Review: Physical and chemical changes
Chemical
reactions
and
chemical
equations
Matter can undergo two types of changes -- physical and chemical
Physical changes are changes in the physical properties of a substance
(e.g., size, shape, density) or changes in the state of matter (solid,
liquid, gas) without an accompanying change in chemical composition
Examples:
• Melting ice (change from solid to liquid state)
• Boiling methanol (change from liquid to vapor state)
• Heating water (increase in volume, decrease in density)
• Hammering a gold nugget into a thin sheet of foil
(change in size, shape)
No new substances are formed in physical changes
Chemical equations
Review: Physical and chemical changes
Matter can undergo two types of changes -- physical and chemical
Chemical changes result in the formation of new substances that
have different properties and composition than the starting materials
In formal terms, a chemical change is referred to as a chemical reaction
chemical equation -- shorthand for expressing a chemical reaction
Example: Combustion of propane
• propane and oxygen are consumed
• carbon dioxide and water are produced
• energy is released as light and heat
Examples:
• Adding vinegar to baking soda (fizzing bubbles indicate acid-base
neutralization reaction)
Energy transitions associated with chemical reactions fall
under the subject of thermodynamics (more on this later)
• Heating a copper wire to form black residue on surface
(conversion of metallic copper to copper oxide)
• Using electricity to split water into hydrogen and oxygen gas
(electrolysis)
Review: Balanced chemical equations
In a balanced chemical equation, the total number of atoms of each
element must be the same on both sides of the equation
-- you can think of this as applying accounting principles to chemistry
Equation:
C3H8(g) + 5 O2(g)
propane
oxygen
Word
equation
propane + oxygen
carbon dioxide + water
Chemical
equation
C3H8 (g) + O2 (g)
CO2 (g) + H2O (g)
Conservation of mass
Mass (matter) can neither be created nor destroyed
-- it can only change from one form to another
3 CO2(g) + 4 H2O(g)
carbon
dioxide
water
Law of conservation of mass
No change is observed in the total mass of the substances
involved in a chemical change (reaction)
In any chemical reaction:
3 carbon atoms, 8 hydrogen atoms,
10 oxygen atoms
3 carbon atoms, 8 hydrogen atoms,
10 oxygen atoms
A balanced chemical equation is an expression of the Law of Conservation of Mass
Matter can not be created nor destroyed -- it can only shift from one form to another
In a chemical reaction, no atoms are created or destroyed
-- they are just recombined to form new substances
mass of reactants = mass of products
The “Mole”?
Sample problem
Consider the following reaction for the electrolysis of water:
2 H 2O
electricity
2 H2
+
What does a small rodent have to do
with chemistry?
O2
If 22.4 g of hydrogen and 177.6 g of oxygen are formed, how
many grams of water reacted?
Conservation of mass
Nothing, actually. When chemists talk about a “mole”, they are
actually referring to a certain number of objects…
mass of reactants = mass of products
mass of H2O = mass of H2 + mass of O2
mass of H2O = 22.4 g + 177.6 g
An extremely LARGE number of objects !
mass of H2O = 200.0 g
The mole is a counting unit
The mole is a counting unit
1 pair = 2 objects
1 pair = 2 objects
1 dozen = 12 objects
1 dozen = 12 objects
Four score and seven years ago…
1 mole = 6.022 x 1023 objects!
(Avogadro’s number)
Old fashioned
1 score = 20 objects
(4 x 20) + 7 = 87
Why is this number so huge?
Because the mole is a counting unit for atoms, and atoms are tiny
-- even a small amount of matter contains an enormous amount of atoms
Different counting units
1 pair of carbon atoms
= 2 carbon atoms
1 dozen carbon atoms
= 12 carbon atoms
Why did they pick 6.022 x 1023?
