Chapter Three
Stoichiometry: Chemical Calculations
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Molecular Masses and
Formula Masses
• Molecular mass: sum of the masses of the atoms
represented in a molecular formula.
• Simply put: it is the mass of a molecule.
• Molecular mass is specifically for molecules.
• Ionic compounds don’t exist as molecules;
for them we use …
• Formula mass: sum of the masses of the atoms or ions
present in a formula unit.
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Example 3.1
Calculate the molecular mass of glycerol (1,2,3-propanetriol).
Solution
In glycerol, one –OH group replaces one H atom on each of the three C
atoms in propane, leading to the condensed structural formula
CH2OH – CHOH - CH2OH
which translates to the molecular formula C3H8O3.
To obtain the molecular mass, we must add together
three times the atomic mass of carbon,
eight times the atomic mass of hydrogen, and
three times the atomic mass of oxygen:
3 x atomic mass of C = 3 x 12.011 u = 36.033 u
8 x atomic mass of H = 8 x 1.00794 u = 8.06352 u
3 x atomic mass of O = 3 x 15.9994 u = 47.9982 u
Molecular mass of C3H8O3 = 92.095 u
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Example 3.2
Calculate the formula mass of ammonium sulfate, a fertilizer
commonly used by home gardeners.
Solution
Ammonium sulfate is an ionic compound consisting of ammonium ions (NH4+) and
sulfate ions (SO42–); its formula is therefore (NH4)2SO4.
To derive a formula mass from a complex formula like this one, we must make
certain that all the atoms in the formula unit are accounted for, which means paying
particular attention to all the subscripts and parentheses in the formula.
Let’s first note the relevant atomic masses and the way in which they must be
combined:
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Calculate the formula mass of ammonium sulfate, a
fertilizer commonly used by home gardeners.
• Summing the atomic masses
•
•
•
(NH4) 2 : [14.0067 u + 4 x 1.00794 u)] x 2 = 36.0769 u
SO4 :
32.066 u + (4 x 15.9994 u) = 96.064 u
Formula mass of (NH4)2 SO4 = 132.141 u
Assessment
As long as every atom in the formula unit is accounted for, we can
check our answer by using a different summation, for example, by
considering each element separately:
Formula mass = (2 x atomic mass N) + (8 x atomic mass H)
+ (1 x atomic mass of S) + (4 x atomic mass O)
= (2 x 14.0067 u) + (8 x 1.00794 u)
+ (1 x 32.066 u) + (4 x 15.9994 u)
= 132.141 u
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3.2 The Mole & Avogadro’s Number
• Mole (mol): amount of substance that contains as many
elementary entities, as there are atoms in exactly 12 g of the
carbon-12 isotope.
• Atoms are small, so this is a BIG number …
• Avogadro’s number (NA) = 6.022 × 1023 /mol
• 1 mol = 6.022 × 1023 “things” (atoms, molecules, ions,
formula units, oranges, etc.)
– A mole of oranges would weigh about as much as the
earth!
• Mole is NOT abbreviated as either M or m (but mol).
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12.011 amu
31.9988 amu
44.010 amu
Two views of
the
combination
of carbon
and oxygen
to form
carbon
dioxide
At the microscopic (molecular) level, chemical reactions occur between atoms
and molecules (top) but we can show only a few atoms and molecules to
represent the enormous numbers that actually make up the samples.
In reality, we usually observe substances at the macroscopic level.
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One Mole of Four
Elements
One mole each of
helium, sulfur, copper, and mercury.
6.022 × 1023 atoms of
each
4,002g
He,
32.066g S,
63,55g
Cu,
200,59
Hg
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Example 3.3
Determine
(a) the mass of a 0.0750-mol sample of Na,
(b) the number of moles of Na in a 62.5-g sample,
(c) the mass of a sample of Na containing 1.00 x 1025 Na atoms, and
(d) the mass of a single Na atom.
Solution
(a) To convert from moles to grams, we need to the conversion factor:
22.99 g Na / 1 mol Na:
(b) This time we write the conversion factor as the inverse 1 mol Na / 22.99 g Na - because we are to convert from grams to moles:
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(c) the mass of a sample of Na containing 1.00 x 1025 Na
atoms
(S1) Here we can use conversion factors, to convert first from number of atoms
to number of moles, and then from moles to the mass in grams:
(S2) Alternatively, we can use the conversion factor 22.99 g Na/6.022 x 1023 Na atoms,
to convert directly from number of atoms to mass in grams:
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(d) the mass of a single Na atom.
(d) The answer must have the unit grams per sodium atom (g/Na atom). We know
these quantities from the molar mass (22.99 g Na / mol Na) and Avogadro’s
number (1 mol Na / 6.022 x 1023 atoms). Our answer is the product of these two
factors:
? g Na atom = 1 Na atom
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(d) the mass of a single Na atom.
(d) The answer must have the unit grams per sodium atom (g/Na atom). We know
these quantities from the molar mass (22.99 g Na / mol Na) and Avogadro’s
number (1 mol Na / 6.022 x 1023 atoms). Our answer is the product of these two
factors:
1 mol Na atom
? g Na atom = 1 Na atom x
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6.022 x 1023 Na atom
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(d) the mass of a single Na atom.
