Download Stoichiometry Objectives

Stoichiometry Objectives
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Identify the quantitative relationships in a balanced chemical
chemical equation.
Determine the mole ratios from from a balanced chemical
equation.
Explain the sequence of steps used in solving stoichiometric
problems.
Use the steps to solve stoichiometric problems.
Identify the limiting reactant in a chemical equation.
Identify the excess reactant and calculate the amount remaining
after the reaction is complete.
Calculate the mass of a product when the amounts of more than
one reactant are given.
Calculate the theoretical yield of a chemical reaction from data.
Determine the percent yield for a chemical reaction.
The Mole (Ch 11)
Chemists need a convenient method for counting accurately
the number of atoms, molecules, or formula units in a
sample of a substance.
-atoms and molecules are extremely small
-even in the smallest sample it’s impossible to actually
count each individual atom.
To fix this problem chemists created their own counting unit
called the mole.
-mole: commonly abbreviated mol, is the SI base unit used
to measure the amount of a substance
The Mole
Through experimentation, it has been established 1 mole =
6.022 136 7 x 1023 representative particles.
-representative particle: any particle such as atoms,
molecules, formula units, electrons, or ions.
-called Avogadro’s number in honor of the Italian
physicist and lawyer Amedeo Avogadro who, in 1811,
determined the volume of one mole of a gas.
-we round Avogadro’s number to three significant
figures— 6.02 x 1023.
- If you write out Avogadro’s number, it looks like this:
- 602 000 000 000 000 000 000 000
One Mole Quantities
Other Mole & Stoichiometry Vocabulary
1. What is stoichiometry?
Study of quantitative relationships among amounts of reactants
and products.
2. What is a mole ratio?
Ratio between the moles of any two substances in a
balanced chemical equation
3. What is molar mass?
Mass, in grams, of one mole of pure substance.
-numerically equal to its atomic mass
-has the units g/mol
Converting Moles to Particles (11.1)
Determine how many particles of sucrose are in 3.50mol
of sucrose.
- write a conversion factor using Avogadro’s number
that relates representative particles to moles of a
substance.
Converting Particles to Moles (11.1)
Now, suppose you want to find out how many moles are
represented by a certain number of representative
particles, such as 4.50 x 1024 atoms of zinc.
-You can use the inverse of Avogadro’s number as a
conversion factor.
Mole Practice 1
Identify and calculate the number of representative particles
in each of the following quantities.
1. 2.15 moles of gold
2. 0.151 mole of nitrogen oxide
3. 11.5 moles of potassium bromide
Calculate the number of moles of the substance that contains
the following number of representative particles.
4. 8.92 x 1023 atoms of barium
5. 5.50 x 1023 molecules of carbon monoxide
6. 2.66 x 1023 formula units of potassium iodide
Practice-Homework
p 311 # 1-3; p 312 # 4a-c
1. Determine the number of atoms in 2.50 mol Zn
2. Determine the number of formula units in 3.25 mol
AgNO3
3. Determine the number of molecules in 11.5 mol H2O
4. Determine the number of moles in
a. 5.75 x 1024 atoms Al
b. 3.75 x 1024 molecules CO2
c. 3.58 x 1023 formula units ZnCl2
Moles to Mass (11.2)
To convert between moles and mass, you need to use the
atomic mass found on the periodic table.
Calculate the mass of 0.625 moles of calcium.
-According to the periodic table, the atomic mass of
calcium is 40.078 amu, so the molar mass of calcium is
40.078 g/mol.
Mass to Moles (11.2)
How many moles of copper are in a roll of copper that has a
mass of 848g?
Practice-Homework
p 316 # 11ab-12ab
11. Determine the mass in grams of
a. 3.57 mol Al
b. 42.6 mol Si
12. Determine the number of moles of
a. 25.5 g Ag
b. 300.0 g S
Mass to Atoms (11.2)
To find the number of atoms in a sample, you must first
determine the number of moles.
Calculate the number of atoms in 4.77 g lead.
1. Determine moles
Mass to Atoms (cont.)
2. Determine atoms
You can also convert from number of particles to mass
by reversing the procedure above and dividing the
number of particles by Avogadro’s number to
determine the number of moles present.
Atoms to Mass
Example problem 11-5, p 318
A party balloon has 5.50x1022 atoms of helium.
What is the mass in grams of the helium?