C
Avogadro’s number is defined as the number of atoms present in
exactly 12 grams of carbon-12 (the isotope of carbon containing 6
protons and 6 neutrons in its nucleus)
C
C
C
C
C
C
C
C
C
C
C
C
C
Mass number = 12 (6 + 6)
Atomic number = 6
carbon-12
6.022 x 1023 atoms
e
1 mol of carbon atoms
12
6
e
e
N
P N P
N
P
N
P
P N
N
P
= 6.022 x 1023 carbon atoms
e
e
e
C
(Avogadro’s number)
12.0000 g
A mole is defined as the amount of any substance
that contains Avogadro’s number (6.022 x 1023)
of atoms or molecules
Different substances require
different counting units
If someone asks you to go to the store for a dozen eggs…
-- no problem!
Different substances require
different counting units
On the other hand, if someone in the lab asks you to weigh out a
dozen iron atoms…
-- good luck!
1 iron atom = 9.27 x 10-23 g
But if someone asks you to go to the store for a mole of eggs…
-- you’d better bring a lot of trucks with you
• even if you could fit one million eggs in each truck, you would
still need over 600 quadrillion (1015) trucks!
12 iron atoms = 12 x (9.274 x 10-23 g) = 1.113 x 10-21 g
But if someone asks you to weigh out a mole of iron atoms…
-- no problem!
1 mole of iron atoms = (6.022 x 1023) x (9.274 x 10-23 g)
= 55.85 g
(easily weighed using a standard lab balance)
One mole of a monoatomic substance (only one atom)
contains 6.022 x 1023 atoms of the substance
One mole of a molecular substance (more than one atom)
contains 6.022 x 1023 molecules of the substance
How many carbon atoms in one mole of carbon?
1 mol x
6.022 x 1023 atoms
6.022 x 1023 atoms
=
1 mol
Compound
Example: Water (H2O)
H
How many carbon atoms in 4.00 moles of carbon?
4.00 mol x
6.022 x 1023 atoms
2.41 x 1024 atoms
=
1 mol
6.022 x 1023 molecules
6.022 x 1023 molecules
=
1 mol
6.022 x 1023 atoms
=
1 mol
1.81 x 1023 atoms
Remember that 1 mole = 6.022 x 1023 objects
(i.e. pay attention to how the question is asked!)
Compound
Example: Water (H2O)
= 6.022 x 1023 molecules
1 mol x
6.022 x 1023 molecules
= 6.022 x 1023 molecules
1 mol
3 atoms
x
H
How many hydrogen atoms in one mole of water?
1 mol
6.022 x 1023 molecules
O
H
H
How many atoms (total) in one mole of water?
6.022 x 1023 molecules
Remember that 1 mole = 6.022 x 1023 objects
(i.e. pay attention to how the question is asked!)
Compound
Example: Water (H2O)
O
H
1 mol x
H
How many water molecules in one mole of water?
1 mol x
How many carbon atoms in 0.300 moles of carbon?
0.300 mol x
O
molecule
= 1.807 x 1024 atoms
2 H atoms
6.022 x 1023 molecules
x
molecule
= 1.204 x 1024 H atoms
Remember that 1 mole = 6.022 x 1023 objects
(i.e. pay attention to how the question is asked!)
Mole questions
How may atoms in one mole of iron?
Compound
Example: Water (H2O)
O
6.022 x 1023
How many atoms in one mole of argon?
H
H
6.022 x 1023
How many molecules in one mole of water (H2O)?
How many oxygen atoms in one mole of water?
6.022 x 1023 molecules
1 mol x
6.022 x 1023
How many atoms in one mole of water (H2O)?
= 6.022 x 1023 molecules
1 mol
3 x (6.022 x 1023) = 1.807 x 1024
How many molecules in one mole of sulfuric acid (H2SO4)?
6.022 x 1023
1 O atom
6.022 x 1023 molecules
x
= 6.022 x
1023
O atoms
molecule
How many atoms in one mole of sulfuric acid (H2SO4)?
