Transcript Molar mass - HCC Learning Web

Chapter 3
Mass Relationships in Chemical
Reactions
1
Micro World
atoms & molecules
Macro World
grams
Atomic mass is the mass of an atom in
atomic mass units (amu)
By definition:
1 atom 12C “weighs” 12 amu
On this scale
1H
= 1.008 amu
16O
= 16.00 amu
2
3
Copper, a metal known since ancient times, is used in
electrical cables and pennies, among other things.
The atomic masses of its two stable isotopes,
(69.09
percent) and (30.91 percent), are 62.93 amu and 64.9278
amu, respectively.
Calculate the average atomic mass of copper. The relative
abundances are given in parentheses.
Each isotope contributes to the average
atomic mass based on its relative
abundance.
(0.6909) (62.93 amu) + (0.3091)
(64.9278 amu) = 63.55 amu
Multiplying the mass of an isotope by its
fractional abundance (not percent) will give
the contribution to the average atomic
mass of that particular isotope.
4
5
The Mole (mol): A unit to count numbers of particles
Dozen = 12
Pair = 2
The mole (mol) is the amount of a substance that
contains as many elementary entities as there
are atoms in exactly 12.00 grams of 12C
1 mol = NA = 6.0221415 x 1023
Avogadro’s number (NA)
6
Avogadro’s number (NA)
6.022 x 1023
= NA
= 1mol
7
eggs
Molar mass is the mass of 1 mole of shoes in grams
marbles
atoms
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li
For any element
atomic mass (amu) = molar mass (grams)
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One Mole of:
S
C
Hg
Cu
Fe
9
1 12C atom
12.00 g
1.66 x 10-24 g
x
=
23
12
12.00 amu
6.022 x 10
C atoms
1 amu
M = molar mass in g/mol
NA = Avogadro’s number
10
3.2
Helium (He) is a valuable gas used in industry, low-temperature
research, deep-sea diving tanks, and balloons.
How many moles of He atoms are in 6.46 g of He?
Thus, there are 1.61 moles of He atoms in 6.46 g of He.
Check
Because the given mass (6.46 g) is larger than the molar mass of He,
we expect to have more than 1 mole of He.
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Moles to g
• How many grams of gold (Au) are in 15.3
moles of Au?
1mole of Au = 197.0 g
15.3 moles of Au = (197.0 g/1mol)x 15.3 mol
= 3014.0 g
=3.01x103 g
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Mass and Moles of a Substance
• Mole calculations
Suppose we have 100.0 grams of iron (Fe). The
atomic weight of iron is 55.8 g/mol. How many
moles of iron does this represent?
100.0 g Fe
moles Fe 
55.8 g/mol
 1.79 moles of Fe
13
Mass and Moles of a Substance
• Mole calculations
Conversely, suppose we have 5.75 moles of
magnesium (atomic wt. = 24.3 g/mol). What is
its mass?
mass Mg  (5.75 moles)  (24.3 g/mol)
 140 grams of Mg
14
Do You Understand Molar Mass?
How many atoms are in 0.551 g of potassium (K) ?
1 mol K = 39.10 g K
1 mol K = 6.022 x 1023 atoms K
1 mol K
6.022 x 1023 atoms K
0.551 g K x
x
=
1 mol K
39.10 g K
8.49 x 1021 atoms K
15
1. What is the mass in grams of 2.5 mol Ca
2. Calculate the number of atoms in 1.7 mol B.
3. Calculate the number of atoms in 5.0 g Al.
4. Calculate the mass of 5,000,000 atoms of Au.
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Molecular Weight and Formula Weight
Formula mass is the sum of the atomic masses
(in amu) in a formula unit of an ionic compound.
1Na
1Cl
NaCl
22.99 amu
+ 35.45 amu
58.44 amu
For any ionic compound
formula mass (amu) = molar mass (grams)
1 formula unit NaCl = 58.44 amu
1 mole NaCl = 58.44 g NaCl
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Do You Understand Formula Mass?
What is the formula mass of Ca3(PO4)2 ?
