Transcript No Slide Title

Chapter 3
Calculations with
Chemical Formulas
and Equations
Dr. S. M. Condren
Molar Mass
Sum
atomic masses
represented by formula
atomic masses => gaw
molar mass => MM
Dr. S. M. Condren
One Mole of each Substance
Clockwise from top left:
1-Octanol, C8H17OH;
Mercury(II) iodide, HgI2;
Methanol, CH3OH; and
Sulfur, S8.
Dr. S. M. Condren
Example
What is the molar mass of ethanol, C2H5O1H1?
MM = 2(gaw)C + (5 + 1)(gaw)H + 1(gaw)O
= 2(12.011)C + 6(1.00794)H + 1(15.9994)O
= 24.022 + 6.04764 + 15.9994
= 46.069 g/mol
Significant figures rule for multiplication
Significant figures rule for addition
Sequence – multiplication then addition,
apply significant figure rules in proper sequence
Dr. S. M. Condren
The Mole
• a unit of measurement, quantity of matter
present
• Avogadro’s Number
6.022 x 1023 particles
• Latin for “pile”
Dr. S. M. Condren
Example How many moles of carbon
dioxide molecules are there in 10.00g
of carbon dioxide? CO2
MM = 1(gaw)C + 2(gaw)O = 44.01 g/mol
#mol CO2 = (10.00g)(1 mol/44.01g)
Dr. S. M. Condren
Example How many moles of carbon
dioxide molecules are there in 10.00g
of carbon dioxide? CO2
MM = 1(gaw)C + 2(gaw)O = 44.01 g/mol
#mol CO2 = (10.00g)(1 mol/44.01g)
Dr. S. M. Condren
Example How many moles of carbon
dioxide molecules are there in 10.00g
of carbon dioxide? CO2
MM = 1(gaw)C + 2(gaw)O = 44.01 g/mol
#mol CO2 = (10.00)(1 mol/44.01)
= 0.2272 mol
Dr. S. M. Condren
Combustion Analysis
Dr. S. M. Condren
Percentage Composition
description of a compound based on the
relative amounts of each element in the
compound
Dr. S. M. Condren
EXAMPLE: What is the percent composition
of chloroform, CHCl3, a substance once used
as an anesthetic?
MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= (12.011 + 1.00797 + 3*35.453)amu
= 119.377amu
Dr. S. M. Condren
EXAMPLE: What is the percent composition
of chloroform, CHCl3, a substance once used
as an anesthetic?
MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= (12.011 + 1.00797 + 3*35.453)amu
= 119.377amu
1(gaw)
%C = ------------ X 100
MM
1(12.011)
%C = -------------- X 100 = 10.061% C
119.377
Dr. S. M. Condren
EXAMPLE: What is the percent composition
of chloroform, CHCl3, a substance once used
as an anesthetic?
MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= (12.011 + 1.00797 + 3*35.453)amu
= 119.377amu
1(1.00797)
%H = ---------------- X 100 = 0.844359% H
119.377
Dr. S. M. Condren
EXAMPLE: What is the percent composition
of chloroform, CHCl3, a substance once used
as an anesthetic?
MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= (12.011 + 1.00797 + 3*35.453)amu
= 119.377amu
3(35.453)
%Cl = -------------- X 100 = 89.095% Cl
119.377
Dr. S. M. Condren
EXAMPLE: What is the percent composition
of chloroform, CHCl3, a substance once used
as an anesthetic?
MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= 119.377amu
%C = 10.061% C
%H = 0.844359% H
%Cl = 89.095% Cl
Dr. S. M. Condren
Simplest (Empirical) Formula
• formula describing a substance based on the
smallest set of subscripts
Dr. S. M. Condren
Acetylene, C2H2, and benzene, C6H6, have
the same empirical formula. Is the correct
empirical formula:
C2H2
CH
C6H6
Dr. S. M. Condren
EXAMPLE: A white compound is formed
when phosphorus burns in air. Analysis shows
that the compound is composed of 43.7% P
and 56.3% O by mass. What is the empirical
formula of the compound?
Element %
43.7
P
56.3
O
Relative Number of Atoms
(%/gaw)
Divide by Smaller
43.7/30.97 = 1.41 1.41/1.41 = 1.00
56.3/15.9994 = 3.52 3.52/1.41 = 2.50
Dr. S. M. Condren
EXAMPLE: A white compound is formed
when phosphorus burns in air. Analysis shows
that the compound is composed of 43.7% P
and 56.3% O by mass. What is the empirical
formula of the compound?
