CHM 101 American University of Sharjah Calculations with Chemical Formulas and Equations Exam The exam will be 10 chapters I will attach them below.The exa

CHM 101 American University of Sharjah Calculations with Chemical Formulas and Equations Exam The exam will be 10 chapters I will attach them below.The exam will be at 21 july ,2020 at (8am uae timing ). CHEMISTRY
The Central Science
Chapter 3
Stoichiometry: Calculations with
Chemical Formulas and Equations
Yehya Elsayed
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Atomic Masses
Micro World
atoms & molecules
Macro World
grams
Atomic mass is the mass of an atom in
atomic mass units (amu)
By definition:
1 atom 12C “weighs” 12 amu
On this scale
1H
= 1.008 amu
16O
= 16.00 amu
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Natural lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
Average atomic mass of lithium:
7.42 x 6.015 + 92.58 x 7.016
= 6.941 amu
100
3.1
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Atomic Masses
Average Atomic Masses
• Atomic mass (weight) of an element is defined as the
weighted average of the masses of the naturally occuring
isotopes of that element
• Average masses of isotopes:
– Naturally occurring C: 98.892 % 12C + 1.108 % 13C.
• Average atomic mass of C:
(0.98892)(12 amu) + (0.0108)(13.00335) = 12.011 amu.
Contribution of Carbon-12
Contribution of Carbon-13
• Atomic weight (AW) is also known as average atomic
mass (atomic weight).
• Atomic weights are listed on the periodic table.
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Atomic Masses
The Atomic Mass Scale
Using atomic mass units (amu) or (u):
• We define: mass of 12C = exactly 12 amu.
mass of 1H = 1 amu
• 1 amu = 1.66054 x 10-24 g
1 g = 6.02214 x 1023 amu
•
16O
weighs 2.6560 x 10-23 g.
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Average atomic mass of an element X is 37.45amu.
Two isotopes exist for X: X-37 and X-38. The X-37
isotope has a fractional abundance of 0.7721 and
isotopic mass of 37.24 amu. What is the isotopic mass
of the other isotope
• (0.7721)(37.24 amu) + (1  0.7721)(x) = 37.45 amu
• x = = 38.161 amu = 38.16 amu
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Average atomic mass (6.941)
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Molecular Masses and Formula
weights
• Molecular mass or molecular weight (MW) is the sum of the masses
of the atoms represented in a molecular formula.
MW(C6H12O6) = 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu)
= 180.0 amu
• Formula weights (FW): sum of masses of the atoms or ions (in amu)
present in a formula unit of the compound.
.
FW (MgCl2) = AW(Mg) + 2AW(Cl)
= (24.3 amu) + 2(35.4 amu)
= 95.1 amu
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
The Mole and Avogadro’s
Number
Avogadro’s number
Mole: convenient measure chemical quantities.
• 1 mole of something = 6.022142  1023 of that thing.
• Avogadro’s number (NA)= 6.022142  1023 mole-1
• Experimentally, 1 mole of 12C has a mass of 12 g.
Molar Mass
• Molar mass: mass in grams of 1 mole of substance (units
g/mol, g.mol-1).
• Mass of 1 mole of 12C = 12 g.
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
The Mole and Avogadro’s
Number
Molar mass is the mass of 1 mole of
eggs in grams
shoes
marbles
atoms
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole lithium atoms = 6.941 g of Li
For any element
atomic mass (amu) = molar mass (grams)
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
The Mole and Avogadro’s
Number
1 12C atom
12.00 g
1.66 x 10-24 g
x
=
23
12
12.00 amu
6.022 x 10
C atoms
1 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
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
The Mole and Avogadro’s Number
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
The Mole and Avogadro’s Number
This photograph shows one
mole of solid (NaCl), liquid
(H2O), and gas (N2).
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
The Mole and Molar Masses
Interconverting Masses, Moles, and
Number of Particles
• Molar mass: sum of the molar masses of the atoms:
molar mass of N2 = 2  (molar mass of N).
• Molar masses for elements are found on the periodic
table.
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Average atomic mass (6.941)
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Do You Understand Formula Mass?
What is the formula mass of Ca3(PO4)2 ?
