Download - Catalyst

Chapter 3 – Stoichiometry
Mole - Mass Relationships in Chemical Systems
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
Atomic Masses
The Mole
Molar Mass
Percent Composition of Compounds
Determining the Formula of a Compound
Chemical Equations
Balancing Chemical equations
Stoichiometric Calculations:
Amounts of Reactants and Products
3.9 Calculations Involving a Limiting Reactant
1
Atomic Definitions
Atomic Mass Unit (amu) = 1/12 the mass of a carbon-12 atom; on
this scale hydrogen has a mass of 1.008 amu
(unified atomic mass unit...“u”)
Dalton (D) = another name for the atomic mass unit; on this scale
12C has a mass of 12.00 Daltons;1 Dalton = 1 amu
Isotopic Mass = The relative mass of an isotope compared 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.
2
How Do We Determine Atomic Mass?
GRAVIMETRY ® weighing and analyzing an element
or compound
MASS SPECTROMETRY ® ionization of elements and
deflection of their path by applied magnetic field;
RATIO of atomic masses determined; accurate
comparison of masses of elements and their
isotopes.
3
4
Operation of a Mass Spectrometer
3.1)
A
E
B
D
C
A.
B.
Sample vaporized (made into a gas)
Electron gun removes an electron, forming a positively charged species
C.
Positive ion accelerated through an electric field
D.
E.
Magnetic deflection ◊ Lighter ions deflected more than heavier ones.
Observe impact on detector
Figure 3.2a: Peaks of neon
from injected sample
THREE “PEAKS”, so
THREE ISOTOPES!
5
Figure 3.2b: Relative
peak areas
This gives
91% 20Ne
the isotopic
0.3% 21Ne
composition
22Ne
9%
of Ne.
6
What is the average atomic mass of the three stable
isotopes of neon: 20Ne ( 90.70%), 21Ne (0.30%), and
22Ne (9.00%)?
1.
2.
3.
4.
5.
20 amu
20.2 amu
21 amu
21.1 amu
22 amu
7
Calculating the Average Atomic Mass of an Element
Problem: Calculate the average atomic mass of the three stable
isotopes of magnesium:
(11.1%).
24Mg
24
Mg
(78.7%)
(78.7%)
(10.2%)
(10.2%)
26Mg (11.1%)
26
Mg (11.1%)
25Mg
25
Mg
24Mg
( 78.7%), 25Mg (10.2%), and 26Mg
23.98504 amu
amu x 0.787 = 18.876226 amu
23.98504
24.98584 amu
amu x 0.102 = 2.548556 amu
24.98584
25.98636 amu
amu x 0.111 = 2.884486 amu
25.98636
Total = 24.309268 amu
With appropriate significant digits ◊ 24.3 amu
8
What is the abundance of the two bromine isotopes,
= 78.918336 amu and 81Br = 80.91629 amu, given
that the average mass of bromine is 79.904 amu?
79Br
1.
79Br
= 50.664% ,
= 49.334%
79Br = 50.67% ,
81Br = 49.33%
79Br = 50.7% ,
81Br = 49.3%
81Br
2.
3.
9
Problem: Calculate the abundance of the two bromine isotopes,
79Br = 78.918336 amu and 81Br = 80.91629 amu, given that the
average mass of bromine is 79.904 amu.
Plan: Let the abundance of 79Br = X and of 81Br = Y and X + Y = 1.0
Solution:
X(78.918336) + Y(80.91629) = 79.904
X + Y = 1.00, therefore X = 1.00 – Y
(1.00 - Y)(78.918336) + Y(80.91629) = 79.904
78.918336 - 78.918336 Y + 80.91629 Y = 79.904
1.997954 Y = 0.985664 ◊ Y = 0.49334
X = 1.00 - Y = 1.00 - 0.4933 = 0.50666
%X = % 79Br = 0.5067 x 100 =
%Y = % 81Br = 0.4933 x 100 =
50.666% = 79Br
49.334% = 81Br
10
Isotopes of Hydrogen
Natural
Abundance
•
•
•
1 H
1
2 H
1
3 H
1
1 Proton
0 Neutrons 99.985 % 1.00782503 amu
(D) 1 Proton
1 Neutron
0.015 % 2.01410178 amu
(T) 1 Proton
2 Neutrons
----------------The average mass of Hydrogen is 1.008 amu
3H
radioactive with a half life of 12 years
H2O normal water “light water “
mass = 18.0 g/mole , BP = 100.000000C
D2O “heavy water”
mass = 20.0 g/mole , BP = 101.42 0C
11
What will happen to the ball
of H2O? D2O?
