1. Determine the oxidation state of the metal in
the compound.
2. Name the metal, put the oxidation state in
ROMAN NUMERALS in parenthesis and end
the nonmetal in –ide.
Review: Find the Formula
Criss-Cross
Pb+4 O-2
Cu+2 (SO4) -2
Sn+2(CO3)
-2
Find Empirical
formula
Work Backwards: Start with the Empirical Formula to determine
the Oxidation state of a Metal with Multiple Oxidation States
Empirical
Formula
Fe1O1
Write in the oxidation
number for the nonmetal
or polyatomic ion you are
sure of and criss-cross.
Fe O
Fe1(SO4)1
Fe (SO4)
Cu1(SO4)1
Cu (SO4)
Non-reduced form with
Oxidation States
Work Backwards: Start with the Empirical Formula to determine
the Oxidation state of a Metal with Multiple Oxidation States
Empirical
Formula
Write in the oxidation
number for the nonmetal
or polyatomic ion you are
sure of and criss-cross.
Sn1(SO3)1
Sn (SO3)
Mn1(SO4)2
Mn (SO4)
Cr1(PO4)2
Cr (PO4)
Non-reduced form with
Oxidation States
Name the following compound using
the Stock System:
1. Fe O ____________________________
2. Fe Cl2 ___________________________
3. Cu SO4 __________________________
4. Pb Cl2 ___________________________
5. Pb O2 ___________________________
6. Cu3(PO4)2 ________________________
7. Cu2 S ___________________________
8. Fe2(CrO4)3 ______________________
19. Sn CO3 _________________________
10. Sn F4 ___________________________
Name each of the following
compounds, use Roman Numerals
only when necessary.
Put a check next to every compound that begins
with a metal with more than 1 oxidation state.
Put parenthesis around all the polyatomic ions.
1. NH4 Cl _________________________
2. Pb SO4 _________________________
3. Co Cl3 __________________________
4. Ba (NO3)2 ______________________
5. Co2 (SO3)3 _____________________
6. KH ____________________________
7. NH4 F _________________________
8. K2Cr2O7 _______________________
9. Cu S __________________________
10. Cu ClO2 ______________________
11. Ag NO3 _______________________
12. Fe Cl3 ________________________
13. Cr F2 _________________________
14. Na Cl _________________________
15. Fe PO4 ______________________
16. Li F _________________________
17. Fe F3 _______________________
18. Al (OH)3 _____________________
19. Mg I2 ________________________
20. Fe Cl3 _______________________
Another Naming System
• For metallic elements that can form two
positive ions, the suffixes –ous and –ic may be
attached to the Latin name of the element:
– Ex: FeCl2 = ferrous chloride
– Ex: FeCl3 = ferric chloride
– Ex: Cu2O = cuprous oxide
– Ex: CuO = cupric oxide
– Ex: Hg2Br2 = mercurous bromide
– Ex: HgBr2 = mercuric bromide
Do Now
• What types of bonds are in molecular
compounds? _____________________
• How can we tell if a formula has a covalent
bond? ___________________________
Review Molecular Compounds
• Molecule = a neutral compound held together
by covalent bonds
• Molecules may consist of identical atoms
bonded together (O2) or different atoms
bonded together (H2O)
Naming Covalent Compounds
*using the IUPAC system
• Covalent compounds are named in a similar
way to ionic compounds
– The first element in the formula is usually written
first in the name
– The second element has an –ide ending
• Ex: SO2 = sulfur oxide
– However, this is not completely correct…
Naming Covalent Compounds
*using the IUPAC system
• Since multiple covalent compounds can be
made from the same elements, the name
must distinguish them as different.
– Prefixes are used to indicate the number of atoms
of each element in the molecule.
• Ex: SO2 = sulfur dioxide
• Ex: SO3 = sulfur trioxide
Naming Covalent Compounds
*using the STOCK SYSTEM
• Most nonmetals have more than 1 oxidation
state, therefore you can use the Stock System
(Roman Numerals) or the Prefix System.
• The Prefix System includes:
MonoDiTriTetraPenta-
HexaHectaOctaNonaDeca-
Naming Molecular Compounds
• According to the number of atoms of each
element, state the prefix for the number of
each atom before the name of the element
and end the nonmetal in –ide.
• For example:
P2O5 = ______________
CO2 = ______________
Naming Molecular Compounds
• Name the first element. Use a prefix ONLY if
there is more than one.
• Name the second element. ALWAYS use a
prefix. Change the ending to –ide.
• The prefix mono only needs to be used for the
second half of the compound NOT the first
element.
Name the following covalent
compounds:
1.
CI4 ________________________________
2.
PCl5 ______________________________
3.
SI6 _______________________________
4.
P2S6 ______________________________
5.
N3O4 ______________________________
6. SO2 ____________________________
7. N2O4 ___________________________
8. CO _____________________________
9. NF3 ____________________________
10. ICl5 ___________________________
11. H2S3 __________________________
12. N2O3 _________________________
13. ClF7 __________________________
14. SO3 ___________________________
15. NI5 ___________________________
16. BN2 __________________________
17. P2O5 _________________________
18. IF7 ___________________________
Do Now
• What is the chemical formula for dihydrogen
monoxide?
• What is the chemical formula for acetate?
Writing Formulas
• The chemical formula of a compound can be
determined from the chemical name!
General Steps to Determining
Chemical Formulas:
• 1. Determine what elements and/or
polyatomic ions are in the compound.
• 2. Write the symbols for each substance with
oxidation states of each substance to the top
right of the symbol.
– If there is a metal with more than one oxidation
state, it will be indicated as roman numerals in
parenthesis after the element.
– If there is a nonmetal with more than one
oxidation state, use the negative number.
General Steps to Determining
Chemical Formulas (continued):
• 3. All oxidation states in a neutral compound add
up to ZERO, so figure out how many of each
substance you need to make all oxidation states
add up to ZERO.
• 4. Put these numbers as subscripts and write the
chemical formula.
• 5. If you have more than one polyatomic ion,
make sure you put the symbols in parenthesis.
• 6. Check to make sure the subscripts are the
lowest ratio.
Polyatomic Ions (Table E)
Writing Formulas for Ionic Compounds
• All ionic compounds must be in empirical form.
(Reduced Form)
• If the compound ends in –ide, most likely it is a
binary compound. Except for cyanide and hydroxide.
