Download - Catalyst

Chapter #3 : Stoichiometry Mole - Mass Relationships in Chemical Systems
1. Atomic Masses
2. The Mole
3. Determining the Formula of an Unknown
Compound
4. Writing and Balancing Chemical Equations
5. Calculating the amounts of Reactant and Product
6. Limiting Reagent Calculations:
MOLE – just a number, like a dozen
• The Mole is based upon the definition: The amount of
substance that contains the same number of elementary
particles (atoms, molecules, ions, or other?) as there are
atoms in exactly 12 grams of carbon-12 (i.e., 12C).
That number (i.e. 1 Mole) = 6.022045 x 1023 particles
(atoms, molecules, ions, electrons, apples, hairs, etc…)
= NA particles (called Avogadro’s Number)
~million x million x million x million
Avogadro’s Number (NA)
NA = 6.022045 x 1023
= # of particles (atoms, molecules, ions,
electrons, or…) in one mole of that thing.
Think of it as a units conversion factor:
NA !
6.022045 x 1023
whatevers
= 1
mol whatevers
Example:
1.00 moles of CO2 molecules x (6.022x1023 CO2 molecules / mol CO2 molecules ) =
6.02x1024 CO2 molecules
2 H2(g) + O2(g) ! 2 H2O(g)
2 dozen H2 molecules react with exactly
1 dozen O2 molecules to give exactly
2 dozen H2O molecules.
2 moles of H2 molecules react with exactly
1 mole of O2 molecules to give exactly
2 moles of H2O molecules.
Why do we do this? Because these last sizes are
in the gram range and were easy to weigh in 1800s.
Conventions: 1 mole of 12C atoms weighs 12 g exactly.
1 atom of 12C weighs 12 amu exactly.
(amu = atomic mass unit = ~mass of a proton or neutron)
2 H2(g) + O2(g) ! 2 H2O(g)
2 H2 molecules react with exactly
1 O2 molecule to give exactly
2 H2O molecules.
2 moles of H2 molecules
react with exactly
1 mole of O2 molecules
4 x 1.008 g = 4.032 g
or 4 mol H x
1.008 g H
mol H
= 4.032 g H
2 x 16.00 g = 32.00 g
to produce exactly
2 moles of H2O molecules
2 x (2x1.008 g + 16.00 g) = 36.03 g
Number – mass relationships
Mole – mass relationships
12 red marbles @ 7g each = 84g
12 yellow marbles @4g each=48g
55.85g Fe = 6.022 x 1023 atoms Fe
32.07g S = 6.022 x 1023 atoms S
Mole - Mass Relationships of Elements
Element
Atomic Mass
1 atom of H = 1.008 amu
Molar Mass Number of Atoms
1 mole of H = 1.008 g = 6.022 x 1023 atoms
1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 1023atoms
1 atom of S =
amu
1 atom of O =
amu
1 mole of S =
Molecular mass:
1 molecule of O2 =
1 mole of O2 =
1 molecule of S8 =
1 mole of S8 =
1 mole of O =
g =
atoms
g=
atoms
amu
g =
molecules
amu
g =
molecules
Mole - Mass Relationships of Elements
Element
Atomic Mass
1 atom of H = 1.008 amu
Molar Mass Number of Atoms
1 mole of H = 1.008 g = 6.022 x 1023 atoms
1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 1023 atoms
1 atom of S = 32.07 amu
1 mole of S = 32.07 g = 6.022 x 1023 atoms
1 atom of O = 16.00 amu
1 mole of O = 16.00 g = 6.022 x 1023 atoms
Molecular mass:
1 molecule of O2 = 16.00 x 2 = 32.00 amu
1 mole of O2 = 32.00 g = 6.022 x 1023 molecule
1 molecule of S8 = 32.07 x 8 = 256.56 amu
1 mole of S8 = 256.56 g = 6.022 x 1023 molecules
Isotopes of Hydrogen
Natural
Abundance
•
•
•
1 Proton
0 Neutrons 99.985 % 1.00782503 amu
2 H (D) 1 Proton
1 Neutron
0.015 % 2.01410178 amu
1
3 H (T) 1 Proton
2 Neutrons
----------------1
The average mass of Hydrogen is 1.008 amu
1
1H
WATER H-ISOTOPE DEMO:
2H O (or D O, >$200 / glass) vs. normal H O
2
2
2
Which is more dense?
Element #8 : Oxygen, Isotopes
•
16
8 Protons
99.759%
8 Neutrons
15.99491462 amu
•
17
8 Protons
0.037%
9 Neutrons
16.9997341 amu
•
18
8 Protons
10 Neutrons
0.204 %
17.999160 amu
H218O WATER >$700 / mL
8O
8O
8O
Atomic Definitions II: AMU, Dalton, 12C Std.
Atomic mass Unit (AMU) = 1/12 the mass of a carbon - 12 atom
on this scale Hydrogen has a mass of 1.008 AMU.
Dalton (D) = The new name for the Atomic Mass Unit,
one dalton = one Atomic Mass Unit
on this scale, 12C has a mass of 12.00 daltons.
Isotopic Mass = The relative mass of an Isotope relative to the
12C isotope standard = 12.00 amu.
Atomic Mass or “Atomic Weight” of an element = the average of the
masses of its naturally occurring isotopes weighted
according to their abundances.
3.1)
Calculating the “Average” Atomic Mass of an
Element
Problem: Calculate the average atomic mass of Magnesium!
Magnesium has three stable isotopes, 24Mg ( 78.7%);
25Mg (10.2%); 26Mg (11.1%).
24Mg
(78.7%) 23.98504 amu
25Mg (10.2%) 24.98584 amu
26Mg (11.1%) 25.98636 amu
Total =
With Significant Digits =
amu
Calculating the “Average” Atomic Mass of an
Element
Problem: Calculate the average atomic mass of Magnesium!
Magnesium Has three stable isotopes, 24Mg ( 78.7%);
25Mg (10.2%); 26Mg (11.1%).
24Mg
(78.7%)
25Mg (10.2%)
26Mg (11.1%)
23.98504 amu x 0.787 = 18.876226 amu
24.98584 amu x 0.102 = 2.548556 amu
25.98636 amu x 0.111 = 2.884486 amu
24.309268 amu
With Significant Digits = 24.3 amu
Calculate the Average Atomic Mass of
Zirconium, Element #40
Zirconium has five stable isotopes: 90Zr, 91Zr, 92Zr, 94Zr, 96Zr.
Isotope (% abd.)
90Zr
91Zr
92Zr
94Zr
96Zr
(51.45%)
(11.27%)
(17.17%)
(17.33%)
(2.78%)
Mass (amu)
fraction
Fractional Mass
89.904703 amu
90.905642 amu
91.905037 amu
93.906314 amu
95.908274 amu
amu
With Significant Digits =
amu
Calculate the Average Atomic Mass of
Zirconium, Element #40
Zirconium has five stable isotopes: 90Zr, 91Zr, 92Zr, 94Zr, 96Zr.
Isotope (% abd.)
