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Stoichiometry
Chapter/Unit 3
Significant Figures Review (1.5)
Rules for counting Significant Figures:
1. Nonzero integers always count!
2. Zeroes (3 classes):
a. Leading zeroes do not count (place holders)
b. Captive zeroes do count
c. Trailing zeroes are if there is a decimal point
3. Exact numbers (counting or from
definitions)
Sig Fig Calculation Review (1.5)
Rules for Significant Figures in calculations:
1. For multiplication or division: the resultant is the same
as the LEAST precise number in the calculation.
2. For addition or subtraction: the resultant is the same
number of decimal places as the least precise
measurement in the calculation.
3. Rounding: DO NOT round until all calculations are
completed. Use only the first number to the right of the
last significant figure.
Stoichiometry
Chemical Stoichiometry: The quantities of materials
consumed and those produced in chemical reactions.
How much of A
How much of B
How much of AB
To understand Stoichiometry it is necessary to
understand how average atomic mass is calculated.
Average Atomic Mass
http://www.tutorvista.com/content/chemistry/chemistry-iii/organic-compounds/molecular-massdetermination.php
Average Atomic Mass
Average Atomic Mass of Carbon
12C
= 98.89 % at 12 amu
13C
= 1.11 % at 13.0034 amu
14C
= Negligibly small at this level
http://harmonyscienceacademy.web.officelive.com/AtomicTheoryandStructure2.aspx
98.89%*(12 amu) + 1.11%*(13.0034 amu) = 12.01 amu
Average Atomic Mass
What does knowing that carbons average atomic mass
is 12.01 amu do for us?
Even though natural carbon does not contain a
single atom with a mass of 12.01, for
Stoichiometry purposes, we can consider
carbon to be composed of only one type of
atom with a mass of 12.01 amu.
So to obtain 1000 atoms of Carbon, with an average
atomic mass of 12.01, we can weigh out 12,010 atomic
mass units. (this would be a mixture of 12C and 13C)
The Mole
Mole: the number equal to the number of carbon
atoms in exactly 12 grams of pure 12C
(abbreviated as mol).
Turns out that number is 6.022 x 1023 which is
referred to Avogrado’s number.
One Mole of something equals 6.022 x 1023 units of that
something.
Ex: 1 mol of students contains 6.022 x 1023 students
Ex: 1 mol of Carbon has 6.022 x 1023 atoms and
has an amu of 12.01.
The Mole
The mole is defined such that a sample of a natural
element with a mass equal to the element’s atomic
mass expressed in grams contains 1 mole of atoms.
1 mol of Helium = 6.022 x 1023 atoms = 1.00794 grams
1 mol of Oxygen = 6.022 x 1023 atoms =15.9994 grams
1 mol of Zirconium = 6.022 x 1023 atoms =91.224 grams
(6.022
x 1023 atoms)(12 amu) = 12 grams
(atom)
6.022 x 1023 amu = 1 g
or
6.022 x 1023 amu
1g
The Mole
Determining the mass of a sample of Atoms
Remember that 1 atom of Carbon = 12.01
amu
If we have 6 atoms of Carbon what is
the mass of that sample?
6 atoms x 12.01 amu x _____1g_____
atom
6.022x1023 amu
= 1.197 x 10-22 g
The Mole
Determining the moles of a sample of
atoms
Aluminum has mass of 26.98 amu
If we have 10.0 grams of Al how many
moles is in the sample?
10.0 grams Al x 1 mol Al __
26.98 g Al
= 0.371 mol Al Atoms
6.022x1023 atoms x 0.371 mol Al
1 mol Al
= 2.23 x 1023 Atoms
The Mole
Calculating the numbers of Atoms in a
sample
Silicon has an amu of 28.09
A silicon ship used in an integrated
circuit has a mass of 5.68 mg. How many
silicon atoms are present in the chip?
5.68 mg
Si x 1 g Si __
1000 mg Si
5.68 x 10-3 g Si x 1 mol Si
28.09 g Si
= 5.68 x 10-3 g Si
= 2.02 x 10-4 mol Si
2.02 x 10-4 mol Si x 6.022 x 1023 atoms = 1.22 x 1020 atoms of Si
1 mol Si
The Mole
Calculating the numbers of Moles and Mass
Cobalt has an amu of 58.93
Calculate both the number of moles in a
sample of cobalt containing 5.00 x 1020
atoms and the mass of the sample?