• If the substance ends in –ate or –ite it contains a
polyatomic ion. Put the polyatomic ion in parenthesis
with oxidation numbers indicated before you crisscross. Put the oxidation states on top and criss-cross
the numbers.
•
For example: Aluminum Sulfide =
Write the formulas for each of the following
compounds:
(Make sure you reduce it to empirical form)
1. Aluminum Chloride ___________________
2. Silver Phosphate _____________________
3. Lithium Hydride ______________________
4. Magnesium Acetate __________________
5. Potassium Sulfite _____________________
6. Zinc Thiosulfate ____________________
7. Strontium Nitride ____________________
8. Calcium Oxide _______________________
9. Gallium Oxalate ______________________
10. Ammonium Hydroxide ________________
Writing formulas for compounds with metals with
more than one oxidations state:
(Roman Numerals will be given)
1.
If there are Roman Numerals with the name, the Roman Numeral is
the charge of the metals.
For example: Iron (II) Oxide= Fe+2O-2=Fe2O2=FeO
2. If the compound ends in –ate or –ite, most likely you should look on
the polyatomic ion chart.
3. Write the metal, then look up the polyatomic ion and place it in
parenthesis. Put the charges on top and criss-cross. Reduce if
necessary.
For example:
Zinc Carbonate=
Gold (III) Thiocyanate =
For each compound listed below, write the
correct formula using the stock system.
1.
Iron (II) Chloride _____________________
2.
Lead (IV) Phosphide __________________
3.
Tin (II) Oxide ________________________
4.
Copper (I) Iodide _____________________
5.
Nickel (III) Sulfide ____________________
6.
Cobalt (II) Thiocyanate ______________
7. Manganese (IV) Oxide _________________
8. Titanium (IV) Chromate _________________
9. Iron (III) Sulfate _______________________
10. Lead (II) Nitrate ______________________
11. Tin (IV) Carbonate ______________
12. Copper (II) Acetate _____________
Writing Formulas for Covalent
Compounds
• Sometimes the name will have prefixes (only
when the two elements in the compound are
nonmetals). Simply use the prefixes to figure
out the formula!
For each compound listed below, write the
correct formula using the prefix system.
1.
Diphosphorous pentoxide _____________________
2.
Silicon tetrafluoride ________________________
3.
Dihydrogen monoxide ________________________
4.
Tetraphosphorous trisulfide __________________
Naming Ionic Compounds using the Formula
Determining subscripts of
elements or polyatomic in
ionic compounds
Use Criss-cross method
Example:
Naming ionic compounds with
metals with only 1 oxidation
state
Binary Ionic Compounds
Ionic compounds with only 2 different elements.
For example: CaBr2 = Calcium
Bromide
Name the metal and end the nonmetal in –ide.
Ternary Ionic Compounds
Ionic Compounds with 3 different elements.
•Name the metal and then name the polyatomic ion.
Naming ionic compounds with
metals with multiple
oxidation state
1. Determine the oxidation state of the metal in the
compound.
2. Name the metal, put the oxidation state in ROMAN
NUMERALS in parenthesis and end the nonmetal in
–ide.
For example: KNO3 = Potassium
Nitrate
Example: FeO= Iron
(II) Oxide
Writing the Formula of Ionic Formulas using the name of the
Ionic Compound
Formula Writing for Ionic
Compounds
Writing formulas for compounds
with metals with more than one
oxidations state:
(Roman Numerals will be given)
All ionic compounds must be in empirical
form. (Reduced Form)
If the compound ends in –ide, most likely it
is a binary compound. Except for cyanide
and hydroxide.
If the substance ends in –ate or –ite it
contains a polyatomic ion. Put the polyatomic
ion in parenthesis before you criss-cross.
Put the oxidation states on top and crisscross the numbers.
If there are Roman Numerals with the
name, the Roman Numeral is the charge of
the metals.
2. If the compound ends in –ate or –ite,
most likely you should look on the
polyatomic ion chart.
3. Write the metal, then look up the
polyatomic ion and place it in parenthesis.
Put the charges on top and criss cross.
Reduce if necessary.
For example:
Zinc Carbonate= Zn+2(CO3) -2 =
Zn2(CO3)2 =Zn(CO3)
Gold (III) Thiocyanate = Au+3 (SCN)
Au(SCN)3
-1
=
For example:
Iron (II) Oxide= Fe+2O-2=Fe2O2=FeO
Quiz
• Short Quiz on Oxidation Numbers and Naming
CHEMICAL REACTIONS
Do Now
• What are some signs that a chemical change
may have taken place?
• Where are the reactants and products in a
reaction? What do they represent?
• Describe the law of conservation of mass.
Chemical Reaction
• A process in which one or more substances
are converted into new substances with
different chemical and physical properties.
Chemical Reaction
• In chemical reactions, existing bonds are
broken, atoms are rearranged, and new bonds
are formed.
• Reactant: substances that enter the reaction
(start)
• Product: substances that come out of reaction
(end)
Evidence
• How do we know that a reaction has taken
place?
– Production or absorption of energy (heat and
light)
– Production of a gas
– Formation of a precipitate (a solid that forms as a
result of mixing 2 solutions).
– Color change
Chemical Equations
• Represent reactions with symbols and
formulas, tells us the identities and relative
amounts of reactants and products in a
chemical reaction.
Word Equations
• Give names of reactants and products
+ = “reacts with,” “add,” or “ added to”
→ = “yields” or “produces”
• Describe the following as a word equation:
Calcium + Oxygen → Calcium Oxide
Formula Equations
• Give symbols or formulas of reactants and
products
• When going from word to formula, be careful
to use the correct formulas. (Remember the
diatomic molecules Br2, I2, N2, Cl2, H2, O2, F2)
• Write the formula for the previous word
equation:
• Ca + O2 → CaO
Additional symbols in equations:
• (s) or ↓ = Solid
• (l) = Liquid
• (g) or ↑ = Gas
• (aq) = Aqueous (dissolved in water)
• Write the word equation for the reaction that
occurs when solid sodium oxide is added to
water and forms sodium hydroxide (dissolved
in water).
• Na2O(s) + H2O → NaOH (aq)
• Write the word equation and formula for
methane reacting with oxygen to produce
carbon dioxide and water.