1.
2.
3.
4.
H2O : sink, D2O : float
H2O : float, D2O : sink
H2O : float, D2O : float
H2O : sink, D2O : sink
12
The mole…
NOT this kind of mole
13
MOLE – just a number, like a dozen
The amount of substance that contains as many
elementary particles as there are atoms in exactly
12 grams of carbon -12 (i.e., 12C).
1 mole = 6.022045 x 1023 particles
(atoms, molecules, ions, marbles, apples, etc…)
~ 100 million x 100 million x 100 million
14
Avogadro’s Number (NA)
NA = 6.022045 x 1023 = # of particles
(atoms, molecules, ions,…) in one mole
of that substance.
Conversion factor:
6.022 x 1023 “whatever” = 1
mol
15
One mole of
common
substances
CaCO3:100.09g
Oxygen gas:
32.00g
Note: one mole
of oxygen gas (O2)
would occupy
22.4 L at room
temperature.
Copper: 63.55g
Water:18.02g
16
Dozen – mass
relationships
12 red marbles (7g each) = 84g
12 yellow marbles (4g each) = 48g
Mole – mass
relationships
6.022 x 1023 atoms Fe = 55.85g Fe
6.022 x 1023 atoms S = 32.07g S
17
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 amounts are
in the gram range and easy to weigh.
Conventions:
1 mole of 12C atoms weighs exactly 12 g
1 atom of 12C weighs exactly 12 amu
(amu = atomic mass unit ≈ mass of a proton or neutron)
18
Mole - Mass Relationships of Elements
Element
Atomic Mass
1 Mole of Atoms
Molar Mass
1 atom of H = 1.008 amu
6.022 x 1023 atoms
= 1.008 g H
1 atom of Fe = 55.85 amu
6.022 x 1023 atoms
= 55.85 g Fe
1 atom of S = 32.07 amu
6.022 x 1023 atoms = 32.07 g S
1 atom of O = 16.00 amu
6.022 x 1023 atoms = 16.00 g O
amu and g/mol are numerically the same!
Molecular mass:
1 molecule of O2 = 16.00 x 2 = 32.00 amu
1 mole of O2 = 6.022 x 1023 molecules = 32.00 g
1 molecule of S8 = 32.07 x 8 = 256.56 amu
1 mole of S8 = 6.022 x 1023 molecules = 256.56 g
19
5
What is the molar mass of
water (H2O)?
1.
2.
3.
4.
18.016 g/mol
18.02 g/mol
10.00 g/mol
10.000 g/mol
20
Molecular Mass - Molar Mass (MM)
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 ) + (1 x atomic mass of O)
= 2 (1.008 amu) + 16.00 amu = 18.02 amu
Mass of one molecule 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
18.02
21
Calculating the Number of Moles and Atoms
in a Given Mass of Element
Problem:
Tungsten (W) is the element used as the filament in incandescent
light bulbs and has the highest melting point of any element,
3695 K (3422oC). How many moles of tungsten, and atoms of the
element, are present in a 35.0 mg sample of the metal?
Plan:
Convert Mass to
Moles
Convert Moles
to Atoms
Molar Mass
Avogadro’s Number
From the Periodic Table, we find that
the molar mass of W is 183.85 g/mol.
22
Calculating the Number of Moles and Atoms
in a Given Mass of Element
Problem:
Tungsten (W) is the element used as the filament in incandescent
light bulbs and has the highest melting point of any element,
3680oC. How many moles of tungsten, and atoms of the element,
are present in a 35.0 mg sample of the metal?
Solution:
Moles of W = 35.0 x 10-3 g W
1 mol W = 0.00019037 mol
183.85 g W
= 1.90 x 10 - 4 mol
No. of W atoms = 1.90 x 10 - 4 mol W 6.022 x 1023 atoms =
1 mol W
1.15 x 1020 atoms of tungsten
23
.
24
Calculating the Moles in a Given
Mass of a Compound
Problem: Sodium phosphate is a component of some detergents.
How many moles 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 subscripts.
25
5
What is the chemical formula for
sodium phosphate?
1.
2.
3.
4.
5.
NaPO3
NaPO4
Na2PO4
Na3PO4
None of the
above
26
5
What is the molecular mass for
sodium phosphate (Na3PO4)?
1.
2.
3.
4.
5.
163.94 g/mol
164 g/mol
164.00 g/mol
80 g/mol
80.00 g/mol
27
Calculating the Moles in a Given
Mass of a Compound
Problem: Sodium phosphate is a component of some detergents.
