Stoichiometry
(Chapter 3)
Antoine Lavoisier experimental observations lead to the development of the Law of
Conservation of Mass (a.k.a. Law of Conservation of Matter). The updated version of the
law of conservation of mass states: During an ordinary (non nuclear) chemical reaction,
atoms are neither created nor destroyed. Stoichiometry is the study of the quantitative
nature of chemical reactions. Chemical equations represent the qualitative and
quantitative nature of chemical reactions.
The equation:
2 H2 (g) + O2(g) à 2 H2O(l)
represents a chemical reaction in which two diatomic hydrogen (gas) molecules react with
one diatomic oxygen (gas) molecule react to yield two water (liquid) molecules. Hydrogen
and oxygen are the reactants (a.k.a. reagents) and water is the product. The “à” symbol
separates the reactant(s) from the product(s) and serves the same purpose as the “=” sign
does in algebra.
The total number of atoms of each element (and the sum of their respective masses) on the
reactant side of the “à” must be equal to the total number of atoms of each element (and
the sum of their respective masses) on the product side.
Chemical equations frequently contain additional symbols to represent the physical state
(phase) of the substance(s). These symbols include: (S) for solid, (l) for liquid, (g) for gas,
and (aq) for aqueous (the substance is dissolved in water).
In the equation: CH4 (g) + 2 O2 (g) à CO2 (g) + 2 H2O (g)
The small “4” (known as the subscript) behind the H in “CH4 (g)” indicates that there are four
atoms of hydrogen in each methane molecule.
The large “2” (known as the coefficient) in front of the O2 (g) indicates that there are two
oxygen molecules. If no coefficient is shown, it is understood to be “one.” Thus we have
one molecule of methane (CH4 (g)) reacting with two oxygen molecules to produce (yield)
one carbon dioxide and two water molecules.
The coefficient is multiplied throughout the entire molecule whereas the subscript only
applies to the element (or polyatomic ion in a parenthesis) it immediately follows.
So in the equation: Na2SO4 (aq) + BaCl2 (aq) à BaSO4 (s) + 2 NaCl (aq)
The “Na2SO4 (aq)” represents one sodium sulfate molecule that contains two sodium atoms,
one sulfur atoms, and four oxygen atoms. The “2 NaCl (aq)” represents two sodium chloride
molecules – each contains one sodium atom and one chlorine atom.
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Checking the equation, we find that we have two sodium atoms, one sulfur atom, four
oxygen atoms, one barium atom and two chlorine atoms on the reactant side of the
equation and one barium atom, one sulfur atom, four oxygen atoms, two sodium atoms,
and two chlorine atoms on the product side. Thus the total number of atoms of each
element is unchanged – they are just combined differently to form different substances.
Since atoms were neither created or nor destroyed, the above balanced equations satisfy
the law of conservation of mass.
Since an atom’s valence electrons are involved in the formation of chemical bonds, we can
make some general predictions regarding an element’s chemical properties based on its
position on the Periodic Table. Elements in the same family (column) typically react in a
similar manner. For example: the Group 1 metal, sodium (Na) reacts vigorously with water
to produce sodium hydroxide and hydrogen:
2 Na (s) + H2O (l) à 2 NaOH (aq) + H2 (g)
We would expect the other Group 1 (alkali metals) to undergo a similar reaction with water.
Reaction Types
A combustion reaction involves a substance reacting rapidly in the presence of oxygen.
For examples, hydrogen burns to form water:
2 H2 (g) + O2(g) à 2 H2O(l)
Hydrocarbons (compounds containing only carbon and hydrogen) burn in the presence of
sufficient oxygen to form carbon dioxide and water. Examples:
CH4 (g) + 2 O2 (g) à CO2 (g) + 2 H2O (g)
C3H8 (g) + 5 O2 (g) à 3 CO2 (g) + 4 H2O (l)
Combination (a.k.a. composition or synthesis) reactions occur when two (or more)
elements combine to form a single product: 2 N2 (g) + 3 H2 (g) à 2 NH3 (g)
Decomposition (a.k.a. analysis) reactions occur when a single substance breaks down
into two (or more) simpler substances: 2 KClO3 (s) à 2 KCl (s) + O2 (g)
See Sections 4.2-4.4 for a discussion of other types of chemical reactions:
Exchange (Metathesis) Reactions.