(d) The answer must have the unit grams per sodium atom (g/Na atom). We know
these quantities from the molar mass (22.99 g Na / mol Na) and Avogadro’s
number (1 mol Na / 6.022 x 1023 atoms). Our answer is the product of these two
factors:
1 mol Na atom
? g Na atom = 1 Na atom x
6.022 x 1023 Na atom
22.99 g Na atom
X
1 mol Na atom
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(d) the mass of a single Na atom.
(d) The answer must have the unit grams per sodium atom (g/Na atom). We know
these quantities from the molar mass (22.99 g Na / mol Na) and Avogadro’s
number (1 mol Na / 6.022 x 1023 atoms). Our answer is the product of these two
factors:
1 mol Na atom
? g Na atom = 1 Na atom x
6.022 x 1023 Na atom
22.99 g Na atom
X
1 mol Na atom
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(d) the mass of a single Na atom.
(d) The answer must have the unit grams per sodium atom (g/Na atom). We know
these quantities from the molar mass (22.99 g Na / mol Na) and Avogadro’s
number (1 mol Na / 6.022 x 1023 atoms). Our answer is the product of these two
factors:
1 mol Na atom
? g Na atom = 1 Na atom x
6.022 x 1023 Na atom
22.99 g Na atom
X
1 mol Na atom
= 3.818 x 10 -23 g / Na atom
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The Mole and Molar Mass
•
Molar mass is the mass of one mole of a substance.
•
Molar mass is numerically equal to atomic mass, molecular mass,
or formula mass. However …
•
… the units of molar mass are grams (g/mol).
•
Examples:
1 atom Na = 22.99 u
1 mol Na = 22.99 g
1 molecule CO2 = 44.01 u
1 mol CO2 = 44.01 g
1 formula unit KCl = 74.56 u
1 mol KCl = 74.56 g
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Conversions involving Mass, Moles,
and Number of Atoms/Molecules
1 mol Na = 6.022 × 1023 Na atoms =
22.99 g Na
We can use these equalities to construct conversion factors,
such as:
1 mol Na
–––––––––
22.99 g Na
22.99 g Na
–––––––––
1 mol Na
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1 mol Na
––––––––––––––––––
6.022 × 1023 Na atoms
22.99 g Na
––––––––––––––––––
6.022 × 1023 Na atoms
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We can read formulas in terms of moles of
atoms or ions.
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Example 3.4
Determine
(a)
the number of NH4+ ions in a 145-g sample of (NH4)2SO4
and
(b)
the volume of 1,2,3-propanetriol (glycerol) (d = 1.261 g/mL)
that contains 1.00 mol O atoms.
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a) Determine the number of NH4+ ions in
a 145-g sample of (NH4)2SO4
Solution
(a) In this calculation, we must use the relationship between the NH4+ ion
and the (NH4)2SO4 formula unit; (132.14 g / mol) that is,
2 mol NH4+ / 1 mol (NH4)2SO4 .
The path to the desired answer is
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b) Determine the volume of 1,2,3-propanetriol (glycerol, d = 1.261 g/mL)
that contains 1.00 mol O atoms.
(b) The conversion factors needed for this calculation are
(1) the relationship between moles of O atoms and C3H8O3 molecules,
(2) the molar mass of C3H8O3, and
(3) the inverse of the density of C3H8O3:
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Example 3.5 An Estimation Example
Which of the following is a reasonable value for the number
of atoms in 1.00 g of helium?
(a) 4.1 × 10–25
(b) 4.0
(c) 1.5 × 1023
(d) 1.5 × 1024
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number of atoms in 1.00 g of helium?
a) 4.1 × 10–25
b)4.0
c) 1.5 × 1023
(d) 1.5 × 1024
(a) is what we get if, in error, we divide the number of moles of He, 0.25, by
Avogadro’s number. Clearly, we can’t have anything less than one atom.
(b) expressed as 4.0 u is the atomic mass of He; expressed as 4.0 g/mol He, it is the
molar mass. In either case, it is far too small or big to be the number of atoms for any
macroscopic sample.
(c) is the correct order of magnitude, and the correct answer.
(d) Order of magnetude is reasonable, however, it is larger than Avogadro’s number.
We would expect a fraction of a mole of helium to have fewer than Avogadro’s
number of atoms.
Response (c) is the correct answer, obtained from the calculation:
0.25 x NA
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=
1.5 X 10 23
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3.4 Mass Percent Composition
from Chemical Formulas
The mass percent composition of a compound refers to the
proportion of the constituent elements, expressed as the number of
grams of each element per 100 grams of the compound. In other
words …
X g element
X % element = –––––––––––––– OR …
100 g compound
g element
% element = ––––––––––– × 100
g compound
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Percentage Composition of Butane
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Example 3.6
Calculate, to four significant figures, the mass percent of
each element in ammonium nitrate
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(NH4 NO3).
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Calculate, to four significant figures, the mass percent of each element in
ammonium nitrate NH4NO3.
Solution
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Example 3.7
How many grams of nitrogen are present
in 46.34 g ammonium nitrate?
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How many grams of nitrogen are present in 46.34 g ammonium nitrate?
Solution
The central factor (shown in red) in the conversion is based on the
chemical formula NH4NO3. The other factors in the following setup are
based on molar masses:
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Example 3.8
An Estimation Example
Without doing detailed calculations, determine which of these
compounds contains the greatest mass of sulfur per gram of
compound:
barium sulfate, (BaSO4)
lithium sulfate, (Li2SO4)
sodium sulfate, (Na2SO4)
lead sulfate.