7 x (6.022 x 1023) = 4.215 x 1024
Atomic mass units and relative atomic mass
Atomic mass and molar mass
The atomic mass of an element is the mass of a single atom of that element
expressed in atomic mass units (amu)
By definition:
The mass of one carbon-12 atom is exactly 12 atomic mass units (amu)
• 1 amu = 1/12 the mass of one carbon-12 atom
Remember that Avogadro’s number is defined as the number of atoms
present in exactly 12 grams of carbon-12
12.000 g carbon-12
6.022 x
1023
atoms carbon-12
1 amu =
= 1.9927 x 10-23 g per atom of carbon-12
Example: Carbon
relative atomic mass = 12.01 amu
atomic
number
element
symbol
1
mass of one
12 carbon-12 atom
=
1 amu =
1
12
6
C
12.01
This is the mass of one carbon atom
(average mass weighted for isotopic abundance)
1.9927 x 10-23 g
1.6606 x 10-24 g
Atomic mass and molar mass
The molar mass of an element is the mass of 1 mole of that element
(6.022 x 1023 atoms) expressed in grams
• the molar mass has the same value in grams as the atomic mass in amu
Sample problems
What is the mass of 3.50 moles of sodium (Na)?
Change moles of Na to grams of Na
Molar mass of Na = 22.99 g / mol
12.01 amu
6.022 x 1023 atoms C
1.6606 x 10-24 g
1 atom C
1 mol C
1 amu
Example: Carbon
atomic
number
element
symbol
= 12.01 g / mol C
molar mass = 12.01 g
6
C
12.01
This is the mass of one mole of carbon
(average mass weighted for isotopic abundance)
3.50 moles Na
x
22.99 g Na
1 mol Na
=
80.5 g Na
Sample problems
Sample problems
How many magnesium atoms are contained in 5.00 g of Mg?
How many moles of iron (Fe) does 25.0 g of Fe represent?
Change grams of Fe to moles of Fe
Change grams of Mg to moles Mg
Then change moles Mg to atoms of Mg
Molar mass of Fe = 55.85 g / mol
Molar mass of Mg = 24.31 g / mol
1 mol Mg
5.00 g Mg x
1 mol Fe
25.0 g Fe x
=
=
0.206 mol Mg
24.31 g Mg
0.448 mol Fe
55.85 g Fe
6.022 x 1023 atoms Mg
= 1.24 x 1023 atoms Mg
0.206 mol Mg x
1 mol Mg
Calculating the molar mass of a compound
H H
H H
H H
H H
H H
H
O
H
H
O
H
O
O
O
H
H H
H
H H
H H
H H
H H
H H
H
H
1 mole O
=
16.00 g
1 mole H2O
=
18.016 g
2.016 g
18.02 g
Molar mass of H2O
18.02 g / mol
Molar mass of compounds
What is the mass of 1 mol of diatomic oxygen (O2)?
Formula of sulfuric acid: H2SO4
2 moles H atoms
1 mole S atoms
4 moles O atoms
# of atoms
H
2
molar mass
1.008 g
Formula of diatomic oxygen: O2
Remember that 1 molecule of O2 contains 2 oxygen atoms
So 1 mole of O2 molecules contains 2 moles of oxygen atoms
total mass
2.016 g
S
1
32.07 g
32.07 g
O
4
16.00 g
64.00 g
element
# of atoms
O
2
molar mass
16.00 g
Molar mass of diatomic oxygen =
Molar mass of sulfuric acid =
O
H
=
1 mol H
6.022 x 1023 H atoms
mass = 1.008 g
What is the mass of 1 mol of sulfuric acid?
element
O
H
2 moles H
2 x (1.008 g)
Molar mass of compounds
=
O
O
H
11.5 g Na
1 mol Na
1 mole of H2SO4
O
O
H
mass = 16.00 g
22.99 g Na
98.09 g
O
O
H
H
1 mol O
6.022 x 1023 O atoms
= 0.500 mol Na
=
O
+
6.022 x 1023 atoms Na
0.500 mol Na x
O
H
3.01 x 1023 atoms Na x
O
H
H
O
1 mol Na
mass = 1.008 g
=
O
H
H
1 mol H2O
6.022 x 1023 H2O molecules
O
H
O
1 mol H
6.022 x 1023 H atoms
Molar mass of Na = 22.99 g / mol
O
O
O
H
Change atoms of Na to moles of Na
Then change moles of Na to grams of Na
H H
What is the mass of 3.01 x 1023 atoms of sodium (Na)?