Multiply
%
(%/gaw)
Divide by Smaller
by Integer
P 43.7 43.7/30.97 = 1.41 1.41/1.41 = 1.00 2*1.00 => 2
O 56.3 56.3/15.9994 = 3.52 3.52/1.41 = 2.50 2*2.50 => 5
Relative Number of Atoms
Empirical Formula => P2O5
Dr. S. M. Condren
EXAMPLE: A sample of a brown-colored
gas that is a major air pollutant is found to
contain 2.34 g of N and 5.34 g of O. What is
the empirical formula of the compound?
2.34
%N = ----------------- X 100 = 30.5% N
2.34 + 5.34
5.34
%O = ----------------- X 100 = 69.5% O
2.34 + 5.34
Dr. S. M. Condren
EXAMPLE: A sample of a brown-colored
gas that is a major air pollutant is found to
contain 2.34 g of N and 5.34 g of O. What is
the empirical formula of the compound?
%N = 30.5% N
%O = 69.5% O
Relative # Atoms
(%/gaw)
Element %
Divide by Smaller
N
30.5 30.5/14.0067 = 2.18 2.18/2.18 = 1.00
O
69.5 69.5/15.9994 = 4.34 4.34/2.18 = 1.99
Dr. S. M. Condren
EXAMPLE: A sample of a brown-colored
gas that is a major air pollutant is found to
contain 2.34 g of N and 5.34 g of O. What is
the empirical formula of the compound?
%N = 30.5% N
%O = 69.5% O
Relative # Atoms
(%/gaw)
Divide by Smaller
Element %
30.5 30.5/14.0067 = 2.18 2.18/2.18 = 1.00
N
69.5 69.5/15.9994 = 4.34 4.34/2.18 = 1.99
O
Empirical Formula => NO2
Dr. S. M. Condren
Multiply
by Integer
1*1.00=>1
1*1.99=>2
Molecular Formula
• the exact proportions of the elements that
are formed in a molecule
Dr. S. M. Condren
Molecular Formula from
Simplest Formula
empirical formula => EF
molecular formula => MF
MF = X * EF
Dr. S. M. Condren
Molecular Formula from
Simplest Formula
formula mass => FM
sum of the atomic weights represented by
the formula
molar mass = MM = X * FM
Dr. S. M. Condren
Molecular Formula from
Simplest Formula
first, knowing MM and FM
X = MM/FM
then
MF = X * EF
Dr. S. M. Condren
EXAMPLE: A colorless liquid used in rocket
engines, whose empirical formula is NO2, has
a molar mass of 92.0. What is the molecular
formula?
FM = 1(gaw)N + 2(gaw)O = 46.0
MM
92.0
X = ------- = -------- = 2
FM
46.0
thus MF = 2 * EF
Dr. S. M. Condren
What is the correct molecular formula for
this colorless liquid rocket fuel?
2NO2
NO2
N2O4
Dr. S. M. Condren
Stoichiometry
stoi·chi·om·e·try noun
1. Calculation of the quantities of
reactants and products in a chemical
reaction.
2. The quantitative relationship between
reactants and products in a chemical
reaction.
Dr. S. M. Condren
The Mole and Chemical Reactions:
The Macro-Nano Connection
2 H2 + O2 -----> 2 H2O
2 H2 molecules
1 O2 molecule
2 H2 moles molecules 1 O2 mole molecules
4 g H2
32 g O2
Dr. S. M. Condren
2 H2O molecules
2 H2O moles molecules
36 g H2O
EXAMPLE
How much H2O, in moles results from
burning an excess of H2 in 3.3 moles of O2?
H2 + O2 -----> H2O
2 H2 + O2 -----> 2 H2O
Dr. S. M. Condren
EXAMPLE
How much H2O, in moles results from
burning an excess of H2 in 3.3 moles of O2?
H2 + O2 -----> H2O
2 H2 + O2 -----> 2 H2O
(3.3 mol O2) (2 mol H2O)
#mol H2O = -----------------------------------(1 mol O2)
Dr. S. M. Condren
EXAMPLE
How much H2O, in moles results from
burning an excess of H2 in 3.3 moles of O2?
H2 + O2 -----> H2O
2 H2 + O2 -----> 2 H2O
(3.3 mol O2) (2 mol H2O)
#mol H2O = -----------------------------------(1 mol O2)
Dr. S. M. Condren
EXAMPLE
How much H2O, in moles results from
burning an excess of H2 in 3.3 moles of O2?