How many moles 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 subscripts.
Solution: The formula is Na3PO4.
Calculating the molar mass:
MM = 3 x 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
28
5
How many moles of sodium phosphate are
in a 38.6 g sample?
MM=163.94 g/mol
1.
2.
3.
4.
0.235 mol
0.24 mol
0.230 mol
0.240 mol
29
Calculating the Moles in a Given
Mass of a Compound
Problem: Sodium phosphate is a component of some detergents.
How many moles 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 subscripts.
Solution: The formula is Na3PO4. The molar mass is 163.94 g/mol.
Calculating the moles in a 38.6 g sample:
moles of Na3PO4 = 38.6 g sample 1 mol
163.94 g sample
= 0.235 mol
30
MM (g/mol)
MM (g/mol)
31
Atoms
6.022 x 1023
mol
Molecular
Formula
Moles
Molecules
6.022 x 1023
mol
32
33
Mass Fraction and Mass %
Mass of Red Balls = 3 balls x 3.0 g/ball = 9.0 g
Fraction Red
Red == 9.0 g / 16.0 g total = 0.56
Mass Fraction
Mass % Red = 0.56 x 100 = 56% red
Fraction Purple
and Mass
Mass Fraction
= % Purple =
2 balls x 2.0 g/ball / (16.0 g total) = 0.25 = 25%
fraction yellow
yellow == 3x1.0/16.0 = 0.19
Similarly, mass fraction
Check: 56% + 25% + 19% = 100%
34
Calculate the Percent Composition of
Sulfuric Acid H2SO4
Molar mass of sulfuric acid =
2(1.008 g) + 1(32.07 g) + 4(16.00 g) = 98.09 g/mol
%H = 2(1.008 g H)
98.09 g
x 100% = 2.06% H
%S = 1(32.07 g S) x 100% = 32.69% S
98.09 g
%O = 4(16.00 g O)
98.09 g
x 100% = 65.25% O
Check = 100.00%
35
Calculating Mass Percentage and Masses of
Elements in a Sample of a Compound
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?
36
What is the mass percent of each element
in sucrose (C12H22O11)?
1.
2.
3.
4.
%C=26.471%,
%H=8.824%,
%O=64.71%
%C=26.47%,
%H=8.83%,
%O=64.71%
%C=42.10%,
%H=6.48%,
%O=51.42%
%C=42.103%,
%H=6.479%,
%O=51.42%
37
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 fraction of C = mass of C per mol sucrose = 144.12 g C/mol
mass of 1 mol sucrose
342.30 g sucrose/mol
= 0.42103
To find mass % of C = 0.4210 x 100% = 42.103%
38
Calculating Mass Percents and Masses of
Elements in a Sample of Compound - II
(a) continued
Mass % of H =
mol H x M of H
x100% = 22 x 1.008 g H x 100%
mass of 1 mol sucrose
342.30 g
= 6.479% H
Mass % of O =
mol O x M of O
x100% = 11 x 16.00 g O x 100%
mass of 1 mol sucrose
342.30 g
= 51.42% O
39
How many grams of carbon are in 24.35 g
of sucrose?
1.
2.
3.
4.
1025 g
1030 g
10.25 g
10.3 g
40
Calculating Mass Percentage and Masses of
Elements in a Sample of a Compound
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 = 42.103%
(b) Determining the mass of carbon in 24.35 g sucrose:
Mass (g) of C = mass of sucrose x ( mass fraction of C in sucrose)
Mass (g) of C = 24.35 g sucrose 0.4210 g C
1 g sucrose
= 10.25 g C
41
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.
42
Some Examples of Compounds with the
Same Elemental Ratios
Empirical Formula
CH2 (unsaturated hydrocarbons)
Molecular Formula
C2H4 C3H6 C4H8
OH or HO
H2O2
S
S8
P
P4
Cl
Cl2
CH2O (carbohydrates)
C6H12O6
43
44
Condensed structural CH CH OH
3
2
formula
CH3OCH3
45
Elemental Analysis
Decomposition or combustion analysis is used to determine
the mass of each type of element present in a compound.
Figure 3.5:
Schematic Diagram of a Combustion Analysis Device
46
Steps for Determining Empirical Formulas
Mass (g) of each element in sample
÷ MM (g/mol ) for that element
Moles of each element
Use # of moles as subscripts.
Preliminary formula
Change to integer subscripts:
÷ by smallest subscript, convert
to whole # subscripts.