Acid-Base Neutralization Reactions
Acid-Base Reactions with Gas Formation
Oxidation-Reduction Reactions
In each type of reaction, we must balance the equation using only coefficients to obtain the
same number of atoms of each elements on both the reactant and products sides of the
raction.
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Atomic and Molecular Masses
Atom, molecules, and their subatomic particles (i.e., electron, proton, and neutron) are
extremely small with negligible mass. Rather than expressing their mass in terms of grams
or kilograms, a special, more convenient, atomic mass unit (u) has been defined. The
atomic mass unit is defined as 1/12th the mass of a single atom of the most abundant isotope
of carbon, 12C. You do not need to memorize these numbers, but
1 u = 1.6605402 x 10-24 g and
1 g = 6.0221421 x 1023 u.
The atomic mass (a.k.a. atomic weight) for an element is expressed in terms of atomic
mass units. Since most elements have multiple naturally occurring isotopes, their atomic
masses are not whole numbers. An element’s atomic mass is an average mass based on
the mass and relative natural abundance of its various naturally occurring isotopes.
The formula mass for a formula unit (ionic substances) or molecule (molecular
substances) equals the sum of the atomic masses of the atoms in the formula. For
example, the formula mass for potassium chlorate, KClO3 would be determined as follows:
(Please refer to your Periodic Table for the atomic masses.)
KClO3 has one potassium atom (AM = 39.0983 u), one chlorine atom (AM = 35.4527 u),
and three oxygen atoms (AM = 15.9994 u), so the formula mass would be find by the
equation
39.0983 u + 35.4527 u + 3 (15.9994 u) = 122.5492 u
Al2(SO4)3 has two aluminum atoms (AM = 26.981538 u), three sulfur atoms
(AM = 32.066 u), and 12 oxygen atoms (AM = 15.9994 u), the formula mass would be
found by:
2 (26.981538 u) + 3 (32.066 u) + 12 ( 15.9994 u) = 342.154 u
Molecular mass refers to the formula mass of a single molecule of a molecular substance.
Please note that the term molecular mass (molecular compounds) is often (incorrectly)
used to refer to the formula mass of an ionic compound. It’s a hard habit to break, so I’ll
probably be guilty of this several times throughout the semester.
Percentage Composition
The elemental percentage composition for compounds typically is determined by
experimental analysis. However, we can obtain some information from the compound’s
chemical formula.
Please refer to the following steps:
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To determine the percent composition of the elements in a compound, follow these steps:
1.
Determine the formula mass for the substance.
2.
Determine the total atomic mass for each element in the substance.
3.
Divide the substance’s formula mass into each of the respective total atomic masses
and multiple each of the result by 100.
Example: Find the percentage composition for aluminum sulfate
1.
The formula mass is 342.154 u (see above).
2.
The total atomic masses of
aluminum atoms: 2 (26.981538 u) = 53.963076 u Al
sulfur atoms: 3 (32.066 u) = 96.198 u S
oxygen atoms: 12 ( 15.9994 u) = 191.993 u O
3.
% Al =
53.963076 u Al
x 1 0 0 = 15.7716% Al
342.154 u Al2(SO4 )3
%S=
96.198 u S
x 1 0 0 = 28.115% S
342.154 u Al2(SO4 )3
%O =
191.993 u O
x 1 0 0 = 56.1130% O
342.154 u Al2(SO4 )3
To check our work, we add the respective percentages:
15.7716% + 28.115% + 56.1130% = 99.9996%
which when rounded to the required three decimal places gives us 100.000%.