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(PbSO4)
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The Periodic table
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the greatest mass of sulfur per gram of compound:
a) barium sulfate, b) lithium sulfate,
c) sodium sulfate, d) lead sulfate.
Analysis and Conclusions
To make this comparison, we need formulas of the compounds, which we can get from
their names:
BaSO4
Li2SO4
Na2SO4
PbSO4
The compound with the greatest mass of sulfur per gram of compound also has the
greatest mass of sulfur per 100 g of compound—in other words, the greatest % S by
mass.
From the formulas, we see that in one mole of each compound there is one mole of sulfur,
which means 32.066 g S. Thus, the compound with the greatest % S is the one with the
smallest formula mass.
Because each formula unit has one SO42– ion, all we have to do is compare some atomic
masses: that of barium to twice that of lithium, and so on. With just a glance at an
atomic mass table: we see the answer must be
lithium sulfate, Li2SO4.
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3.5 Chemical Formulas from Mass Percent
Composition
•
We can “reverse” the process of finding percentage composition.
• First we use the percentage or mass of each element to find
moles of each element.
• Then we can obtain the empirical formula by finding the
smallest whole-number ratio of moles.
•
Find the whole-number ratio by dividing each number of moles by
the smallest number of moles.
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Example 3.9
Phenol, a general disinfectant, has the composition
76.57% C,
6.43% H, and
17.00% O by mass.
Determine its empirical formula.
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the composition 76.57% C, 6.43% H, and 17.00% O by mass. Determine its
empirical formula.
Solution
Step 1: A 100.00-g sample of phenol contains
76.57 g C,
6.43 g H, and
17.00 g O.
Step 2: We convert the masses of C, H, and O to amounts in moles:
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Step 3:Now we use the numbers of moles in Step 2 as subscripts in a tentative
formula:
C6.375H6.38O1.063
Step 4:Next we divide each subscript by the smallest (1.063) to try to get integral
subscripts:
Step 5:The subscripts in Step 4 are all integers. We need do nothing further.
The empirical formula of phenol is C6H6O.
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Example 3.10
Diethylene glycol, used in antifreeze, as a softening agent for
textile fibers and some leathers, and as a moistening agent for
glues and paper, has the composition 45.27% C, 9.50% H, and
45.23% O by mass.
Determine its empirical formula.
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the composition 45.27% C, 9.50% H, and 45.23% O by mass.
Determine its empirical formula.
Solution
Step 1: A 100.00-g sample of diethylene glycol contains
45.27 g C, 9.50 g H, and 45.23 g O.
Step 2: We convert the masses of C, H, and O to amounts in moles:
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the composition 45.27% C, 9.50% H, and 45.23% O by mass.
Determine its empirical formula.
Step 3:Now we use the numbers of moles in Step 2 as subscripts in a tentative formula:
C3.769H9.42O2.827
Step 4:Next we must divide each subscript by the smallest (2.827) in an attempt to get
integral subscripts:
Step 5:Finally we multiply all the subscripts from Step 4 by a common factor to
convert them all to integers. By recognizing that 1.333 = 4/3 and
3.33 = 10/3, we can see that the common factor we need is 3:
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f.p.dep. for Water 1.86oC/mol
Diethylene glycol m.w. = 106.12g
(Na+)(Cl-) m.w. = 58.8
Which is better to use as Antifreeze = ?
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Relating Empirical Formulas
to Molecular Formulas
•
A molecular formula is a simple integer multiple of the empirical
formula.
•
That is, an empirical formula of CH2 means that the molecular
formula is CH2, or C2H4, or C3H6, or C4H8, etc.
•
So: we find the molecular formula by:
molecular formula mass
= integer (nearly)
empirical formula mass
We then multiply each subscript in the empirical formula by the integer.
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Example 3.11
The empirical formula of hydroquinone, a chemical
used in photography, is C3H3O, and its molecular
mass is 110 u.
What is its molecular formula?
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The empirical formula of hydroquinone, is C3H3O, and its molecular mass is
110 u.
What is its molecular formula?
Solution
The empirical formula mass is (3 x 12.0 u) + (3 x 1.0 u) + 16.0 u = 55.0 u.
The multiplier we need to convert the subscripts in the empirical formula
to those in the molecular formula is the integral factor
The molecular formula is
(C3H3O) x 2 ,
or C6H6O2.
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3.6 Elemental Analysis …
•
… is one method of determining empirical formulas in the laboratory.
•
This method is used primarily for simple organic compounds (that contain
carbon, hydrogen, oxygen).
– The organic compound is burned in oxygen.
– The products of combustion (usually CO2 and H2O) are weighed.
– The amount of each element is determined from the mass of products.
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Elemental Analysis
… H2O formed, is
absorbed by
MgClO4, and …
The sample is
burned in a
stream of
oxygen gas,
producing …
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… CO2 formed, is
absorbed by
NaOH.
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Example 3.12
Burning a 0.1000-g sample of a carbon–hydrogen–oxygen
compound in oxygen yields 0.1953 g CO2 and 0.1000 g H2O.
A separate experiment shows that the molecular mass of the
compound is 90 u.
Determine
(a)
the mass percent composition,
(b)
the empirical formula, and
(c)
the molecular formula of the compound.