5.00 x 1020 atoms Co x _____1 mol Co_____
6.022x 1023 atoms Co
8.30 x 10-4 mol Co x ____58.93 grams Co
1 mol Co
= 8.30 x 10-4 mol Co
= 4.89 x 10-2 grams Co
The Mole
Calculating Molar Mass
Molar Mass: the mass in grams of one mole of
the compound
Calculating the molar mass of a molecule
requires that we sum the molar masses
of the individual elements. Example: The
molar mass of methane (CH4)?
Mass of 1 mol of C =
12.011 g
Mass of 1 mol of H = 4 x 1.00794 g
Mass of 1 mol of CH4 =
16.043 g
The Mole
Calculating Molar Mass
Example: Juglone, C10H6O3, is a natural herbicide.
Calculate the molar mass and how many moles are
present in 1.56 x10-2 g of the pure compound.
Mass of 1 mol of C (x10) = 10 x 12.011 g
Mass of 1 mol of H (x6)=
6 x 1.00794 g
Mass of 1 mol of O (x3)=
3 x 15.9994 g
Mass of 1 mol of C10H6O3 =
174.156 g
1.56 x 10-2 g C10H6O3 x _1 mol of C10H6O3__
174.156 g
= 8.96x 10-5 mols of C10H6O3
The Mole
Calculating Molar Mass
Example: Bees release 1 x10-6 grams of Isopentyl
acetate (C7H14O2) when they sting. How many
molecules are present per sting?
Mass of 1 mol of C (x7) =
Mass of 1 mol of H (x14)=
Mass of 1 mol of O (x2)=
7 x 12.011 g
14 x 1.00794 g
2x 15.9994 g
Mass of 1 mol of C7H14O2 =
130.187 g
1 x 10-6 g C7H14O2 x _1 mol of C7H14O2__
130.187 g
8 x 10-9 mol C7H14O2 x _6.022 x 1023 molecules__
1 mol of C7H14O2
= 8 x 10-9 mols of C7H14O2
= 5 x 1015 molecules C7H14O2
The Mole
Calculating Percent Composition
Percent composition: comparison of the mass of
individual elements to its parent molecules mass.
Example: What percent mass is each element in
ethanol, C2H5OH?
Mass of 1 mol of C (x2) =
Mass of 1 mol of H (x6)=
Mass of 1 mol of O (x1)=
Mass of 1 mol of C2H5OH =
2x 12.011 g
6 x 1.00794 g
1x 15.9994 g
2x12.011 g = 24.022 g C
6 x1.00794 g = 6.04764 g H
1x15.9994 g = 15.9994 g O
46.069 g
____24.022g C___
X 100%
46.069 g C2H5OH
____6.04674 g H___ X 100%
46.069 g C2H5OH
= 52.14 % Carbon
= 13.13 % Hydrogen
= 34.73 % Oxygen
The Mole
Calculating Percent Composition
Example: What percent mass is each element in
Penacillin C14H20N2SO4?
Mass of 1 mol of C (x14) =
Mass of 1 mol of H (x20)=
Mass of 1 mol of O (x4)=
Mass of 1 mol of N (x2)=
Mass of 1 mol of S (x1)=
Mass of 1 mol of C14H20N2SO4 =
____168.154 g C___
312.38988 g C14H20N2SO4
____20.1588 g H __
312.38988 g C14H20N2SO4
14x 12.011 g
20x 1.00794 g
4x 15.9994 g
2x 14.00674 g
1x 32.066 g
= 168.154 g C
= 20.1588 g H
= 63.9976 g O
= 28.01348 g N
= 32.066 g S
312.38988 g
X 100%
= 53.83 % Carbon
X 100% = 6.45 % Hydrogen
= 20.49 % Oxygen
= 10.27 % Sulfur
= 8.96 % Nitrogen
The Mole
Calculating empirical and Molecular formula
Example: A compound contains 71.65% Cl, 24.27%
C and 4.07% H. The molar mass is known to be
98.95916 g/mol. What is the empirical and
molecular formula?