1 formula unit of Ca3(PO4)2
3 Ca
3 x 40.08
2P
2 x 30.97
8O
+ 8 x 16.00
310.18 amu
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Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
1S
SO2
2O
SO2
32.07 amu
+ 2 x 16.00 amu
64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
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Mass and Moles of a Substance
• Mole calculations
This same method applies to compounds.
Suppose we have 100.0 grams of H2O
(molecular weight = 18.0 g/mol). How many
moles does this represent?
100.0 g H 2O
moles H 2O 
18.0 g/mol
 5.56 moles of H 2O
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Mass and Moles of a Substance
• Mole calculations
Conversely, suppose we have 3.25 moles of
glucose, C6H12O6 (molecular wt. = 180.0 g/mol).
What is its mass?
mass C6 H12O6  ( 3.25 moles)  (180.0 g/mol)
 585 grams of C6 H12O6
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Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol C3H8O molecules = 8 mol H atoms
1 mol H = 6.022 x 1023 atoms H
1 mol C3H8O 8 mol H atoms 6.022 x 1023 H atoms
72.5 g C3H8O x
x
x
=
1 mol C3H8O
1 mol H atoms
60 g C3H8O
5.82 x 1024 atoms H
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23
Heavy
Light
Heavy
Light
24
Determining Chemical Formulas
• The percent composition of a compound is the
mass percentage of each element in the
compound.
We define the mass percentage of “A” as the
parts of “A” per hundred parts of the total, by
mass. That is,
mass of " A" in whole
mass % " A" 
 100%
mass of the whole
25
Mass Percentages from Formulas
• Let’s calculate the percent composition of
butane, C4H10.
First, we need the molecular mass of C4H10.
4 carbons @ 12.0 amu/atom  48.0 amu
10 hydrogens @ 1.00 amu/atom  10.0 amu
1 molecule of C4 H10  58.0 amu
Now, we can calculate the percents.
amu C
% C  5848.0
.0 amu total  100%  82.8%C
amu H
% H  5810.0
.0 amu total  100%  17.2%H
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Percent composition of an element in a compound =
n x molar mass of element
x 100%
molar mass of compound
n is the number of moles of the element in 1 mole
of the compound
2 x (12.01 g)
x 100% = 52.14%
46.07 g
6 x (1.008 g)
%H =
x 100% = 13.13%
46.07 g
1 x (16.00 g)
%O =
x 100% = 34.73%
46.07 g
%C =
C2H6O
52.14% + 13.13% + 34.73% = 100.0%
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Determining Chemical Formulas
• Determining the formula of a compound
from the percent composition.
The percent composition of a compound leads
directly to its empirical formula.
An empirical formula (or simplest formula) for
a compound is the formula of the substance
written with the smallest integer (whole
number) subscripts.
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Percent Composition and Empirical Formulas
Determine the empirical formula of a
compound that has the following
percent composition by mass:
K 24.75, Mn 34.77, O 40.51 percent.
1 mol K
nK = 24.75 g K x
= 0.6330 mol K
39.10 g K
nMn = 34.77 g Mn x
1 mol Mn
= 0.6329 mol Mn
54.94 g Mn
nO = 40.51 g O x
1 mol O
= 2.532 mol O
16.00 g O
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Percent Composition and Empirical Formulas
nK = 0.6330, nMn = 0.6329, nO = 2.532
0.6330 ~
K:
~ 1.0
0.6329
Mn :
0.6329
= 1.0
0.6329
2.532 ~
O:
~ 4.0
0.6329
KMnO4
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Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
g CO2
mol CO2
mol C
gC
6.0 g C = 0.5 mol C
g H2O
mol H2O
mol H
gH
1.5 g H = 1.5 mol H
g of O = g of sample – (g of C + g of H)
4.0 g O = 0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
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Determining Chemical Formulas
• Determining the empirical formula from
the percent composition.
Benzoic acid is a white, crystalline powder
used as a food preservative. The
compound contains 68.8% C, 5.0% H, and
26.2% O by mass. What is its empirical
formula?
In other words, give the smallest wholenumber ratio of the subscripts in the
formula
Cx HyOz
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Determining Chemical Formulas
Determining the empirical formula from
the percent composition.