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Burning a 0.1000-g sample of a carbon–hydrogen–oxygen compound in oxygen
yields 0.1953 g CO2 and 0.1000 g H2O.
A separate experiment shows that the molecular mass of the compound is 90 u.
Determine a) the mass percent composition,
Solution
(a) First, we do the conversions outlined to calculate the mass of
carbon in the CO2 produced:
This mass of carbon originated from the 0.1000-g sample. Thus,
the mass percent carbon in the compound is
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Burning a 0.1000-g sample of a carbon–hydrogen–oxygen compound in oxygen
yields 0.1953 g CO2 and 0.1000 g H2O.
A separate experiment shows that the molecular mass of the compound is 90 u.
Determine a) the mass percent composition,
We can use similar calculations to determine first the mass of hydrogen and
then the mass percent of hydrogen in the compound:
Finally, we can calculate the mass percent oxygen by subtracting the
mass percents of C and H from 100.00%:
% O = 100.00% – 53.30% – 11.19% = 35.51%
Therefore the mass percent composition is
53.30% C,
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11.19% H
and
35.51% O
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Burning a 0.1000-g sample of a carbon–hydrogen–oxygen compound in
oxygen yields 0.1953 g CO2 and 0.1000 g H2O. A separate experiment
shows that the molecular mass of the compound is 90 u. Determine
b) the empirical formula,
(b) Here we apply the method of Examples 3.9 and 3.10, but we need only the
first four steps.
Step 1:
A 100.00-g sample of the compound contains 53.30 g C, 11.19 g H,
and 35.51 g O.
Step 2: We convert the masses of C, H, and O to numbers of moles:
C4.438H11.10O2.220
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Burning a 0.1000-g sample of a carbon–hydrogen–oxygen compound in
oxygen yields 0.1953 g CO2 and 0.1000 g H2O. A separate experiment
shows that the molecular mass of the compound is 90 u. Determine
b) the empirical formula,
.
Step 3: Next, we use the numbers of moles in Step 2 as subscripts in a tentative
formula:
C4.438H11.10O2.220
Step 4: We divide all subscripts by the smallest subscript (2.220) to get
integral subscripts:
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Burning a 0.1000-g sample of a carbon–hydrogen–oxygen compound in
oxygen yields 0.1953 g CO2 and 0.1000 g H2O. A separate experiment
shows that the molecular mass of the compound is 90 u. Determine
c) the molecular formula of the compound.
c) The empirical formula mass of C2H50 is
(2 x 12.0 u) + (5 x 1.0 u) + 16.0 u = 45.0 u.
The multiplier we need to convert the subscripts in the empirical formula to those in
the molecular formula is the integral factor
The molecular formula is
(C2H5O)x2 ,
or C4H10O2.
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Animation 1: Reactions
(Sodium, Potassium)
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3.7 Writing Chemical Equations
•
A chemical equation is a shorthand description of a chemical
reaction, using symbols and formulas to represent the elements
and compounds involved.
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Writing Chemical Equations
• Sometimes
additional
information about
the reaction is
shown in the
equation.
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Balancing Equations
How can we tell that
the equation is not
balanced?
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… not by
changing the
equation …
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Balancing Equations
How can we tell that
the equation is not
balanced?
… not by
changing the
equation …
… and not by
changing the
formulas.
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Balancing Equations
How can we tell
that the
equation is not
balanced?
… not by changing
the equation …
… and not by
changing the
formulas.
The equation is
balanced by
changing the
coefficients …
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Guidelines for Balancing
Chemical Equations
•
If an element is present in just one compound on each side of the
equation, try balancing that element first.
•
Balance any reactants or products that exist as the free element last.
•
In some reactions, certain groupings of atoms (such as polyatomic
ions) remain unchanged. In such cases, treat these groupings as a
unit.
•
At times, an equation can be balanced by first using a fractional
coefficient(s). The fraction is then cleared by multiplying each
coefficient by a common factor.
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Example 3.13
Balance the equation
Fe + O2  Fe2O3 (not balanced)
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Fe + O2  Fe2O3
(not balanced)
Solution
It seems that the easiest place to begin is with the iron atoms. There is one on the left (Fe)
and two on the right (Fe2O3), and we can balance them by placing the coefficient 2 on the
left:
2 Fe + O2  Fe2O3 (Fe balanced, O not balanced)
To balance the oxygen atoms, we begin by noting that there are two of them on the left
and three on the right. An easy way to get three O atoms on the left, thereby balancing
the equation, is to use the fractional coefficient 3/2 before O2 on the left {3/2 x 2 = 3}:
2 Fe + 3/2 O2  Fe2O3 (balanced, fractional coefficient)
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Fe + O2  Fe2O3
(not balanced)
Solution
Fractional coefficients are not only acceptable in equations, sometimes they are
desirable. If we don’t want them, however, we can easily clear an equation of them
by multiplying every coefficient by the smallest integer required to clear the
fractions, in this case, 2:
2 x {2 Fe + 3/2 O2  Fe2O3}
becomes
4 Fe + 3 O2  2 Fe2O3 (balanced, coefficients integral)
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Example 3.14
Balance the equation
C2H6 + O2  CO2 + H2O
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Solution
C2H6 + O2  CO2 + H2O
Oxygen appears as the free element on the left, so let’s leave it for last and
balance the other two elements first.