H
Sample problems
total mass
32.00 g
32.00 g
Stoichiometry
STOICHIOMETRY
Quantitative
relationships
between
reactants and
products in
chemical
reactions
stoichiometry -- the quantitative relationships between
reactants and products in a chemical reaction
You can think of stoichiometry as a kind of accounting
system for chemistry
• empirical formulas and mass percent compositions
of compounds
• moles of substances consumed and produced in
a chemical reaction
• masses of substances consumed and produced
in a chemical reaction
• limiting reactants
• theoretical yields
Percent composition of compounds
Percent composition of compounds
The percent composition of a compound refers to the mass percent
of each of the elements in the compound
The percent composition of a compound can be determined from
experimental data without knowing the formula of the compound
H H H H H H H H H
H H H H H H H H H H
O
1 mol H
O
O
O
O
O
O
O
O
O
O
O
O
H H H H H H H H H
H
H
H
H
H
H
O
H
O
H
H
O
H
H H H H H H H H H H
18.02 g
O
6.022 x 1023 H2O molecules
+
O
H
1 mol H2O
O
mass = 1.008 g
=
O
O
6.022 x 1023 H atoms
1 mol O
6.022 x 1023 O atoms
mass = 16.00 g
1 mol H
Zn
1.63 g
6.022 x 1023 H atoms
mass = 1.008 g
H 2O
18.02 g (100%)
H
1.008 g (5.6%)
H
2.016 g (11.2%)
H
1.008 g (5.6%)
Example: When heated in air, 1.63 g of zinc combines with 0.40 g of
oxygen to form zinc oxide. Calculate the percent composition of the zinc
oxide product.
O
16.00 g (88.8%)
O
16.00 g (88.8%)
Empirical formula versus molecular formula
There are two types of formulas that can be used to describe a
compound: the empirical formula and the molecular formula
+
zinc oxide
(formula: ZnO? ZnO2? other?)
O2
0.40 g
mass of the reactants
2.03 g
=
mass of the products
Calculate the mass percent of each element in the compound
mass % of zinc = 100 x ( 1.63 g / 2.03 g ) = 80.3% Zn
mass % of oxygen = 100 x ( 0.40 g / 2.03 g ) = 20.% O
Example of empirical formula determination
When heated in air, 1.63 g of zinc combines with 0.40 g of oxygen to form
zinc oxide. What is the empirical formula of the zinc oxide product?
empirical formula (simplest formula) -- gives the smallest whole-number
ratio of atoms of each element in the compound
-- i.e., the relative number of atoms of each element in the compound
The word empirical has a similar meaning to the word experimental
-- empirical formulas are based on results from laboratory techniques
(known as quantitative analysis) that determine the amount of a given
element in a substance
Note: The empirical formula does not give any information about;
• the exact number of atoms present in a molecule of the compound
• the structural arrangement of the atoms in a molecule of the compound
Zn
1.63 g
+
O2
0.40 g
zinc oxide
(formula: ZnxOy)
Convert grams to moles
1.63 g Zn x ( 1 mol Zn / 65.39 g Zn ) = 0.0249 moles of Zn atoms
0.40 g O x ( 1 mol O / 16.00 g O ) = 0.025 moles of O atoms
Divide by smallest number of moles
Zinc: ( 0.0249 mol / 0.0249 mol ) = 1.0
Oxygen: ( 0.025 mol / 0.0249 mol ) = 1.0
Assign subscripts
Zn1O1
ZnO
H
Calculating empirical formulas
To calculate the empirical formula of a compound, you need to know
the following:
-- the elements that combine to form the compound
-- the molar masses for each of the elements
Calculating empirical formulas
Example: A compound is formed by combining 2.233 g of iron and 1.926
g of sulfur. What is the empirical formula of the compound?