1 formula unit of Ca3(PO4)2
3 Ca
3 x 40.08
2P
2 x 30.97
😯
+ 8 x 16.00
310.18 amu
3.3
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
The Mole and Molar Masses
Interconverting Masses, Moles, and
Number of Particles
1 mol of (NH4)3PO4 contains
EXAMPLE:
3 X (6.02X1023)
molecules of NH4
3 mol of NH4
3 mol of N
atoms
12 mol of
H atoms
1 mol of PO4
1 mol of P
atoms
4 mol of O
atoms
3 X (6.02X1023) 12 X (6.02X1023)
atoms of nitrogen atoms of hydrogen
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
The Mole and Avogadro’s
Number
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
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
1 mol K
6.022 x 1023 atoms K
0.551 g K x
x
=
1 mol K
39.10 g K
8.49 x 1021 atoms K
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Do You Understand Molar Mass?
How many H atoms are in 72.5 g of C3H8O ?
Molar mass C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g/mol C3H8O
1 mol C3H8O molecules = 8 mol H atoms
1 mol H = 6.022 x 1023 atoms H
1 mol C3H8O 8 mol H atoms 6.022 x 1023 H atoms
72.5 g C3H8O x
x
x
=
1 mol C3H8O
1 mol H atoms
60 g C3H8O
5.82 x 1024 atoms H
3.3
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Example
• Calculate the number of sulfate ions in 86.2 g of barium
sulfate:
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Mass Percent Composition
from Chemical Formulas
Percentage Composition from
Formulas
Percent composition of an element in a compound =
n is the number of moles of the element in 1 mole
of the compound
OR
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
EXAMPLE
• Calculate the mass percent of chlorine in
Ba(ClO3)2
• Answer:
Formula Weight
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Mass Percent Composition
from Chemical Formulas
Percent composition of an element in a compound =
n x molar mass of element
x 100%
molar mass of compound
2 x (12.01 g)
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 =
C2H6O
52.14% + 13.13% + 34.73% = 100.0%
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Determining Empirical Formulas
from Mass percent Composition
• Start with mass % of elements (i.e. empirical data) and
calculate a formula, or
• Start with the formula and calculate the mass % elements.
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Percent Composition and Empirical Formulas
Determine the empirical formula of a compound that
has the following percent composition by mass:.
K 24.75, Mn 34.77, O 40.51 percent
1 mol K
nK = 24.75 g K x
= 0.6330 mol K
39.10 g K
1 mol Mn
nMn = 34.77 g Mn x
= 0.6329 mol Mn
54.94 g Mn
nO = 40.51 g O x
1 mol O
= 2.532 mol O
16.00 g O
nK = 0.6330, nMn = 0.6329, nO = 2.532
0.6330 ~
0.6329
2.532 ~
=
1.0
K:
Mn
:
1.0
O
:
4.0
~
~
0.6329
0.6329
0.6329
KMnO4
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
EXAMPLE
• Determine the empirical formula of a compound that is
81.32% carbon, 5.12% hydrogen and 13.56% oxygen by
mass
6.77 5.12 0.848
C8H6O
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Empirical Formulas from
Analyses
Molecular Formula from Empirical
Formula
• Once we know the empirical formula, we need the MW to
find the molecular formula.
(Molecular Weight) / (Empirical Formula Weight) = whole number
• Subscripts in the molecular formula are always wholenumber multiples of subscripts in the empirical formula
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
EXAMPLE
• Determine the Molecular formula of a compound of
molar mass of 236 g/mol that is 81.32% carbon, 5.12%
hydrogen and 13.56% oxygen by mass
6.77 5.12 0.848
Empirical Formula: C8H6O
Empirical Formula weight = 118 g/mol
MFW/EFM= 236/118= 2
Molecular formula= C16H12O2
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Experimental Determination of
Mass Percent Composition
Combustion Analysis
• Empirical formulas are determined by combustion analysis:
•g CO2
mol CO2
•g H2O
mol H2O
mol C
mol H
%C
%H
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Given:
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
Empirical formula???
g CO2
g H2O
÷44
÷18
mol CO2
mol H2O
1:1
1:2
x12
mol C
gC
0.5 mol C = 6.0 g C
mol H
x1
gH
1.5 mol H= 1.5 g H
g of O = g of sample – (g of C + g of H)
0.25 mol O = 4.0 g O
= 11.5g – 7.5 g = 4. g
0.5 mol C 1.5 mol H
0.25 mol O
Divide by smallest subscript (0.25)
Empirical formula C2H6O
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Chemical Reactions &
Chemical Equations
• Chemical equations: Symbolic representation of a
chemical reactions.