Our 100.0 grams of benzoic acid would
contain:
1 mol C
68.8 g C 
 5.73 mol C
12.0 g
1 mol H
5.0 g H 
 5.0 mol H
1.0 g
1 mol O
26.2 g O 
 1.63mol O
16.0 g
This isn’t quite a
whole number
ratio, but if we
divide each
number by the
smallest of the
three, a better
ratio might
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emerge.
Determining Chemical Formulas
• Determining the empirical formula from
the percent composition.
Our 100.0 grams of benzoic acid would
contain:
now it’s not too
5.73 mol C  1.63  3.50
difficult to see that
the smallest whole
5.0 mol H  1.63  3.0
number ratio is
1.63mol O  1.63  1.00 7:6:2.
The empirical
formula is C7H6O2 .
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Determining Chemical Formulas
• Determining the “true” molecular formula
from the empirical formula.
For example, suppose the empirical formula of a
compound is CH2O and its “true” molecular weight
is 60.0 g/mol.
The molar weight of the empirical formula (the
“empirical weight”) is only 30.0 g/mol.
This would imply that the “true” molecular
formula is actually the empirical formula doubled,
or
C2H4O2
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3.11
A sample of a compound contains 30.46 percent nitrogen and
69.54 percent oxygen by mass, as determined by a mass
spectrometer.
In a separate experiment, the molar mass of the compound is
found to be between 90 g and 95 g.
Determine the molecular formula and the accurate molar mass
of the compound.
36
3.11
Strategy
To determine the molecular formula, we first need to determine
the empirical formula. Comparing the empirical molar mass to
the experimentally determined molar mass will reveal the
relationship between the empirical formula and molecular
formula.
Solution
We start by assuming that there are 100 g of the compound.
Then each percentage can be converted directly to grams; that
is, 30.46 g of N and 69.54 g of O.
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3.11
Let n represent the number of moles of each element so that
Thus, we arrive at the formula N2.174O4.346, which gives the
identity and the ratios of atoms present. However, chemical
formulas are written with whole numbers.
Try to convert to whole numbers by dividing the subscripts by
the smaller subscript (2.174). After rounding off, we obtain NO2
as the empirical formula.
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3.11
The molecular formula might be the same as the empirical
formula or some integral multiple of it (for example, two, three,
four, or more times the empirical formula).
Comparing the ratio of the molar mass to the molar mass of the
empirical formula will show the integral relationship between the
empirical and molecular formulas.
The molar mass of the empirical formula NO2 is
empirical molar mass = 14.01 g + 2(16.00 g) = 46.01 g
39
3.11
Next, we determine the ratio between the molar mass and the
empirical molar mass
The molar mass is twice the empirical molar mass. This means
that there are two NO2 units in each molecule of the compound,
and the molecular formula is (NO2)2 or N2O4. The actual molar
mass of the compound is two times the empirical molar mass,
that is, 2(46.01 g) or 92.02 g, which is between 90 g and 95 g.
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Chemical Reactions and Chemical Equations
A process in which one or more substances is changed into one
or more new substances is a chemical reaction
A chemical equation uses chemical symbols to show what
happens during a chemical reaction
3 ways of representing the reaction of H2 with O2 to form H2O
reactants
products
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How to “Read” Chemical Equations
2 Mg + O2
2 MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
IS NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
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The Experimental Basis of a Chemical Equation
We know that a chemical equation represents a
chemical change. The following is evidence for a
reaction:
• Release of a gas.
– CO2 is released when acid is placed in a
solution containing CO32- ions.
– H2 is released when Na is placed in
water.
• Formation of a solid (precipitate.)
– A solution containing Ag+ ions is mixed
with a solution containing Cl- ions.
43
Chemical Reactions: Equations
• Writing chemical equations
A chemical equation is the symbolic representation
of a chemical reaction in terms of chemical formulas.
For example, the burning of sodium and chlorine to
produce sodium chloride is written
2Na  Cl 2  2NaCl
The reactants are starting substances in a chemical
reaction. The arrow means “yields.” The formulas on
the right side of the arrow represent the products.
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Chemical Reactions: Equations
• Writing chemical equations
In many cases, it is useful to indicate the states of
the substances in the equation.
When you use these labels, the previous equation
becomes
2Na(s )  Cl 2 (g )  2NaCl(s )
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Chemical Reactions: Equations
• Writing chemical equations
The law of conservation of mass dictates that the
total number of atoms of each element on both
sides of a chemical equation must match. The
equation is then said to be balanced.