To balance carbon, we place the coefficient 2 in front of CO2:
C2H6 + O2  2 CO2 + H2O (C balanced, H and O not balanced)
To balance hydrogen, we need the coefficient 3 before H2O:
C2H6 + O2  2 CO2 + 3 H2O (C and H balanced, O not balanced)
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C2H6 + O2  CO2 + H2O
Solution
Now, if we count oxygen atoms, we find two on the left and seven on the right. We
can get seven on each side by using the fractional coefficient 7/2
(7/2 x 2 = 7):
C2H6 + 7/2 O2  2 CO2 + 3 H2O (balanced)
To obtain integral coefficients, we multiply each coefficient by 2:
2 x {C2H6 + 7/2 O2  2 CO2 + 3 H2O}
That leads to
2 C2H6 + 7 O2  4 CO2 + 6 H2O (balanced)
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Example 3.15
Balance the equation
H3PO4 + NaCN  HCN + Na3PO4
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H3PO4 + NaCN  HCN + Na3PO4
Solution
Notice that the PO4 and CN groups remain unchanged in the reaction. For
purposes of balancing equations, we can often treat such groups as a whole, rather
than breaking them down into their constituent atoms.
To balance hydrogen atoms, we place a 3 before HCN:
H3PO4 + NaCN  3 HCN + Na3PO4 (not balanced)
To balance cyanide ions, we put a 3 in front of the NaCN. Doing this also balances
the sodium ions:
H3PO4 + 3 NaCN  3 HCN + Na3PO4 (balanced)
Because the PO4 was balanced to begin with and we did nothing to upset that
balance, we are finished; the equation is balanced.
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Example 3.16
A Conceptual Example
Write a reasonable chemical equation for the reaction between water
and a liquid molecular chloride of phosphorus to form an aqueous
solution of hydrochloric acid and phosphorus acid (H3PO3).
The phosphorus-chlorine compound is 77.45% Cl by mass.
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chemical equation for the reaction between water and a liquid molecular chloride of
phosphorus to form an aqueous solution of hydrochloric acid and phosphorus acid.
The phosphorus-chlorine compound is 77.45% Cl by mass.
Analysis and Conclusions
Establishing the formula of the chloride of phosphorus.
We can form a compound that is 22.55% P and 77.45% Cl.
A 100.00-g sample of the compound consists of 22.55 g P and 77.45 g Cl, corresponding
to 0.728 mol P and 2.185 mol Cl. The formula P0.728Cl2.185 reduces to PCl3.
Establishing the formulas of hydrochloric and phosphorus acids.
Hydrochloric acid, a binary acid, has the formula HCl. Table 2.5 gives H3PO4 as the
formula of phosphoric acid. Phosphorus acid should have one O atom fewer per
molecule than the “-ic” acid, giving the formula H3PO3.
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chemical equation for the reaction between water and a liquid molecular chloride of
phosphorus to form an aqueous solution of hydrochloric acid and phosphorus acid.
The phosphorus-chlorine compound is 77.45% Cl by mass.
Analysis and Conclusions
Writing and balancing the equation.
The unbalanced equation, including an indication of the physical form of each
substance, is
PCl3(l) + H2O(l)  HCl(aq) + H3PO3(aq)
The balanced equation requires the coefficient 3 for H2O(l) and for HCl(aq):
PCl3(l) + 3 H2O(l)  3 HCl(aq) + H3PO3(aq)
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3.8 Stoichiometric Equivalence
and Reaction Stoichiometry
• A stoichiometric factor or mole ratio is a
conversion factor obtained from the stoichiometric
coefficients in a chemical equation.
• In the equation: CO(g) + 2 H2(g)  CH3OH(l)
– 1 mol CO is chemically equivalent to 2 mol H2
– 1 mol CO is chemically equivalent to 1 mol CH3OH
– 2 mol H2 is chemically equivalent to 1 mol CH3OH
1 mol CO
–––––––––
2 mol H2
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1 mol CO
–––––––––––––
1 mol CH3OH
2 mol H2
–––––––––––––
1 mol CH3OH
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Outline of Simple Reaction Stoichiometry
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Example 3.17
When 0.105 mol propane is burned in an excess of
oxygen, how many moles of oxygen are consumed?
The reaction is
C3H8 + 5 O2  3 CO2 + 4 H2O
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When 0.105 mol propane is burned in an excess of oxygen, how many moles of
oxygen are consumed? The reaction is
C3H8 + 5 O2  3 CO2 + 4 H2O
Solution
The stoichiometric coefficients in the equation allow us to write the equivalence
From this equivalence, we can derive two conversion factors:
Because we are given the number of moles of propane (substance A in Figure
3.8) and are seeking the number of moles of O2 consumed (substance B in
Figure 3.8), we need the factor that has the unit “mol O2” in the numerator and
the unit “mol C3H8 ” in the denominator. This is the conversion factor shown
in red above:
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Outline of Stoichiometry Involving Mass
… we’ve added a
conversion from
mass at the
beginning …
To our simple
stoichiometry
scheme …
Substances A and B
may be two reactants,
two products, or
reactant and product.
… and a
conversion to
mass at the end.
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ARA
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Example 3.18
The final step in the production of nitric acid involves the
reaction of nitrogen dioxide with water; nitrogen monoxide is
also produced.