5.50x1022 atoms He x
1 mol He
= 0.0914 mol He
6.02 x1023 atoms He
0.0914 mol He x 4.00 g He = 0.366 g He
1 mol He
Mole Practice 2
How many atoms are in the following samples?
1. 1.24 g cobalt
2. 0.575 g cesium
How many grams are in the following samples?
3. 4.16 x1023 atoms of radium
4. 1.50 x 1020 atoms of cadmium
Practice-Homwork
p 316 # 11ab-12ab, p 318 # 13ab-14ab
Moles of Compounds (11.3)
A mole of a compound contains as many moles of each
element as are indicated by the subscripts in the formula.
-For example, a mole of ammonia (NH3) consists of one
mole of nitrogen atoms and three moles of hydrogen
atoms.
-the molar mass of the compound is found by adding the
molar masses of all of the atoms in the representative
particle.
Molar mass of NH3 = 1(molar mass of N) + 3(molar mass of H)
Molar mass of NH3 = 1(14.007 g) + 3(1.008 g) = 17.031 g/mol
Practice
p 322 # 25
P 318 # 13ab-14ab
p322
25. Determine the molar mass of each of the following:
NaOH, CaCl2, Sr(NO3)2
p318
13. How many atoms are in
a. 55.2 g Li
b. 0.230 g Pb
14. What is the mass of
a. 6.02x1023 atoms Bi
b. 1.00x1024 atoms Mn
-Mole relationships from a formula (p 321)
Determine the number of moles of aluminum ions
in 1.25 moles of aluminum oxide (Al2O3).
First we need the ratio of Al ions to Al2O3.
2 mol Al ions
1 mole Al2O3
1.25 mol Al2O3 x 2 mol Al ions = 2.50 mol Al ions
1 mole Al2O3
-Mole relationships from a formula (p 321)
P321# 20-21
20. Determine the number of moles of chloride ions
in 2.50 mol ZnCl2.
21. Calculate the number of moles of each element
in 1.25 mole glucose (C6H12O6).
-Mole to Mass for compounds ( p 323)
What is the mass of 2.50 moles of allyl sulfide,
(C3H5)2S?
1. Calculate the mass of allyl sulfide.
6(12.01 g/mol) = 72.06 g/mol
10(1.01 g/mol) = 10.10 g/mol
1(32.07 g/mol) = 32.07 g/mol
114.23 g/mol
-Mole to Mass for compounds ( p 323)
What is the mass of 2.50 moles of allyl sulfide,
(C3H5)2S?
2. Convert the moles to mass.
2.50mol (C3H5)2S x 114.23g (C3H5)2S
1 mol (C3H5)2S
= 286g (C3H5)2S
Mass of Compound to Moles
Calculate the number of moles of water that are in 1.000 kg
of water?
1. Before you can calculate moles, you must determine the
molar mass of water (H2O).
molar mass H2O = 2(molar mass H) + molar mass O
2. Now you can use the molar mass of water as a conversion
factor to determine moles of water.
-Notice 1.000 kg is converted to 1.000 x 103 g
Practice
p 323 # 27-28,
p 324 # 30a&b
27. What is the mass of 3.25 moles H2SO4?
28. What is the mass of 4.35x10-2 moles of ZnCl2?
30. Determine the number of moles present in each of the
following:
a. 22.6 g AgNO3
b. 6.50 g ZnSO4
Mole Practice 3
Calculate the molar mass of the following:
1. C2H5OH
2. HCN
What is the mass of the following:
3. 2.25 moles of KMnO4
4. 1.56 moles of H2O
Determine the number of moles in the following:
5. 35.0 g HCl
6. 254 g PbCl4
What is the mass in grams of one molecule of the following:
7. H2SO4
Percent Composition, Molecular & Empirical
Formulas
Recall that every chemical compound has a definite
composition—a composition that is always the same
wherever that compound is found.
The composition of a compound is usually stated as the
percent by mass of each element in the compound, using
the following process.
Percent Composition
Example: Determine the percent composition of calcium
chloride (CaCl2).
1. Determine mass of each ion in CaCl2.
-1mol CaCl2 consists of 1mol Ca+2 ions and 2mol Cl- ions.
1mol Ca+2 ions x 40.08g Ca+2 ions = 40.08g Ca+2 ions
1mol Ca+2 ions
2mol Cl- ions x 35.45g Cl- ions = 70.90g Cl- ions
1mol Cl- ions
Percent Composition
Example: Determine the percent composition of calcium
chloride (CaCl2).