1.) Convert the mass percents to masses in grams (use 100 grams
as your conversion)
71.65 g Cl x 1 mol Cl__ = 2.079662842 mol Cl
34.4527 g
24.27 g C x 1 mol C__
12.011 g
4.07 g H x 1 mol H__
1.00794 g
= 2.02064774 mol C
= 4.037938766 mol H
The Mole
Calculating empirical and Molecular formula
Example: A compound contains 71.65% Cl, 24.27%
C and 4.07% H. The molar mass is known to be
98.95916 g/mol. What is the empirical and
molecular formula?
2.) Divide each mole value by the smallest mole number present
2.079662842 Cl
2.02064774
= 1.029206032 Chlorines
2.02064774__C
2.02064774
= 1 Carbons
4.037938766_H
2.02064774
= 1.998338793 Hydrogens
Empirical Formula = ClCH2
The Mole
Calculating empirical and Molecular formula
Example: A compound contains 71.65% Cl, 24.27%
C and 4.07% H. The molar mass is known to be
98.95916 g/mol. What is the empirical and
molecular formula?
3.) Determine the molar mass of the empirical formula and compare
to the given molar mass.
Empirical Formula = ClCH2
Empirical Formula molar mass = 49.47958 g/mol
98.95916 g/mol
49.47958 g/mol
=2
Molecular Formula = (ClCH2)2
Molecular Formula = Cl2C2H4
Physical vs. Chemical
Changes in physical properties
• Melting
• Boiling
• Condensation
No change occurs in the identity of the
substance
Example: Ice , rain, and steam are all
water
Physical vs. Chemical
In a chemical Reaction:
Atoms in the reactants are rearranged to
form one or more different substances.
Old bonds are broken; new bonds form
Examples:
Fe and O2 form rust (Fe2O3)
Ag and S form tarnish (Ag2S)
Chemical Reactions
http://www.mikeblaber.org/oldwine/chm1045/notes/Stoich/Equation/Stoich01.htm
Balancing Chemical Reactions
States within a chemical reaction:
Solid: symbolized using a (s)
Liquid: symbolized using a (l)
Gas: symbolized using a (g)
Dissolved in water: symbolized using a
(aq) for aqueous
HCl(aq) + NaHCO3(s)
CO2(g) + H2O(l) +NaCl(aq)
Balancing Chemical Reactions
Rules for balancing chemical equations:
1. Determine what reaction is occurring. What
are the reactants and what are the products?
What are the physical states involved?
2. Write the unbalanced equation that
summarizes the reaction.
3. Balance the equation by inspection, starting
with the most complicated molecules. Determine
the coefficients necessary. DO NOT change the
formulas of the reactants or products.
Balancing Chemical Reactions
Balancing by inspection
C2H5OH(l) + O2(g)
CO2(g) + H2O(g)
Make a list of what atoms are present on each side:
Reactants
Products
C x 2 = 2C
H x 6 = 6H
O x 3 = 3O
C x 1 = 1C
H x 2 = 2H
O x 3 = 3O
Start with the most complicated molecule
Since C2H5OH contains 2 carbons then we need to have
2 carbons on the reactant side. We place a coefficient of
2 in front of carbon dioxide
C2H5OH(l) + O2(g)
2CO2(g) + H2O(g)
Balancing Chemical Reactions
Balancing by inspection
C2H5OH(l) + O2(g)
2CO2(g) + H2O(g)
Change the list of what atoms are present on each side after
the change:
Reactants
Products
C x 2 = 2C
H x 6 = 6H
O x 3 = 3O
C x 2 = 2C
H x 2 = 2H
O x 5 = 5O
Since C2H5OH contains 6 H’s we can place a coefficient of 3
in front of water. Which will give us a change in the number of
H’s and O’s
C2H5OH(l) + O2(g)
2CO2(g) + 3H2O(g)
Balancing Chemical Reactions
Balancing by inspection
C2H5OH(l) + O2(g)
2CO2(g) + 3H2O(g)
Reactants
Products
C x 2 = 2C
H x 6 = 6H
O x 3 = 3O
C x 2 = 2C
H x 6 = 6H
O x 7 = 7O
All that’s left is to balance the oxygens.