NOTE: Due to rounding errors, the total percent may not exactly equal 100%.
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Masses in the laboratory
Since atoms and molecules are so incredibly small, it is impossible to count them out
individual. Instead we take advantage of the relationship that exists between the number of
particles and their formula mass.
By definition, the mass of a single atom of 12C is 12 u.
Since 1 g = 6.0221421 x 1023 u, 6.0221421 x 1023 12C atoms have a mass of 12 g.
This relationship between the number of particles and mass is defined has follows:
The number of 12C atom in exactly 12.0000 g of 12C is the mole.
The number of particles in a mole, 6.0221421 x 1023, is known as Avogadro’s number.
The mole (abbreviated as mol) is the bridge between the number of particles and their
mass in grams.
Examples:
Example 1: Convert 25.00 g H2O to mol
2 5 . 0 0 g H 2O x
1 mol H 2O
= 1.388 mol H 2 O
18.01528 g H 2O
Example 2: Convert 1.388 mol H2O to molecules
1 . 3 8 8 mol H 2O x
6.0221421 x 10 23 molecules H 2 O
= 8 . 3 5 8x 10 23 molecules H 2O
1 mol H 2 O
Example 3: Convert 25.00 g H2O to molecules
25.00 g H 2 O x
1 mol H 2 O
18.01528 g H 2O
23
x
6.0221421 x 1 0 molecules H 2 O
1 mol H 2 O
= 8 . 3 5 9x 10
23
molecules H 2O
Note the slightly different answers for examples two and three. The round-off error
occurred when the rounded answer from example one were used in example two.
Example three combined the steps into one calculation – without the intermediate rounding.
Empirical and Molecular Formulas
The molecular formula represents the actual number of the atoms of the various elements
in a molecule of a molecular compound. The empirical formula represents the simplest
ratio of the atoms of the various elements in the compound.
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Octane, the primary ingredient in gasoline, has a molecular formula of C8H18. Since both
“8” and “18” are divisible by “2”, octane’s empirical formula would be C4H9.
Ionic compounds do not consist of molecules, so their empirical formulas are used to
represent them.
A compound’s empirical and/or molecular formula(s) can be determined from
experimentally determined percentage composition data.
DETERMINING A COMPOUND’S EMPIRICAL FORMULA:
1.
Assume a 100.00 g sample so that the mass percents may be converted to grams.
63.15% C, 5.30% H, and the rest is O
63.15 g C. 5.30 g H, and 31.55 g O
2.
For each element, convert its mass to moles.
63.15 g C x 1 mol C/12.011 g C = 5.258 mol C
5.30 g H x 1 mol H/1.00794 g H = 5.26 mol H
31.55 g O x 1 mol O/15.9994 g O = 1.972 mol O
3.
Divide the number of moles of each element by the smallest number of moles from
above to get the mole ratio.
5.258 mol C/1.972 mol O = 2.666 = 8/3
5.26 mol H/1.972 mol O = 2.67 = 8/3
1.972 mol O/1.972 mol O = 1 = 3/3
3a.
If the above ratios contain fractions, multiply each term by some whole-number to
eliminate that fraction.
Multiple by 3, the common denominator, to get:
C8H8O3
If molecular mass data is also available, one may determine a compound’s molecular
formula.
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DETERMINING A COMPOUND’S MOLECULAR FORMULA:
Method 1
1.
Determine the compound’s empirical formula.
Example: Given: empirical formula C4H9, MM = 114.2309
2.
Determine the empirical formula’s formula mass
4 * 12.011 + 9 * 1.00794 = 57.11546
3.
Divide the empirical formula’s formula mass by the compound’s molar mass.
114.2309/57.11546 = 2
4.
Multiply the empirical formula’s subscripts by the number obtained in Step 3.
(C4H9)2 --> C8H18
Method 2
1.
For each element, multiple the mass percent (as a decimal) for that element by the
compound’s molar mass and divide by the element’s atomic mass.
