ed. Brad Collins
Mass Relationships in
Chemical Reactions
Chapter 4
Some images copyright © The McGraw-Hill Companies, Inc.
Sunday, August 18, 13
Dozen = 12
Pair = 2
The mole (mol) is the amount of a substance that
contains as many elementary entities as there
are atoms in exactly 12.00 grams of 12C
1 mol = NA = 6.0221367 x 1023
Avogadro’s number (NA)
Sunday, August 18, 13
4.1
eggs
shoes
Molar mass is the mass of 1 mole of marbles in grams
atoms
(M)
molecules
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li
For any element
atomic mass (amu) = molar mass (grams)
4.1
Sunday, August 18, 13
One Mole of:
S
C
Hg
Cu
Fe
4.1
Sunday, August 18, 13
12.00 g
1 12C atom
1.66 x 10-24 g
x
=
23 12C atoms
1 amu
6.022
x
10
12.00 amu
1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu
M = molar mass in g/mol
NA = Avogadro’s number
4.1
Sunday, August 18, 13
Molar Mass Conversion Factors
We will use two equalities in molar mass calculations
For mass to moles and moles to mass:
1 mol X = molar mass X (g)
For # atoms to moles and moles to # atoms:
1 mol X = 6.022 x 1023 X atoms
4.1
Sunday, August 18, 13
Do You Understand Molar Mass?
How many atoms are in 0.551 g of potassium (K) ?
1 mol K = 39.10 g K
1 mol K = 6.022 x 1023 atoms K
6.022 x 1023 atoms K
1 mol K
x
0.551 g K x
=
1 mol K
39.10 g K
8.49 x 1021 atoms K
4.1
Sunday, August 18, 13
Practice
• How many sulfur atoms are in 16.3 grams of sulfur?
4.1
Sunday, August 18, 13
Percent composition of an element in a compound =
n x molar mass of element
x 100%
molar mass of compound
n is the number of moles of the element in 1 mole
of the compound
2 x (12.01 g)
%C =
x 100% = 52.14%
46.07 g
6 x (1.008 g)
%H =
x 100% = 13.13%
46.07 g
1 x (16.00 g)
%O =
x 100% = 34.73%
46.07 g
C 2H 6O
52.14% + 13.13% + 34.73% = 100.0%
4.2
Sunday, August 18, 13
Chemical composition of a lead bullet.
4.2
Sunday, August 18, 13
Elemental Analysis
• Determines the percent composition to determine the empirical formula
• Example: Formaldehyde
• 40.00 % carbon
• 6.714% hydrogen
• 53.29% oxygen
• Assume percentages are grams in a 100-gram sample
• Convert grams to moles:
• 40.00 g C x 1 mol C/12.01 g C = 3.33 mol C
• 6.714 g H x 1 mol H/1.008 g H = 6.66 mol H
• 53.29 g O x 1 mol O/16.00 g O = 3.33 mol O
• Write formula with mol of each element as subscripts: C3.33H6.66O3.33
• Convert to whole numbers by dividing each subscript by the smallest subscript: CH2O
4.2
Sunday, August 18, 13
Practice
• Determine the empirical formula of vitamin C
(ascorbic acid) given the following elemental
analysis data:
• Carbon, 40.92%
• Hydrogen, 4.58%
• Oxygen, 54.50%
4.2
Sunday, August 18, 13
Molecular Formulas from Percent
Composition Data
Formulas determined from percent composition data are always
empirical formulas
To determine a molecular formula you must know its molar mass.
Example Ascorbic Acid: molar mass = 176.12
First: Calculate Empirical Formula (see previous example)
Second: Calculate empirical molar mass
Third: Determine whole number multiple to calculate the
accurate molar mass and determine the molecular formula.
4.2
Sunday, August 18, 13
A process in which one or more substances is changed into
one or more new substances is a chemical reaction
A chemical equation uses chemical symbols to show what
happens during a chemical reaction
3 ways of representing the reaction of H2 with O2 to form H2O
reactants
products
4.3
Sunday, August 18, 13
How to “Read” Chemical Equations
2 Mg + O2
2 MgO
IS:
2 atoms Mg + 1 molecule O2 makes 2 formula units
MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
IS NOT:
2 grams Mg + 1 gram O2 makes 2 g MgO
4.3
Sunday, August 18, 13
Balancing Chemical Equations
1. Write the correct formula(s) for the reactants on
the left side and the correct formula(s) for the
product(s) on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2
CO2 + H2O
2. Change the numbers in front of the formulas
(coefficients) to make the number of atoms of
each element the same on both sides of the
equation. Do not change the subscripts.
2C2H6
NOT
C4H12
4.3
Sunday, August 18, 13
Balancing Chemical Equations
3. Start by balancing those elements that appear in
only one reactant and one product.
C2H6 + O2
2 carbon
on left
C2H6 + O2
6 hydrogen
on left
C2H6 + O2
Sunday, August 18, 13
CO2 + H2O
start with C or H but not O
1 carbon
on right
multiply CO2 by 2
2CO2 + H2O
2 hydrogen
on right
2CO2 + 3H2O
multiply H2O by 3
4.3
Balancing Chemical Equations
4. Balance those elements that appear in two or
more reactants or products.