90Zr
91Zr
92Zr
94Zr
96Zr
(51.45%)
(11.27%)
(17.17%)
(17.33%)
(2.78%)
Mass (amu)
89.904703 amu
90.905642 amu
91.905037 amu
93.906314 amu
95.908274 amu
fraction
X
X
X
X
X
0.5145
0.1127
0.1717
0.1733
0.0278
Fractional Mass
=
=
=
=
=
46.2560 amu
10.2451 amu
15.7801 amu
16.2740 amu
2.6663 amu
91.2215 amu
With Significant Digits = 91.22 amu
Problem: Estimate the abundance of the two Bromine isotopes,
given that the average mass of Br is 79.904 amu. Since exact
masses of isotopes not given, estimate from: mass in amu = #p+ + #n:
79Br = 79 g/mol and 81Br = 81 g/mol (approximately).
Plan: Let the abundance of 79Br = X and of 81Br = Y and X + Y = 1.0
Solution:
Problem: Estimate the abundance of the two Bromine isotopes,
given that the average mass of Br is 79.904 amu. Since exact
masses of isotopes not given, estimate from: mass in amu = #p+ + #n:
79Br = 79 g/mol and 81Br = 81 g/mol (approximately)
Plan: Let the abundance of 79Br = X and of 81Br = Y and X + Y = 1.0
Solution:
X(79) + Y(81) = 79.904
X + Y = 1.00 therefore X = 1.00 - Y
(1.00 - Y)(79) + Y(81) = 79.904
79 – 79 Y + 81Y = 79.904
2 Y = 0.904 so Y = 0.452,
or 0.5 w/ sig. figs.
X = 1.00 - Y = 1.00 - 0.5 = 0.5
%X = % 79Br = 0.5 x 100% = 50% (Actual: 50.67% = 79Br)
%Y = % 81Br = 0.5 x 100% = 50% (Actual: 49.33% = 81Br)
Calculating the Number of Moles and Atoms in a
Given Mass of Element
Problem: Tungsten (W) is the element used as the filament in light
bulbs, and has the highest melting point of any element,
3680oC. How many moles of tungsten, and atoms of the
element, are contained in a 35.0 mg sample of the metal?
Plan: Convert mass into moles by dividing the mass by the atomic
weight of the metal, then calculate the number of atoms by
multiplying by Avogadro’s number!
Solution: Converting from mass of W to moles:
Moles of W =
No. of W atoms =
Calculating the Number of Moles and Atoms in a
Given Mass of Element
Problem: Tungsten (W) is the element used as the filament in light
bulbs, and has the highest melting point of any element
3680oC. How many moles of tungsten, and atoms of the
element, are contained in a 35.0 mg sample of the metal?
Plan: Convert mass into moles by dividing the mass by the atomic
weight of the metal, then calculate the number of atoms by
multiplying by Avogadro’s number!
Solution: Converting from mass of W to moles:
Moles W = 35.0 mg W x
1 g W x 1 mol W
= 0.00019032 mol
1000 mg W
183.9 g W 1.90 x 10 - 4 mol
NO. of W atoms = 1.90 x
10 - 4
mol W x
6.022 x 1023 atoms W
=
1 mole of W
= 1.15 x 1020 atoms of Tungsten
Molecular Mass - Molar Mass ( M )
The Molecular mass of a compound expressed in amu is
numerically the same as the mass of one mole of the
compound expressed in grams, called its molar mass.
For water: H2O
Molecular mass = (2 x atomic mass of H ) + atomic mass of O
= 2(
amu) +
amu =
amu
Mass of one molecules of water =
amu
Molar mass = ( 2 x molar mass of H ) + (1 x molar mass of O)
=2(
g)+
g =
g
g H2O = 6.022 x 1023 molecules of water = 1 mole H2O
Molecular Mass - Molar Mass ( M )
The Molecular mass of a compound expressed in amu is
numerically the same as the mass of one mole of the
compound expressed in grams , called its molar mass.
For water: H2O
Molecular mass = (2 x atomic mass of H ) + atomic mass of O
= 2 ( 1.008 amu) + 16.00 amu = 18.02 amu
Mass of one molecules of water = 18.02 amu
Molar mass = ( 2 x molar mass of H ) + (1 x molar mass of O)
= 2 ( 1.008 g ) + 16.00 g = 18.02 g per mole H2O
18.02 g H2O = 6.022 x 1023 molecules of water = 1 mole H2O
One mole
of
common
substances
CaCO3
100.09 g
Oxygen
32.00 g
Copper
63.55 g
Water
18.02 g
Demo: Mole Board
.
.
From this, what is the mass fraction of O in glucose? Ans.: 96.00 / 180.16
Calculate the Molecular Mass of
Glucose: C6H12O6
• Carbon
• Hydrogen
• Oxygen
Calculate the Molecular Mass of
Glucose: C6H12O6
• Carbon
6 x 12.01 amu =
72.06 amu
• Hydrogen 12 x 1.008 amu = 12.096 amu
• Oxygen
6 x 16.00 amu =
96.00 amu
180.16 amu
Calculate the Molar Mass of Glucose:
C6H12O6
• Carbon
6 x 12.01 g/mol = 72.06 g/mol
• Hydrogen 12 x 1.008 g/mol = 12.096 g/mol
• Oxygen
6 x 16.00 g/mol =
96.00 g/mol
180.16 g/mol
Calculating the Molecular Mass of a Compound
Problem: Using the data in the periodic table, calculate the molecular
mass of the following compounds:
a) Tetraphosphorous decoxide b) Ammonium sulfate
Plan: We first write the formula, then multiply the number of atoms
(or ions) of each element by its atomic mass, and find the sum.
Solution:
a) Tetraphosphorous decoxide
b) Ammonium sulfate
DEMO: P4O10 makes phosphoric acid
Calculating the Molecular Mass of a Compound
Problem: Using the data in the periodic table, calculate the molecular
mass of the following compounds:
a) Tetraphosphorous decoxide b) Ammonium sulfate
Plan: We first write the formula, then multiply the number of atoms
(or ions) of each element by its atomic mass, and find the sum.
Solution:
a) The formula is P4O10.
Molecular mass = (4 x atomic mass of P ) +(10 x atomic mass of O )
= ( 4 x 30.97 amu) + ( 10 x 16.00 amu)
= 283.88 = 283.9 amu
b) The formula is (NH4)2SO4
Molecular mass =
Calculating the Molecular Mass of a Compound
Problem: Using the data in the periodic table, calculate the molecular
mass of the following compounds:
a) Tetraphosphorous decoxide b) Ammonium sulfate
Plan: We first write the formula, then multiply the number of atoms
(or ions) of each element by its atomic mass, and find the sum.
Solution:
a) The formula is P4O10.
Molecular mass = (4 x atomic mass of P ) +(10 x atomic mass of O )
= ( 4 x 30.97 amu) + ( 10 x 16.00 amu)
= 283.88 = 283.9 amu
b) The formula is (NH4)2SO4
Molecular mass = ( 2 x atomic mass of N ) + ( 8 x atomic mass of H)
+ ( 1 x atomic mass of S ) + ( 4 x atomic mass of O)
= ( 2 x 14.01 amu) + ( 8 x 1.008 amu) +
( 1 x 32.07 amu) + ( 4 x 16.00 amu)
= 132.154 amu = 132.15 amu
Calculating the Moles and Number of
Formula Units in a given Mass of Cpd.
