Transcript Stoichiometry: Calculations with Chemical Formulas and Equations

Chemistry, The Central Science,
8 or 9 or 10 or 11th edition
Theodore L. Brown, H. Eugene LeMay, Jr.,
and Bruce E. Bursten
Dr.Imededdine Arbi Nehdi
Chemistry Department, Science College, King
Saud University
Box 2A93, B5
Stoichiometry
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Chapter 2
Stoichiometry:
Calculations with Chemical
Formulas and Equations
Dr.Imededdine Arbi Nehdi
Chemistry Department, Science College, King
Saud University
Stoichiometry
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Chemical Equations
Chemical reactions are represented in a
concise way by chemical equations.
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
-The starting substances, called reactants (left side of the equation)
- The substances produced in the reaction, called products
(appear on the right side of the equation)
- The number in front of the chemical formulas are
Stoichiometry
coefficients.
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Law of Conservation of Mass
(Antoine Lavoisier, 1789)
This law states that matter can neither be created nor
destroyed; it can only change from one form to another.
- According to this Law: during any chemical reaction, the
total mass of the reactants or starting materials must equal
the mass of the products.
- According to this Law :because atoms are neither created
nor destroyed in any reaction, a chemical equation must
have an equal number of atoms of each element on each
side of the narrow. When this condition is met, the equation
is said to be balanced.
Stoichiometry
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Subscripts and Coefficients Give
Different Information
• Subscripts tell the number of atoms of
each element in a molecule.
Stoichiometry
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Subscripts and Coefficients Give
Different Information
• Subscripts tell the number of atoms of each
element in a molecule
• Coefficients tell the number of molecules.
• Subscripts should never be changed in
balancing an equation.
Stoichiometry
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Reaction
Types
Stoichiometry
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Combination Reactions
• In this type of
reaction two
or more
substances
react to form
one product.
• Examples:
– 2 Mg (s) + O2 (g)  2 MgO (s)
– N2 (g) + 3 H2 (g)  2 NH3 (g)
– C3H6 (g) + Br2 (l)  C3H6Br2 (l)
Stoichiometry
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Decomposition Reactions
• In a decomposition
one substance breaks
down into two or more
substances.
• Examples:
– CaCO3 (s)  CaO (s) + CO2 (g)
– 2 KClO3 (s)  2 KCl (s) + O2 (g)
– 2 NaN3 (s)  2 Na (s) + 3 N2 (g)
Stoichiometry
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Combustion Reactions
• These are generally
rapid reactions that
produce a flame.
• Most often involve
hydrocarbons
reacting with oxygen
in the air.
• Examples:
– CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
– C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
Stoichiometry
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Formulas
Weights
Stoichiometry
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CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
• Chemical formulas and chemical equations both are
a quantitative significance; the subscripts in formulas
and the coefficients in equations represent precise
quantities.
• But how do to relate the numbers of atoms or
molecules to the amounts we measure out in the
laboratory?
• So, we must examine the masses of atoms and
molecules.
Stoichiometry
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ATOMIC MASS UNIT (amu)
• The masses of individual atoms are
extremely very small.
• One atom of hydrogen (1H) has a mass of
1.6735 x 10-24 g
• One atom of oxygen (16O) has a mass of
2.6560 x 10-23 g
• It is convenient to use a unit called the atomic
mass unit (amu):
• 1 amu = 1.66054 x 10-24 g
• The amu is defined by assigning a mass of
exactly 12 amu to the 12C.
Stoichiometry
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ATOMIC MASS UNIT (amu)
• In these units the atomic weight :
• Hydrogen (1H) atom is 1.0078 amu,
• Oxygen (16O) = 15.9949 amu
• Iron (56Fe) = 55.845 amu
Stoichiometry
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AVERAGE ATOMIC MASS (ATOMIC WEIGHT)
• Most elements occur in nature as mixture of isotopes
• Atoms of a given element that differ in the number of
neutrons, and consequently in mass, are called
isotopes.