How many grams of nitric acid are produced for every 100.0 g
of nitrogen dioxide that reacts?
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How many grams of nitric acid are produced for
every 100. 0 g of nitrogen dioxide that reacts?
Solution 3.18
Because no equation is given, we must first write a chemical equation
from the description of the reaction.
To balance the equation, let’s first balance H atoms because H appears
in one reactant and one product. Then we can balance N atoms because
they appear on the reactant side only in NO2 . At this point, O atoms
will also be balanced.
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How many grams of nitric acid are produced for every 100. 0 g of nitrogen
dioxide that reacts?
Stepwise Solution 3.18
Next, we use coefficients from the balanced equation to establish the
stoichiometric equivalence of NO2 and HNO3
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How many grams of nitric acid are produced for every 100. 0 g of nitrogen dioxide
that reacts?
Stepwise Solution 3.18 continued
Now, we convert the mass of the given substance, NO2, to an
amount in moles, using the molar mass .
Because we need to convert from moles of NO2 to moles of HNO3, we should use
this equivalence to write the stoichiometric factor needed for the conversion.
Finally, we can convert from moles of HNO3 to grams of HNO3 using its molar mass.
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How many grams of nitric acid are produced for every 100. 0 g of nitrogen
dioxide that reacts?
Solution of question 3.18 in one equation
? g HNO3 = 100.0 g NO2  convert to moles NO2
convert to moles HNO3 (2/3)
 convert to g HNO3
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Example 3.19
Ammonium sulfate, a common fertilizer used by gardeners, is
produced commercially
By passing gaseous ammonia into an
aqueous solution that is 65% H2SO4 by mass and
has a density of 1.55 g/mL.
How many milliliters of this sulfuric acid solution are
required to convert 1.00 kg NH3 to (NH4)2SO4 ?
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Ammonium sulfate produced by passing gaseous ammonia into an aqueous
solution that is 65% H2SO4 by mass, and has a density of 1.55 g/mL.
How many milliliters of this sulfuric acid solution are required to convert 1.00 kg
NH3 to (NH4)2SO4?
Solution 3.19
First, we must write the balanced equation for the reaction:
2 NH3(g) + H2SO4(aq)  (NH4)2SO4(aq)
The required setup has 1.00 kg NH3 —the given substance—as its starting
point.
Because we are seeking a volume of H2SO4(aq), the setup has the general
form
? mL H2SO4(aq) = 1.00 kg NH3 x conversion factors
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Ammonium sulfate produced by passing gaseous ammonia into an aqueous
solution that is 65% H2SO4 by mass, and has a density of 1.55 g/mL.
How many milliliters of this sulfuric acid solution are required to convert 1.00 kg
NH3 to (NH4)2SO4?
Solution 3.19 continued
We begin by converting kilograms of NH3 to grams of NH3. Then we can apply
the flow chart below to find grams of H2SO4.
Finally, we use the percent composition and then the density of the H2SO4(aq) to
convert from grams of H2SO4 to milliliters of H2SO4(aq):
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Ammonium sulfate produced by passing gaseous ammonia into an aqueous
solution that is 65% H2SO4 by mass, and has a density of 1.55 g/mL.
How many milliliters of this sulfuric acid solution are required to convert 1.00 kg
NH3 to (NH4)2SO4?
Solution 3.19 continued
In the combined setup, the conversions are done in the order in
which they are described above:
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3.9 Limiting Reactants
• Many reactions are carried out with a limited amount of one
reactant and a plentiful amount of the other(s).
• The reactant that is completely consumed in the reaction,
limits the amounts of products and is called the limiting
reactant, or limiting reagent.
• The limiting reactant is not necessarily the one present in
smallest amount.
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Limiting Reactant Analogy
If we have
10 sandwiches, 18 cookies, and 12 oranges …
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… how many
packaged meals
can we make?
86
Molecular View of the Limiting Reactant Concept
1. Why is ethylene left
over, when we started
with more bromine
than ethylene? (Hint:
count no. of moles.)
2. What mass of ethylene
is left over after
reaction is complete?
(Hint: it’s an easy
calculation; why?)
When 28 g
(1.0 mol) ethylene
reacts with …
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… 128 g (0.80 mol)
bromine, we get …
… 150 g of
1,2-dibromoethane,
and leftover is
ethylene!
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Recognizing and Solving Limiting Reactant
Problems
•
We recognize limiting reactant problems by the fact that
amounts of two (or more) reactants are given.
•
One way to solve them is to perform a normal stoichiometric
calculation of the amount of product obtained, starting with each
reactant.
•
The reactant that produces the smallest amount of product is
the limiting reactant.
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Example 3.20
Magnesium nitride Mg3N2 can be formed by the reaction of
magnesium metal with nitrogen gas.
(a)
How many grams of magnesium nitride can be made in the
reaction of 35.00 g of magnesium and 15.00 g of nitrogen?
(b)
How many grams of the excess reactant remain after the
reaction?
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How many grams of magnesium nitride can be made in the reaction of
35.00 g of magnesium and 15.00 g of nitrogen?
How many grams of the excess reactant remain after the reaction?
Solution 3.20
(a) As usual, we must first write a balanced equation .
We can identify the limiting reactant by finding the number of
moles of Mg3N2(s) produced by assuming
- first one reactant completely consumed,
- and then the other
- the one with the less product, is the limiting reactant.