Empirical formula
47
Determining Empirical Formulas from
Measured Masses of Elements - I
Problem: The elemental analysis of a sample 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:
Convert mass to moles
Construct preliminary
formula
Moles = mass/MM
Convert to empirical
formula
Divide each moles by smallest
number of moles
48
Determining Empirical Formulas from
Measured Masses of Elements - I
Problem: The elemental analysis of a sample 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?
Solution: Finding the moles of the elements:
= 0.2469 mol Na
1 mol Na
22.99 g Na
Moles of Cr = 6.420 g Cr 1 mol Cr
= 0.1235 mol Cr
52.00 g Cr
Moles of O = 7.902 g O 1 mol O
= 0.4939 mol O
16.00 g O
Moles of Na = 5.677 g Na
49
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)
Na1.999 Cr1.000 O3.999
Rounding off to whole numbers:
Na2CrO4
sodium chromate
50
Determining the Molecular Formula from
Elemental Composition and Molar Mass - I
Problem: The sugar burned for energy in cells of the body is glucose
(MM = 180.16 g/mol), elemental analysis shows that it contains 40.00
mass % C, 6.729 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 we have 100 g of the compound.
Convert mass to moles
Construct preliminary
formula
Moles = mass/MM
Convert to empirical
formula
Divide each moles by smallest
number of moles
51
Determining the Molecular Formula from
Elemental Composition and Molar Mass - I
Problem: The sugar burned for energy in cells of the body is glucose
(MM = 180.16 g/mol), elemental analysis shows that it contains 40.00
mass % C, 6.729 mass % H, and 53.27 mass % O.
(a) Determine the empirical formula of glucose.
(b) Determine the molecular formula.
Remember: We are only given mass % and no weight of the
compound so we will assume 100 g of the compound.
Solution:
Mass carbon = 40.00% x 100 g/100% = 40.00 g C
Mass hydrogen = 6.729% x 100 g/100 = 6.729 g H
Mass oxygen = 53.27% x 100 g/100% = 53.27 g O
99.999 g cmpd
52
What is the empirical formula of glucose?
40.00 mass % C, 6.729 mass % H, and 53.27 mass % O
1.
2.
3.
4.
C3H6O3
CH2O
C3H7O3
CH7O
53
Determining the Molecular Formula from
Elemental Composition and Molar Mass - II
Converting from grams of elements to moles:
1 mole C = 3.331 moles C
12.01 g C
Moles of H = 6.719 g H 1 mol H = 6.666 moles H
1.008 g H
Moles of O = 53.27 g O 1 mol O = 3.330 moles O
16.00 g O
Moles of C = 40.00 g C
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
Empirical Formula!!
54
Determining the Molecular Formula from
Elemental Composition and Molar Mass - I
Problem: The sugar burned for energy in cells of the body is glucose
(MM = 180.16 g/mol), elemental analysis shows that it contains 40.00
mass % C, 6.729 mass % H, and 53.27 mass % O.
(a) Determine the empirical formula of glucose: CH2O
(b) Determine the molecular formula.
Plan: We are only given mass %, and no weight of the compound so we
will assume we have 100 g of the compound.
Determine the
empirical formula
Determine molar mass
of empirical formula
MOLECULAR FORMULA
Multiply empirical
formula by “n”
Divide molecular
formula mass by
empirical formula
mass
n
55
What is the molecular formula of glucose?
CH2O, MM=180.16 g/mol
1.
2.
3.
4.
CH2O
C3H6O3
C6H12O6
C9H18O9
56
Determining the Molecular Formula from
Elemental Composition and Molar Mass - III
(b) Determining the molecular formula:
The molar mass of the empirical formula is:
(1*C) + (2*H) + (1*O) = (1*12.01) + (2*1.008) + (1*16.00)
= 30.03 g/mol
whole-number multiple =
MM of the compound
empirical molar mass
Stated in the
problem
180.16 g/mol
= 6.00 = 6
30.03 g/emp. mol
Therefore the molecular formula is:
C1x6H2x6O1x6 =
C6H12O6
57
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 the empirical formula of ascorbic acid.
Plan:
Calculate mass of C
and H in CO2 and
H2O
Generic:
Calculate mass of all
elements (except
common element) in
respective compounds
Subtract masses of C and
H from compound mass
to determine mass of O
Subtract known masses of all
elements from compound
mass to determine mass of
common element
Calculate
empirical formula
Calculate
empirical formula
58
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 the empirical formula of ascorbic acid.