• An atom of a specific isotope is called a nuclide.
• We can determine the average atomic mass of an
element by using the masses of its various isotopes and
their relative abundances.
• For example, naturally occuring carbon is composed of
98.892 percent of 12C and 1.108 % of 13C.
• The masses of these nuclides are 12 amu and
13.00355 amu, respectively.
• The average atomic mass (atomic weight) of carbon is:
Stoichiometry
0.98892 *12 + 0.01108 * 13.00335 = 12.011 amu
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Formula
Weights
Stoichiometry
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Formula Weight (FW)
• A formula weight is the sum of the
atomic weights for the atoms in a
chemical formula.
• So, the formula weight (FW) of calcium
chloride, CaCl2, would be
Ca: 1(40.1 amu)
+ Cl: 2(35.5 amu)
111.1 amu
• Formula weights are generally reported
for ionic compounds.
Stoichiometry
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Molecular Weight (MW)
• If the chemical formula of a substance is its molecular
formula, then the formula weight is also called the
molecular weight.
• A molecular weight is the sum of the atomic weights
of the atoms in a molecule.
• For the molecule ethane, C2H6, the molecular weight
(MW) would be
C:
2(12.0 amu)
+ H:
6(1.0 amu)
30.0 amu
Stoichiometry
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Percent Composition from Formulas
One can find the percentage of the
mass of a compound that comes from
each of the elements in the compound
by using this equation:
(number of atoms of that element)(atomic weight of element)
x 100
% element =
(FW of the compound)
Stoichiometry
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Percent Composition
So the percentage of carbon in ethane
(C2H6) is…
(2)(12.0 amu)
%C =
(30.0 amu)
24.0 amu
x 100
=
30.0 amu
= 80.0%
Stoichiometry
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Avogadro’s Number and the Mole
Stoichiometry
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The Mole
Even the smallest samples that we deal with in
the laboratory contain enormous numbers of
atoms, ions, or molecules. For example, a
teaspoon of water (about 5 mL) contains 2 X 10
23 water
molecules? It is convenient to have a
special unit for describing such large numbers
of objects, like dozen (12 objects) and gross
(144 objects).
Stoichiometry
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The Mole is a convenient measure of chemical quantities.
- A mole is defined as the amount of matter that contains as
many objects (atoms, ions, or molecules) as the number of
atoms in exactly 12 g of 12C.
- The number of atoms in 12 g of 12C is 6.0221421 x 1023.
- This number(6.0221421 x 1023) is called Avogadro’s number.
- Thus, a mole of ions, a mole of molecules, or a mole of
anything else all contain Avogadro’s number of these objects:
1 mole of carbon atoms = 6.02 x 1023 carbon atoms.
1 mol H2O molecules = 6.02 x 1023 H2O molecules.
1 mol NO3- ions = 6.02 x 1023 NO3- ions .
Stoichiometry
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Molar Mass
• By definition, a molar mass is the mass of 1
mol of a substance (i.e., g/mol).
– The mass of a single atom of an element ( in amu)
is numerically equal to the mass (in grams) of 1
mol of atoms of that element.
One 12C atom weighs 12 amu
1 mol 12C weighs 12 g.
One 24Mg atom weighs 24 amu 1 mol 24Mg weighs 24 g.
One 197Au atom weighs 197 amu 1 mol 197Au weighs 197 g.
Stoichiometry
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Molar Mass
• The molar mass (in g/mol) of any substance is
always numerically equal to its formula weight
(in amu).
One H2O molecule weighs 18.0 amu
One NO3- ion weighs 62.0 amu
One NaCl unit weighs 58.5 amu
1 mol H2O weighs 18 g.
1 mol NO3- weighs 62.0 g.
1 mol NaCl weighs 58.5 g.
Stoichiometry
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Mole Relationships
• One mole of atoms, ions, or molecules contains
Avogadro’s number of those particles.