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How many grams of magnesium nitride can be made in the reaction of
35.00 g of magnesium and 15.00 g of nitrogen?
How many grams of the excess reactant remain after the reaction?
-) Assuming Mg is the limiting reactant and N2 is in excess:
-) Assuming N2 is the limiting reactant and Mg is in excess:
Because the amount of product from 35.00 g Mg (0.4800 mol
Mg3N2) is smaller than that formed from 15.00 g N2 (0.5355 mol
Mg3N2), we know that magnesium is the limiting reactant.
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How many grams of magnesium nitride can be made in the reaction of
35.00 g of magnesium and 15.00 g of nitrogen?
How many grams of the excess reactant remain after the reaction?
Solution 3.20 continued
When 0.4800 mol Mg3N2 has been formed, the Mg is completely consumed
and the reaction stops, and the amount of Mg3N2 produced is :
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How many grams of the excess reactant remain after the reaction?
Solution 3.20 continued
(b) Having found that the amount of product is 0.4800 mol Mg3N2, we can
now calculate how much N2 must have been consumed.
From this, we calculate the mass of excess N2.
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3.10 Yields of Chemical Reactions
• The theoretical yield of a chemical reaction is the calculated
quantity of product in the reaction.
• The actual yield is the amount you actually get when you
carry out the reaction.
• Actual yield will be less than the theoretical yield, for many
reasons … can you name some?
actual yield
Percent yield = ––––––––––––– × 100
theoretical yield
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Actual Yield of ZnS Is Less than the
Theoretical Yield
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Example 3.21
Ethyl acetate is a solvent used as fingernail polish remover.
What mass of acetic acid is needed to prepare 252 g ethyl acetate if the
expected percent yield is 85.0%?
Assume that the other reactant, ethanol, is present in excess and water is
the by-product.
The equation for the reaction, carried out in the presence of H2SO4, is
CH3COOH + HOCH2CH3
Acetic acid
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Ethanol

CH3COOCH2CH3
+
H2O
Ethyl acetate
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What mass of acetic acid is needed to prepare 252 g ethyl acetate if the expected
percent yield is 85.0%?
Solution
First, we can solve Equation (3.8) for the theoretical yield and substitute known
quantities for the actual and percent yields.
Then, we can determine the mass of acetic acid required to produce 296 g
ethyl acetate.
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Example 3.22
A Conceptual Example
What is the maximum yield of CO(g) obtainable from
725 g of C6H14(l), assuming no other carbon-containing
reactant or product is formed?
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What is the maximum yield of CO(g) obtainable from 725 g of C6H14 (l),
Analysis and Conclusions
We can start the calculation by converting grams of C6H14 to moles of C6H14, and
we can end it by converting moles of CO to grams of CO.
The critical link between these two “ends” of the calculation are the factors that
relate moles of C to moles of C6H14 on the one hand, and moles of C to moles of
CO on the other.
In the solution equation, conversion factors are shown in red .
The solution equation is:
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3.11 Solutions and
Solution Stoichiometry
• Solute: the substance being dissolved.
• Solvent: the substance doing the dissolving.
• Solution: A mixture of solute and solvent.
• Concentration of a solution: the quantity of a solute in a given
quantity of solution (or solvent).
– A concentrated solution contains a relatively large amount of
solute vs. the solvent (or solution).
– A dilute solution contains a relatively small concentration of
solute vs. the solvent (or solution).
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Molar Concentration
Molarity (M), or molar concentration, is the amount of solute,
in moles, per liter of solution:
moles of solute
Molarity = ––––––––––––––
liters of solution
•
A solution that is 0.35 M sucrose contains 0.35 moles of
sucrose in each liter of solution.
•
Keep in mind that molarity signifies moles of solute per
liter of solution, not liters of solvent.
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Preparing 0.01000 M KMnO4
Mol weight of
KMnO4 is 158
g. How much
do we need?
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Weigh
0.01000 mol
(1.580 g)
KMnO4.
Dissolve in water. How
much water? Doesn’t
matter, as long as we
don’t go over a liter.
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After dissolving
solid add more
water to reach the
1.000 liter mark.
102
Animation 3 : Preparation of a Solution
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Example 3.23
What is the molarity of a solution in which 333 g
potassium hydrogen carbonate ( KHCO3 ) is dissolved in
enough water to make 10.0 L of solution?
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molarity of a solution in which 333 g potassium hydrogen carbonate is dissolved in enough water to
make 10.0 L of solution?
Solution
First, let’s prepare the setup that converts mass to number of moles of
KHCO3.
Now, without solving this expression, let’s use it (number of moles) as the
numerator in the defining equation for molarity.
The solution volume, 10.0 L, is the denominator.
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Example 3.24
We want to prepare a 6.68 molar solution of NaOH (6.68 M
NaOH).
(a) How many moles of NaOH are required to prepare
0.500 L of 6.68 M NaOH?
(b) How many liters of 6.68 M NaOH can we prepare with
2.35 kg NaOH?
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(a) How many moles of NaOH are required to prepare 0.500 L of
6.68 M NaOH?
Solution 3.24
(a) This calculation requires only a one-step conversion from liters of
solution to moles of NaOH, with the molarity of the solution as the
conversion factor:
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(b) How many liters of 6.68 M NaOH can we prepare with 2.35 kg NaOH?