Convert grams to moles
2.233 g Fe x ( 1 mol Fe / 55.85 g Fe ) = 0.03998 moles of Fe atoms
-- the mass percentages for each of the elements
1.926 g S x ( 1 mol S / 32.07 g S ) = 0.06006 moles of S atoms
1) Convert the mass of each element (grams) into moles
2) Divide the number of moles of each element by the smallest
number of moles
Divide by smallest number of moles (multiply by 2 to get whole numbers)
iron: ( 0.03998 mol / 0.03998 mol ) = 1.000
x 2 = 2.000
3) If the values from the preceding step are not whole numbers,
multiply them by the smallest factor that will result in whole
numbers
sulfur: ( 0.06006 mol / 0.03998 mol ) = 1.502
x 2 = 3.004
Assign subscripts to elements in the empirical formula
4) Use the whole numbers as subscripts for the elements in the
empirical formula
Fe2S3
Empirical formula versus molecular formula
Calculating molecular formulas
empirical formula (simplest formula) -- gives the relative number of atoms
of each element in the compound
Remember that the molecular formula of a compound will either be the
same as its empirical formula or a whole number multiple of its empirical
formula
molecular formula (true formula) -- gives the exact number of atoms of
each element present in one molecule of the compound
Compound
empirical formula
molecular formula
CH2O
CH2O
formaldehyde
acetic acid
CH2O
glucose
= CH2O
C2H4O2
CH2O
= (CH2O)2
C6H12O6 = (CH2O)6
Note: The molecular formula is either the same
as the empirical formula or a whole number
multiple of the molecular formula
Molecular formula = (empirical formula)n
Compound
water
empirical formula
H2O
[ n = 1, 2, 3, etc. ]
molecular formula
H 2O
acetylene
CH
C2H2
= (CH)2
benzene
CH
C6H6
= (CH)6
The molecular formula can be calculated from the empirical formula if the
molar mass of the compound is known
molecular mass of compound
Number of empirical formula units =
molecular mass of empirical formula
Calculating molecular formulas
Calculating molecular formulas
Example (trivial): We saw that the empirical formula for water is H2O. The
Example: A sample of polypropylene contains 14.3 g of hydrogen and
85.7 g of carbon. Polypropylene has a molar mass of 42.08 g/mol. What
is the molecular formula for polypropylene?
molecular mass of water is 18.02 amu. What is the molecular formula for
water?
Step 1: Calculate the empirical formula
Molecular formula = (empirical formula)n = (H2O)n
Convert grams to moles
mass of empirical formula = 2 x (molar mass of H) + molar mass of O
14.3 g H x ( 1 mol H / 1.008 g H ) = 14.2 moles of H atoms
!
!
!
= 2 x ( 1.008 amu ) + ( 16.00 amu )
85.7 g C x ( 1 mol C / 12.01 g C ) = 7.14 moles of C atoms
!
!
!
= 18.02 amu
Divide by smallest number of moles
n =
molecular mass of compound
molecular mass of empirical formula
=
18.02 amu
18.02 amu
= 1.000
hydrogen: ( 14.2 mol / 7.14 mol ) = 1.989 ≈ 2
carbon: ( 7.14 mol / 7.14 mol ) = 1.000 ≈ 1
Assign subscripts to elements in the empirical formula
Molecular formula = (H2O)1 =
H 2O
(same as empirical formula)
C 1H 2
CH2
Calculations from chemical equations
Calculating molecular formulas
Example: A sample of polypropylene contains 14.3 g of hydrogen and
85.7 g of carbon. Polypropylene has a molar mass of 42.08 g/mol. What
is the molecular formula for polypropylene?