Mass of C and H in the CO2 and H2O, respectively:
9.74 x10-3g CO2
= 2.65 x 10-3 g C
12.01 g C
44.01 g CO2
2.64 x10-3g H2O
2.016 g H
18.02 g H2O
= 2.95 x 10-4 g H
Mass of O:
6.49 x10-3 g sample - 2.65 x10-3 g C - 2.95 x10-4g H = 3.55 x 10-3 g O
59
What is the empirical formula
for ascorbic acid?
2.65 x 10-3 g C, 2.95 x 10-4 g H, 3.55 x 10-3 g O
1. CH1.33O
2. C2HO
3. C3H4O3
60
Vitamin C combustion - II
Now we convert to moles:
-3
C: 2.65 x 10 g C 1 mol C
12.011g C
H: 2.95 x 10-3 g H 1 mol H
1.008g H
-3
O: 3.55 x 10 g O 1 mol O
16.00g O
-4
= 2.21 x 10 mol C
= 2.93 x 10-4 mol H
= 2.22 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 = 1.33
(to get ~integers)
H = 3.99 = 4.0
O = 1.00
O = 3.00 = 3.0
C3H4O3
61
If the empirical formula of ascorbic acid is C3H4O3,
and the molecular mass of ascorbic acid is
176 g/mol, what is the molecular formula?
1.
2.
3.
4.
5.
C3H4O3
C6H8O6
C9H12O9
C12H16O12
None of the above
62
Chemical Equations
Qualitative information about a chemical reaction:
Reactants
States of Matter:
Products
(s) solid
(l) liquid
(g) gaseous
(aq) aqueous
2 H2 (g) + O2 (g)
2 H2O (g)
But also Quantitative Information!
63
(Like Table 3.2)
64
How to Balance Equations
Mass Balance (or Atom Balance)- same number of each element
on each side of the equation:
(1) start with largest/most complicated molecule
(2) progress to other elements, leaving lone elements for last
(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)
1 CH4 (g) + 2 O2 (g)
1 CO2 (g) + 2 H2O (g)
65
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 compounds,
with blanks before each compound. Begin element balance,
putting 1 on the most complex compound first, and save oxygen
until last!
Solution:
C6H14 (l) +
O2 (g)
CO2 (g) +
H2O(g) + Energy
Begin with one C6H14 molecule which says that we will get 6 CO2’s!
1 CH
6 14 (l) +
O2 (g)
6 CO
2 (g) +
H2O(g) + Energy
66
Balancing Chemical Equations - II
The H atoms in the hexane will end up as H2O, and we have 14
H atoms. Since each water molecule has two H atoms, we will get
a total of 7 water molecules.
1 CH
6 14 (l) +
O2 (g)
6 CO
7 H O + Energy
2 (g) +
2 (g)
Since oxygen atoms only come as diatomic molecules (two O atoms,
O2),we must have even numbers of oxygen atoms on the product side.
We do not since we have 7 water molecules! Therefore, multiply
everything by 2, giving a total of 2 hexane molecules, 12 CO2
molecules, and 14 H2O molecules.
2 CH
6 14 (l) +
O2 (g)
12 CO
14 H O + Energy
2 (g) +
2 (g)
This now gives 24 O from the carbon dioxide, and 14 O atoms from the
water, which will be a total of 24+14 = 38 O or 19 O2 !
2 C6H14 (l) + 19 O2 (g)
12 CO2 (g) + 14 H2O(g) + Energy
67
Which of the following is always true for a
balanced chemical equation?
1.
2.
3.
4.
5.
Number of atoms in
reactants equal number
of atoms in products.
Volume of reactants
equals volume of
products.
Moles of reactants equal
moles of products.
Mass of reactant equal
mass of product.
More than one of the
above.
68
Chemical Equations &
Calculations
Atoms (Molecules)
Avogadro’s
Number
6.022 x
mol
Reactants
1023
Moles
Molar
Mass
Mass
g/mol
Products
69
MM (g/mol) of
compound A
MM (g/mol) of
compound B
70
Calculating Reactants and Products
in a Chemical Reaction
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) + _ H2O(l)
Plan:
Balance the
chemical reaction
Calculate the
moles of
aluminum sulfide
Convert to masses
using molar mass of
compounds
_ Al(OH)3 (s) + _ H2S(g)
Calculate moles of water
needed, using the
numbers in front of the
compounds
Calculate moles of H2S
and Al(OH)3 produced,
using the numbers in
front of the compounds
71
What are the coefficients in the
balanced equation?
_ Al2S3 (s) + _ H2O(l)
1.
2.
3.
4.