• One mole of molecules or formula units contains
Avogadro’s number times the number of atoms or
ions of each element in the compound.
Stoichiometry
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Empirical
Formulas
from analyses
Stoichiometry
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Calculating Empirical Formulas
The empirical formula for a substance tells us the relative
number of atoms of each element it contains. Thus, the
formula H2O indicates that the water contains two H atoms
for each O atom. This ratio also applies on the molar level;
Thus 1 mol of H2O contains 2 mol of H atoms and 1 mol of
O atoms. Conversely, the ratio of the number of moles of
each element in a compound gives the substcripts in a
compounds empirical formula. Thus the mole concept
provides a way of calculating the empirical formulas of
Stoichiometry
chemical substances.
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Example of calculating empirical formula
The compound para-aminobenzoic acid (you may have
seen it listed as PABA on your bottle of sunscreen) is
composed of carbon (61.31%), hydrogen (5.14%),
nitrogen (10.21%), and oxygen (23.33%). Find the
empirical formula of PABA.
Stoichiometry
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Calculating Empirical Formulas
Assuming 100.00 g of para-aminobenzoic acid,
C:
H:
N:
O:
1 mol
12.01 g
1 mol
5.14 g x
1.01 g
1 mol
10.21 g x
14.01 g
1 mol
23.33 g x
16.00 g
61.31 g x
= 5.105 mol C
= 5.09 mol H
= 0.7288 mol N
= 1.456 mol O
Stoichiometry
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Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number
of moles:
C:
5.105 mol
0.7288 mol
= 7.005  7
H:
5.09 mol
0.7288 mol
= 6.984  7
N:
0.7288 mol
0.7288 mol
= 1.000
O:
1.458 mol
0.7288 mol
= 2.001  2
Stoichiometry
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Calculating Empirical Formulas
These are the subscripts for the empirical formula:
C7H7NO2
Stoichiometry
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Procedure used to Calculate
Empirical Formulas
Stoichiometry
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Calculating Molecular Formula from
Empirical Formula
We can obtain the molecular formula from the empirical
formula if we known the molecular weight of the
compound. The subscripts in the molecular formula
of a compound are whole-number multiple
of the subscripts in its empirical formula.
Procedure of calcul
Example: Mesitylene, a hydrocarbon that occurs in small amounts in crude oil,
has an empirical formula of C3H4. The experimentally determined molecular
weight of this substance is 121 amu.
What is the molecular formula of mesitylene?
Stoichiometry
Solution:
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Plan The subscripts in the molecular formula of a compound are
whole-number multiple of the subscripts in its empirical formula.
To find the appropriate multiple, we must compare the molecular
weight with the formula weight of the empirical formula.
Solve First, we calculate the formula weight of the empirical
formula, C3H4:
3(12.0 amu) + 4(1.0 amu) = 40.0 amu
Next, we divide the molecular weight by the empirical formula
weight to obtain the multiple used to multiply the subscripts in
C 3H 4:
Only whole-number ratios make physical sense because we
must be dealing with whole atoms. The 3.02 in this case could
result from a small experimental error in the molecular weight.
We therefore multiply each subscript in the empirical formula by
3 to give the molecular formula: C9H12.
Stoichiometry
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Practice Exercise
Ethylene glycol, the substance used in automobile antifreeze, is composed of
38.7% C, 9.7% H, and 51.6% O by mass. Its molar mass is 62.1 g/mol. (a)
What is the empirical formula of ethylene glycol? (b) What is its molecular
formula?
Answers: (a) CH3O, (b) C2H6O2
Stoichiometry
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Combustion Analysis
• Compounds containing C, H and O are routinely
analyzed through combustion in an apparatus like
this.