If you know the amount of any reactant or product involved in the reaction:
•
you can calculate the amounts of all the other reactants and products
that are consumed or produced in the reaction
Step 2: Calculate the molecular formula
C3H8(g) + 5 O2(g)
molecular formula = (empirical formula)n = (CH2)n
mass of empirical formula = (molar mass of C) + 2 x molar mass of H
= 12.01 g/mol + ( 2 x 1.008 g/mol )
= 14.03 g/mol
n =
molar mass of compound
42.08 g
=
molar mass of empirical formula
molecular formula = (CH2)3 =
BUT REMEMBER!
The coefficients in a chemical equation provide information ONLY about
the proportions of MOLES of reactants and products
•
given the number of moles of a reactant/product involved in a
reaction, you CAN directly calculate the number of moles of other
reactants and products consumed or produced in the reaction
•
given the mass of a reactant/product involved in a reaction, you can
NOT directly calculate the mass of other reactants and products
consumed or produced in the reaction
= 2.999 ≈ 3
14.03 g
C 3H 6
Mole - mole calculations
Mole - mole calculations
Given: • A balanced chemical equation
• A known quantity of one of the reactants/product (in moles)
Calculate: The quantity of one of the other reactants/products (in moles)
Example:
How many moles of ammonia are produced from 8.00 mol of hydrogen
reacting with nitrogen?
Equation:
3 H2 + N2
2 NH3
Mole ratio between unknown
substance (ammonia) and
known substance (hydrogen):
Use ratio between coefficients
of substances A and B from
balanced equation
Moles of
substance A
3 CO2(g) + 4 H2O(g)
Moles of
substance B
( 8.00 moles H2 )
n
8.00 moles H2
2 moles NH3
3 moles H2
=
2 moles NH3
3 moles H2
( 8.00 moles H2 )
n = 5.33 moles NH3
Mole - mole calculations
Mole - mole calculations
Given the balanced equation:
K2Cr2O7 + 6 KI + 7 H2SO4
Given the balanced equation:
Cr2(SO4)3 + 4 K2SO4 + 3 I2 + 7 H2O
K2Cr2O7 + 6 KI + 7 H2SO4
Cr2(SO4)3 + 4 K2SO4 + 3 I2 + 7 H2O
Calculate:
Calculate:
a) The number of moles of potassium dichromate (K2Cr2O7) required
to react with 2.0 mol of potassium iodide (KI)
b) The number of moles of sulfuric acid (H2SO4 ) required to produce
2.0 moles of iodine (I2 )
Mole ratio between the unknown
substance (potassium dichromate) and
the known substance (potassium iodide):
( 2.0 mol KI )
n
2.0 mol Kl
=
Mole ratio between the unknown
substance (sulfuric acid) and the
known substance (iodine):
1 mol K2Cr2O7
6 mol Kl
1 mol K2Cr2O7
6 mol Kl
n = 0.33 mol K2Cr2O7
( 2.0 mol KI )
( 2.0 mol l2 )
n
2.0 mol l2
=
7 mol H2SO4
3 mol l2
7 mol H2SO4
3 mol l2
n = 4.7 mol H2SO4
( 2.0 mol l2 )
Mass - mass calculations
Given: • A balanced chemical equation
• A known mass of one of the reactants/product (in grams)
Calculate: The mass of one of the other reactants/products (in grams)
Mass - mass calculations
How many grams of nitric acid are required to produce 8.75 g of
dinitrogen monoxide (N2O)?
The balanced equation is:
4 Zn (s) + 10 HNO3 (aq)
Grams of
substance A
Grams of
substance B
Use molar mass
of substance A
Use molar mass
of substance B
Use ratio between coefficients
of substances A and B from
balanced equation
Moles of
substance A
Step 1: Convert the amount of known substance (N2O) from
grams to moles
Molar mass N2O: ( 2 x 14.01 g/mol ) + 16.00 g/mol = 44.02 g/mol
8.75 g N2O ( 1 mol N2O / 44.02 g N2O )
=
0.199 mol N2O
Moles of
substance B
Mass - mass calculations
Mass - mass calculations
How many grams of nitric acid are required to produce 8.75 g of
dinitrogen monoxide (N2O)?
