Download Stoichiometry - coercingmolecules

We measure ordinary objects either
by counting or weighing them,
depending on which method is more convenient
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
http://weyume.com/wp-content/uploads/2011/04/rice.jpg
http://farm1.static.flickr.com/21/90994367_5613e69fd9.jpg
Certain nouns can be used to define
a collection of objects
Dozen = 12
Pair = 2
The mole
The mole (n or mol) is the amount of matter that
contains as many entities (atoms, molecules,
ions, or other particles) as there are atoms in
exactly 12 g of the carbon-12 isotope (12C)
• The actual number of atoms in 12 g of
carbon-12 was determined experimentally
• Avogadro’s number (NA)
NA = 6.02 x 1023
Brown, , E. LeMay, and B. Bursten. 2000. Chemistry: The Central Science. 8th ed. Phils: Pearson Education Asia Pte. Ltd.
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
Just as 1 dozen of oranges contains 12 oranges,
1 mole of matter contains 6.02 x 1023 entities
1 mole 12C atoms = 6.02 x 1023
12C
atoms
1 mole H2O molecules = 6.02 x 1023 H2O molecules
1 mole NO3- ions = 6.02 x 1023 NO3- ions
Brown, , E. LeMay, and B. Bursten. 2000. Chemistry: The Central Science. 8th ed. Phils: Pearson Education Asia Pte. Ltd.
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
Each of these contains one mole of the
substance
carbon
sulfur
mercury
copper
iron
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
One mole (or an Avogadro’s number)
is an extremely big number
• One mole of softdrink cans would cover
the surface of the earth to a depth of over
300 kilometers
• If we were able to count the number of
atoms at a rate of 10 million per second, it
would take about 2 billion years to count a
mole of atoms
Molar mass
The molar mass (M) of a substance is the mass
of one mole of its entities (atoms, molecules,
ions, or other particles) in units of g/mol
MC = 12.01 g/mol
(one mole of C atom weighs 12.01 g)
MH2O = 18.0 g/mol
(one mole of H2O molecule weighs 18.0 g)
MNO3- = 62.0 g/mol
(one mole of NO3- ion weighs 62.0 g)
Brown, T., E. LeMay, and B. Bursten. 2000. Chemistry: The Central Science. 8th ed. Phils: Pearson Education Asia Pte. Ltd.
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
The periodic table is indispensable for
calculating the molar mass of a substance
• Elements
– M is the numerical value from the
periodic table
MH = 1.008 g/mol
MO = 16.00 g/mol
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
The periodic table is indispensable for
calculating the molar mass of a substance
• Compounds
– M is the sum of the molar masses of the
atoms of the elements in the formula
MSO2 = MS + (2 x MO)
= 32.07 g/mol + (2 x 16.00 g/mol)
= 64.07 g/mol
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
The periodic table is indispensable for
calculating the molar mass of a substance
• Compounds
– M is the sum of the molar masses of the
atoms of the elements in the formula
MK2S = (2 x MK) + MS
= (2 x 39.10 g/mol) + 32.07 g/mol
= 110.27 g/mol
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
Interconverting moles, mass,
and chemical entities
(atoms, molecules, ions, or other particles)
The factor-label method is used to convert
from one unit to another
1 peso = 4 25-centavos
1 peso
4 25-centavos
unit factor
4 25-centavos
1 peso
=1
=1
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
Alexa bought 18 fresh chicken’s eggs. How
many dozens of egg did she buy?
unit factor
dozens of egg =
18 eggs
x
1 dozen egg
12 eggs
= 1.5 dozens of egg
In order to convert between moles, mass, and
chemical entities, the factor label method is used
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
Methane (CH4) is the principal component of
natural gas. How many moles of methane
are present in 6.07 g of CH4?
MCH4 = MC + (4 x MH)
= 12.01 g/mol + (4 x 1.01 g/mol)
= 16.05 g/mol
nCH4 =
6.07 g CH4
x
1 mol CH4
16.05 g CH4
= 0.378 mol CH4
Report final answer with the correct number of significant
figures!