C2H5OH(l) + 3O2(g)
2CO2(g) + 3H2O(g)
Reactants
Products
C x 2 = 2C
H x 6 = 6H
O x 7 = 7O
C x 2 = 2C
H x 6 = 6H
O x 7 = 7O
Chemical Stoichiometry
Rules for calculating masses of reactants
and products in chemical equations:
1. Write the equation and balance
2. Convert the known mass of the reactant or product to
moles of that substance
3. Use the balanced equation to set up the appropriate
mole ratios
4. Use the mole ratio to calculate the number of moles of
the desired reactant/product
5. Convert from moles back to grams if the desired by
question
Chemical Stoichiometry
Example: Solid lithium hydroxide is used in space
vehicles to remove exhaled carbon dioxide forming solid
lithium carbonate and liquid water. What mass of
carbon dioxide can be absorbed by 1.00 kg of lithium
hydroxide?
Step 1: Write the equation and balance
2 LiOH(s) + CO2(g)
Li2CO3(s) + H2O(l)
Step 2: Convert the mass of LiOH to moles
1.00 kg LiOH 1000 g LiOH
1 mol LiOH
x
x
= 41.8 moles of
1
1.00 kg LiOH 23.948 g LiOH
LiOH
Chemical Stoichiometry
Example: Solid lithium hydroxide is used in space
vehicles to remove exhaled carbon dioxide forming solid
lithium carbonate and liquid water. What mass of
carbon dioxide can be absorbed by 1.00 kg of lithium
hydroxide?
Step 3: Write the appropriate mole ratio
1 mol CO2_
2 mol LiOH
Step 4: Calculate the moles of CO2 needed to react with the
moles of LiOH
41.8 mol LiOH
1 mol CO2_
x
1
2 mol LiOH
= 20.9 moles of CO2
Chemical Stoichiometry
Example: Solid lithium hydroxide is used in space
vehicles to remove exhaled carbon dioxide forming solid
lithium carbonate and liquid water. What mass of
carbon dioxide can be absorbed by 1.00 kg of lithium
hydroxide?
Step 5: Convert the moles of the desired product
to mass using the molar mass of the product
20.9 mol CO2
1
x
44.010 g CO2_ = 9.20 x 102 g of CO
2
1 mol CO2
Limiting Reactants
Limiting Reactant: the reactant that is consumed first
and therefore limits the amount of products that can
be formed.
The rules for determining Limiting reactants are
the same as for mass to mass calculations, but you
are given the masses of both reactants.
Limiting Reactants
Example: Nitrogen gas can be prepared by passing
gaseous ammonia over solid Copper (II) Oxide at high
temperatures producing solid copper and water vapor.
If a sample containing 18.1 grams of NH3 is reacted
with 90.4 grams of Copper (II) Oxide, which is the
limiting reactant? How many grams of Nitrogen gas are
produced?
Step 1: Write the equation and balance
3 CuO(s) + 2 NH3(g)
N2(g) + 3 Cu(s) + 3 H2O(l)
Limiting Reactants
3 CuO(s) + 2 NH3(g)
N2(g) + 3 Cu(s) + 3 H2O(l)
Step 2: Convert the mass of CuO and NH3 to moles
18.1 g NH3
x
1
90.4 g CuO
1
x
1 mol NH3
= 1.062795 moles of NH3
17.03056 g NH3
1 mol CuO
= 1.136457 moles of CuO
79.5454 g CuO
Limiting Reactants
3 CuO(s) + 2 NH3(g)
N2(g) + 3 Cu(s) + 3 H2O(l)
Step 3: Write the a mole ratio for the two
reactants
Required
Actual
3 mol CuO_
2 mol NH3
= 3/2 = 1.5
1.14 mol CuO_ = 1.14/1.06 = 1.08
1.06 mol NH3
Since the actual ratio is smaller than the
required CuO will be limiting
Limiting Reactants
3 CuO(s) + 2 NH3(g)
N2(g) + 3 Cu(s) + 3 H2O(l)
Step 4: Calculate the number of moles that are produced for
your target product using a your limiting reactant mole ratio.
In this case the ratio between N2 and CuO.
1.14 moles of CuO 1 mol N2_
x
1
3 mol CuO
= 0.380 moles of N2
Step 5: Convert the moles of the desired product
to mass using the molar mass of the product
0.380 moles of N2 x 28.01348 g N2_ = 10.6 grams of N
2
1
1 mol N2