Problem: Sodium phosphate is a component of some detergents.
How many moles and formula units are in a 38.6 g sample?
Plan: We need to determine the formula, and the molecular mass from
the atomic masses of each element multiplied by the coefficients.
Solution: The formula is Na3PO4. Calculating the molar mass:
M=
Convert mass to moles:
# Formula units = moles x NA =
Calculating the Moles and Number of
Formula Units in a given Mass of Cpd.
Problem: Sodium Phosphate is a component of some detergents.
How many moles and formula units are in a 38.6 g sample?
Plan: We need to determine the formula, and the molecular mass from
the atomic masses of each element multiplied by the coefficients.
Solution: The formula is Na3PO4. Calculating the molar mass:
M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen =
= 3 x 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol
= 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/mol
Converting mass to moles:
Moles Na3PO4 = 38.6 g Na3PO4 x (1 mol Na3PO4)
163.94 g Na3PO4
= 0.23545 mol Na3PO4
# Formula units = 0.23545 mol Na3PO4 x 6.022 x 1023 formula units
1 mol Na3PO4
23
= 1.46 x 10 formula units
Mass Fraction and Mass %
Mass of Red Balls =
Mass Fraction Red =
Mass % Red =
Mass Fraction Purple =
Similarly, mass fraction yellow =
Check:
Mass Fraction and Mass %
Mass of Red Balls = 3 balls x 3.0 g/ball = 9.0 g
Mass Fraction Red = 9.0 g / 16.0 g total = 0.56
Mass % Red = 0.56 x 100% = 56% red
Mass Fraction Purple =
2 balls x 2.0 g/ball / (16.0 g total) = 0.25 = 25%
Similarly, mass fraction yellow = 3x1.0/16.0 = 0.19
Check: 56% + 25% + 19% = 100%
Calculate M and % composition of NH4NO3.
•
•
•
2 mol N x
4 mol H x
3 mol O x
Molar mass = M =
Calculate M and % composition of NH4NO3.
•
•
•
2 mol N x 14.01 g/mol = 28.02 g N
4 mol H x 1.008 g/mol = 4.032 g H
3 mol O x 15.999 g/mol = 48.00 g O
80.05 g/mol
%N =
x 100%
=
%H =
x 100% =
%O =
x 100% =
Haldor Topsoe
Calculate M and % composition of NH4NO3.
•
•
•
2 mol N x 14.01 g/mol = 28.02 g N
4 mol H x 1.008 g/mol = 4.032 g H
3 mol O x 15.999 g/mol = 48.00 g O
80.05 g/mol
%N = 28.02g N x 100%
80.05g
= 35.00%
%H =
4.032g H x 100% =
80.05g
%O =
48.00g O x 100% = 59.96%
80.05g
99.997%
5.037%
Haldor Topsoe
Mass Percent Composition of Na2SO4
Na2SO4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen
Elemental masses (g)
2 x Na = 2 x 22.99 = 45.98
1xS=
4xO=
Check
% Na + % S + % O = 100%
Percent of each Element
% Na = Mass Na / Total mass x 100%
% Na =
% S = Mass S / Total mass x 100%
%S=
%O=
%O=
Mass Percent Composition of Na2SO4
Na2SO4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen
Elemental masses (g)
2 x Na = 2 x 22.99 = 45.98
1 x S = 1 x 32.07 = 32.07
4 x O = 4 x 16.00 = 64.00
142.05
Check
Percent of each Element
% Na = Mass Na / Total mass x 100%
% Na = (45.98 / 142.05) x 100% =32.37%
% S = Mass S / Total mass x 100%
% S = (32.07 / 142.05) x 100% = 22.58%
% O = Mass O / Total mass x 100%
% O = (64.00 / 142.05) x 100% = 45.05%
% Na + % S + % O = 100%
32.37% + 22.58% + 45.05% = 100.00%
Calculate the Percent Composition
of Sulfuric Acid H2SO4
Molar Mass of Sulfuric acid =
2(1.008g) + 1(32.07g) + 4(16.00g) = 98.09 g/mol
%H = 2(1.008g H) x 100% =
98.09g
2.06% H
%S = 1(32.07g S) x 100% =
98.09g
32.69% S
%O = 4(16.00g O) x 100% =
98.09 g
65.25% O
Check =
100.00%
Flow Chart of Mass Percentage Calculation
Moles of X in one
mole of Compound
Multiply by M (g / mol of X)
Mass (g) of X in one
mole of compound
Divide by mass (g) of one mole
of compound
Mass fraction of X
Multiply by 100 %
Mass % of X
Calculating Mass Percentage and Masses of
Elements in a Sample of a Compound - I
Problem: Sucrose (C12H22O11) is common table sugar.
( a) What is the mass percent of each element in sucrose?
( b) How many grams of carbon are in 24.35 g of sucrose?
(a) Determining the mass percent of each element:
mass of C per mole sucrose =
mass of H / mol =
mass of O / mol =
total mass per mole =
Finding the mass fraction of C in Sucrose & % C :
mass of C per mole
Mass Fraction of C =
=
mass of 1 mole sucrose
=
To find mass % of C =
Calculating Mass Percentage and Masses of
Elements in a Sample of a Compound - I
Problem: Sucrose (C12H22O11) is common table sugar.
( a) What is the mass percent of each element in sucrose?
( b) How many grams of carbon are in 24.35 g of sucrose?
(a) Determining the mass percent of each element:
mass of C per mole sucrose = 12 x 12.01 g C/mol = 144.12 g C/mol
mass of H / mol = 22 x 1.008 g H/mol =
22.176 g H/mol
mass of O / mol = 11 x 16.00 g O/mol =
176.00 g O/mol
total mass per mole = 342.296 g/mol
Finding the mass fraction of C in Sucrose & % C :
mass of C per mole
144.12 g C/mol
Mass Fraction of C =
=
mass of 1 mole sucrose
342.30 g Cpd/mol
= 0.4210
To find mass % of C = 0.4210 x 100% = 42.10%
Calculating Mass Percents and Masses of
Elements in a Sample of Compound - II
(a) continued
Mass % of H =
mol H x M of H x 100% =
mass of 1 mol sucrose
Mass % of O =
mol O x M of O x 100% =
mass of 1 mol sucrose
(b) Determining the mass of carbon:
Mass (g) of C = mass of sucrose x ( mass fraction of C in sucrose)
Mass (g) of C =
Calculating Mass Percents and Masses of
Elements in a Sample of Compound - II
(a) continued
Mass % of H =
mol H x M of H x 100% = 22 x 1.008 g Hx 100%
mass of 1 mol sucrose
342.30 g
= 6.479% H
Mass % of O =
mol O x M of O x 100% = 11 x 16.00 g Ox 100%
mass of 1 mol sucrose
342.30 g
= 51.417% O
(b) Determining the mass of carbon:
Mass (g) of C = mass of sucrose x ( mass fraction of C in sucrose)
Mass (g) of C = 24.35 g sucrose x 0.421046 g C
1 g sucrose
= 10.25 g C
Calculating the Mass of an Element in a
Compound: Ammonium Nitrate
How much Nitrogen is in 455 kg of Ammonium Nitrate?