How many molecules of methane are
present 6.07 g of CH4?
molecules CH4 =
6.07 g CH4
x
1 mol CH4
16.05 g CH4
x
6.02 x 1023 molecules CH4
mol CH4
= 2.28 x 1023 molecules CH4
Glucose (C6H12O6), also known as blood sugar,
is used by the body as energy source.
How many moles of glucose are present in
1.75 x 1022 molecules of glucose?
nC6H12O6 =
1.75 x 1022 molecules C6H12O6
x
1 mol C6H12O6
6.02 x 1023
molecules
C6H12O6
= 0.0291 mol C6H12O6
How many grams of glucose are present in
1.75 x 1022 molecules of glucose?
MC6H12O6 = (6 x MC) + (12 x MH) + (6 x MO)
= (6 x 12.01 g/mol) + (12 x 1.01 g/mol)
+ (6 x 16.00 g/mol)
= 180.18 g/mol
How many grams of glucose are present in
1.75 x 1022 molecules of glucose?
nC6H12O6 =
1.75 x 1022 molecules C6H12O6
x
1 mol C6H12O6
6.02 x 1023
molecules
C6H12O6
x
180.18 g C6H12O6
1 mol C6H12O6
= 5.24 g C6H12O6
1. Urea [(NH2)2CO] is used in animal
feeds, as fertilizer and in the
manufacture of polymers.
a. Draw the Lewis structure of area. C is
surrounded by O and the N’s. (Where are the
H’s connected?)
b. Calculate its molar mass.
c. Consider 25.6 g of urea. How many moles of
urea present?
d. How many moles of N are present?
e. How many moles of C are present?
f. How many molecules of urea are present?
g. How many atoms of N are present?
2. Vitamin C, ascorbic acid, is often
sold as sodium ascorbate.
OH
HO
O
O
a. Calculate its molar mass.
Na+ O
OH
b. Consider a 500.-mg tablet. How many moles
of sodium ascorbate are present?
c. How many moles of C are present?
d. How many moles of Na are present?
e. How many formula units of sodium
ascorbate are present?
f. How many atoms of Na are present?
Chemical reactions and
chemical equations
A chemical reaction shows the process in
which a substance (or substances) is
changed into one or more new substances
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
A chemical equation uses chemical symbols to
show what happens during a chemical reaction
(g)
(g)
reactants
(l)
product
“Two molecules of hydrogen react with one molecule of
oxygen to yield two moles of water”
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
The Law of Conservation of Mass states that
matter is neither created nor destroyed
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
To conform with the Law of Conservation of Mass, there
must be the same number of each type of atom on both
sides of the arrow. Hence, we balance the equation by
adding coefficients before each chemical symbol
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
Calculating the amounts of
reactant and product
Stoichiometry a double
cheeseburger
2 bun slices + 2 cheese slices
+ 2 burger patties =
In a balanced equation, the number of moles of
one substance is equivalent to the number of
moles of any of the other substances
2CO(g) + O2(g)  2CO2(g)
2 mol CO = 1 mol O2
2 mol CO
1 mol O2
=1
1 mol O2
=1
2 mol CO
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
In a balanced equation, the number of moles of
one substance is equivalent to the number of
moles of any of the other substances
2CO(g) + O2(g)  2CO2(g)
2 mol CO = 2 mol CO2
2 mol CO
2 mol CO2
=1
2 mol CO2
=1
2 mol CO
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
In a balanced equation, the number of moles of
one substance is equivalent to the number of
moles of any of the other substances
2CO(g) + O2(g)  2CO2(g)
1 mol O2 = 2 mol CO2
1 mol O2
2 mol CO2
=1
2 mol CO2
=1
1 mol O2
Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
The amount of one substance in a reaction
is related to that of any other
Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.
All alkali metals react with water to produce
hydrogen gas and the corresponding
alkali metal hydroxide
2Li(s) + 2H2O(l)
2LiOH(aq) + H2(g)
How many moles of H2 will be formed by the
complete reaction of 6.23 moles of Li with water?