.84117 C * 114.2309 / 12.011 = 8
.158827 H * 114.2309 / 1.00794 = 18
thus, C8H18 NOTE: We can determine the empirical formula for the molecular since
both subscripts are divisible by a common factor (2) to yield C4H9
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Combustion Analysis: Determing Empirical and/or Molecular Formulas
The combustion analysis of a 5.982 g sample of a compound in the vitamin D family
(containing C, H, and O) yielded 18.490 g of CO2 and 6.217 g of H2O. If the compound’s
molar mass is 398.6726, what are its empirical and molecular formulas?
The general reaction is:
CxHyOz(s) + O2(g) --> CO2(g) + H2O(g)
18.490 g CO2 x 12.011 g C/44.0098 g CO2 = 5.0462 g C
6.217 g H2O x 2(1.00794 g H)/18.01528 g H2O = 0.69569 g H
Why must we solve for the mass of oxygen indirectly? Since some of the oxygen is from
the atmosphere, we must determine the amount of oxygen in the original compound by
subtracting the mass(es) of the other element(s) from the initial mass of the compound.
g O = g sample - (g C + g H) = 5.982 g - (5.0462 + 0.69569) g = 0.2401 g O
5.0462 g C x 1 mol C/12.011 g C = 0.4201 mol C
0.69569 g H x 1 mol H/1.00794 g H = 0.6902 mol H
0.2401 g O x 1 mol O/15.9994 g O = 0.015007 mol O
0.4201 mol C/0.015007 mol O = 28
0.6902 mol H/0.015007 mol O = 46
empirical formula = C28H46O
formula mass = 398.6728
molar mass/formula mass = 398.6728/398.6728 = 1
molecular formula = C28H46O
Empirical and Molecular Formula Practice Problems
1.
Serotonin (MM = 176) is a compound that conducts nerve impulses in the brain and
muscles. It is 68.2% C, 6.86% H, 15.9% N and 9.08% O. What is serotonin’s
molecular formula?
2.
A sample of a certain inorganic insecticide contains 5.969 g Pb, 0.029 g H, 2.158 g
As, and 1.844 g O. What is the empirical formula for this insecticide?
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3.
A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68 mg
of the compound yields 16.01 mg CO2 and 4.37 mg H2O. The compound’s molar
mass is 176.1 g/mol. What are the compound’s empirical and molecular formulas?
4.
Coniine, a toxic substance isolated from poison hemlock, contains only carbon,
hydrogen, and nitrogen. Combustion analysis of a 5.024 mg sample yields 13.90
mg of CO2 and 6.048 mg of H2O. What is coniine’s empirical formula?
5.
Ibuprofen, a headache remedy, is 75.69% C, 15.51% O, and 8.80% H. What is
ibuprofen’s empirical formula?
6.
Ethylene glycol (MM = 62.0), commonly used as automotive antifreeze, contains
only carbon, hydrogen, and oxygen. Combustion analysis of a 23.46 mg sample
yields 20.42 mg of H2O and 33.27 mg of CO2. What are ethylene glycol’s empirical
and molecular formulas?
7.
Menthol (MM = 156.3), a strong-smelling substance used in cough drops, is a
compound of carbon, hydrogen, and oxygen. When 0.1595 g menthol was
subjected to combustion analysis, it produced 0.449 g CO2 and 0.184 g H2O. What
is menthol’s empirical formula?.
8.
An analysis of nicotine, a poisonous compound found in tobacco leaves, shows that
it is 74.0348% C, 8.6980% H, and 17.2672 % N. Its molar mass is 162.2346 g/mol.
What are the empirical and molecular formulas for nicotine?
9.
Hexachlorophene, a compound made up of atoms of carbon, hydrogen, chlorine,
and oxygen, is an ingredient in germicidal soaps. Combustion of a 1.000 g sample
yields 1.407 g of carbon dioxide, 0.134 g of water, and 0.5228 g of chlorine gas.