• CuO (copper oxide) helps to oxide traces of carbon
and carbon monoxide to carbon dioxide and to oxide
hydrogen to water
Stoichiometry
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Calculating Empirical Formula from
Combustion Analysis
When a compound containing carbon and hydrogen is completely
combusted in this apparatus, all the carbon and hydrogen
are converted to CO2 and H2O, respectively. From the masses of
CO2 and H2O we can calculate the number of moles of C and H
in the compound and thereby the empirical formula.
If the compound contains a third element, its mass can be
determined by subtracting the masses of C and H from the
compound's original mass.
Stoichiometry
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Calculating Empirical Formula from
Combustion Analysis
Example: Isopropyl alcohol, a substance sold as rubbing alcohol, is composed of C, H,
and O. Combustion of 0.255 g of isopropyl alcohol produces 0.561 g of CO2 and 0.306
g of H2O. Determine the empirical formula of isopropyl alcohol.
Solution
Solve To calculate the number of grams of C, we first use the molar mass of CO 2, 1 mol CO2 = 44.0 g CO2, to
convert grams of CO2 to moles of CO2. Because each CO2 molecule has only 1 C atom, there is 1 mol of C atoms per mole of
CO2 molecules. This fact allows us to convert the moles of CO2 to moles of C. Finally, we use the molar mass of C, 1 mol C
=1 2.0 g C, to convert moles of C to grams of C. Combining the three conversion factors, we have
The calculation of the number of grams of H from the grams of H2O is similar, although we must remember that there are
2 mol of H atoms per 1 mol of H2O molecules:
The total mass of the sample, 0.255 g, is the sum of the masses of the C, H, and O. Thus, we can calculate the
mass of O as follows:
Stoichiometry
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Calculating Empirical Formula from
Combustion Analysis
We then calculate the number of moles of C, H, and O in the sample:
To find the empirical formula, we must compare the relative number of moles of
each element in the sample. The relative number of moles of each element is
found by dividing each number by the smallest number, 0.0043. The mole ratio
of C:H:O so obtained is 2.98:7.91:1.00. The first two numbers are very close to
the whole numbers 3 and 8, giving the empirical formula C3H8O.
Stoichiometry
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Calculating Empirical Formula from
Combustion Analysis
Practice Exercise
a) Caproic acid, which is responsible for the foul odor of dirty socks, is
composed of C, H, and O atoms. Combustion of a 0.225-g sample of
this compound produces 0.512 g CO2 and 0.209 g H2O. What is the
empirical formula of caproic acid? (b) Caproic acid has a molar mass
of 116 g/mol. What is its molecular formula?
Answers: (a) C3H6O, (b) C6H12O2
Stoichiometry
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Quantitative information from
balanced equations
aA + bB
cC + dD
If a reaction occurs stoichiometrically, the relationship
between the quantities of the substances are:
nA/a = nB/b = nC/c = nD/d
nA, nB, nC, nD, quantities (mol) of formed products and
reacted reactants
Stoichiometry
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Limiting
Reactants
Stoichiometry
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Limiting Reactants
The number of moles of O2 needed to react with all the 10 moles of H2
is : nH2/2 = 10/2 = nO2/1
nO2= 5
Because 7 mol O2 was available at the start of the reaction,
7 -5 = 2 mol O2 will be present when the H2 is all consumed.
• The reactant that is completely consumed in the reaction is called the
limiting reactant because it determines, or limits, the amount of
product formed. The other reactant is called excess reactant.
Stoichiometry
• In our example, H2 is the limiting reactant, and O2 is the excess
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reactant or excess reagent.
Limiting Reactants
• The limiting reactant is the reactant present in the smallest
stoichiometric amount.
• In other words, it’s the reactant you’ll run out of first
– The quantity of limiting reactant is used to
calculate the quantity of any product formed
In our case, the number of moles of H2O formed is :
nH2/2 = 10/2 = nH2O/2
nH2O= 10
Stoichiometry
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Theoretical and Actual Yields
• The quantity of product that is
calculated to form when all of the
limiting reactant reacts is called the
theoretical yield.
• The amount of product actually obtained
in the reaction is called the actual yield.