_ Al(OH)3 (s) + _ H2S(g)
1,3,2,3
1,3,2,6
1,6,2,3
2,6,4,3
72
Calculating Reactants and Products
in a Chemical Reaction
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)
Solution:
a) molar mass of aluminum sulfide = 150.17 g/mol
moles Al2S3 = 65.80 g Al2S3
mol Al2S3
150.17 g Al2S3
= 0.4382 moles Al2S3
Now think back to unit conversion and the information the
chemical equation provides……..
73
Calculating Reactants and Products
in a Chemical Reaction
a) cont.
0.4382 moles Al2S3 6 moles H2O
1 mole Al2S3
b) 0.4382 moles Al2S3 3 moles H2S
1 mole Al2S3
molar mass of H2S = 34.09 g/mol
= 2.629 moles H2O
= 1.314 moles H2S
mass H2S = 1.314 moles H2S 34.09 g H2S
= 44.81 g H2S
1 mole H2S
0.4382 moles Al2S3 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 78.00 g Al(OH)3 =
1 mole Al(OH)3
74
= 68.36 g Al(OH)3
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - I
Problem: Calcium phosphate could be prepared from phosphorus 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)
HISSSSS
4 H3PO4 (aq)
6 H2O(aq) + Ca3(PO4)2 (s)
Given 15.50 g P4 and sufficient KClO3 , H2O and Ca(OH)2. What mass
of calcium phosphate could be formed?
Plan:
(1) Calculate moles of P4 provided.
(2) Use molar ratios to get moles from P4 to Ca3(PO4)2.
(3) Convert the moles of product back into mass using the
molar mass of calcium phosphate.
75
How many moles of P4 are in
15.50 g? MM P = 30.97 g/mol
1.
2.
3.
4.
0.1251 mol
0.1251 g
0.5005 mol
0.5005 g
76
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - II
Solution:
moles of phosphorous = 15.50 g P4 1 mole P4
= 0.1251 mol P4
123.88 g P4
For Reaction #1: 3 P4 (s) + 10 KClO3 (s)
3 P4O10 (s) + 10 KCl (s)
For Reaction #2: 1 P4O10 (s) + 6 H2O (l)
For Reaction #3: 2 H3PO4 + 3 Ca(OH)2
4 H3PO4 (aq)
1 Ca3(PO4)2 + 6 H2O
0.1251 moles P4 3 moles P4O10 4x moles H3PO4 1 mole
x Ca3(PO4)2
3 moles P4
1 mole P4O10 2 moles H3PO4
= 0.2502 moles Ca3(PO4)2
Molar mass of Ca3(PO4)2 = 310.18 g/mol
mass of product = 0.2502 moles Ca3(PO4)2 310.18 g Ca3(PO4)2
1 mole Ca3(PO4)2 77
= 77.61 g Ca (PO )
3
4 2
3 P4(s) + 10 KClO3(s) ® 3 P4O10(s) + 10 KCl(s)
Heat of reaction = - 9,425 kJ
The head of "strike anywhere" matches contain an oxidizing agent
such as potassium chlorate together with tetraphosphorus trisulfide (P4S3),
glass and binder. The phosphorus sulfide is easily ignited, the potassium
chlorate decomposes to give oxygen, which in turn causes the phosphorus
sulfide to burn more vigorously.
The head of safety matches are made of an oxidizing agent such as
potassium chlorate, mixed with sulfur, fillers and glass powder. The side of
the box contains red phosphorus, binder and powdered glass. The heat
generated by friction when the match is struck causes a minute amount of
red phosphorus to be converted to white phosphorus, which ignites
spontaneously in air. This sets off the decomposition of potassium chlorate
to give oxygen and potassium chloride. The sulfur catches fire and ignites
the wood.
78
LIMITING REACTANT
The amount of reagent that limits the amount of product that
can be formed is the limiting reactant.
When chemicals are mixed in exact stoichiometric quantities,
none of the reagents is limiting.
So far, we have been assuming that most of the reactants are
available in excess of the amount needed to react with a fixed
amount of one of the reactants.
Read about the HABER process for producing ammonia (section
3.9 in the book) as an example of a limiting reactant.
79
LIMITING REAGENT
3N2 (g) + 9 H2 (g) ® 6 NH3 (g)
80
Figure 3.9: Hydrogen and nitrogen reaction to form ammonia
81
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 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, this is a limiting reactant problem!
Calculate the moles of each reactant. Pick one of the reactants and figure
out how much of the other you would need. If you have more than
enough of the second reagent, then the first is limiting. Use the moles of
the limiting reactant to calculate the moles of nitrogen gas that will be
formed. Finally, calculate the mass using the molecular weight of
nitrogen gas.