Methane + oxygen  carbon dioxide + water
CH4 + O2  CO2 + H2O
LAW OF CONSERVATION OF MATTER
(MASS)
• Formula equations must be written in
accordance with the LAW OF CONSERVATION
OF MATTER (MASS) = atoms cannot be
created or destroyed in a chemical reaction.
They can only be rearranged!
*If you have 50g of reactants, you must end up with 50g of products.
Law of Conservation of Mass
Mass is neither created nor
destroyed during chemical
or physical reactions.
Total mass of reactants = Total mass of products
Formula equations must be written in
accordance with the LAW OF
CONSERVATION OF MATTER (MASS)
Antoine Lavoisier
Law of Conservation of Mass
• The same number and kind of atoms must
appear on both sides of the arrow!!!
• COEFFICIENTS= small whole numbers that
appear in front of a formula in a chemical
equation.
• The coefficient multiplies the number of
atoms of each element in the formula that
follows.
– Ex: 2 H2O = 4 H and 2 O
Trial and Error is the best way to
balance equations.
Steps to Writing Balanced Equations:
1. Read the word equation that describes a
reaction:
zinc + hydrochloric acid → zinc chloride + hydrogen
2. Replace the words with symbols.
Zn +HCl → ZnCl2 + H2
• 3. Count the number of atoms on each side of
the arrow. (Keep polyatomic ions that exist on
both sides of the equation as one!)
Zn +HCl → ZnCl2 + H2
1 Zn
1 Zn
1H
2H
1 Cl
2 Cl
• 4. Balance using trial and error.
Zn +2HCl → ZnCl2 + H2
• 5. Check
1 Zn
2H
2 Cl
1 Zn
2H
2 Cl
1mol of Zn will react with 2mol HCl to produce 1mol ZnCl2 and 1mol H2
Practice
• Na + H2O  NaOH + H2
1 Na
1 Na
2H
3H
1O
1O
• 2Na + 2H2O  2NaOH + H2
2 Na
2 Na
4H
4H
2O
2O
Practice
• Iron (III) oxide and hydrogen are the reactants.
Iron and Water are the products. Find the
balanced equation.
Fe2O3 + H2  Fe + H2O
2 Fe
1 Fe
3O
1O
2H
2H
Practice
Fe2O3 + H2  Fe + H2O
2 Fe
1 Fe
3O
1O
2H
2H
Fe2O3 + 3H2  2Fe + 3H2O
2 Fe
2 Fe
3O
3O
6H
6H
Balanced Particle Diagrams
• What are the chemical formulas for the
reactants and products?
What did you learn today?
What did you learn today?
• In all chemical reactions there is a conservation of
mass, energy, and charge.
• A balanced chemical equation represents
conservation of atoms. The coefficients in a
balanced chemical equation can be used to
determine mole ratios in the reaction.
Do Now
• What do you notice about the arrangement of
elements in the reactants and products?
• 2Al + Fe2O3  2Fe + Al2O3
• HCl + NaOH  H2O + NaCl
TYPES OF CHEMICAL REACTIONS
• Thousands of chemical reactions occur in
nature. Reactions can be classified into one of
several categories. Classification is important
for predicting products of reactions.
– 1. SYNTHESIS REACTIONS
– 2. DECOMPOSITION REACTIONS
– 3. SINGLE REPLACEMENT REACTIONS
– 4. DOUBLE REPLACEMENT REACTIONS
– 5. COMBUSTION
1. SYNTHESIS REACTIONS
• Two or more substances combine to form a
new compound (composition or direct
combination).
• Examples:
• A + X → AX
• 2 Mg +O2 → 2 MgO
• 2 CO + O2 → 2 CO2 (car exhaust)
• CO2 + H2O  H2CO3 (carbonic acid)
2. DECOMPOSITION REACTIONS
• A single compound undergoes a reaction that
produces two or more simpler substances
(breakdown).
• Examples:
• AX → A + X
• 2 H2O2(l) → 2 H2O(l) +O2(g)
• 2 H2O(l) → 2 H2(g) +O2(g)
– Electrolysis- the decomposition of a substance by
electric current.
Electrolysis of Water
3. SINGLE REPLACEMENT REACTIONS
• One element replaces a similar element in a
compound (displacement).
• A + BX → B + AX
• Mg(s) + 2HCl(aq) → MgCl2(aq) + H2 ↑
• 2Al(s)+3Pb(NO3)2(aq) → 2Al(NO3)3(aq)+3Pb(s)
• This type of reaction occurs only under certain
conditions!
What does “certain conditions” mean?
• A + BX → B + AX
• A (metal or nonmetal)
• B (metal)
• X (nonmetal)
Using the Activity Series
(Table J)
• Some elements are considered to be very
reactive, some are considered unreactive. The
ability of an element to react is called the
element’s “activity.”
• Higher metals are more reactive than lower
metals. A metal will only replace another
metal in a compound if it is HIGHER on Table J.
Using the Activity Series
• Metals: the higher the activity, the greater its
tendency is to lose electrons and therefore more
reactive.
• Nonmetals: the higher the reactivity, the greater
its tendency to gain electrons and therefore more
reactive.
• The activity series is used to predict whether or
not a chemical reaction will occur (single
replacement reactions).
• If it is lower no reaction (NR) will occur!
Using the Activity Series
• Single Replacement reactions only occur when
element A is more active than the element
being replaced.
• Element A must be more active on the Activity
Series (Table J) than the element being
replaced!!
• Metals replace metals, nonmetals replace
nonmetals.
METALS: Single-replacement
• If element A lies further up on the activity
chart (Table J) than element B, element A is
more active, therefore the reaction will occur.
• For example:
• Al + CuCl2 → AlCl3 + Cu
• A BX
AX B
• Aluminum is more active than copper, it lies
further up on the chart, therefore the reaction
will occur!
In each of the following, circle which
metal is more active:
• 1. Li or Al
• 2. Au or Mg
• 3. Na or Pb
*notice
precious
metals at the
bottom =
less reactive
and more
valuable
In each of the following, predict whether
there will be a reaction, then balance the
reactions:
• Mg + CuSO4 →
• Al + LiCl →
• Sn + Ba3PO4 →
NONMETALS: Single-replacement
• If element A is a stronger nonmetal than
element X (nonmetals replace nonmetals)
then Element A is more active and the
reaction will occur!