• The actual yield is always less than the
theoretical yield.
Stoichiometry
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Percent Yield
One finds the percent yield by
comparing the amount actually obtained
(actual yield) to the amount it was
possible to make (theoretical yield).
Actual Yield
Percent Yield =
Theoretical Yield
x 100
Stoichiometry
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Percent Yield
Consider the following reaction:
2 NaPO4(aq) + 3 Ba(NO3)2 (aq)
Ba3(PO4)2 (s) + 6 NaNO3 (aq)
4.92 g of Ba3(PO4)2 should form when 3.50 g of NaPO4 is mixed
with 6.40 g of Ba(NO3)2 .
If the actual yield turned out to be 4.70 g, calculate the percent
yield?
The theoretical yield is 4.92 g, then the percent yield is:
(4.70g/4.92 g ) x 100 % = 95.5 %
Stoichiometry
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Solution
Stoichiometry
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* A solution is a homogeneous mixture of two
or more substances
* One of these substances is called the
solvent; it is present in greater quantity
* The other substances in the solution are the
solutes
Stoichiometry
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Ways of
Expressing
Concentrations
of Solutions
Stoichiometry
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Molarity
• Two solutions can contain the same
compounds but be quite different because the
proportions of those compounds are different.
• Molarity is one way to measure the
concentration of a solution.
Molarity (M) =
moles of solute
volume of solution in liters
M = n(mol)/V (L)
n (mol) = M x V (L)
•Since volume is temperature-dependent, molarity
Stoichiometry
can change with temperature.
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Molality (m)
m=
mol of solute
kg of solvent
Since both moles and mass do not
change with temperature, molality
(unlike molarity) is not temperaturedependent.
Stoichiometry
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Mole Fraction (X)
moles of A
XA =
total moles in solution
• In some applications, one needs the
mole fraction of solvent, not solute —
make sure you find the quantity you
need!
Stoichiometry
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Mass Percentage
mass of A in solution
 100
Mass % of A =
total mass of solution
Stoichiometry
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Parts per Million and
Parts per Billion
Parts per Million (ppm)
mass of A in solution
 106
ppm =
total mass of solution
Parts per Billion (ppb)
mass of A in solution
 109
ppb =
total mass of solution
-The mass units of the solute and solution
should be the same.
Stoichiometry
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Changing Molarity to Molality
If we know the
density of the
solution, we can
calculate the
molality from the
molarity and vice
versa.
Stoichiometry
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Mixing a Solution
• To create a solution of a
known molarity, one
weighs out a known mass
(and, therefore, number of
moles) of the solute.
• The solute is added to a
volumetric flask, and
solvent is added to the line
on the neck of the flask.
Stoichiometry
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Dilution
• One can also dilute a more concentrated
solution by
– Using a pipet to deliver a volume of the solution to
a new volumetric flask, and
– Adding solvent to the line on the neck of the new
flask.
Stoichiometry
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Dilution
The molarity of the new solution can be determined
from the equation
Mc  Vc = Md  Vd,
where Mc and Md are the molarity of the concentrated and dilute
solutions, respectively, and Vc and Vd are the volumes of the
two solutions.
Stoichiometry
Vc
Vd
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Practice exercise
• What is the volume (in ml) required to
dilute 220 ml of 1.75 M HCl to exactly
1.25 M
Stoichiometry
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Solution
•
•
•
•
•
Mc x Vc = Md x Vd
1.75 x 220 = 1.25 x Vd
Vd = (1.75 x 220)/1.25 = 308 mL
The volume required of water is
V = 308 – 220 = 88 mL
Stoichiometry
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Using Molarities in
Stoichiometric Calculations
Stoichiometry
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One finds the percent yield by
comparing the amount actually obtained
(actual yield) to the amount it was
possible to make (theoretical yield).
Actual Yield
Percent Yield =
Theoretical Yield
x 100
Stoichiometry
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