Chapter 3
Stoichiometry
Chapter 3 - Stoichiometry
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
Reaction of zinc and sulfur.
Figure 3.1: Mass spectrometer
Chemists using a mass spectrometer to
analyze for copper in blood plasma.
Source: USDA Agricultural Research Service
A herd of savanna-dwelling elephants
Source: Corbis
Figure 3.2: Relative intensities of the
signals recorded when natural neon is
injected into a mass spectrometer.
Figure 3.3: Mass spectrum of natural
copper
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
Isotope 12C the chosen standard.
Atomic Mass = “Atomic Weight” of an element is the average of the
masses of its naturally occurring isotopes weighted
according to their abundances.
Isotopes of Hydrogen
•
•
•
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 is 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
Element #8 : Oxygen, Isotopes
16
8O
8 Protons
99.759%
8 Neutrons
15.99491462 amu
•
17 O
8
8 Protons
0.037%
9 Neutrons
16.9997341 amu
•
18 O
8
8 Protons
0.204 %
10 Neutrons
17.999160 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
___________ 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
(%)
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
____________ amu
With Significant Digits = ___________ amu
Problem: Calculate the abundance of the two Bromine isotopes:
79Br = 78.918336 g/mol and 81Br = 80.91629 g/mol , given that
the average mass of Bromine is 79.904 g/mol.
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
or Y = 0.4933
X = 1.00 - Y = 1.00 - 0.4933 = 0.5067
%X = % 79Br = 0.5067 x 100% = 50.67% = 79Br
%Y = % 81Br = 0.4933 x 100% = 49.33% = 81Br
LIKE SAMPLE PROBLEM 3.2
During a perplexing dream one evening you come across
200 atoms of Einsteinium. What would be the total mass
of this substance in grams?
SOLUTION:
200 Atoms of Es X 252 AMU/Atom = 5.04 x 104 AMU
5.04 x 104 AMU x (1g / 6.022 x 1023 AMU) =
______________ g of Einsteinium
MOLE
• The Mole is based upon the following definition:
• The amount of substance that contains as many
elementary parts (atoms, molecules, or other?) as
there are atoms in exactly
• 12 grams of carbon -12.
• 1 Mole = 6.022045 x 1023 particles
Figure 3.4: One-mole samples of copper,
sulfur, mercury, and carbon
One mole
of
common
Substances
CaCO3
100.09 g
Oxygen
32.00 g
Copper
63.55 g
Water
18.02 g
Atoms
Molecular
Formula
Avogadro’s
Number
Molecules
6.022 x 1023
Moles
Moles
Mole - Mass Relationships of Elements
Element Atom/Molecule Mass
1 atom of H = 1.008 amu
Mole 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
1 molecule of O2 = 32.00 amu
1 mole of O2 = 32.00 g = 6.022 x 1023 molecule
1 molecule of S8 = 256.56 amu
1 mole of S8 = _______ g = 6.022 x 1023 molecules
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.
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 atomic mass of H ) + atomic mass of O
= 2 ( 1.008 g ) + 16.00 g = __________ g
18.02 g H2O = 6.022 x 1023 molecules of water = 1 mole H2O
LIKE SAMPLE PROBLEM 3.4
How many carbon atoms are present in a 2.0 g tablet of
Sildenafil citrate (C28H38N6O11S) ?
SOLUTION:
MW of Sildenafil citrate = 28 X 12 amu (C) +
38 X 1 amu (H) +
6 X 14 amu (N) +
11 X 16 amu (O) +
1 X 32 amu (S) =
666 AMU
LIKE SAMPLE PROBLEM 3.4 (cont…)
2.0 g (C28H38N6O11S) X 1 mol/666g =
3.0 X 10-3 mol (C28H38N6O11S)
3.0 X 10-3 mol (C28H38N6O11S)
X
6.022 X 1023 molecules / 1 mol (C28H38N6O11S) =
1.8 X 1021 molecules of C28H38N6O11S
1.8 X 1021 molecules of C28H38N6O11S X
28 atoms of C / 1 molecules of C28H38N6O11S =
Carbon 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 of W = 35.0 mg W x
1 mol W
183.9 g W
NO. of W atoms = 1.90 x 10 - 4 mol W x
= 0.00019032 mol
1.90 x 10 - 4 mol
6.022 x 1023 atoms
=
1 mole of W
= __________________ atoms of Tungsten
Calculating the Moles and Number of
Formula Units in a given Mass of Cpd.