Ammonium Nitrate = NH4NO3 The Formula Mass of Cpd is:
4 x H = 4 x 1.008 = 4.032 g
2 x N = 2 X 14.01 = 28.02 g
Therefore mass fraction N:
3 x O = 3 x 16.00 = 48.00 g
28.02 g Nitrogen
80.052 g
=
80.052 g Cpd
Mass N in sample =
Calculating the Mass of an Element in a
Compound: Ammonium Nitrate
How much Nitrogen is in 455 kg of Ammonium Nitrate?
Ammonium Nitrate = NH4NO3 The Formula Mass of Cpd is:
4 x H = 4 x 1.008 = 4.032 g
2 x N = 2 X 14.01 = 28.02 g
Therefore mass fraction N:
3 x O = 3 x 16.00 = 48.00 g
28.02 g Nitrogen
80.052 g
= 0.35002249 g N / g Cpd
80.052 g Cpd
455 kg x 1000g / kg = 455,000 g NH4NO3
455,000 g Cpd x 0.35002249 g N / g Cpd = 1.59 x 105 g Nitrogen
or: 455 kg NH4NO3
x 28.02 kg Nitrogen = 159 kg Nitrogen
80.052 kg NH4NO4
Empirical and Molecular Formulas
Empirical Formula - The simplest formula for a compound
that agrees with the elemental analysis! The
smallest set of whole numbers of atoms.
Molecular Formula - The formula of the compound as it
really exists. It must be a multiple of the
empirical formula.
Some Examples of Compounds or Molecules
with the same Elemental Ratios
Empirical Formula
CH2 (unsaturated HydroCarbons)
CH2O (carbohydrates)
OH or HO
Molecular Formula
C2H4 , C3H6 and C4H8
C6H12O6 and C4H8O4
H2O2 and OH
S
S8 and S
P
P4 and P
Cl
Cl2 and Cl
Empirical and Molecular Formulas
Name
Molecular
water
H2O
hydrogen
peroxide
H 2 O2
ethane
C2 H 6
sulfur
S8
acetic acid
CH3COOH
Empirical
DEMO: Genie in a bottle
Empirical and Molecular Formulas
Name
Molecular
Empirical
water
H2O
H2O
hydrogen
peroxide
H 2 O2
HO
ethane
C2 H 6
CH3
sulfur
S8
acetic acid
CH3COOH
S
COH2
Steps to Determine Empirical Formulas
Mass (g) of Element in sample
÷ M (g/mol ) for that element
Moles of Element
Use no. of moles as subscripts.
Preliminary Formula
Change to integer subscripts:
÷ smallest, conv. to whole #.
Empirical Formula
Determining Empirical Formulas from
Measured Masses of Elements - I
Problem: The elemental analysis of a sample compound gave the
following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the
empirical formula and name of the compound?
Plan: First we have to convert mass of the elements to moles of the
elements using the molar masses. Then we construct a preliminary
formula and name of the compound.
Solution: Finding the moles of the elements:
Moles of Na =
Moles of Cr =
Moles of O =
Determining Empirical Formulas from
Measured Masses of Elements - I
Problem: The elemental analysis of a sample compound gave the
following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the
empirical formula and name of the compound?
Plan: First we have to convert mass of the elements to moles of the
elements using the molar masses. Then we construct a preliminary
formula and name of the compound.
Solution: Finding the moles of the elements:
1 mol Na
Moles of Na = 5.678 g Na x
= 0.2469 mol Na
22.99 g Na
Moles of Cr = 6.420 g Cr x 1 mol Cr = 0.12347 mol Cr
52.00 g Cr
1 mol O
Moles of O = 7.902 g O x
= 0.4939 mol O
16.00 g O
Determining Empirical Formulas from
Masses of Elements - II
Constructing the preliminary formula:
Converting to integer subscripts (dividing all by smallest subscript):
Rounding off to whole numbers:
Determining Empirical Formulas from
Masses of Elements - II
Constructing the preliminary formula:
Na0.2469 Cr0.1235 O0.4939
Converting to integer subscripts (dividing all by smallest subscript):
Na1.99 Cr1.00 O4.02
Rounding off to whole numbers:
Na2CrO4
Sodium Chromate
Determining the Molecular Formula from
Elemental Composition and Molar Mass - I
Problem: The sugar burned for energy in cells of the body is Glucose
(M = 180.16 g/mol), elemental analysis shows that it contains
40.00 mass % C, 6.719 mass % H, and 53.27 mass % O.
(a) Determine the empirical formula of glucose.
(b) Determine the Molecular formula.
Plan: We are only given mass %, and no weight of the compound so we
will assume 100g of the compound, and % becomes grams, and
we can do as done previously with masses of the elements.
Solution:
Mass Carbon =
Mass Hydrogen =
Mass Oxygen =
Determining the Molecular Formula from
Elemental Composition and Molar Mass - I
Problem: The sugar burned for energy in cells of the body is Glucose
(M = 180.16 g/mol), elemental analysis shows that it contains
40.00 mass % C, 6.719 mass % H, and 53.27 mass % O.
(a) Determine the empirical formula of glucose.
(b) Determine the Molecular formula.
Plan: We are only given mass %, and no weight of the compound so we
will assume 100g of the compound, and % becomes grams, and
we can do as done previously with masses of the elements.
Solution:
Mass Carbon = 40.00% x 100g/100% = 40.00 g C
Mass Hydrogen = 6.719% x 100g/100% = 6.719 g H
Mass Oxygen = 53.27% x 100g/100% = 53.27 g O
99.989 g Cpd
Determining the Molecular Formula from
Elemental Composition and Molar Mass - II
Converting from Grams of Elements to moles:
Moles of C =
Moles of H =
Moles of O =
Constructing the preliminary formula:
Converting to integer subscripts, ÷ all subscripts by the smallest:
Determining the Molecular Formula from
Elemental Composition and Molar Mass - II
Converting from Grams of Elements to moles:
Moles of C = Mass of C x 1 mole C = 3.3306 moles C
(40.00 g C)
12.01 g C
Moles of H = Mass of H x 1 mol H = 6.6657 moles H
(6.719 g H)
1.008 g H
Moles of O = Mass of O x 1 mol O = 3.3294 moles O
(53.27 g O)
16.00 g O
Constructing the preliminary formula C 3.33 H 6.67 O 3.33
Converting to integer subscripts, ÷ all subscripts by the smallest:
C 3.33/3.33 H 6.667 / 3.33 O3.33 / 3.33 = C1H2O1 = CH2O
Determining the Molecular Formula from
Elemental Composition and Molar Mass - III
(b) Determining the Molecular Formula:
The formula weight of the empirical formula (CH2O) is:
Whole-number multiple =
M compound
empirical formula mass
Therefore the Molecular Formula is:
=
Determining the Molecular Formula from
Elemental Composition and Molar Mass - III
(b) Determining the Molecular Formula:
The formula weight of the empirical formula (CH2O) is:
1xC + 2xH + 1xO = 1x12.01 + 2x1.008 + 1x16.00 = 30.03 g/mol
Whole-number multiple =
=
M of the compound
empirical formula mass
=
180.16 g/mol
emp.mol
= 6.00 = 6
30.03 g/emp. mol
mol
Therefore the Molecular Formula is:
C1x6H2x6O1x6 =
C6H12O6
Adrenaline is a very Important
Compound in the Body - I
• Analysis gives mass % s:
•
C = 56.8 %
•
H = 6.50 %
•
O = 28.4 %
•
N = 8.28 %
• Calculate the Empirical
Formula !