Solution:
2 N2H4 (l) + N2O4 (l)
3 N2 (g) + 4 H2O (g) + Energy
82
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
1.00 x 102 g
Moles N2H4 =
32.05 g/mol
Limiting !
= 3.12 moles N2H4
2
Moles N2O4 = 2.00 x 10 g = 2.17 moles N2O4
92.02 g/mol
For 3.12 moles N2H4 we would need:
3.12 mol N2H4 1 mol N2O4 = 1.56 moles N2O4 …we have 2.17moles
2 mol N2H4
Nitrogen yielded = 3.12 mol N2H4 3 mole N2 = 4.68 moles N2
2 mol N2H4
Mass of nitrogen = 4.68 moles N2 x 28.02 g N2/mol = 131 g N2 83
Which of these is always true for
limiting reagent questions?
1.
2.
3.
4.
It will say in the question
that it is a limiting
reactant question.
The equation provided
will be balanced.
There will be quantitative
information about more
than one of the reactants.
There will be information
about only products.
84
Acid - Metal Reaction
2 Al(s) +
HCl(g)
2 AlCl3(s) + H2(g)
Consider the reaction above.
Is it balanced? (An unbalanced reaction won’t
help you much.)
2 Al(s) + 6 HCl(g)
2 AlCl3(s) + 3 H2(g)
If we react 30.0 g Al and 20.0 g HCl, how many
moles of aluminum chloride will be formed?
85
Is this a Limiting Reagent
question?
1. Yes
2. No
86
Acid - Metal Limiting Reactant - I
2 Al(s) + 6 HCl(g)
→
2 AlCl3(s) + 3 H2(g)
If we react 30.0 g Al and 20.0 g HCl, how many moles of aluminum
chloride will be formed?
A shortcut…calculate the moles of each reactant and divide each # of
moles by the reactant coefficient in the balanced reaction to find
which reactant is limiting. Use the moles of that reactant to calculate
the moles of product formed.
30.0 g Al * (mol Al/26.98 g Al) = 1.11 mol Al
1.11 mol Al / 2 = 0.555 equivalents of Al
20.0 g HCl * (mol HCl/36.5 g HCl) = 0.548 mol HCl
0.548 mol HCl / 6 = 0.0913 equivalents of HCl
Fewer equivalents of HCl, so HCl is the limiting reactant!
87
Acid - Metal Limiting Reactant - II
2 Al(s) + 6 HCl(g) → 2 AlCl3(s) + 3 H2(g)
Since 6 moles of HCl yield 2 moles of AlCl3,
0.548 moles of HCl will yield:
0.548 mol HCl 2 moles of AlCl3 = 0.183 mol of AlCl3
6 mol HCl
88
Limiting Reactant Problems
aA + bB + cC
dD + eE + f F
Steps for solving:
1) Identify it as a limiting reactant problem - Information about the mass,
number of moles, number of molecules, or volume and molarity of a
solution is given for more than one reactant!
2) Calculate moles of each reactant!
3) Determine which reactant is limiting: either 1) calculate the amount of
other reactants needed and compare with what is available or
2) divide the moles of each reactant by stoic. coefficient (a, b, etc...)
and whichever is smallest, that reactant is the limiting reactant!
4) Use the limiting reactant to calculate the moles of product(s) that will
form or the moles of other reactants that will be used up, then convert
to the units needed (moles, mass, volume, number of atoms, etc....)!
89
Solving for Masses of Reactants and Products
Start here
90
Titrations as Examples of
Limiting Reactant Problems
• Before the endpoint the titrant is limiting and the
sample is in excess.
• At the endpoint there are stoichiometric quantities
(equivalent amounts) of both reactants.
• Past the endpoint the sample is limiting and the titrant
is in excess.
(endpoint ≈ equivalence point)
You’ll see this titration terminology in
Lab #2 and we’ll revisit it in Chapter 7.
91
Chemical Reactions in Practice:
Theoretical, Actual, and Percent Yields
Theoretical yield: The amount of product expected based on amounts of
reactants and stoichiometry (molar ratios) in the balanced equation.
Actual yield: The actual amount of product that is obtained.
Side Reactions: These form smaller amounts of different products that
reduce the actual yield of the product of interest.
Percent yield (%Yield):
% Yield =
Actual Yield (mass or moles)
Theoretical Yield (mass or moles)
x 100%
92
93
What is the actual yield of a chemical reaction if
the percent yield was 50%, and the theoretical
yield was 8 g product?
1.
2.