Problem: Trisodium 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
= ______________ formula units
Flow Chart of Mass Percentage Calculation
Moles of X in one
mole of Compound
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 = 12 x 12.01 g C/mol =
144.12 g C/mol
mass of H = 22 x 1.008 g H/mol =
22.176 g H/mol
mass of O = 11 x 16.00 g O/mol =
176.00 g O/mol
342.296 g/mol
Finding the mass fraction of C in Sucrose & % C :
Mass Fraction of C = Total mass of C
= 144.12 g C
mass of 1 mole of sucrose
342.30 g Cpd
= 0.421034 To find mass % of C = 0.421034 x 100% = ______%
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 O x 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 0.421034 g C
1 g sucrose
=
C
Mol wt 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 N2 x 100%
80.05g
= 35.00% N
%H =
4.032g H2 x 100% =
80.05g
%O =
48.00g O2 x 100% = 59.96% O
80.05g
99.997%
5.037% H
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 H2) 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%
Penicillin is isolated from a mold
Source: Getty Images
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
exists, it may be a multiple of the Empirical
formula.
Figure 3.6: Examples of substances whose
empirical and molecular formulas differ.
Steps to Determine Empirical Formulas
Mass (g) of Element
M (g/mol )
Moles of Element
use no. of moles as subscripts
Preliminary Formula
change to integer subscripts
Empirical Formula
Some Examples of Compounds with the same
Elemental Ratios
Empirical Formula
Molecular Formula
CH2(unsaturated Hydrocarbons)
C2H4 , C3H6 , C4H8
OH or HO
H 2O2
S
S8
P
P4
Cl
Cl2
CH2O (carbohydrates)
C6H12O6
Figure 3.7: Structural Formula of P4O10
Determining Empirical Formulas from
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
= _________ mol Na
22.99 g Na
1 mol Cr
Moles of Cr = 6.420 g Cr x
= ___________ mol Cr
52.00 g Cr
1 mol O
Moles of O = 7.902 g O x
= ____________ mol O
16.00 g O
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 = 40.00% x 100g/100% = 40.00 g C
Mass Hydrogen = 6.719% x 100g/100% = 6.719g 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 = Mass of C x 1 mole C = 3.3306 moles C
12.01 g C
1 mol H
Moles of H = Mass of H x
= 6.6657 moles H
1.008 g H
Moles of O = Mass of O x 1 mol O = 3.3294 moles O
16.00 g O
Constructing the preliminary formula C 3.33 H 6.67 O 3.33
Converting to integer subscripts, divide all subscripts by the smallest:
C 3.33/3.33 H 6.667 / 3.33 O3.33 / 3.33 = CH2O
Determining the Molecular Formula from
Elemental Composition and Molar Mass - III
(b) Determining the Molecular Formula:
The formula weight of the empirical formula is:
1 x C + 2 x H + 1 x O = 1 x 12.01 + 2 x 1.008 + 1 x 16.00 = 30.03
Whole-number multiple =
=
M of Glucose
empirical formula mass
180.16
= 6.00 = 6
30.03
Therefore the Molecular Formula is:
C 1 x 6 H 2 x 6 O 1 x 6 = C6H12O6
=
Adrenaline is a very Important
Compound in the Body - I
• Analysis gives :
•
C = 56.8 %
•
H = 6.50 %
•
O = 28.4 %
•
N = 8.28 %
• Calculate the
Empirical Formula !
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 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
Figure 3.5: Combustion device
Ascorbic acid ( Vitamin C ) - I
contains 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!