What is hexachlorophene’s empirical formula?
Balanced Equations and Masses of Reactants and Products
Once again, according to the Law of Conservation of Mass, atoms can neither be created
nor destroyed during a chemical reaction. We use balanced chemical equations to
represent the quantitative relationships that exist between the reactants and products.
Remember that the mole is the “bridge” that links the reactants to products. The general
format is:
Mass reactant à Moles reactant à Moles product à Mass reactant
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A chemical reaction can continue to produce products only as long as some of each
reactant is available. Once one of the reactants is completely consumed, the reaction
stops. The reactant that is consumed first – the limiting reactant – determines the
maximum amount of product(s) – the theoretical yield – that can be produced under ideal
conditions.
Example:
What mass of water can be produced when 10.00 g of hydrogen is reacted
with 50.00 g of oxygen?
Step 1:
Write the balanced chemical equation for the reaction.
2 H2 (g) + O2 (g) à 2 H2O (g)
Step 2:
Convert the given mass(es) of the reactant(s) to moles
Step 3:
10.00g H2 x
1 mol H 2
= 4 . 9 6 1mol H 2
2.01588 g H 2
5 0 . 0 0 g O2 x
1 mol O2
= 1.563 mol O2
31.9988 g O2
Identify the limiting reactant.
Separately, for each reactant, we can convert mass of the reactant to moles,
use the coefficient ratio to get moles of product and then convert to grams of
product. The smallest calculated yield is the “theoretical yield” and the
starting reactant is “limiting.”
10.00g H2 x
50.00g O x
1 mol H 2
2 mol H 2 O
18.01528 g H 2O
x
x
= 8 9 . 3 7g H 2 O
2.01588 g H 2
2 mol H 2
1 mol H 2O
1 mol O
2 mol H 2 O
18.01528 g H 2 O
x
x
= 5 6 . 3 0g H 2O
31.9988 g O
1 mol O2
1 mol H 2O
Since 56.30 g < 89.37 g, oxygen was the limiting reactant and 56.30 g O2 is the theoretical
yield.
Percent Yield =
actual yield
x 1 0 0=
theoretical yield
What would be the percent yield if the above reaction only produced an actual yield of
37.95 g H2O?
% yield =
37.95 g
x 1 0 0 = 67.41%
56.30 g
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An alternate approach
From the balanced equation, we see that two moles of hydrogen are needed for each mole
of oxygen. To determine which reactant is consumed first, we can do the following –
compare the number of moles of each reactant divided by their respective coefficient from
the balanced equation
4.961 mol H
1.563 mol O
>
2 mol H
1 mol O
2.481 > 1.563
so oxygen would be consumed first
When we compare the two fractions, the “>” or “<” symbol always points to the limiting
reactant … in this case the oxygen.
Use the coefficient ratios from the balanced equation and formula masses to complete the
calculations:
50.00g O x
1 mol O
2 mol H 2 O
18.01528 g H 2 O
x
x
= 5 6 . 3 0g H 2O
31.9988 g O
1 mol O2
1 mol H 2O
What would be the percent yield if the above reaction only produced an actual yield of
37.95 g H2O?
% yield =
37.95 g
x 1 0 0 = 67.41%
56.30 g
PLEASE NOTE:
Chemical reactions rarely (if ever) occur under “ideal” conditions, so
the actual yield normally is always less than 100%. Many reactions
may have very small percent yields.
If your calculated percent yield exceeds 100%, then some combination
of experimental and/or calculation error(s) has occurred.
Answers for the Empirical and Molecular Formula Practice Problems on Pages 8-9:
1.
2.
3.
4.
5.
6.
7.
8.
9.
C10H12N2O
PbHAsO4
C3H4O3
C8H17N
C13H18O2
CH3O
C10H20O
C 5H 7N
C13H6Cl6O2
C6H8O6
C2H6O2
C10H14N2
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