• For example:
• F2 + 2LiBr → Br2 + 2LiF
• A
BX
X BA
• Fluorine is more active than Bromine,
therefore the reaction will occur!!
In each of the following, circle which
nonmetal is more active:
• 4.
F2 or I2
• 5.
Br2 or H2
• 6.
Cl2 or I2
In each of the following, predict whether
there will be a reaction, then balance the
reactions:
• I2 + NaF →
• F2 + MgBr2 →
• Cl2 + BaH2 →
Rules for completing and writing single
replacement reactions.
1. Determine whether the reaction will occur or not
Li + CuSO4 →
Yes a reaction will occur (Lithium is higher up on the chart)
2. Determine the products of the reaction by switching the
appropriate elements.
Li +CuSO4 → LiSO4 + Cu
3. Place the charges on the elements and balance each of the
new compounds formed.
Li + CuSO4 → Li2(SO4) + Cu
4. Balance the entire equation.
2 Li + CuSO4 → Li2(SO4) + Cu
4. DOUBLE REPLACEMENT REACTIONS
• The ions of 2 compounds exchange places in an
AQUEOUS solution to form 2 new compounds.
• Examples:
• AB + CD → AD + CB
• Pb(NO3)2(aq) +KI(aq) → KNO3(aq) + PbI2(s)
• Double replacement reactions do not always
occur.
• We use the term No Reaction (NR) when no
significant change occurs in a reaction.
Conditions when Double Replacement
Reactions occur
• 1. If the product is insoluble. (A precipitate is
formed) ↓
– Insoluble: cannot be dissolved in water
– Precipitate: an insoluble product (solid)
• 2. If water is a product
• 3. If the product decomposes into a gas. ↑
– H2CO3 →
– H2SO3 →
– NH4OH →
H2O + CO2 (g)
H2O + SO2 (g)
H2O + NH3 (g)
PRODUCT IS INSOLUBLE
• Use the Solubility Chart (Table F) on your
reference table to figure out the solubility for
the problems on the next slide:
• Soluble = dissolvable in water, therefore it is
aqueous (aq)
• Insoluble = cannot dissolve in water, therefore
it forms a precipitate (s)
Solubility Chart (Table F)
Determine whether the compound is
soluble or insoluble
a. NH4OH
b. PbCl2
c. BaCO3
d. Al(OH)3
e. Na2S
f. K2SO4
g. CaSO4
h. Mg3(PO4)2
i. Zn(OH)2
Now complete each of the following double replacement
equations. For each of the products use Table F to assign a
notation of (s) or (aq) below each formula.
1. BaCl2 (aq) + Na2CO3 (aq) →
2. Zn(NO3)2 (aq) + KOH (aq) →
3. CaI2 (aq) + Na3PO4 (aq) →
4. CuSO4 (aq) + (NH4)2CO3 (aq) →
5. COMBUSTION REACTIONS
• A substance combines with oxygen (O2) to
release large amounts of energy in the form of
light and heat.
• CH4 +O2 → CO2 +2H2O
http://www.chem.uiuc.edu/clcwebsite/meth.html
Type of reaction
Definition
Example
Synthesis
Two or more substances
combine to form a new
compound.
2Mg + O2→ 2MgO
Decomposition
A single compound undergoes a
reaction that produces two or
more simpler substances
(breakdown).
H2O2(l) → 2H2O(l) + O2 (g)
Single Replacement
One element replaces a similar
element in a compound.
Mg (s) + HCl (aq) → MgCl2 (aq) + H2 (g)
Double Replacement
The ions of 2 compounds
exchange places in an AQUEOUS
solution to form 2 new
compounds.
The metals in the compounds
switch places.
2 Li + CuSO4 → Li2(SO4) + Cu
A substance combines with
oxygen (O2) to release large
amounts of energy in the form
of light and heat.
CH4 +O2 → CO2 +2H2O
Combustion Reaction
When will the reaction take place
•Single Replacement reactions only take
occur when one element is more active than
the element being replaced.
•Use Table J Activity Series
•Metals replace metals in a compound.
•Nonmetals replace nonmetals in a
compound.
1. If the product is insoluble. (A precipitate is
formed) ↓
2. If water is a product
3. If the product decomposes into a gas. ↑
Lab
• Balancing Reactions web lab
Lab
• Single Replacement lab
Lab
• Double Replacement lab
Quiz
• Quiz on Balancing and Types of Reactions
Chemical formulas and equations
Do Now
• What is a mole in chemistry?
• What was Avogadro's number? What did it
represent?
• What is molar mass?
Do Now
• What is a mole in chemistry?
• Remember: Mole (mol) = the SI unit for amount
• What was Avogadro's number? What did it
represent?
• Remember: Avogadro's number = the number of
particles in a mole = 6.02 x 1023
• What is molar mass?
• Remember: Molar mass = mass in grams of one
mole of an element or compound
Remember
• You can convert from moles to particles, or
moles to mass, and vice versa.
• Ex: How many particles do you have if you
have 2.5 mol of sulfur?
• Ex: How many grams of carbon do you have if
you have 2.44 x 1022 atoms?
* You must use conversion factors to convert from number of atoms
to moles and then to grams.
Formula Mass
• The molar mass is numerically equal to the
atomic mass of monatomic elements and the
formula mass of compounds and diatomic
elements.
• Ex: Find the formula mass of KBr
• Ex: Find the formula mass of H2O
Chemical Formula
• Chemical formulas indicate the relative
number of atoms of each kind in a chemical
compound.
• Identify the number of atoms in each of the
following compounds:
1. KCl= _____________________
2. C6H12O6=__________________
3. NH3=__________________
Chemical Formulas
• Formulas for covalent compounds show both
the elements and the number of atoms of
each element in a molecule.
• Formulas for ionic compounds do not show
numbers of atoms, but show the simplest
ratio of cations and anions.
Chemical Formulas
• The meaning of formulas do not change when
polyatomic ions are involved.
• Polyatomic ions = a group of covalently bond
elements that behave as a single ion
Finding the number of atoms in a
compound with polyatomic ions
Compound
1. MgCO3
2. Al(ClO4)3
What does it look like
expanded?
Numbers of
atoms
Compound
3. Zn3(PO4)2
4. (NH4)2S
What does it look like
expanded?