Solution 3.24 continued
(b) The central conversion factors in this calculation are the inverse of the molar mass
of NaOH to convert from grams of NaOH to moles of NaOH and the inverse of the
molarity—1 L soln/6.68 NaOH—to convert from moles of NaOH to liters of
solution.We must also convert from kilograms of NaOH to grams of NaOH. In all,
the required conversions are
kg NaOH  g NaOH  mol NaOH  L soln
which are set up as follows:
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Example 3.25
The label of a stock bottle of aqueous ammonia
indicates that the solution is 28.0% NH3 by mass
and has a density of 0.898 g/mL.
Calculate the molarity of the solution.
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-) aqueous ammonia indicates that the solution is 28.0% NH3 by mass
and has a density of 0.898 g/mL. Calculate the molarity of the solution.
Solution 3.25
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Dilution of Solutions
•
Dilution is the process of preparing a more dilute solution by
adding solvent to a more concentrated one.
•
Addition of solvent does not change the amount of solute in a
solution but does change the solution concentration.
•
It is very common to prepare a concentrated stock solution of a
solute, then dilute it to other concentrations as needed.
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Visualizing the Dilution of a Solution
We start and
end with the
same amount of
solute.
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Addition of
solvent has
decreased the
concentration.
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Dilution Calculations …
• Number of moles of solute does not change on dilution.
• Moles of solute = M × V
• Therefore …
Mconc × Vconc = Mdil × Vdil
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Example 3.26
How many milliliters of a 2.00 M CuSO4 stock solution
are needed to prepare 0.250 L of 0.400 M CuSO4?
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How many milliliters of a 2.00 M CuSO4 stock solution are needed to prepare
0.250 L of 0.400 M CuSO4?
Solution 3.26
This can also be easily solved by using the dilution equation:
Mdil = 0.400 M; Vdil = 0.250 L;
Mconc = 2.00 M; Vconc = ?
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Assessment
As expected, the two methods yield the same result: 50.0 mL 2.00 M CuSO4(aq).
To prepare the dilute solution, we should measure out 50.0 mL of 2.00 M CuSO4 and
add it to enough water to make 0.250 L of solution.
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Animation 4: Dilution
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Solutions in Chemical Reactions
• Molarity provides an additional tool in
stoichiometric calculations based on chemical
equations.
• Molarity provides factors for converting between
moles of solute (either reactant or product) and
liters of solution.
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Example 3.27
A chemical reaction familiar to geologists is that used to identify
limestone.
The reaction of hydrochloric acid with limestone, which is largely calcium
carbonate, is seen through an effervescence —a bubbling due to the
liberation of gaseous carbon dioxide:
CaCO3(s) + 2 HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g)
How many grams of CaCO3(s) are consumed in a reaction with 225 mL of
3.25 M HCl?
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CaCO3(s) + 2 HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g)
How many grams of CaCO3(s) are consumed in a reaction with 225
mL of 3.25 M HCl?
Solution
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Cumulative Example
The combustion in oxygen of 1.5250 g of an alkane-derived compound
composed of carbon, hydrogen, and oxygen yields 3.047 g CO2 and
1.247 g H2O.
The molecular mass of this compound is 88.1 u.
Draw a plausible structural formula for this compound.
Is there more than one possibility? Explain.
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The combustion in oxygen of 1.5250 g of an alkane-derived compound composed of
carbon, hydrogen, and oxygen yields 3.047 g CO2 and 1.247 g H2O. The molecular
mass of this compound is 88.1 u.
Solution
We begin by finding the moles of carbon in 3.047 g of CO2 and the
moles of hydrogen in 1.247 g of H2O.
We will need these numbers of moles for the empirical formula
determination later. Next, we convert moles of each element to mass.
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The combustion in oxygen of 1.5250 g of an alkane-derived compound composed of
carbon, hydrogen, and oxygen yields 3.047 g CO2 and 1.247 g H2O. The molecular
mass of this compound is 88.1 u.
We find the mass of oxygen by subtracting the masses of carbon and
hydrogen from the sample mass.
Then we convert the mass of oxygen to moles.
Now we use the moles of carbon, hydrogen, and oxygen to
construct a tentative formula.
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The combustion in oxygen of 1.5250 g of an alkane-derived compound composed of
carbon, hydrogen, and oxygen yields 3.047 g CO2 and 1.247 g H2O. The molecular
mass of this compound is 88.1 u.
Then we divide each subscript by the subscript for oxygen (because this
is the smallest of the three subscripts).
The empirical formula mass is 44.053 u.
Next, we calculate the ratio of the molecular mass to the empirical
formula mass.
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Draw a plausible structural formula for this compound. Is there more than one
possibility? Explain.
We multiply each subscript by the factor 2 to obtain the molecular
formula.
A carboxylic acid has two oxygen atoms. Butanoic acid fits the
molecular formula (I).
However, there are many other possibilities. The three-carbon chain
of butanoic acid could be replaced by a branched group, as shown in
structure (II).
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
125
Is there more than one possibility? Explain.
The compounds (III) and (IV) also fit the molecular formula.
Another interesting observation is that alkane-based alcohols
(ROH) and ethers [(ROR´),where R and R´ represent alkyl groups]
are not possible structures for the formula C4H8O2. There are not
enough hydrogen atoms in the molecule for this to be the case.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
126