Adrenaline - II
• Assume 100g!
• C=
• H=
• O=
• N=
Divide by smallest
•
•
•
•
C=
H=
O=
N=
=>
Adrenaline - II
•
•
•
•
•
•
•
•
•
•
Assume 100g!
C = 56.8 g C/(12.01 g C/ mol C) = 4.73 mol C
H = 6.50 g H/( 1.008 g H / mol H) = 6.45 mol H
O = 28.4 g O/(16.00 g O/ mol O) = 1.78 mol O
N = 8.28 g N/(14.01 g N/ mol N) = 0.591 mol N
Divide by smallest (0.591) =>
C = 8.00 mol C = 8.0 mol C
or
H = 10.9 mol H = 11.0 mol H
O = 3.01 mol O = 3.0 mol O C8H11O3N
N = 1.00 mol N = 1.0 mol N
Fig. 3.5
m
CnHm + (n+ )O2(g)
2
m
n CO2(g) +
H2O(g)
2
Ascorbic acid ( Vitamin C ) - I
contains only C , H , and O LINUS PAULING
• Upon combustion in excess oxygen, a 6.49 mg
sample yielded 9.74 mg CO2 and 2.64 mg H2O.
• Calculate its Empirical formula!
• Mass C: (from CO2)
• Mass H: from (H2O)
• Mass Oxygen = (subtract C+H from total)
Ascorbic acid ( Vitamin C ) - I
contains only C , H , and O
• Upon combustion in excess oxygen, a 6.49 mg
sample yielded 9.74 mg CO2 and 2.64 mg H2O.
• Calculate its Empirical formula!
• Mass C: 9.74 x10-3g CO2 x (12.01 g C / 44.01 g CO2)
= 2.65 x 10-3 g C
• Mass H: 2.64 x10-3g H2O x (2.016 g H / 18.02 g H2O)
= 2.95 x 10-4 g H
• Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg
= 3.54 mg O = 3.54x10-3 g O
Vitamin C combustion - II
• Moles C =
• Moles H =
• Moles O =
• Divide each by smallest:
• Moles C =
• Moles H =
• Moles O =
Vitamin C combustion - II
• C = 2.65 x 10-3 g C / ( 12.01 g C / mol C ) =
= 2.21 x 10-4 mol C
• H = 0.295 x 10-3 g H / ( 1.008 g H / mol H ) =
= 2.92 x 10-4 mol H
• O = 3.54 x 10-3 g O / ( 16.00 g O / mol O ) =
= 2.21 x 10-4 mol O
•
•
•
•
Divide each by smallest (2.21 x 10-4 ):
C = 1.00
Multiply each by 3: C = 3.00 = 3.0
H = 1.32
(to get ~integers)
H = 3.96 = 4.0
O = 1.00
O = 3.00 = 3.0
C3H4O3
Determining a Chemical Formula from
Combustion Analysis - I
Problem: Erythrose (M = 120 g/mol) is an important chemical
compound used often as a starting material in chemical
synthesis, and contains Carbon, Hydrogen, and Oxygen.
Combustion analysis of a 700.0 mg sample yielded:
1.027 g CO2 and 0.4194 g H2O.
Plan: We find the masses of Hydrogen and Carbon using the mass
fractions of H in H2O, and C in CO2. The mass of Carbon and
Hydrogen are subtracted from the sample mass to get the mass
of Oxygen. We then calculate moles, and construct the empirical
formula, and from the given molar mass we can calculate the
molecular formula.
Determining a Chemical Formula from
Combustion Analysis - II
Calculating the mass fractions of the elements:
Mass fraction of C in CO2 =
Mass fraction of H in H2O =
Calculating masses of C and H:
Mass of Element = mass of compound x mass fraction of element
Determining a Chemical Formula from
Combustion Analysis - II
Calculating the mass fractions of the elements:
mol C x M of C
Mass fraction of C in CO2 =
=
mass of 1 mol CO2
= 1 mol C x (12.01 g C/ 1 mol C) =
44.01 g CO2
0.2729 g C / 1 g CO2
mol H x M of H
=
mass of 1 mol H2O
2 mol H x (1.008 g H / 1 mol H)
=
= 0.1119 g H / 1 g H2O
18.02 g H2O
Mass fraction of H in H2O =
Calculating masses of C and H
Mass of Element = mass of compound x mass fraction of element
Determining a Chemical Formula from
Combustion Analysis - III
Mass (g) of C =
Mass (g) of H =
Calculating the mass of O:
Calculating moles of each element:
C=
H=
O=
Determining a Chemical Formula from
Combustion Analysis - III
Mass (g) of C = 1.027 g CO2 x 0.2729 g C = 0.2803 g C
1 g CO2
0.1119 g H
Mass (g) of H = 0.4194 g H2O x
= 0.04693 g H
1 g H2 O
Calculating the mass of O:
Mass (g) of O = Sample mass -( mass of C + mass of H )
= 0.700 g - 0.2803 g C - 0.04693 g H = 0.37277 g O
Calculating moles of each element:
C = 0.2803 g C / 12.01 g C/ mol C = 0.02334 mol C
H = 0.04693 g H / 1.008 g H / mol H = 0.04656 mol H
O = 0.37277 g O / 16.00 g O / mol O = 0.02330 mol O
C0.02334H0.04656O0.02330 = CH2O
Formula wt = 30 g / formula unit
120 g /mol / 30 g / formula = 4 formula units / cpd = C4H8O4
Chemical Equations
Qualitative Information:
Reactants
Products
States of Matter: (s) solid
(l) liquid
(g) gaseous
(aq) aqueous
2 H2 (g) + O2 (g)
But also Quantitative Information!
2 H2O (g)
How to Balance Equations
• Mass Balance (or Atom Balance)- same number of
each element on each side of the equation:
(1) start with simplest element (or largest molecule)
(2) progress to other elements
(3) make all whole numbers
(4) re-check atom balance
1 CH4 (g) + O2 (g)
1 CO2 (g) + H2O (g)
1 CH4 (g) + O2 (g)
1 CO2 (g) + 2 H2O (g)
1 CH4 (g) + 2 O2 (g)
1 CO2 (g) + 2 H2O (g)
• Make charges balance. (Remove “spectator” ions.)