3.
4.
5.
6.
0.5 g product
1 g product
2 g product
4 g product
16 g product
None of the above
94
Percent Yield Calculation
Problem: Given the chemical reaction between iron and water to form
the iron oxide Fe3O4 and hydrogen gas, if 4.55 g of iron is reacted with
sufficient water to convert all of the iron to rust, what is the percent yield
if only 6.02 g of the oxide forms?
Plan: Balance the chemical reaction, calculate the theoretical yield and
use it and the actual yield to calculate the percent yield.
Solution:
_ Fe(s) + _ H2O(l)
_ Fe3O4 (s) + _ H2 (g)
95
What are the coefficients of the
balanced chemical reaction?
_ Fe(s) + _ H2O(l)
1.
2.
3.
4.
_ Fe3O4 (s) + _ H2 (g)
1,4,1,2
3,2,1,2
3,4,1,2
3,4,1,4
96
Percent Yield Calculation
Problem: Given the chemical reaction between iron and water to form
the iron oxide Fe3O4 and hydrogen gas, if 4.55 g of iron is reacted with
sufficient water to convert all of the iron to rust, what is the percent yield
if only 6.02 g of the oxide forms?
Plan: Balance the chemical reaction, calculate the theoretical yield and
use it and the actual yield to calculate the percent yield.
Solution:
3 Fe(s) + 4 H2O(l)
Fe3O4 (s) + 4 H2 (g)
97
What is the theoretical yield of
Fe3O4(s)?
Fe (s) = 4.55 g, MM Fe = 55.85 g/mol, MM O = 16.00 g/mol
1.
2.
3.
4.
0.0272 g
0.0272 mol
6.30 g
6.30 mol
98
Percent Yield Calculation
Problem: Given the chemical reaction between iron and water to form
the iron oxide Fe3O4 and hydrogen gas, if 4.55 g of iron is reacted with
sufficient water to convert all of the iron to rust, what is the percent yield
if only 6.02 g of the oxide forms?
Plan: Balance the chemical reaction, calculate the theoretical yield and
use it and the actual yield to calculate the percent yield.
Solution:
3 Fe(s) + 4 H2O(l)
Fe3O4 (s) + 4 H2 (g)
4.55 g Fe
= 0.081468 mol Fe = 0.0815 mol Fe
55.85 g Fe/mol
0.0815 mol Fe 1 mol Fe3O4 = 0.0272 mol Fe3O4
3 mol Fe
0.0272 mol Fe3O4 231.55 g Fe3O4 = 6.30 g Fe3O4
1 mol Fe3O4
6.02 g Fe3O4
Actual
Yield
Percent Yield =
x 100% =
x100%= 95.6 %
6.30 g Fe3O4
Theoretical Yield
99
Percent Yield Problem - I
Problem: Ammonia is produced by the Haber process using nitrogen
and hydrogen gas. If 85.90 g 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 finally the percent yield.
100
What is the limiting reactant?
N2 (g) = 85.92 g, H2 (g) = 21.66 g,
MM N = 14.01 g/mol, MM H = 1.008 g/mol
1.
2.
3.
4.
5.
N2 (g)
H2 (g)
NH3 (g)
N2 (l)
H2 (l)
101
Percent Yield Problem - I
Problem: Ammonia is produced by the Haber process using nitrogen
and hydrogen gas. If 85.90 g 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 finally the percent yield.
Solution:
Divide by coefficient
to get eqs. of each:
85.90 g N2
moles N2 =
= 3.066 mol N2 3.066 mol N2
= 3.066 eqs.
28.02 g N2/mol
1
LR
21.66 g H2
10.74
mol
H
2= 3.582 eqs.
moles H2 =
= 10.74 mol H2
2.016 g H2/mol
3
102
What is the theoretical yield of
NH3 (g)?
N2 (g)= 3.066 mol, H2 (g) =10.74 mol,
MM N = 14.01 g/mol, MM H = 1.008 g/mol
N2 (g) + 3 H2 (g)
1.
2.
3.
4.
2 NH3 (g)
6.132 g
6.132 mol
104.427 g
104.427 mol
103
Percent Yield 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:
3.066 mol N2
2 mol NH3 = 6.132 mol NH3
1 mol N2
6.132 mol NH3 17.03 g NH = 104.427 g NH3
3
1 mol NH3
Percent Yield =
Percent Yield =
(Theoretical Yield)
(Theoretical Yield)
Actual Yield
x 100%
Theoretical Yield
98.67 g NH3
104.427 g NH3
x 100% =
94.49 %
104