The balanced equation is:
4 Zn (s) + 10 HNO3 (aq)
How many grams of nitric acid are required to produce 8.75 g of
dinitrogen monoxide (N2O)?
The balanced equation is:
4 Zn(NO3)2 (aq) + N2O (g) + 5 H2O (l)
Step 2: Determine the number of moles of the unknown substance (HNO3) required to
produce the number of moles of the known substance (0.199 mol N2O)
Mole ratio between the unknown
substance (nitric acid) and the known
substance (dinitrogen monoxide):
( 0.199 mol N2O )
4 Zn(NO3)2 (aq) + N2O (g) + 5 H2O (l)
n
0.199 mol N2O
=
10 mol HNO3
1 mol N2O
4 Zn(NO3)2 (aq) + N2O (g) + 5 H2O (l)
Step 3: Convert the amount of unknown substance (1.99 moles HNO3)
from moles to grams
Molar mass HNO3: 1.008 g/mol + 14.01 g/mol + ( 3 x 16.00 g/mol )
1 mol N2O
10 mol HNO3
4 Zn (s) + 10 HNO3 (aq)
= 63.02 g/mol
( 0.199 mol N2O )
1.99 mol HNO3 ( 63.02 g HNO3 / 1 mol HNO3 )
=
125 g HNO3
n = 1.99 mol HNO3
Mass - mass calculation: Another example
Mass - mass calculation: Another example
How many grams of carbon dioxide are produced by the complete
combustion of 100. g of pentane (C5H12)?
How many grams of carbon dioxide are produced by the complete
combustion of 100. g of pentane (C5H12)?
The balanced equation is:
The balanced equation is:
C5H12 (g)
+
8 O2(g)
5 CO2 (g)
+
6 H2O(g)
Step 1: Convert the amount of known substance (C5H12) from
grams to moles
Molar mass C5H12: ( 5 x 12.01 g/mol ) + ( 12 x 1.008 g/mol )
= 72.15 g/mol
100. g C5H12 ( 1 mol C5H12 / 72.15 g C5H12 ) = 1.39 mol C5H12
C5H12 (g)
+
8 O2(g)
5 CO2 (g)
+
6 H2O(g)
Step 2: Determine the number of moles of the unknown substance (CO2) required to
produce the number of moles of the known substance (1.39 mol C5H12)
Mole ratio between the unknown
substance (carbon dioxide) and the
known substance (pentane):
( 1.39 mol C5H12 )
n
1.39 mol C5H12
5 mol CO2
1 mol C5H12
=
5 mol CO2
1 mol C5H12
n = 6.95 mol CO2
( 1.39 mol C5H12 )
Mass - mass calculation: Another example
How many grams of carbon dioxide are produced by the complete
combustion of 100. g of pentane (C5H12)?
Mole - mass calculations
Given: • A balanced chemical equation
• A known quantity of one of the reactants/product (in moles)
The balanced equation is:
Calculate: The mass of one of the other reactants/products (in grams)
C5H12 (g)
+
8 O2(g)
5 CO2 (g)
+
6 H2O(g)
Grams of
substance B
Step 3: Convert the amount of unknown substance ( 6.95 moles CO2 )
from moles to grams
Molar mass CO2: 12.01 g/mol + ( 2 x 16.00 g/mol ) = 44.01 g/mol
n = 6.95 mol CO2 ( 44.01 g CO2 / 1 mol CO2 )
= 306 g CO2
Moles of
substance A
Mole - mass calculations
Example:
What mass of hydrogen is produced by reacting 6.0 mol of aluminum with
hydrochloric acid?