Ca2+ (aq) + 2 OH- (aq) + Na+(aq)
Ca(OH)2 (s) + Na+(aq)
Methane/oxygen reaction
CH4(g) + 2 O2(g) ! CO2(g) + 2 H2O(g)
DEMO: Methane Bubbles
Table 3.2 (P 66) Information Conveyed by the
Balanced Equation for the Combustion of Methane
Reactants
Products
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
1 molecule CH4 +
2 molecules of O2
1 molecule CO2 +
2 molecules H2O
1 mol CH4 molecules +
2 mol O2 molecules
1 mol CO2 molecules +
2 mol H2O molecules
6.022 x 1023 CH4 molecules +
2 x (6.022 x 1023) O2 molecules
16g CH4 + 2 (32g) O2
80g reactants
6.022 x 1023 CO2 molecules +
2 x (6.022 x 1023) H2O molecules
44g CO2 + 2 (18g) H2O
80g products
Information Contained in a Balanced Equation
Viewed in
terms of:
Reactants
Products
2 C2H6 (g) + 7 O2 (g) = 4 CO2 (g) + 6 H2O(g) + Energy
Molecules
Amount (mol)
Mass (amu)
Mass (g)
Total Mass (g)
Information Contained in a Balanced Equation
Viewed in
terms of:
Reactants
Products
2 C2H6 (g) + 7 O2 (g) = 4 CO2 (g) + 6 H2O(g) + Energy
Molecules 2 molecules of C2H6 + 7 molecules of O2 =
4 molecules of CO2 + 6 molecules of H2O
Amount (mol) 2 mol C2H6 + 7 mol O2 = 4 mol CO2 + 6 mol H2O
Mass (amu)
Mass (g)
60.14 amu C2H6 + 224.00 amu O2 =
176.04 amu CO2 + 108.10 amu H2O
60.14 g C2H6 + 224.00 g O2 = 176.04 g CO2 + 108.10 g H2O
Total Mass (g)
284.14g = 284.14g
Balancing Chemical Equations - I
Problem: The hydrocarbon hexane is a component of Gasoline that
burns in an automobile engine to produce carbon dioxide and
water as well as energy. Write the balanced chemical
equation for the combustion of hexane (C6H14).
Plan: Write the skeleton equation, converting the words into chemical
compounds, with blanks before each compound. Begin the
element balance, putting 1 on the most complex compound first,
and save oxygen until last!
Solution:
Write down unbalanced reaction.
Set coeff. of biggest molecule = 1. Balance any elements that this forces:
Balancing Chemical Equations - I
Problem: The hydrocarbon hexane is a component of Gasoline that
burns in an automobile engine to produce carbon dioxide and
water as well as energy. Write the balanced chemical
equation for the combustion of hexane (C6H14).
Plan: Write the skeleton equation, converting the words into chemical
compounds, with blanks before each compound. Begin the
element balance, putting 1 on the most complex compound first,
and save oxygen until last!
Solution:
1 C6H14 (l) +
O2 (g)
CO2 (g) +
H2O(g) + Energy
One C6H14 molecule says that we must get 6 CO2’s!
1 C6H14 (l) +
O2 (g)
6 CO2 (g) +
H2O(g) + Energy
Balancing Chemical Equations - II
Next balance H atoms:
1 C6H14 (l) +
O2 (g)
6 CO2 (g) +
H2O(g) + Energy
Balance O atoms last:
C6H14 (l) +
C6H14 (l) +
O2 (g)
O2 (g)
CO2 (g) +
H2O(g) + Energy
CO2 (g) +
H2O(g) + Energy
Balancing Chemical Equations - II
The H atoms in the hexane will end up as H2O, and we have 14
H atoms, and since each water molecule has two H atoms, we will get
a total of 7 water molecules.
1 C6H14 (l) +
1 C6H14 (l) +
O2 (g)
9.5
O2 (g)
6 CO2 (g) + 7 H2O(g) + Energy
6 CO2 (g) +
7 H2O(g) + Energy
But must have even numbers of molecules: Double all!
2 C6H14 (l) + 19 O2 (g)
2 C6H14 (l) + 19 O2 (g)
12 CO2 (g) +14 H2O(g) + Energy
12 CO2 (g) +14 H2O(g) + Energy
Molecular model: Balanced equation
C2H6O(liquid) + 3 O2(g)
2 CO2 (g) + 3 H2O(g) + Energy
Thermal decomposition of ammonium dichromate:
(NH4)2Cr2O7(s) ! _ Cr2O3(s) + _ N2(g) + _ H2O(g)
(NH4)2Cr2O7(s) ! 1 Cr2O3(s) + 1 N2(g) + 4 H2O(g)
These numbers are called the
“stoichiometric coefficients”
Sample Problem: Calculating Reactants and
Products in a Chemical Reaction - I
Problem: Given the following chemical reaction between Aluminum
Sulfide and water, if we are given 65.80 g of Al2S3: a) How many moles
of water are required for the reaction? b) What mass of H2S & Al(OH)3
would be formed?
Al2S3 (s) + 6 H2O(l)
2 Al(OH)3 (s) + 3 H2S(g)
Plan: Calculate moles of Aluminum Sulfide using its molar mass, then
from the equation, calculate the moles of Water, and then the moles of
Hydrogen Sulfide, and finally the mass of Hydrogen Sulfide using
it’s molecular weight.
Solution:
a) molar mass of Aluminum Sulfide =
moles Al2S3 =
Sample Problem: Calculating Reactants and
Products in a Chemical Reaction - I
Problem: Given the following chemical reaction between Aluminum
Sulfide and water, if we are given 65.80 g of Al2S3: a) How many moles
of water are required for the reaction? b) What mass of H2S & Al(OH)3
would be formed?
Al2S3 (s) + 6 H2O(l)
2 Al(OH)3 (s) + 3 H2S(g)
Plan: Calculate moles of Aluminum Sulfide using its molar mass, then
from the equation, calculate the moles of Water, and then the moles of
Hydrogen Sulfide, and finally the mass of Hydrogen Sulfide using
it’s molecular weight.
Solution:
a) molar mass of Aluminum Sulfide = 150.17 g / mol
moles Al2S3 = 65.80 g Al2S3 x 1 mol Al2S3 = 0.4382 moles Al2S3
150.17 g Al2S3
Calculating Reactants and Products in a
Chemical Reaction - II
a) cont.:
Al2S3 (s) + 6 H2O(l)
!
H2O: 0.4382 moles Al2S3 x
2 Al(OH)3 (s) + 3 H2S(g)
___ moles H2O
1 mole Al2S3
___ moles H S
b)H2S: 0.4382 moles Al2S3 x 1 mole Al S 2
2 3
molar mass of H2S =
mass H2S =
Al(OH)3: 0.4382 moles Al2S3 x
molar mass of Al(OH)3 =
mass Al(OH)3 =
Calculating Reactants and Products in a
Chemical Reaction - II
a) cont. Al2S3 (s) + 6 H2O(l) ! 2 Al(OH)3 (s) + 3 H2S(g)
0.4382 moles Al2S3 x
6 moles H2O = 2.629 moles H2O
1 mole Al2S3
b) 0.4382 moles Al2S3 x 3 moles H2S = 1.314 moles H2S
1 mole Al2S3
molar mass of H2S = 34.09 g / mol
mass H2S = 1.314 moles H2S x 34.09 g H2S = 44.81 g H2S
1 mole H2S
0.4382 moles Al2S3 x 2 moles Al(OH)3 = 0.8764 moles Al(OH)3
1 mole Al2S3
molar mass of Al(OH)3 = 78.00 g / mol
mass Al(OH)3 = 0.8764 moles Al(OH)3 x 78.00 g Al(OH)3 =
1 mole Al(OH)3
= 68.36 g Al(OH)3
Reactions in Aqueous Solutions
often give precipitates (solids) which
can be weighed.