H2 + O2 -----> H2O
2 H2 + O2 -----> 2 H2O
(3.3) (2 mol H2O)
#mol H2O = ------------------------ = 6.6 mol H2O
(1)
Dr. S. M. Condren
Combination Reaction
PbNO3(aq) + K2CrO4(aq)  PbCrO4(s) + 2 KNO3(aq)
Colorless
yellow
yellow
Dr. S. M. Condren
colorless
Stoichiometric Roadmap
Dr. S. M. Condren
EXAMPLE How much iron(III) oxide
and aluminum powder are required to field
weld the ends of two rails together?
Assume that the rail is
132 lb/yard and that
one inch of the rails
will be covered by an
additional 10% mass
of iron.
http://www.cbu.edu/~mcondren/Demos/Thermite-Welding.ppt
Dr. S. M. Condren
EXAMPLE How much iron(III) oxide
and aluminum powder are required to field
weld the ends of two rails together?
Assume that the rail is 132 lb/yard and that
one inch of the rails will be covered by an
additional 10% mass of iron.
The mass of iron in 1 inch of this rail is:
#g/in = (132 #/yard) (1 yard/36 in) (454 g/#)
= 1.67 X 103 g/in
Dr. S. M. Condren
EXAMPLE How much iron(III) oxide
and aluminum powder are required to field
weld the ends of two rails together?
Assume that the rail is 132 lb/yard and that
one inch of the rails will be covered by an
additional 10% mass of iron.
The mass of iron in 1 inch of this rail is:
#g/in = (132) (1/36 in) (454 g)
= 1.67 X 103 g/in
The mass of iron in a weld adding 10% mass:
#g = (1.67 X 103 g) (0.10) = 167 g
Dr. S. M. Condren
EXAMPLE How much iron(III) oxide and
aluminum powder are required to field weld the ends of two
rails together? Assume that the rail is 132 lb/yard and that
one inch of the rails will be covered by an additonal
10% mass of iron.
The mass of iron in 1 inch of this rail is:
#g/in = (132 #/yard) (1 yard/36 in) (454 g/#)
= 1.67 X 103 g/in
The mass of iron in a weld adding 10% mass:
#g = (1.67 X 103 g) (0.10) = 167 g
Balanced chemical equation:
Fe2O3 + 2 Al ---> 2 Fe + Al2O3
Dr. S. M. Condren
EXAMPLE How much iron(III) oxide and
aluminum powder are required to field weld the ends of two
rails together? Assume that the rail is 132 lb/yard and that
one inch of the rails will be covered by an additonal
10% mass of iron.
The mass of iron in 1 inch of this rail is:
#g/in = (132 #/yard) (1 yard/36 in) (454 g/#)
= 1.67 X 103 g/in
The mass of iron in a weld adding 10% mass:
#g = (1.67 X 103 g) (0.10) = 167 g
Balanced chemical equation:
Fe2O3 + 2 Al ---> 2 Fe + Al2O3
What mass of Fe2O3 is required for the
thermite process?
Dr. S. M. Condren
EXAMPLE How much iron(III) oxide and
aluminum powder are required to field weld the ends of two
rails together? Assume that the rail is 132 lb/yard and that one
inch of the rails will be covered by an additonal
10% mass of iron.
The mass of iron in a weld adding 10% mass:
#g = (1.67 X 103 g) (0.10) = 167 g
Balanced chemical equation:
Fe2O3 + 2 Al ---> 2 Fe + Al2O3
What mass of Fe2O3 is required for the
thermite process?
mol Fe) (1 mol Fe2O3) (159.7 g Fe2O3)
#g Fe2O3 = (167 g Fe) *(1
------------------------------- ------------------(55.85 g Fe) (2 mol Fe) (1 mol Fe2O3)
= 238 g Fe2O3
Dr. S. M. Condren
EXAMPLE How much iron(III) oxide and
aluminum powder are required to field weld the ends of two
rails together? Assume that the rail is 132 lb/yard and that
one inch of the rails will be covered by an additonal
10% mass of iron.
The mass of iron in a weld adding 10% mass: #g Fe = 167 g Fe
Balanced chemical equation:
Fe2O3 + 2 Al ---> 2 Fe + Al2O3
What mass of Fe2O3 is required for the thermite process?
#g Fe2O3 = 238 g Fe2O3
What mass of Al is required for the thermite process?