Equation:
2 Al (s) + 6 HCl (aq)
2 AlCl3 (aq) + 3 H2 (g)
Mole ratio between unknown
substance (hydrogen) and
known substance (aluminum):
n
( 6.0 mol Al )
6.0 mol Al
Mole - mass calculations
How many grams of potassium chloride and oxygen are produced from
3.00 moles of potassium chlorate?
Molar mass KCl
Equation:
2 KClO3
3 mol H2
( 6.0 mol Al )
2 mol Al
( 3.00 mol KClO3 )
( 2.016 g / mol H2 )
n
3.00 mol KClO3
= 74.55 g
=
2 mol KCl
2 mol KClO3
2 mol KCl
2 mol KClO3
( 3.00 mol KClO3 )
n = 223.65 g = 224 g KCl
Mass - mole calculations
Mole - mass calculations
How many grams of potassium chloride and oxygen are produced from
3.00 moles of potassium chlorate?
Molar mass O2
2 KClO3
39.10 g + 35.45 g
n = 3.00 mol KCl ( 74.55 g / mol KCl )
n = 18 g H2
Equation:
2 KCl + 3 O2
Mole ratio between the unknown
substance (potassium chloride) and the
known substance (potassium chlorate):
2 mol Al
n = 9.0 mol H2
Moles of
substance B
Step 1: Determine the mass of potassium chloride produced
3 mol H2
=
Use molar mass
of substance B
Use ratio between coefficients
of substances A and B from
balanced equation
2 KCl + 3 O2
2 x 16.00 g
Given: • A balanced chemical equation
• A known mass of one of the reactants/product (in grams)
Calculate: The quantity of one of the other reactants/products (in moles)
= 32.00 g
Step 2: Determine the mass of oxygen produced
Mole ratio between the unknown
substance (oxygen) and the known
substance (potassium chlorate):
( 3.00 mol KClO3 )
n
3.00 mol KClO3
=
n = 4.50 mol O2
Grams of
substance A
3 mol O2
2 mol KClO3
3 mol O2
2 mol KClO3
Use molar mass
of substance A
( 3.00 mol KClO3 )
( 32.00 g / mol O2 )
n = 144 g O2
Use ratio between coefficients
of substances A and B from
balanced equation
Moles of
substance A
Moles of
substance B
Mass - mole calculations
How many moles of water can be produced by burning 325 g of octane
(C8H18)?
Mass - mole calculations
How many moles of silver nitrate (AgNO3) are required to produce
100.0 g of silver sulfide (Ag2S)?
The balanced equation is:
2 C8H18 (g)
+
25 O2 (g)
16 CO2 (g)
+
18 H2O (g)
325 g C8H18 ( 1 mol C8H18 / 114.2 g C8H18 ) = 2.85 mol C8H18
Mole ratio between unknown
substance (water) and
known substance (octane):
( 2.85 mol C8H18 )
18 mol H2O
2 mol C8H18
n
=
2.85 mol C8H18
18 mol H2O
2 mol C8H18
2 AgNO3 +
H 2S
Ag2S + 2 HNO3
Step 1: Convert the amount of known substance (Ag2S) from
grams to moles
100.0 g Ag2S ( 1 mol Ag2S / 247.87 g Ag2S )
=
0.403 mol Ag2S
( 2.85 mol C8H18 )
n = 25.7 mol H2O
Mass - mole calculations
Homework
How many moles of silver nitrate (AgNO3) are required to produce
100.0 g of silver sulfide (Ag2S)?
2 AgNO3 +
H 2S
Homework Assignment 3: (Due Oct 7 at 9:00 pm)
Ag2S + 2 HNO3
Step 2: Determine the number of moles of the unknown substance
(AgNO3) required to produce the number of moles of the known
substance (0.403 mol Ag2S)
Mole ratio between the unknown
substance (silver nitrate) and the
known substance (silver sulfide):
( 0.403 mol Ag2S )
n
0.403 mol Ag2S
2 mol AgNO3
1 mol Ag2S
=
2 mol AgNO3
1 mol Ag2S
n = 0.806 mol AgNO3
( 0.403 mol Ag2S )