• Example:
Pb(NO3)2(aq) + 2 NaI(aq) ! PbI2(s) + 2 NaNO3(aq)
Lead iodide is a Bright yellow precipitate
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - I
Problem: Calcium Phosphate could be prepared in the following
reaction sequence:
POW!!
3 P4 (s) + 10 KClO3 (s)
3 P4O10 (s) + 10 KCl (s)
P4O10 (s) + 6 H2O (l)
2 H3PO4 (aq) + 3 Ca(OH)2 (aq)
4 H3PO4 (aq)
6 H2O(aq) + Ca3(PO4)2 (s)
Given: 15.5 g P4 and sufficient KClO3 , H2O and Ca(OH)2. What mass
of Calcium Phosphate could be formed?
Plan: (1) Calculate moles of P4.
(2) Use molar ratios to get moles of Ca3(PO4)2.
(3) Convert the moles of product back into mass by using the
molar mass of Calcium Phosphate. DEMO: Match head reaction- Somorjai
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - II
Solution:
moles of P4 =
For Reaction #1 [ 3 P4 (s) + 10 KClO3 (s)
For Reaction #2 [ 1 P4O10 (s) + 6 H2O (l)
For Reaction #3 [ 2 H3PO4 + 3 Ca(OH)2
moles Ca3(PO4)2 =
moles P4 x
3 P4O10 (s) + 10 KCl (s) ]
4 H3PO4 (aq) ]
1 Ca3(PO4)2 + 6 H2O]
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - II
Solution:
1 mole P4
moles of Phosphorous = 15.50 g P4 x 123.88 g P4 = 0.1251 mol P4
For Reaction #1 [ 3 P4 (s) + 10 KClO3 (s)
For Reaction #2 [ 1 P4O10 (s) + 6 H2O (l)
For Reaction #3 [ 2 H3PO4 + 3 Ca(OH)2
3 P4O10 (s) + 10 KCl (s) ]
4 H3PO4 (aq) ]
1 Ca3(PO4)2 + 6 H2O]
0.1251 moles P4 x 3 moles P4O10 x 4 moles H3PO4 x 1 mole Ca3(PO4)2
3 moles P4
1 mole P4O10
2 moles H3PO4
= 0.2502 moles Ca3(PO4)2
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - III
Molar mass of Ca3(PO4)2 = 310.18 g mole
Mass of Ca3(PO4)2 product =
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - III
Molar mass of Ca3(PO4)2 = 310.18 g mole
310.18 g Ca3(PO4)2
mass of product = 0.2502 moles Ca3(PO4)2 x
=
1 mole Ca3(PO4)2
= 77.61 g Ca3(PO4)2
Balanced
reaction!
Defines
stoichiometric
ratios!
Unbalanced (i.e., non-stoichiometric)
mixture!
Limited by syrup!
Molecular model: N2 molecules require
3H2 molecules for the reaction
N2 (g) + 3 H2 (g)
2 NH3 (g)
Figure 3.9: Hydrogen and Nitrogen reacting to
form Ammonia, the Haber process
N2 (g) + 3 H2 (g)
!
2 NH3 (g)
Limiting Reactant Problem:
A Sample Problem
Problem: A fuel mixture used in the early days of rocketry is composed
of two liquids, hydrazine (N2H4) and dinitrogen tetraoxide (N2O4). They
ignite on contact ( hypergolic!) to form nitrogen gas and water vapor.
How many grams of nitrogen gas form when exactly 1.00 x 102 g N2H4
and 2.00 x 102 g N2O4 are mixed?
Plan: First write the balanced equation. Since amounts of both reactants
are given, it is a limiting reactant problem. Calculate the moles of each
reactant, and then divide by the equation coefficient to find which is
limiting and use that one to calculate the moles of nitrogen gas, then
calculate mass using the molecular weight of nitrogen gas.
Solution: first write reaction from words above and balance it:
2 N2H4 (l) + N2O4 (l)
3 N2 (g) + 4 H2O (g) + Energy
Sample Problem cont.
Now get moles of each reactant:
molar mass N2H4 = ( 2 x 14.01 + 4 x 1.008 ) = 32.05 g/mol
molar mass N2O4 = ( 2 x 14.01 + 4 x 16.00 ) = 92.02 g/mol
2g
1.00
x
10
Moles N2H4 =
32.05 g/mol
=
2.00 x 102 g
Moles N2O4 =
=
92.02 g/mol
Divide by coefficients to see how many moles “per mole as written”,
to decide which is limiting reagent.
N2 yield in moles =
Mass of N2 =
Sample Problem cont.
2 N2H4 (l) + N2O4 (l) !
3 N2 (g) + 4 H2O (g)
molar mass N2H4 = ( 2 x 14.01 + 4 x 1.008 ) = 32.05 g/mol
molar mass N2O4 = ( 2 x 14.01 + 4 x 16.00 ) = 92.02 g/mol
2g
1.00
x
10
Moles N2H4 =
32.05 g/mol
= 3.12 moles N2H4
2.00 x 102 g
Moles N2O4 =
= 2.17 moles N2O4
92.02 g/mol
Divide by coefficients: 3.12 mol / 2 = 1.56 mol N2H4
2.17 mol / 1 = 2.17 mol N2O4
3 mol N2
Nitrogen yielded = 3.12 mol N2H4 x 2 mol N H
2 4
Limiting !
= 4.68 moles N2
Mass of Nitrogen = 4.68 moles N2 x 28.02 g N2 / mol = 131 g N2
used extensively to treat bacterial
infections since 1961.
Acid - Metal Limiting Reactant - I
• 2Al(s) + 6HCl(g)
2AlCl3(s) + 3H2(g)
Consider the reaction above. If we react 30.0 g Al and
20.0 g HCl, how many moles of aluminum chloride
will be formed?
• 30.0 g Al
• 20.0g HCl
• Limiting reactant = the one w/ fewest “equivalents” =
Acid - Metal Limiting Reactant - I
• 2Al(s) + 6HCl(g)
2AlCl3(s) + 3H2(g)
Consider the reaction above. If we react 30.0 g Al and
20.0 g HCl, how many moles of aluminum chloride
will be formed?
• 30.0 g Al / (26.98 g Al / mol Al) = 1.11 mol Al
1.11 mol Al / 2 = 0.555 equivalents Al
• 20.0g HCl / (36.5gHCl / mol HCl) = 0.548 mol HCl
O.548 mol HCl / 6 = 0.0913 equivalents HCl
• HCl is smaller therefore the Limiting reactant!