2Li(s) + 2H2O(l)
6.23 mol Li
nH2 =
x
2LiOH(aq) + H2(g)
1 mol H2
2 mol Li
= 3.12 mol H2
How many grams of H2 will be formed by the
complete reaction of 80.57 g of Li with water?
2Li(s) + 2H2O(l)
80.57 g Li
mH2 =
2LiOH(aq) + H2(g)
1 mol Li
x
x
6.941 g Li
x
2.016 g H2
1 mol H2
= 11.70 g H2
1 mol H2
2 mol Li
In a lifetime, the average American uses about
794 kg of copper in coins, plumbing, and wiring.
Copper is obtained from sulfide ores (such as
Cu2S) by a multistep process. After an initial
grinding, the first step is to “roast” the ore (heat it
strongly with O2) to form Cu2O and SO2
2Cu2S(s) + 3O2(g)
2Cu2O(s) + 2SO2(g)
How many moles of oxygen are required
to roast 10.0 mol of Cu2S?
2Cu2S(s) + 3O2(g)
nO2 =
10.0 mol Cu2S
2Cu2O(s) + 2SO2(g)
x
3 mol O2
2 mol Cu2S
= 15.0 mol O2
How many grams of SO2 are formed when
10.0 mol of Cu2S is roasted?
2Cu2S(s) + 3O2(g)
mSO2 =
10.0 mol Cu2S
2Cu2O(s) + 2SO2(g)
x
2 mol SO2
2 mol Cu2S
= 641 g SO2
x
64.07 g SO2
1 mol SO2
Ch 2 F
•
•
•
•
No meeting this Friday
Lab discussion moved to March 2
1:30-3:30 pm
SOM 201
How many grams of O2 are required to form
2.86 kg of Cu2O?
2Cu2S(s) + 3O2(g)
mO2 =
2.86 kg Cu2O
x
2Cu2O(s) + 2SO2(g)
1000 g Cu2O
x
1 kg Cu2O
x
3 mol O2
2 mol Cu2O
= 960 g O2
x
32.00 g O2
1 mol O2
1 mol Cu2O
143.10 g Cu2O
Limiting Reactants
The reactant used up first in a chemical
reaction is called the limiting reactant. Excess
reactants are present in quantities greater than
necessary to react with the quantity of the
limiting reactant.
A + B --- C + D
Given the amounts of A and B, which is the
limiting reactant?
Urea is prepared by reacting ammonia with
carbon dioxide:
2NH3(g) + CO2(g) -- (NH2)2CO(aq) + H2O(l)
In one process, 637.2 g of NH3 are allowed to
react with 1142 g of CO2.
(a) Which is the limiting reactant?
(b) How much urea (in grams) is produced?
(c) How much of the excess reactant (in grams)
is left at the end of the reaction?
Strategy
• Convert mass of each reactant to moles
• Calculate the amount of product formed
from the each of the reactants.
• The reactant the produces the less
amount is the limiting reactant.
1. The reaction between aluminum and iron (III) oxide can generate
temperatures around 3000⁰C and is used in welding metals:
2Al + Fe2O3 -- Al2O3 + 2Fe
In one process, 124 g of Al are reacted with 601 g of ferric oxide.
(a)Which is the limiting reactant?
(b)How much Al2O3 (in grams) is produced?
(c)How much of the excess reactant (in grams) is left at the end of
the reaction?
2. Titanium is a strong & light metal used in rockets & aircrafts. It is
prepared by the reaction between titanium (IV) chloride with molten
magnesium at around 1000⁰C:
TiCl4 + 2Mg -- Ti + 2MgCl2
In a certain industrial operation, 3.54 x 107g of TiCl4 are reacted with
1.13 x 107 g of magnesium.
(a)Which is the limiting reactant?
(b)How much Ti (in grams) is produced?
(c)How much of the excess reactant (in grams) is left at the end of
the reaction?