Numbers of
atoms
Molar Mass
• Formulas can be used to calculate molar mass
• Ex: ZnCl2
• Ex: ZnSO4
Do Now
• Do you see a pattern in the following
formulas?
NH4NO2
NH2O
Law of Definite Proportions
• Law of Definite Proportions states that every
pure substance always contains the same
elements combined in the same proportions
by weight.
• For example: H2O, will always have the same
percent by weight (11.2% H and 88.8% O)
Percentage Composition
• Percentage composition = the percentage by
mass of each element in a compound
• Percentage composition helps verify a
substance’s identity and can be used to
compare the ratio of masses contributed by
the elements in two substances
• Ex: Fe2O3
FeO
69.9% Fe
77.7% Fe
30.1% O
22.3% O
Review Formulas
• Molecular Formula = represents the number
and kind of atoms in a molecule (not
necessarily the smallest whole number ratio).
• Structural Formula = indicates twodimensional arrangement of the bonds and
lone pairs of electrons in a molecule.
Empirical Formulas
• Empirical formula = a chemical formula that
shows the composition of a compound in
terms of the relative numbers and kinds of
atoms in the simplest ratio.
Empirical Formula for Molecules (have
covalent bonds)
• The chemical formula that represents the simplest (lowest) atomic ratio in
which elements can combine.
• Formulas for molecular compounds are NOT NECESSARILY the empirical
formula
Molecular Formula
Empirical Formula
C2H6
C6H12O6
C4H8
C3H8
C6H10
C4H6
C2H4
C5H10
Empirical Formula for Ionic Compounds
(Always written in empirical form)
• The chemical formula that represents the simplest (lowest) atomic ratio in
which elements can combine.
• Formulas for ionic compounds are the empirical formula because they are
ALWAYS written in the expressed as the lowest possible ratios.
Empirical Form
Ca2O2
Fe2O2
Pb2O4
Mg2O2
Empirical Formula for Ionic Compounds
with Polyatomic Ions
Empirical Form
Fe2(CO3)2
Fe2(HSO4)2
Zn2(SO4)2
Pb2(SO3)2
Empirical or Molecular
Formula
C6H12O6
LiNO3
H2O2
C11H22O11
If written in molecular
formula what is the
empirical formula?
Type of Bond
Finding Empirical Formulas
• You can find the empirical formula from
percentage composition!
– Step 1: Convert to mass in grams (assume you have
100g of the given substance)
– Step 2: Convert from grams to moles (using the molar
mass conversion factor)
– Step 3: Reduce the molar ratio to the simplest wholenumber ratio by dividing by the smaller amount
– Step 4: Round to whole numbers and insert subscripts
Finding Empirical Formulas
• Ex: A given liquid has 60.0% C, 13.4% H, and
26.6% O by mass. Calculate the empirical
formula.
– Step 1: Convert to mass in grams (assume you
have 100g of the given substance)
60.0% C x 100g = 60.0g C
13.4% H x 100g = 13.4g H
26.6% O x 100g = 26.6g O
Finding Empirical Formulas
– Step 2: Convert from grams to moles (using the molar
mass conversion factor)
60.0g C x 1mol = 5.00mol C
12.01g C
13.4g H x 1mol = 13.3mol H
1.01g H
26.6g O x 1mol = 1.66mol O
16.00g O
C5H13.3O1.66 ?
Finding Empirical Formulas
– Step 3: Reduce the molar ratio to the simplest wholenumber ratio by dividing by the smaller amount
C5H13.3O1.66 ?
5.00mol C / 1.66 = 3.01mol C
13.3mol H / 1.66 = 8.01mol H
1.66mol O / 1.66 = 1.00mol O
– Step 4: Round to whole numbers and insert subscripts
C3H8O
Practice
• What is the empirical formula of a compound
that is 78.6% B and 21.4% H?
Practice
• What is the empirical formula of a compound
containing 32.38% Na, 22.65% S, and 44.99%
O?
Practice
• What is the empirical formula of a compound
containing 26.56% potassium, 35.41%
chromium, and the remainder is oxygen?
Practice
• In a 10.150g sample, 4.433g are phosphorous
and the rest is oxygen. What is the empirical
formula for this compound?
Practice
• A compound contains 0.606g nitrogen and
1.390g oxygen. What is the empirical formula
of the compound?
What did you learn today?
What did you learn today?
• A compound is a substance composed of two or more different elements
that are chemically combined in a fixed proportion. A chemical compound
can be broken down by chemical means. A chemical compound can be
represented by a specific chemical formula.
• Types of chemical formulas include empirical, molecular, and structural.
• The empirical formula of a compound is the simplest whole-number ratio
of atoms of the elements in a compound. It may be different from the
molecular formula, which is the actual ratio of atoms in a molecule of that
compound.
Do Now
• What was the difference between the
empirical formula and molecular formula?
Do Now
• What was the difference between the
empirical formula and molecular formula?
• Empirical formula = smallest possible whole
number ratio of elements
• Molecular formula = the actual formula of a
molecule
Molecular Formulas
• For ionic compounds, the molecular formula is
the same as the empirical formula.
• For molecular compounds, the molecular
formula is a whole number multiple of the
empirical formula.
*Both formulas are just different ways of representing the
composition of the same molecule.
Molar Mass
• In order to determine the molecular formula,
you must know the molecular mass!
• The molar mass of a compound is equal to the
molar mass of the empirical formula time a
whole number, n.
n(empirical formula) = molecular formula
Molecular Formula Examples
• Formaldehyde, acetic acid, and glucose each
have the same empirical formula, CH2O
– For formaldehyde, n = 1
– For acetic acid, n = 2
– For glucose, n = 6
Determining Molecular Formula
• Step 1: Find the molar mass of the empirical
formula using the molar masses of the elements
from the periodic table
• Step 2: Solve for n, the factor multiplying the
empirical formula to get the molecular formula
n = experimental molar mass of compound
molar mass of the empirical formula
• Step 3: Multiply the empirical formula by this
factor to get the molecular formula
Determining Molecular Formula
• The empirical formula of a compound is P2O5.
The experimental molar mass is 284 g/mol.
Determine the molecular formula.
– Step 1: Find the molar mass of the empirical
formula using the molar masses of the elements
from the periodic table
2 x molar mass of P = 2(30.97) = 61.94g/mol
5 x molar mass of O = 5(16.00) = 80.00g/mol
molar mass of P2O5 = 141.94g/mol
Determining Molecular Formula
• The empirical formula of a compound is P2O5. The
experimental molar mass is 284 g/mol.