• Two parts to an equation: reactants and products:
• Reactants
 Products
Starting material
Resulting material
• It is important to indicate the physical states of the
reactants and products when we write a chemical
equation
• (g)=gas; (s)=solid; (l) =liquid and (aq)= aqueous (water)
solution
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Writing and Balancing
Chemical Equations
• Two parts to an equation:
reactants and products:
2H
H2 (g) + O2 (g) 
 2H
H2O (l)
Not Balanced
Balanced
Physical states
• Balancing takes place by placing stoichiometric
coefficients in front of formulas in a chemical equation:
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
How to “Read” Chemical
Equations
2 Mg + O2
2 MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
IS NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
3.7
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Chemical Equations
• Law of conservation of mass: matter cannot be lost in any
chemical reactions.
O
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
EXAMPLE
• Balance the following chemical equation:
• C2H6 + 7/2O2 → CO
2 2 + H23O
• Balanced but fractional coefficients : Not acceptable.
• Multiply both sides of the equation by 2 to clear the
fraction
• 2C2H6 + 7 O2 → 4CO2 + 6H2O
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
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
3.7
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
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
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
3.7
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Balancing Chemical Equations
4. Balance those elements that appear in two or
more reactants or products.
C2H6 + O2
2 oxygen
on left
2CO2 + 3H2O
multiply O2 by 7
2
4 oxygen + 3 oxygen = 7 oxygen
(3×1)
on right
(2×2)
C2H6 + 7 O2
2
2CO2 + 3H2O
2C2H6 + 7O2
4CO2 + 6H2O
remove fraction
multiply both sides by 2
3.7
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
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
4CO2 + 6H2O
4 C (2 x 2)
4C
12 H (2 x 6)
12 H (6 x 2)
14 O (7 x 2)
14 O (4 x 2 + 6)
Reactants
4C
12 H
14 O
Products
4C
12 H
14 O
3.7
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Amounts of Reactants and Products
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
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Reaction Stoichiometry
• C2H6 +
O2 → CO2 + H2O
• 2C2H6 + 7 O2 → 4CO2 + 6H2O
• 2 mol C2H6 reacts with 7 mol of O2 to produce 4
mol of CO2 and 6 moles of H2O
• Stoichiometric conversion factor used for
calculation: Examples
6 mol H 2 O
2 mol C 2 H 6
7 mol O2
7 mol O2
2 mol C 2 H 6
4 mol CO 2
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
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
H2O
chemical equation
4 mol H2O
18.0 g H2O
1 mol CH3OH
=
x
x
209 g CH3OH x
32.0 g CH3OH
2 mol CH3OH
1 mol H2O
235 g H2O
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Some Simple Patterns of
Chemical Reactivity
Combination and Decomposition
Reactions
have fewer products than reactants
have fewer reactants than products:
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Some Simple Patterns of
Chemical Reactivity
Combustion in Air
Combustion is the burning of a
substance in oxygen from air:
C3H8(g) + 5O2(g)  3CO2(g) +
4H2O(l)
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Limiting Reactants
• If the reactants are not present in stoichiometric amounts,
at end of reaction some reactants are still present in
excess.
• Limiting Reactant: one reactant that is consumed
completely in a chemical reaction
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Limiting Reactants
2H2 (g) + O2 (g) 
2H2O (l)
H2 is the limiting reagent and O2 is in excess
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3
Al2O3 + 2Fe
Which reactant is the limiting reagent.Calculate the mass of
Al2O3 formed.
g Al
mol Al
mol Fe produced
g Fe2O3
mol Fe2O3
WHILE
mol Fe produced
2 moles Fe
1 mol Al
=4.592 moles Fe
x
124 g Al x
27.0 g Al
2 mol Al
2 mol Fe
1 mol Fe2O3
601g Fe2O3 x
x
= 7.550 moles Fe
159.7 g Fe2O3
1 mol Fe2O3
Fe2O3 (601 g) produced more Fe so Al is limiting reagent
References: General Chemistry, Ebbing, Gammon 9th edition, 2007, Cengage; “Chemistry: The Central Science” by T. L. Brown, 9th edition, Prentice Hall, 2003
3.9
Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3
Al2O3 + 2Fe
Which reactant is the limiting reagent.Calculate the mass of
Al2O3 formed.
g Al
mol Al
mol Fe2O3 needed
g Fe2O3 needed
g Fe2O3
124 g Al x
OR
mol Al needed
mol Fe2O3
1 mol Al
27.0 g Al
x
1 mol Fe2O3
2 mol Al
Start with 124 g Al
160. g Fe2O3
=
x
1 mol Fe2O3
g Al needed
367 g Fe2O3
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting re…
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