• C: 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2)
= 2.66 x 10-3 g C
• H: 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O)
= 2.95 x 10-4 g H
• Mass Oxygen = 6.49 mg - 2.66 mg - 0.30 mg
= 3.53 mg O
Vitamin C combustion - II
• C = 2.66 x 10-3 g C / ( 12.01 g C / mol C ) =
= 2.21 x 10-4 mol C
• H = 2.95 x 10-4 g H / ( 1.008 g H / mol H ) =
= 2.93 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 2.21 x 10-4
• C = 1.00
• H = 1.33
• O = 1.00
Multiply each by 3
C3H4O3
= 3.00 = 3.0
= 3.99 = 4.0
= 3.00 = 3.0
Computer generated molecule:
Caffeine, C8H10N4O2
Determining a Chemical Formula from
Combustion Analysis - I
Problem: Erthrose (M = 120 g/mol) is an important chemical
compound 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:
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 = 0.2729 g C / 1 g CO2
44.01 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 = 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 H2O
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 weight = 30 g / formula
120 g /mol / 30 g / formula = 4 formula units / cpd = C4H8O4
Like Example 3.7 (P 65) - I
Sucrose is the common sugar used in all homes, and chemical
analysis tells us that the chemical composition is 42.14% carbon,
6.48% hydrogen and 51.46% oxygen. What is the molecular formula
of sucrose if its molecular mass is approximately 340 g/mol?
First determine the mass of each element in 1 mole (342.3g) of the
compound, sucrose.
42.14g C
342.3g
144.24g C
=
100.0g sucrose X
mol
mol
6.48g H
100.0g sucrose
342.3g
X
mol
51.46g O
100.0g sucrose
342.3g
mol
X
=
22.18g H
mol
176.15g O
=
mol
Like Example 3.7 (P 65) - II
Now we convert to moles:
144.24g C
1 mol C
C: mol sucrose X 12.011g C
H:
22.18g H
mol sucrose
X 1 mol Hg
1.008g H
=
12.01g C
mol sucrose
=
22.00g H
mol sucrose
176.15g O
1 mol O
11.01g O
X
=
Mol sucrose
16.00g O
mol sucrose
Divide by the smallest number:
C: 1.09
Therefore Empirical Formula = CH2O
since the molecular mass = 340g/mol,
H: 2.00
we must divide the formula mass into the
molecular mass or 340/30 = 11.3 = 11
O: 1.00
therefore the molecular formula is C11H22O11 !!!
O:
Chemical Equations
Qualitative Information:
Reactants
Products
Phases (States of Matter): (s) solid
(l) liquid
(g) gaseous
(aq) aqueous
2 H2 (g) + O2 (g)
2 H2O (g)
Balanced Equations
• mass balance (atom balance)- same number of each
element
(1) start with simplest element
(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)
•charge balance (no “spectator” ions)
Ca2+ (aq) + 2 OH- (aq) + Na+
Ca(OH)2 (s) + Na+
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)
Mass (amu)
Mass (g)
2 mol C2H6 + 7 mol O2 = 4 mol CO2 + 6 mol H2O
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
Flare in a
natural gas
field
Source: Stock Boston
Figure 3.8: Methane/oxygen reaction
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
Molecular model: Balanced equation
C2H6O(aq) + 3 O2(g)
2 CO2 (g) + 3 H2O(g) + Energy
Decomposition of ammonium
dichromate
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 from the words into chemical
compounds with blanks before each compound. begin the
balance with the most complex compound first, and save oxygen
until last!
Solution:
C6H14 (l) +
O2 (g)
CO2 (g) +
H2O(g) + Energy
Begin with one Hexane molecule which says that we will get 6 CO2’s!
1 C6H14 (l) +
O2 (g)
6 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) +
O2 (g)
6 CO2 (g) + 7 H2O(g) + Energy
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 the hexane by 2, giving a total of 12 CO2 molecules, and
14 H2O molecules.
2 C6H14 (l) +
O2 (g)
12 CO2 (g) + 14 H2O(g) + Energy
This now gives 12 O2 from the carbon dioxide, and 14 O atoms from the
water, which will be another 7 O2 molecules for a total of 19 O2 !
2 C6H14 (l) + 19 O2 (g)
12 CO2 (g) +14 H2O(g) + Energy
Chemical Equation Calc - I
Atoms (Molecules)
Avogadro’s
Number
Reactants
6.02 x 1023
Molecules
Products
Chemical Equation Calc - II
Mass
Atoms (Molecules)
Avogadro’s
Number
Reactants
6.02 x 1023
Molecules
Moles
Molecular
g/mol
Weight
Products
Sulfuric Acid
Plant
S(s) + O2 (g)
SO2 (g)
SO2 (g) + O2 (g)
SO3 (g)
SO3 (g) + H2O(l)
H2SO4 (aq)
Source: Southern States Chemical
The process for finding the mass of carbon
dioxide produced from 96.1 grams of propane
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 it’s 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
= ____________ Al2S3
150.17 g Al2S3/ mol Al2S3
Calculating Reactants and Products in a
Chemical Reaction - II
a) cont.