Determine the molecular formula.
– Step 2: Solve for n, the factor multiplying the empirical
formula to get the molecular formula
n = experimental molar mass
molar mass of empirical formula
n = 284 g/mol
141.94 g/mol
n=2
Determining Molecular Formula
• The empirical formula of a compound is P2O5.
The experimental molar mass is 284 g/mol.
Determine the molecular formula.
– Step 3: Multiply the empirical formula by this
factor to get the molecular formula
n(empirical formula) = 2(P2O5) = P4O10
Determining Molecular Formula
• You can verify your answer by finding the
molar mass of the molecular formula and
compare it to the experimental molar mass
4 x molar mass of P = 4(30.97g/mol) = 123.88g/mol
10 x molar mass of O = 10(16.00g/mol) = 160.0g/mol
molar mass of P4O10 = 283.88g/mol
Practice
• What is the molecular formula of a compound
with the empirical formula BH3 and molecular
mass of 28g?
Practice
• What is the molecular formula of a compound
with a molecular mass of 34g that consists of
0.44g H and 6.92g O?
Practice
• What is the empirical formula of a compound
that contains 65.5% carbon, 5.5% hydrogen,
and 29.0% oxygen? What is the molecular
formula if the molecular mass is 110g?
Do Now
• What does percentage mean?
• What does a percentage represent?
Review
• Chemical formulas allow scientists to calculate
a number of characteristics values for a given
compound.
• Chemical formulas represent the number and
kind of atoms in a molecule.
• If you know the chemical formula, then you
can calculate the percentage composition.
Review
• Formula mass = the sum of the average
atomic masses of all the atoms represented in
the formula
• Formula mass is numerically equal to the
molar mass or gram formula mass
• Ex: Find the formula mass of KClO3
• Ex: What is the molar mass of Ba(NO3)2?
Percent Composition
• Percentage composition = the percentage by
mass of each element in a compound
Finding Percent Composition
• From the subscripts, you can determine the
mass contributed by each element and add
these to get the molar mass.
• Divide the mass of each element by the molar
mass.
• Multiply by 100 to find the percentage
composition of that element
Finding Percent Composition
• CO2
1mol x 12.01 g/mol = 12.01 g C
+2mol x 16.00 g/mol = 32.00 g O
mass of 1mol CO2 = 44.01 g
Finding Percent Composition
% C = 12.01 g C x 100 = 27.29%
44.01 g CO2
% O = 32.00 g O x 100 = 72.71%
44.01 g CO2
Finding Percent Composition
• CO
1mol x 12.01 g/mol = 12.01 g C
+1mol x 16.00 g/mol = 16.00 g O
mass of 1mol CO = 28.01 g
Finding Percent Composition
% C = 12.01 g C x 100 = 42.88%
28.01 g CO
% O = 16.00 g O x 100 = 57.71%
28.01 g CO
Percent Composition
• Percentage composition helps verify a
substance’s identity and can be used to
compare the ratio of masses contributed by
the elements in two substances
• Ex: Fe2O3
FeO
69.9% Fe
77.7% Fe
30.1% O
22.3% O
Practice
• Calculate the percentage composition of Cu2S,
a copper ore called chalcocite.
Practice
• Calculate the percent of both elements in
sulfur dioxide
Practice
• Calculate the percentage composition of
ammonium nitrate, NH4NO3
Hydrates
• Hydrates = salts that have crystallized from
water solution
• In the process of crystallization, the water
molecules bind to the salt to form hydrates.
• Hydrates are represented as follows:
Na2CO3 ● 10H2O
(sodium carbonate decahydrate)
Practice
• Calculate the molar mass of
Na2CO3 ● 10H2O
Practice
• Calculate the percent composition of water in
Na2CO3 ● 10H2O
What did you learn today?
What did you learn today?
• The formula mass of a substance is the sum of the
atomic masses of its atoms. The molar mass (gram
formula mass) of a substance equals one mole of
that substance.
• The percent composition by mass of each element in
a compound can be calculated mathematically.
Lab
• Percent Composition of Sugar in Gum
Lab
• Hydrate lab
Quiz
• Quiz on Formulas and Percent Composition
Stoichiometry
Do Now
• A recipe calls for one cup of milk and three
eggs per serving. You quadruple the recipe
because you are expecting guests. How much
milk and eggs do you need?
Stoichiometry
• Stoichiometry = the proportional relationship
between two or more substances during a
chemical reaction.
• Reaction stoichiometry = problems involving
the amount of products in relation to the
amount of reactants
– In order to correctly solve stoichiometry
problems, you MUST be able to correctly balance
equations!
• Stoichiometry problems are solved by using
ratios from the balanced equation.
Balanced Equations
• Balanced equations show proportions
• A balanced chemical equation is very similar
to a recipe in that the coefficients show the
proportions of the reactants and products
involved in the reaction.
– For example: 2 H2 + O2  2 H2O
So, 2 moles of hydrogen react with 1 mole of
oxygen to produce 2 moles of water.
Mole Ratio
• You can use mole ratios to determine how
much of a reactant is needed to produce a
quantity of product, and vice versa.
• Since the coefficients in a balanced equation
show the relative numbers of moles of the
substances in a reaction, ALWAYS convert
between amounts in moles!
– For example:
Al2O3 (l) 
Al (s) +
O2 (g)
Moles to Moles
• Steps to converting between amounts in
moles:
1. Identify the amount in moles that you know
from the problem.
2. Using coefficients from the balanced equation,
set up the mole ratio with the known substance
on the bottom and the unknown substance on
top.
3. Multiply the original amount by the mole ratio.
Moles to Moles Setup
Amount of known moles x unknown moles = unknown moles
known moles
Moles to Moles Example
• How many moles of hydrogen are needed to
prepare 312 moles of ammonia?
N2 + 3H2  2NH3
Moles to Moles Example
• How many moles of hydrogen are needed to
prepare 312 moles of ammonia?
N2 + 3H2  2NH3
• Amount of NH3 = 312 mol
• Amount of H2 = unknown
• From the equation: 3 mol H2 = 2 mol NH3
Moles to Moles Example
• How many moles of hydrogen are needed to
prepare 312 moles of ammonia?