2. Calculate molar mass of CaCl2.
- 40.08g Ca+2 ions + 70.90g Cl- ions = 110.98 g CaCl2
1 mole CaCl2
1 mole CaCl2
3. Determine percent by mass of each element.
Percent Composition
Example: Determine the percent composition of calcium
chloride (CaCl2).
3. Determine percent by mass of each element.
% Ca = 40.08 g Ca+2
x 100 = 36.11 % Ca+2
110.98 g CaCl2
% Cl = 70.90 g Cl-
x 100 = 63.89 % Cl-
110.98 g CaCl2
4. Make sure your percent compositions equal 100%.
36.11% Ca+2 + 63.98% Cl- = 100%
Practice
p 331 # 43, 45
43. Calculate the percent composition of sodium
sulfate (Na2SO4).
45. What is the percent composition of phosphoric
acid (H3PO4).
Empirical Formula
You can use percent composition data to help identify an
unknown compound by determining its empirical formula.
-empirical formula-simplest whole-number ratio of atoms
of elements in the compound.
~In many cases, the empirical formula is the actual
formula for the compound.
the empirical formula of sodium chloride is Na1Cl1,
or NaCl, which is the true formula
~sometimes, the empirical formula is not the actual
formula of the compound.
the empirical formula for N2O4 (the actual) is NO2.
Empirical Formula
Example: The percent composition of an unknown
compound is found to be 38.43% Mn, 16.80% C, and
44.77% O. Determine the compound’s empirical formula.
- Because percent means “parts per hundred parts,” assume
that you have 100 g of the compound.
1. Calculate the number of moles of each element in the 100
g of compound.
Empirical Formula
Example: The percent composition of an unknown
compound is found to be 38.43% Mn, 16.80% C, and
44.77% O. Determine the compound’s empirical formula.
1. Calculate the number of moles of each element in the 100
g of compound.
Empirical Formula
Example: The percent composition of an unknown compound
is found to be 38.43% Mn, 16.80% C, and 44.77% O.
Determine the compound’s empirical formula.
2. The results show the following relationship
3. Obtain the simplest whole-number ratio of moles:
-divide each number of moles by the smallest number of
moles.
0.6995 mol Mn : 1.339 mol C : 2.798 mol O
0.6995 mol
1
0.6995 mol
:
2
0.6995 mol
:
4
Empirical Formula
Example: The percent composition of an unknown compound
is found to be 38.43% Mn, 16.80% C, and 44.77% O.
Determine the compound’s empirical formula.
3. Obtain the simplest whole-number ratio of moles:
Mn
:
C
:
O
1
:
2
:
4
4. Determine the empirical formula.
MnC2O4
Practice
p 333 # 46-47
46. A blue solid is found to contain 36.84% N
and 63.16% O. What is the empirical
formula?
47. Determine the empirical formula for a
compound that contains 35.98% Al and
64.02% S.
Molecular Formula
For many compounds, the empirical formula is not the true
formula.
-Chemists have learned, though, that acetic acid is a
molecule with the formula C2H4O2, which is the molecular
formula for acetic acid.
-molecular formula tells the exact number of atoms of
each element in a molecule or formula unit of a compound.
Notice the molecular formula for acetic acid (C2H4O2) has
exactly twice as many atoms of each element as the
empirical formula (CH2O).
-The molecular formula is always a whole-number
multiple of the empirical formula.
Molecular Formula
Example: Determine the molecular formula for maleic acid,
which has a molar mass of 116.1g/mol.
1. empirical formula of the compound
-composition of maleic acid is 41.39% C, 3.47% H, and
55.14% O (change the % to g)
Molecular Formula
Example: Determine the molecular formula for maleic acid.
1. empirical formula of the compound
-the ratio of C:H:O is 1:1:1, making the empirical formula
CHO
2. calculate the molar mass of CHO (empirical formula).
-29.01g/mol
3. Determine the molecular formula for maleic acid,
Molecular Formula
Example: Determine the molecular formula for maleic acid.
3. Determine the molecular formula for maleic acid,
-shows the molar mass of maleic acid is 4x that of CHO.
4. Multiply CHO by 4 to get C4H4O4
Practice
p 335 # 51, 53
51. A substance has a chemical composition of 65.45% C,
5.45% H and 29.09% O. The molar mass of the
molecular formula is 110.0 g/mol. Determine the
molecular formula.
53. A compound contains 46.68 % N and 53.32 % O. It
has a molar mass of 60.01 g/mol. What is the
molecular formula?