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
= ________________ g Al(OH)3
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - I
Problem: Calcium Phosphate could be prepared in the following
reaction sequence:
4 P4 (s) + 10 KClO3 (s)
P4O10 (s) + 6 H2O (l)
2 H3PO4 (aq) + 3 Ca(OH)2 (aq)
4 P4O10 (s) + 10 KCl (s)
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.
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 [ 4 P4 (s) + 10 KClO4 (s)
For Reaction #2 [ 1 P4O10 (s) + 6 H2O (l)
For Reaction #3 [ 2 H3PO4 + 3 Ca(OH)2
4 P4O10 (s) + 10 KCl (s) ]
4 H3PO4 (aq) ]
1 Ca3(PO4)2 + 6 H2O]
0.1251 moles P4 x 4 moles P4O10 x 4 moles H3PO4 x 1 mole Ca3(PO4)2
4 moles P4
1 mole P4O10
2 moles H3PO4
= _______ 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 product = 0.2502 moles Ca3(PO4)2 x
=
g Ca3(PO4)2
310.18 g Ca3(PO4)2
1 mole Ca3(PO4)2 =
Hydrochloric acid reacts with solid sodium
hydrogen carbonate
Two antacid tablets
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
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 the 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....)!
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:
2 N2H4 (l) + N2O4 (l)
3 N2 (g) + 4 H2O (g) + Energy
Sample Problem cont.
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
dividing 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 =
2 mol N2H4
Limiting !
= 4.68 moles N2
Mass of Nitrogen = 4.68 moles N2 x 28.02 g N2 / mol =
g N2
Acid - Metal Limiting Reactant - I
• 2Al(s) + 6HCl(g)
2AlCl3(s) + 3H2(g)
• Given 30.0g Al and 20.0g HCl, how many moles of
Aluminum Chloride will be formed?
• 30.0g Al / 26.98g Al/mol Al = 1.11 mol Al
•
1.11 mol Al / 2 = 0.555
• 20.0g HCl / 36.5gHCl/mol HCl = 0.548 mol HCl
•
O.548 mol HCl / 6 = 0.0913
• HCl is smaller therefore the Limiting reactant!
Acid - Metal Limiting Reactant - II
• since 6 moles of HCl yield 2 moles of AlCl3
• 0.548 moles of HCl will yield:
• 0.548 mol HCl / 6 mol HCl x 2 moles of
•
AlCl3 = ______________ mol of AlCl3
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?
• 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 mol NH3
• 40.0g O2 / 32.0g O2 /mol O2 = 1.25 mol O2
1.25 mol O2 / 5 = 0.25 mol O2
• 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
=
g NO
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)
% Yield =
Actual Yield
x 100
Theoretical Yield
Percent Yield Problem:
Problem: Given the chemical reaction between Iron and water to form
the iron oxide, Fe3O4 and Hydrogen gas given below. If 4.55g of Iron is
reacted with sufficient water to react all of the Iron to form rust, what is
the percent yield if only 6.02g 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(s) + 4 H2O(l)
Fe3O4 (s) + 4 H2 (g)
4.55 g Fe = 0.081468 mol = 0.0815 mol
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
____ %
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 limiting:
28.02 g N2
3.066 g N2
= 3.066
1 mole N2
1
moles H2 = 21.66 g H2 = 10.74 mol H2
2.016 g H2
10.74 g H2 = 3.582
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:
2 mol NH3 = 6.132 mol NH
3.066 mol N2 x
3
1 mol N2
(Theoretical Yield)
17.03 g NH3
= 104.428 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
104.428 g NH3
x 100% =
%
Flowchart : Solving a stoichiometry
problem involving masses of reactants
Mass Percent Composition of Na2SO4
Na2SO4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen
Elemental masses
Percent of each Element
2 x Na = 2 x 22.99 = 45.98 % Na = Mass Na / Total mass x 100%
1 x S = 1 x 32.07 = 32.07 % Na = (45.98 / 142.05) x 100% =32.37%
4 x O = 4 x 16.00 = 64.00
% S = Mass S / Total mass x 100%
142.05 % S = (32.07 / 142.05) x 100% = 22.58%
Check
% 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%
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 gm Nitrogen/ gm Cpd
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