N2 + 3H2  2NH3
• 312 mol NH3 x 3 mol H2 = ?
2 mol NH3
= 468 mol H2
Mole to Mole Problems
Al2O3 (l) 
Al (s) +
O2 (g)
• How many moles of Al(s) will be produced from
the decomposition of 13.0 mol of Al2O3(l)?
Mole to Mole Problems
Al2O3 (l) 
Al (s) +
O2 (g)
• How many moles of O2(g) will be produced
when 36 mol of Al(s) are produced from the
decomposition of Al2O3(l)?
Mole to Mole Problems
Al2O3 (l) 
Al (s) +
O2 (g)
• How many moles of O2(g) are produced when
15 mol of Al2O3(l) react?
Problems
• Complete the additional problems in your
notes packet…
Do now
• How would you get from Van Cortlandt park
to Flushing Meadows park?
Do Now
• Take out your Reference Table and turn to the
back page!
• List two conversion factors that relate to the
mole.
Remember to Use Table T
Mass Calculations
• Substances are usually measured by mass, so
before using the mole ratio, you will need to
convert from mass to moles!
• The conversion factor for converting between
mass and moles is the molar mass of the
substance.
• Molar mass = the sum of the atomic masses of
the elements using the chemical formula
The Mole Map
Mole Triangle
Multiply by
atomic/molar
mass from
periodic table
Atoms or Molecules A
Divide by 6.02 X 1023
Multiply by 6.02 X 1023
Divide by
atomic/molar mass
from periodic table
1 mole A
↔
Use ratio
of moles
1 mole
B
Multiply by 6.02 X 1023
Atoms or
Molecules B
Divide by 6.02 X
1023
Mass Grams A
Divide by
atomic/molar
mass from
periodic table
Multiply by
atomic/molar mass
from periodic table
Mass
(grams) B
Mole-Mass Conversions
• Most of the time in chemistry, the amounts
are given in grams instead of moles
• We still go through moles and use the mole
ratio, but now we also use molar mass to get to
grams
• Example: How many grams of chlorine are
required to react completely with 5.00 moles of
sodium to produce sodium chloride?
2 Na + Cl2  2 NaCl
5.00 moles Na
mol Cl2
mol Na
g Cl2
1 mol Cl2
=
g Cl2
Practice
• Calculate the mass in grams of Iodine
required to react completely with 0.50
moles of aluminum.
Mass-Mole
• We can also start with mass and convert to
moles of product or another reactant
• We use molar mass and the mole ratio to
get to moles of the compound of interest
– Calculate the number of moles of ethane (C2H6)
needed to produce 10.0 g of water
– 2 C2H6 + 7 O2  4 CO2 + 6 H20
10.0 g H2O
mol H2O
g H2O
mol C2H6 =
mol H20
mol C2H6
Practice
• Calculate how many moles of oxygen are
required to make 10.0 g of aluminum
oxide
Mass to Mass
• Steps to converting between amounts in
mass:
1. Convert you given amount from mass to
moles
2. Use the mole ratio to determine moles of
the unknown substance (from the balanced
equation)
3. Convert the moles calculated into the units
required to solve the problem (grams)
Mass to Mass Setup
mass of known x 1 mol = mol of known
molar mass
mol of known x mol of unknown = mol of unknown
mol of known
mol of unknown x molar mass = grams of unknown
1 mol
Mass to Mass Example
• N2 + 3H2  2NH3
• What mass of NH3 can be made from 21 g H2
and excess N2?
Mass to Mass Example
• N2 + 3H2  2NH3
• What mass of NH3 can be made from 21 g H2
and excess N2?
• Given:
mass of H2 = 21 g
molar mass of H2 = 2.02 g/mol
mass of NH3 = unkwown
molar mass of NH3 = 17.04 g/mol
mole ratio: 3 mol H2 = 2 mol NH3
Mass to Mass Example
• N2 + 3H2  2NH3
• What mass of NH3 can be made from 21 g H2
and excess N2?
21 g H2 x 1mol H2 x 2mol NH3 x 17.04 g NH3 =
2.02 g H2 3mol H2 1 mol NH3
= 118 g NH3
Mass-Mass Conversions
• Most often we are given a starting mass
and want to find out the mass of a
product we will get (called theoretical
yield) or how much of another reactant
we need to completely react with it (no
leftover ingredients!)
• Now we must go from grams to moles,
mole ratio, and back to grams of
compound we are interested in
Mass-Mass Conversion
• Ex. Calculate how many grams of
ammonia are produced when you react
2.00g of nitrogen with excess hydrogen.
N2 + 3 H2  2 NH3
g N2
mol N2
mol NH3
g NH3
g N2
mol N2
mol NH3
=
g NH3
N2 + 3H2 → 2NH3
1. Convert grams to moles.
2. Solve as a mole-mole problem to get # moles of NH3.
3. Convert answer back to grams NH3.
Practice
• How many grams of calcium nitride are
produced when 2.00 g of calcium reacts
with an excess of nitrogen?
Mass to Mass Practice
• Fe2O3 + Al  Fe + Al2O3
• How many grams of Al are needed to
completely react with 135 g Fe2O3?
Mass to Mass Practice
• Fe2O3 + Al  Fe + Al2O3
• How many grams of Al2O3 can form when 23.6
g Al react with excess Fe2O3?
Mass to Mass Practice
• Fe2O3 + Al  Fe + Al2O3
• How many grams of Fe2O3 react with excess Al
to make 475 g Fe?
Mass to Mass Practice
• Fe2O3 + Al  Fe + Al2O3
• How many grams of Fe will form when 97.6 g
Al2O3 form?
Conversions
liters to moles
22.4 Liters of gas = 1 mole of gas
22.4 Liters gas
1 mole of gas
=
1 mole of gas
22.4 Liters gas
↔
22.4 Liters of gas
6.022x1023
(CO2) or
formula units (NaCl)
atoms (Ne) or molecules
↔
1 mole
Grams compound or
element
↔
(Found on Periodic
Table)
How many liters of O2 are present if you have 15 moles O2?
↔
22.4 Liters of gas
6.022x1023
(CO2) or
formula units (NaCl)
atoms (Ne) or molecules
↔
1 mole
Grams compound or
element
↔
(Found on Periodic
Table)
How many moles are there in 30 liters of O2 gas?