C2H6 + O2
2 oxygen
on left
2CO2 + 3H2O
7
multiply O2 by
2
4 oxygen + 3 oxygen = 7 oxygen
(3x1)
(2x2)
on right
7
C 2H 6 +
O2
2
2CO2 + 3H2O
2C2H6 + 7O2
4CO2 + 6H2O
remove fraction
multiply both sides by 2
4.3
Sunday, August 18, 13
Balancing Chemical Equations
5. Check to make sure that you have the same
number of each type of atom on both sides of the
equation.
2C2H6 + 7O2
12
(2
6)
(7xxx2)
2)
414CHO(2
4CO2 + 6H2O
1412OH4(4(6
+ 6)
Cx x2 2)
Reactants
4C
12 H
14 O
Products
4C
12 H
14 O
4.3
Sunday, August 18, 13
Practice: Balancing Equations
Fe (s) + O2 (g)
Fe2O3 (s)
C2H4 (g) + O2 (g)
CO2 (g) + H2O (g)
Al (s) + HCl (aq)
AlCl3 (aq) + H2 (g)
4.3
Sunday, August 18, 13
Stoichiometry
Definition: The quantitative study of reactants and products in a
chemical reaction.
Uses the coefficients in the chemical equation to stand for the
number of moles of reactants and products.
2 CO (g) + O2 (g)
2 molecules
1 molecule
2 * 6.022E23 molecules
6.022E23 molecules
2 mol
2 CO2 (g)
2 molecules
2 * 6.022E23 molecules
1 mol
2 mol
Mole ratio: The ratio of moles of a reactant or product to any
other reactant or product.
• Mole ratio of CO and CO2 is 2:2
• Mole ratio of CO and O2 is 2:1
Stoichiometric calculations use mole ratios to calculate the
amount of a product produced or reactant consumed.
Sunday, August 18, 13
4.4
4.4
Sunday, August 18, 13
Mass Changes in Chemical Reactions
1. Write balanced chemical equation
2. Convert quantities of known substances into moles
3. Use coefficients in balanced equation to calculate the
number of moles of the sought quantity
4. Convert moles of sought quantity into desired units
Sunday, August 18, 13
4.4
Methanol burns in air according to the equation
2CH3OH + 3O2
2CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH
moles CH3OH
molar mass
CH3OH
moles H2O
grams H2O
molar mass
coefficients
H 2O
chemical equation
4 mol H2O
18.0 g H2O
1 mol CH3OH
209 g CH3OH x
x
x
32.0 g CH3OH
1 mol H2O
2 mol CH3OH
=
235 g H2O
4.4
Sunday, August 18, 13
Yields in Chemical Reactions
• Yield is the amount of product that is produced in a
chemical reaction.
• Several definitions:
• Theoretical Yield is the calculated amount of product
that would result from the amount of reactants present.
• Actual Yield is the amount of product that is actually
obtained from a reaction
• % Yield = Actual Yield ÷ Theoretical Yield x 100
4.4
Sunday, August 18, 13
Limiting Reagents
Actual chemical reactions rarely involve exact
stoichiometric amounts.
If an large amount of one reagent is added, some of
it may be left over after the reaction is complete.
Reactant used up first is the Limiting Reagent
Reactant left over is the Excess Reagent
4.5
Sunday, August 18, 13
Limiting Reagents
66green
red left
used
4.5
Sunday, August 18, 13
Solving Limiting Reagent
Problems
Use the reaction coefficients to calculate the amount of each
reactant needed to react completely with the other reactant.
2 CO (g) + O2 (g)
2 CO2 (g)
Example:
If we have 125.0 g of CO and 62.5 g of O2, which one is the
excess reagent? CO
Which is the limiting reagent? O2
Once you know the limiting reagent, use that amount to
calculate the amount of product that can be produced.
62.5 g O2 * 1 mol O2/32.00 g O2 * 2 mol CO2/1 mol O2 = 3.91 mol CO2
4.5
Sunday, August 18, 13
Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3
Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
g Al
mol Al
mol Fe2O3 needed
g Fe2O3 needed
OR
g Fe2O3
mol Al needed
mol Fe2O3
124 g Al x
1 mol Al
27.0 g Al
x
1 mol Fe2O3
2 mol Al
Start with 124 g Al
g Al needed
160. g Fe2O3
= 367 g Fe2O3
x
1 mol Fe2O3
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
Sunday, August 18, 13
4.5
Use limiting reagent (Al) to calculate amount of product that
can be formed.
g Al
mol Al
mol Al2O3
2Al + Fe2O3
124 g Al x
1 mol Al
27.0 g Al
x
1 mol Al2O3
2 mol Al
g Al2O3
Al2O3 + 2Fe
102. g Al2O3
=
x
1 mol Al2O3
234 g Al2O3
4.5
Sunday, August 18, 13
Combustion Reactions
• Reactions can be classified into different types depending on the
reactants involved
• Combustion reactions are any reaction between oxygen and another
substance.
• Burning fuels (butane, ethanol, benzene)
• Cellular respiration
• Reactions of non-metals with oxygen (rusting)
• If reactant contains: carbon, hydrogen, another nonmetal, a metal
• Product is: CO2, H2O, a nonmetal oxide (P4O10), a metal oxide (Fe2O3)
Sunday, August 18, 13
Combustion Analysis
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
g CO2
mol CO2
mol C
g C 6.0 g C = 0.5 mol C
g H2O
mol H2O
mol H
g H 1.5 g H = 1.5 mol H
g of O = g of sample – (g of C + g of H)
4.0 g O = 0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
Sunday, August 18, 13
3.6