Empirical Formula from Mass
You can also calculate the empirical formula of a compound
from mass of individual elements.
Example: Determine the empirical formula for ilmenite,
which contains 5.41g Fe, 4.64g Ti and 4.65g O.
1. Multiply the mass by molar mass to get moles
5.41g Fe x 1 mol Fe = 0.0969 mol Fe
55.85 g Fe
4.64g Ti x 1 mol Ti = 0.0969 mol Ti
47.88g Ti
4.65g O x 1 mol O = 0.291 mol O
16.00g O
Empirical Formula from Mass
Example: Determine the empirical formula for ilmenite,
which contains 5.41g Fe, 4.64g Ti and 4.65g O.
2. Multiply by the smallest number to get the mole ratio.
0.0969 mol Fe : 0.0969 mol Ti : 0.291 mol O
0.0969 mol
0.0969 mol
0.0969 mol
1
:
1
:
3
3. Calculate the empirical formula.
FeTiO3
Practice
p 337 # 54-55
TEST!!!!
Stoichiometry Practice – Conservation of Mass
For the following balanced chemical equations, determine
all possible mole ratios.
1. HCl(aq) + KOH(aq)  KCl(aq) + H2O(l)
2. 2Mg(s) + O2(g)  2MgO(s)
3. 2HgO(s)  2Hg(l) + O2(g)
Stoichiometric Calculations
Many times we need to determine a certain amount of product
from a reaction or want to know how much product will
form from a given amount of reactant.
To do this you need:
1. Balanced chemical equations
2. Mole ratios
Moles of known
x
moles of unknown =
moles of known
Stoichiometric Calculations: mole-mole
Example: If you put 0.0400 mol of K into water, how
much hydrogen gas will be produced?
2K(s) + 2H2O(l) → 2KOH(aq) +H2(g)
Mole ratio between K and KOH is
2 mol K or 1 mol H2
1 mol H2
moles of known
x
moles of unknown
moles of known
0.0400 mol K x 1 mol H2 = 0.0200 mol H2
2 mol K
2 mol K
Stoichiometric Calculations Practice
1. How many moles of carbon dioxide are produced
when 10.0 moles of propane (C3H8) are burned in
excess oxygen in a gas grill. Water is also a product.
2.
Sulfuric acid is formed when sulfur dioxide reacts with
oxygen and water. Write the balanced chemical
equation for the reaction. If 12.5 mol SO2 reacts, how
many moles H2SO4 can be produced? How many mole
O2 is needed?
Practice
p 357 #3a , p 359 #10
3a. Balance the following and determine the possible mole
ratios: ZnO(s) + HCl(aq)  ZnCl2(aq) + H2O(l)
10. CH4(g) + S8(s)  CS2(l) + H2S(g)
a. Balance the equation.
b. Calculate moles of CS2 produced when 1.50 mol S8 is
used.
c. How many mol H2S is produced?
Stoichiometric Calculations: mole-mass
Example: If you put 0.0400 mol of K into water, how
many grams of hydrogen gas will be produced?
2K(s) + 2H2O(l) → 2KOH(aq) +H2(g)
Mole ratio between K and KOH is
2 mol K or 1 mol H2
1 mol H2
0.0400 mol K x 1 mol H2 = 0.0200 mol H2
2 mol K
0.0200 mol H2 x 2.02 g H2 = 0.0404 g H2
1 mol H2
2 mol K
Practice
p 360 #11-12
11. If you begin with 1.25 mol TiO2, what mass of Cl2 is
needed? TiO2 + C + Cl2  TiCl4 + CO2
12. Sodium chloride is decomposed into the elements
sodium and chlorine by means of electrical energy. How
many grams of chlorine gas are produced from 2.50 mol
sodium chloride?
Stoichiometric Calculations: mass-mass
Example: If you put 15g of K into water, how many grams of
hydrogen gas will be produced?
2K(s) + 2H2O(l) → 2KOH(aq) +H2(g)
15 g K x 1 mol K
= 0.384 mol K
39.098 g K
Mole ratio between K and KOH is
2 mol K or 1 mol H2
1 mol H2
0.384 mol K x 1 mol H2 = 0.192 mol H2
2 mol K
0.192 mol H2 x 2.02 g H2 =
1 mol H2
0.388 g H2
2 mol K
Practice
p 361 #13-14
13. Determine the mass of N2 produced if 100.0g NaN3 is
decomposed. NaN3(s)  Na(s) + N2(g)
14. If 2.50 g sulfur dioxide reacts with excess oxygen and
water, how many grams of sulfuric acid are produced?