↔
22.4 Liters of gas
6.022x1023
(CO2) or
formula units (NaCl)
atoms (Ne) or molecules
↔
1 mole
Grams compound or
element
↔
(Found on Periodic
Table)
How many liters of H2O are present in 1.5 moles of H2O?
Do Now (Honors)
• A bicycle mechanic has 10 frames and 16
wheels in the shop. How many complete
bicycles can he assemble using these parts?
Limiting Reactants
• If you have the following available ingredients,
how many peanut butter and jelly sandwiches
can you make?
– 4 slices of bread
– 1 jar of peanut butter
– 1/2 jar of jelly
Definitions
• Limiting Reactant
• Excess Reactant
Limiting Reactants
1. Write a balanced equation.
2. For each reactant, calculate the
amount of product formed.
3. Smaller answer indicates the limiting
reactant
Limiting Reactants
• 79.1 g of zinc react with 0.90 L of
2.5M HCl. Identify the limiting
and excess reactants. How many
liters of hydrogen are formed at
STP?
Zn + 2HCl

ZnCl2 + H2
Limiting Reactants
How many liters of hydrogen are formed if we are given 79.1g of
zinc?
Zn + 2HCl 
79.1 g 0.90 L
2.5M
g Zn
ZnCl2 + H2
?L
mol
Zn
mol
H2
L H2
g Zn
mol
Zn
mol
H2
=
H2
L
Limiting Reactants
How many liters of hydrogen are formed when we are given 0.90
Liters of HCl?
Zn + 2HCl 
79.1 g 0.90 L
2.5M
L HCl
ZnCl2 + H2
?L
mol
HCl
mol
H2
1L
mol
HCl
L H2 =
mol
H2
L
H2
Limiting Reactant is the one that produces the
least amount of Hydrogen
Zn:____L H2
HCl:_____L H2
Limiting reactant:
Excess reactant:
Product Formed: L H2
left over ____
• Solid silicon dioxide (quartz) is usually quite
unreactive but reacts readily with hydrogen
fluoride gas to form silicon tetrafluoride gas and
liquid water. Write the balanced chemical
reaction for this. If 2.0mol of HF are exposed to
4.5mol of SiO2, which is the limiting reactant?
Limiting Reactants
• Limiting reactants are not present in excess
and they are entirely consumed in a reaction.
• First, you need to determine which reactant is
present in a limiting quantity by comparing
moles.
• Then, you can calculate the mole ratio.
• Lastly, you can determine the amount of
product produced.
Practice
• N2 + 3H2  2NH3
• If you have 2 moles of N2 and 2 moles of H2,
which of the reactants is the limiting agent?
• Mole ratio from problem: 2mol N2 = 1
2mol H2
• Mole ratio from equation: 1mol N2 = 0.33
3mol H2
Practice
Larger
• Mole ratio from problem: 2mol N2 = 1
Limiting reactant
2mol H2
• Mole ratio from equation: 1mol N2 = 0.33
3mol H2
• If the mole ratio from the problem is larger, then the
reactant in the numerator is present in excess.
• If the mole ratio from the problem is smaller, then
the reactant in the denominator is present in excess.
• The reactant not present in excess is the limiting
reactant.
Practice
• N2 + 3H2  2NH3
• If H2 is the limiting reactant, then all of it will
be consumed in the reaction.
• So, how much N2 will react with 2mol of H2?
2mol H2 x 1mol N2 = 0.67mol N2
3mol H2
Practice Problem
• C3H8 + 5O2  3CO2 + 4H20
• Determine the limiting reactant and the
amount of excess reactant that reacts if you
have 1.5mol C3H8 and 15mol O2.
Practice Problem
• C3H8 + 5O2  3CO2 + 4H20
• Mole ratio from problem: 1.5mol C3H8 = 0.1
Smaller
15mol O2
• Mole ratio from equation: 1mol C3H8 = 0.2
5mol O2
• O2 is present in excess
• C3H8 is the limiting reactant
Practice Problem
• C3H8 + 5O2  3CO2 + 4H20
1.5mol C3H8 x 5mol O2 = 7.5mol O2
1mol C3H8
Practice Problem
• 2H2 + O2  2H2O
• How much water is produced if 100g H2 and
160g O2 react?
• You must convert
mass  moles  moles  mass
Practice Problem
• 2H2 + O2  2H2O
• 100g H2 x 1mol H2
2g H2
= 50mol H2
• 160g O2 x 1mol O2
32g O2 = 5mol O2
• Mole ratio from problem: 50mol H2 =10
5mol O2
• Mole ratio from equation: 2mol H2 = 2
1mol O2
Limiting
reactant
Practice Problem
• 2H2 + O2  2H2O
5mol O2 x 2mol H2O x 18g H20 =
1mol O2
1mol H2O
= 180g H2O
Check
• 2H2 + O2  2H2O
• 10H2 + 5O2  10H2O
• Conservation of mass:
10H2 x 2g H2 = 20g H2
1mol H2
Sum of reactants
5O2 x 32g O2 = 160g O2 = Sum of products
1mol O2
10H2O x 18g H2O = 180g H2O
1mol H2O
Percent Yield
measured in lab
actual yield (g)
% yield 
 100
theoretical yield(g)
calculated on paper
Percent Yield
• When 45.8 g of K2CO3 react with
excess HCl, 46.3 g of KCl are
formed. Calculate the theoretical
and % yields of KCl.
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
?g
actual: 46.3 g
Percent Yield
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
?g
actual: 46.3 g
Theoretical Yield:
g
mol
K2CO3 K2CO3
mol
KCl g KCl
g
mol
K2CO3 K2CO3
=
mol g KCl
KCl
Percent Yield
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
actual: 46.3 g
Theoretical Yield =
% Yield =
46.3 g
g
g KCl
 100 =
%
• Methanol (CH3OH) can be produced through the
reaction of CO and H2 in the presence of a
catalyst.
CO(g) + 2H2(g) catalyst > CH3OH(l)
If 75.0g of CO reacts to produce 68.4g CH3OH,
what is the percent yield of methanol?
Quiz
• Quiz on Double Replacement Reactions and
Mole Calculations (Stoichiometry)
Lab
• Hydrate lab
Lab
• Stoichiometry lab
Lab
• Mole lab
Test
• Test on Chemical Formulas, Naming,
Reactions, and Moles/Stoichiometry