#g Al =
(1 mol Fe) (2 mol Al) (26.9815 g Al)
(167 g Fe) * ----------------------------------------------(55.85 g Fe) (2 mol Fe) (1 mol Al)
= 80.6 g Al
Dr. S. M. Condren
EXAMPLE How much iron(III) oxide and
aluminum powder are required to field weld the ends of two
rails together? Assume that the rail is 132 lb/yard and that
one inch of the rails will be covered by an additonal
10% mass of iron.
The mass of iron in a weld adding 10% mass:
#g Fe = 167 g Fe
#g Fe2O3 = 238 g Fe2O3
#g Al = 80.6 g Al
Dr. S. M. Condren
Limiting Reactant
reactant that limits the amount of product
that can be produced
Dr. S. M. Condren
EXAMPLE
What is the number of moles of Fe(OH)3 (S) that
can be produced by allowing 1.0 mol Fe2S3, 2.0
mol H2O, and 3.0 mol O2 to react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
Dr. S. M. Condren
EXAMPLE
What is the number of moles of Fe(OH)3 (S)
that can be produced by allowing 1.0 mol
Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to
react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
balanced equation relates:
2Fe2S3(S) <=> 6H2O(l) <=> 3O2(g)
Dr. S. M. Condren
EXAMPLE
What is the number of moles of Fe(OH)3 (S)
that can be produced by allowing 1.0 mol
Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to
react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
balanced equation relates:
2Fe2S3(S) <=> 6H2O(l) <=> 3O2(g)
have only:
1Fe2S3(S) <=> 2H2O(l) <=> 3O2(g)
Dr. S. M. Condren
EXAMPLE What is the number of
moles of Fe(OH)3 (S) that can be produced by
allowing 1.0 mol Fe2S3, 2.0 mol H2O, and
3.0 mol O2 to react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
balanced equation relates:
2Fe2S3(S) <=> 6H2O(l) <=> 3O2(g)
have only:
1Fe2S3(S) <=> 2H2O(l) <=> 3O2(g)
not enough H2O to use all Fe2S3
plenty of O2
Dr. S. M. Condren
EXAMPLE What is the number of
moles of Fe(OH)3 (S) that can be produced by
allowing 1.0 mol Fe2S3, 2.0 mol H2O, and
3.0 mol O2 to react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
if use all Fe2S3:
(1.0 mol Fe2S3) (4 mol Fe(OH)3)
#mol Fe(OH)3 = -----------------------------------------(2 mol Fe2S3)
= 2.0 mol Fe(OH)3
Dr. S. M. Condren
EXAMPLE: What is the number of
moles of Fe(OH)3 (S) that can be produced by
allowing 1.0 mol Fe2S3, 2.0 mol H2O, and
3.0 mol O2 to react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
if use all H2O:
(2.0 mol H2O) (4 mol Fe(OH)3)
#mol Fe(OH)3 = ----------------------------------------(6 mol H2O)
= 1.3 mol Fe(OH)3
Dr. S. M. Condren
EXAMPLE: What is the number of
moles of Fe(OH)3 (S) that can be produced by
allowing 1.0 mol Fe2S3, 2.0 mol H2O, and
3.0 mol O2 to react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
if use all O2
(3.0 mol O2) (4 mol Fe(OH)3)
#mol Fe(OH)3 = --------------------------------------(3 mol O2)
= 4.0 mol Fe(OH)3
Dr. S. M. Condren
EXAMPLE: What is the number of
moles of Fe(OH)3 (S) that can be produced by
allowing 1.0 mol Fe2S3, 2.0 mol H2O, and
3.0 mol O2 to react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
1.0 mol Fe2S3 => 2.0 mol Fe(OH)3
2.0 mol H2O => 1.3 mol Fe(OH)3
3.0 mol O2 => 4.0 mol Fe(OH)3
Dr. S. M. Condren
EXAMPLE: What is the number of
moles of Fe(OH)3 (S) that can be produced by
allowing 1.0 mol Fe2S3, 2.0 mol H2O, and
3.0 mol O2 to react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
1.0 mol Fe2S3 => 2.0 mol Fe(OH)3
2.0 mol H2O => 1.3 mol Fe(OH)3
3.0 mol O2 => 4.0 mol Fe(OH)3
Since 2.0 mol H2O will produce only 1.3 mol
Fe(OH)3, then H2O is the limiting reactant. Thus the
correct number of moles of Fe(OH)3 is 1.33 moles.