Stoichiometric Review
1. Why is a balanced chemical equation needed in solving
stoichiometric calculations?
2. When solving stoichiometric problems, how is the correct
mole ratio expressed?
3. List the steps in solving stoichiometric equations.
4. How many grams of zinc (II) chloride can be obtained in
50.0 g of zinc reacts completely in the following reaction:
Zn + 2HCl → ZnCl2 + H2
5. How many grams of KCl is produced when 0.500 mol of
KClO3 decomposes in the following reaction:
2KClO3 + heat → 2KCl + 3O2
Extra Practice
p 379-380 #61, 64, 70
QUIZ!!!!
Limiting Reactants
Rarely are the reactants in a chemical reaction present in the
exact mole ratios specified in the balanced equation.
-usually, one or more of the reactants are present in excess,
and the reaction proceeds until all of one reactant is used
up.
-the reactant that is used up is called the limiting reactant
The limiting reactant limits the reaction and, thus, determines
how much of the product forms.
- The left-over reactants are called excess reactants
Limiting Reactants
In the reaction below, 40.0 g of sodium hydroxide (NaOH)
reacts with 60.0 g of sulfuric acid (H2SO4).
a. Calculate the limiting reactant
b. Determine the reactant in excess
c. Calculate the mass of water produced
d. Calculate the mass of reactant in excess.
Limiting Reactants
In the reaction below, 40.0 g of sodium hydroxide (NaOH)
reacts with 60.0 g of sulfuric acid (H2SO4).
a. Calculate the limiting reactant
1. calculate the actual number of moles of reactants
Limiting Reactants
In the reaction below, 40.0 g of sodium hydroxide (NaOH)
reacts with 60.0 g of sulfuric acid (H2SO4).
a. determine the limiting reactant
2. look at the actual ratio to the available ratio.
~actual ratio = 1.00 mol NaOH
0.612 mol H2SO4
~available ratio = 2.00 mol NaOH = 1.00 mol NaOH
1.00 mol H2SO4
0.500 mol H2SO4
You can see that when 0.500 mol H2SO4 has reacted, all of
the 1.00 mol of NaOH would be used up.
-NaOH is the limiting reactant
Limiting Reactants
In the reaction below, 40.0 g of sodium hydroxide (NaOH)
reacts with 60.0 g of sulfuric acid (H2SO4).
b. determine the reactant in excess
-since NaOH is the limiting reactant, H2SO4 is the reactant
in excess
c. calculate the mass of sodium sulfate produced
-use the limiting reactant
1.00mol NaOH x 1.00mol Na2SO4 x 142g Na2SO4 = 71.0g Na2SO4
2.00mol NaOH
1.00mol Na2SO4
Limiting Reactants
In the reaction below, 40.0 g of sodium hydroxide (NaOH)
reacts with 60.0 g of sulfuric acid (H2SO4).
d. calculate the mass of reactant in excess
-use NaOH, the limiting reactant to determine moles and
mass of H2SO4 used
~1.00mol NaOH x 1.00mol H2SO4 x 98.09g H2SO = 49.05g H2SO4
2.00mol NaOH
1.00mol H2SO4
-subtract the mass needed from the mass available.
60.0g H2SO4 – 49.05g H2SO4 = 11.0g H2SO4 in excess
Practice
p 368 # 20-21
Percent Yield
Most reactions never succeed in producing the predicted
amount of product.
-not every reaction goes cleanly or completely
●liquids may stick to surfaces of containers
●liquids may vaporize/evaporate
●solids may be left behind on filter paper
●solids may be lost in the purification process
●sometimes unintended products form
The amount you have been calculating so far has been the
theoretical yield, the maximum amount of product that can
be produced from a given amount of reactant.
Percent Yield
A chemical reaction rarely produces the theoretical yield.
-actual yield is the amount of product produced in the
chemical reaction
We can measure the efficiency of the reaction by calculating
the percent yield.
-percent yield (% yield) of the product is the ratio of actual
yield to the theoretical yield, expressed as a percent.
% yield = actual yield (from the experiment) x 100
theoretical yield (from calculations)
Percent Yield
When potassium chromate is added to a solution containing
0.500 g of silver nitrate, solid silver chromate is formed.
a. Determine the theoretical yield of silver chromate
b. If 0.455 g of silver chromate is actually obtained,
calculate the percent yield.