Acid - Metal Limiting Reactant - II
• Since 6 moles of HCl yield 2 moles of AlCl3
________ moles of HCl will yield:
Acid - Metal Limiting Reactant - II
• Coefficients tell us that 6 moles of HCl yield 2 moles of AlCl3
Thus, 0.548 moles of HCl will yield:
0.548 mol HCl x (2 moles of AlCl3 / 6 mol HCl)
= 0.183 mol of AlCl3
Limiting Reactant Problems
aA + bB + cC
dD + eE + f F
Steps to solve
1) Identify it as a limiting reactant problem - Information on the:
mass, number of moles, number of molecules, volume and
molarity of a solution is given for more than one reactant!
2) Calculate moles of each reactant!
3) Divide the moles of each reactant by stoic. coefficient (a,b,c etc...)!
4) Whichever is smallest, that reactant is the limiting reactant!
5) Use the limiting reactant to calculate the moles of product
desired then convert to the units needed (moles, mass, volume,
number of atoms etc....)!
Ostwald Process Limiting Reactant Problem
• What mass of NO could be formed by the reaction 30.0 g of
ammonia gas and 40.0 g of oxygen gas w/ the rxn below?
4NH3 (g) + 5 O2 (g)
4NO(g) + 6 H2O(g)
• 30.0g NH3
• 40.0g O2
• _______ fewer, therefore ______ is the Limiting Reagent!
• Moles NO formed =
• Mass NO =
Ostwald Process Limiting Reactant Problem
• What mass of NO could be formed by the reaction 30.0g of
ammonia gas and 40.0g of oxygen gas w/ the rxn below?
4NH3 (g) + 5 O2 (g)
4NO(g) + 6 H2O(g)
• 30.0g NH3 / (17.0g NH3/mol NH3) = 1.76 mol NH3
1.76 mol NH3 / 4 = 0.44 eq. NH3
• 40.0g O2 / (32.0g O2 /mol O2) = 1.25 mol O2
1.25 mol O2 / 5 = 0.25 eq O2
• Oxygen fewer, therefore oxygen is the Limiting Reagent!
• 1.25 mol O2 x 4 mol NO = 1.00 mol NO
5 mol O2
• mass NO = 1.00 mol NO x 30.0 g NO
1 mol NO
= 30.0 g NO
Flowchart : Solving a stoichiometry problem
involving masses of reactants
This mole ratio is called the
ratio of “stoichiometric coefficients”
Chemical Reactions in Practice: Theoretical,
Actual, and Percent Yields
Theoretical yield: The amount of product indicated by the
stoichiometrically equivalent molar ratio in the balanced equation.
Side Reactions: These form smaller amounts of different products that
take away from the theoretical yield of the main product.
Actual yield: The actual amount of product that is obtained.
Percent yield (%Yield): definition:
% Yield =
or
Actual Yield (mass or moles)
x 100%
Theoretical Yield (mass or moles)
Actual yield = Theoretical yield x (% Yield / 100%)
Percent Yield Problem:
Problem: Given the chemical reaction between iron and water to form
the iron oxide, Fe3O4 and hydrogen gas given below. If 4.55 g of iron is
reacted with sufficient water to react all of the iron to form rust, what is
the percent yield if only 6.02 g of the oxide are formed?
Plan: Calculate the theoretical yield and use it to calculate the percent
yield, using the actual yield.
Solution:
3 Fe + 4 H O
Fe O + 4 H
(s)
2
(l)
3
Moles Fe =
Theoretical moles Fe3O4 =
Theoretical mass Fe3O4 =
Percent Yield = Actual Yield x 100% =
Theoretical Yield
4 (s)
2 (g)
Percent Yield Problem:
Problem: Given the chemical reaction between Iron and water to form
the iron oxide, Fe3O4 and hydrogen gas given below. If 4.55 g of iron is
reacted with sufficent water to react all of the iron to form rust, what is
the percent yield if only 6.02 g of the oxide are formed?
Plan: Calculate the theoretical yield and use it to calculate the percent
yield, using the actual yield.
Solution:
3 Fe + 4 H O
Fe O + 4 H
(s)
2
(l)
3
4 (s)
2 (g)
4.55 g Fe = 0.081468 mol Fe = 0.0815 mol Fe
55.85 g Fe
mol Fe
0.0815 mol Fe x 1 mol Fe3O4 = 0.0272 mol Fe3O4
3 mol Fe
0.0272 mol Fe3O4 x 231.55 g Fe3O4 = 6.30 g Fe3O4
1 mol Fe3O4
Percent Yield = Actual Yield x 100% = 6.02 g Fe3O4 x 100% =
Theoretical Yield
6.30 g Fe3O4
95.6 %
Percent Yield / Limiting Reactant Problem - I
Problem: Ammonia is produced by the Haber process using nitrogen
and hydrogen Gas. If 85.90g of nitrogen are reacted with
21.66 g hydrogen and the reaction yielded 98.67 g of
ammonia what was the percent yield of the reaction.
N2 (g) + 3 H2 (g)
2 NH3 (g)
Plan: Since we are given the masses of both reactants, this is a limiting
reactant problem. First determine which is the limiting reagent
then calculate the theoretical yield, and then the percent yield.
Solution:
moles N2 =
moles H2 =
eqs N2 =
eqs H2 =
Percent Yield / Limiting Reactant Problem - I
Problem: Ammonia is produced by the Haber process using nitrogen
and hydrogen Gas. If 85.90g of nitrogen are reacted with
21.66 g hydrogen and the reaction yielded 98.67 g of
ammonia what was the percent yield of the reaction.
N2 (g) + 3 H2 (g)
2 NH3 (g)
Plan: Since we are given the masses of both reactants, this is a limiting
reactant problem. First determine which is the limiting reagent
then calculate the theoretical yield, and then the percent yield.
Solution: Moles of Nitrogen and Hydrogen:
Divide by coefficient
moles N2 = 85.90 g N2 = 3.066 mol N2 to get eqs. of each:
28.02 g N2
3.066 g N2 = 3.066eq
1 mole N2
1
moles H2 = 21.66 g H2 = 10.74 mol H2
2.016 g H2
10.74 g H2 = 3.582eq
1 mole H2
3
Percent Yield/Limiting Reactant Problem - II
Solution Cont.
N2 (g) + 3 H2 (g)
2 NH3 (g)
We have 3.066 moles of Nitrogen, and it is limiting, therefore the
theoretical yield of ammonia is:
mol NH3 = 3.066 mol N2 x
mass NH3 =
Percent Yield =
Percent Yield =
Actual Yield x 100%
Theoretical Yield
98.67 g NH3
g NH3
x 100% =
Percent Yield/Limiting Reactant Problem - II
Solution Cont.
N2 (g) + 3 H2 (g)
2 NH3 (g)
We have 3.066 moles of Nitrogen, and it is limiting, therefore the
theoretical yield of ammonia is:
2 mol NH3 = 6.132 mol NH
3.066 mol N2 x
3
1 mol N2
(Theoretical Yield)
17.03 g NH3
= 104.427 g NH3
1 mol NH3
(Theoretical Yield)
Actual Yield x 100%
Percent Yield =
Theoretical Yield
6.132 mol NH3 x
Percent Yield =
98.67 g NH3
104.427 g NH3
x 100% =
94.49 %