Dr. S. M. Condren
EXAMPLE: What is the number of
moles of Fe(OH)3 (S) that can be produced by
allowing 1.0 mol Fe2S3, 2.0 mol H2O, and
3.0 mol O2 to react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
1.0 mol Fe2S3 => 2.0 mol Fe(OH)3
2.0 mol H2O => 1.3 mol Fe(OH)3 least amount
3.0 mol O2 => 4.0 mol Fe(OH)3
Since 2.0 mol H2O will produce only 1.3 mol
Fe(OH)3, then H2O is the limiting reactant.
Dr. S. M. Condren
EXAMPLE: What is the number of
moles of Fe(OH)3 (S) that can be produced by
allowing 1.0 mol Fe2S3, 2.0 mol H2O, and
3.0 mol O2 to react?
2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)
1.0 mol Fe2S3 => 2.0 mol Fe(OH)3
2.0 mol H2O => 1.3 mol Fe(OH)3 least amount
3.0 mol O2 => 4.0 mol Fe(OH)3
Since 2.0 mol H2O will produce only 1.3 mol Fe(OH)3,
then H2O is the limiting reactant. Thus the maximum
number of moles of Fe(OH)3 that can be produced by
this reaction is 1.3 moles.
Dr. S. M. Condren
Theoretical Yield
the amount of product produced by a
reaction based on the amount of the limiting
reactant
Dr. S. M. Condren
Actual Yield
amount of product actually produced in a
reaction
Dr. S. M. Condren
Percent Yield
actual yield
% yield = --------------------- * 100
theoretical yield
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the
net equation for the overall process
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation for
the overall process
(1.00 kg Cl2)
#kg N2H4 = ---------------------
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation for
the overall process
(1.00 kg Cl2) (1000 g Cl2)
#kg N2H4 = ----------------------------------(1 kg Cl2)
metric conversion
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation for
the overall process
(1.00) (1000 g Cl2) (1 mol Cl2)
#kg N2H4 = ----------------------------------------(1) (70.9 g Cl2)
molar mass
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation for
the overall process
(1.00)(1000)(1 mol Cl2)
#kg N2H4 = ----------------------------------------(1)(70.9)
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation for
the overall process
(1.00)(1000)(1 mol Cl2)(1 mol N2H4)
#kg N2H4 = ------------------------------------------------(1) (70.9)
(1 mol Cl2)
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation for
the overall process
(1.00)(1000)(1)(1 mol N2H4)
#kg N2H4 = ------------------------------------------------(1) (70.9)(1)
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation for
the overall process
(1.00)(1000)(1)(1 mol N2H4) (32.0 g N2H4)
#kg N2H4 = -------------------------------------------------------(1)(70.9) (1)
(1 mol N2H4)
molar mass
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation for
the overall process
(1.00)(1000)(1)(1) (32.0 g N2H4)(1 kg N2H4)
#kg N2H4 = ---------------------------------------------------------(1)(70.9)(1)(1)
(1000 g N2H4)
metric conversion
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation for
the overall process
(1.00)(1000)(1)(1) (32.0)(1 kg N2H4)
#kg N2H4 = ---------------------------------------------------------(1)(70.9)(1)(1)(1000)
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation for
the overall process
(1.00)(1000)(1)(1) (32.0)(1 kg N2H4)
#kg N2H4 = ---------------------------------------------------------(1)(70.9)(1)(1)(1000)
= 0.451 kg N2H4
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) theoretical yield #kg N2H4 = 0.451 kg N2H4
(b) actual yield
(0.299 kg product)
# kg N2H4 = --------------------------
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) theoretical yield #kg N2H4 = 0.451 kg N2H4
(b) actual yield
(0.299 kg product) (98.0 kg N2H4)
# kg N2H4 = -------------------------------------------(100 kg product)
purity factor
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) theoretical yield #kg N2H4 = 0.451 kg N2H4
(b) actual yield
(0.299 kg product) (98.0 kg N2H4)
# kg N2H4 = -------------------------------------------(100 kg product)
= 0.293 kg N2H4
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) theoretical yield
(b) actual yield
#kg N2H4 = 0.451 kg N2H4
# kg N2H4 = 0.293 kg N2H4
Dr. S. M. Condren
EXAMPLE
A chemical plant obtained
0.299 kg of 98.0% N2H4 for every 1.00 kg of
Cl2 that is reacted with excess NaOH and
NH3. What are the: (a) theoretical, (b) actual,
and (c) percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O
(a) theoretical yield #kg N2H4 = 0.451 kg N2H4
(b) actual yield
# kg N2H4 = 0.293 kg N2H4
(c) percent yield
0.293 kg
% yield = -------------- X 100 = 65.0 % yield
0.451kg
Dr. S. M. Condren