CH 4 
O 2  CO 2  H 2O
Consider the combustion of methane to produce
carbon dioxide and water.
46
Chemical Reactions: Equations
• Writing chemical equations
For this equation to balance, two molecules of
oxygen must be consumed for each molecule of
methane, producing one molecule of CO2 and two
molecules of water.
CH 4  2 O 2  CO 2  2 H 2O
Now the equation is “balanced.”
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Balancing Chemical Equations
1. Write the correct formula(s) for the reactants on
the left side and the correct formula(s) for the
product(s) on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2
CO2 + H2O
2. Change the numbers in front of the formulas
(coefficients) to make the number of atoms of
each element the same on both sides of the
equation. Do not change the subscripts.
2C2H6
NOT
C4H12
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Balancing Chemical Equations
3. Start by balancing those elements that appear in
only one reactant and one product.
C2H6 + O2
2 carbon
on left
C2H6 + O2
6 hydrogen
on left
C2H6 + O2
CO2 + H2O
start with C or H but not O
1 carbon
on right
multiply CO2 by 2
2CO2 + H2O
2 hydrogen
on right
2CO2 + 3H2O
multiply H2O by 3
49
Balancing Chemical Equations
4. Balance those elements that appear in two or
more reactants or products.
C2H6 + O2
2 oxygen
on left
2CO2 + 3H2O
multiply O2 by 7
2
4 oxygen + 3 oxygen = 7 oxygen
(3x1)
on right
(2x2)
C2H6 + 7 O2
2
2CO2 + 3H2O
2C2H6 + 7O2
4CO2 + 6H2O
remove fraction
multiply both sides by 2
50
Balancing Chemical Equations
5. Check to make sure that you have the same
number of each type of atom on both sides of the
equation.
2C2H6 + 7O2
4CO2 + 6H2O
4 C (2 x 2)
4C
12 H (2 x 6)
12 H (6 x 2)
14 O (7 x 2)
14 O (4 x 2 + 6)
Reactants
4C
12 H
14 O
Products
4C
12 H
14 O
51
Chemical Reactions: Equations
• Balance the following equations.
O2 
P4 
PCl 3 
N 2O 
As2S 3 
Ca3 (PO4 )2 
O2 
POCl 3
P4O6 
As2O 3 
H 3 PO 4 
N2
SO 2
Ca(H2 PO 4 )2
52
Chemical Reactions: Equations
• Balance the following equations.
O 2  2 PCl 3  2 POCl 3
P4  6 N 2O 
P4O6  6 N 2
2 As2S 3  9 O 2  2 As2O 3  6 SO 2
Ca3 (PO4 )2  4 H 3 PO 4  3Ca(H2 PO 4 )2
53
Stoichiometry: Quantitative Relations in
Chemical Reactions
• Stoichiometry is the calculation of the quantities
of reactants and products involved in a chemical
reaction.
It is based on the balanced chemical equation
and on the relationship between mass and
moles.
Such calculations are fundamental to most
quantitative work in chemistry.
Amounts of Reactants and Products
1. Write balanced chemical equation
2. Convert quantities of known substances into moles
3. Use coefficients in balanced equation to calculate the
number of moles of the sought quantity
4. Convert moles of sought quantity into desired units
Molar Interpretation of a Chemical
Equation
• The balanced chemical equation can be interpreted in
numbers of molecules, but generally chemists
interpret equations as “mole-to-mole” relationships.
For example, the Haber process for producing
ammonia involves the reaction of hydrogen
and nitrogen.
N 2 (g )  3 H 2 (g )  2 NH 3 (g )
Molar Interpretation of a Chemical
Equation
• This balanced chemical equation shows that one
mole of N2 reacts with 3 moles of H2 to produce 2
moles of NH3.
N 2 (g)
1 molecule N2
1 mol N 2

3H 2 (g)
+


3 molecules H2
3 mol H 2

2 NH 3 (g )
2 molecules NH3
2 mol NH 3
Because moles can be converted to mass, you
can also give a mass interpretation of a
chemical equation.
Calculations Using the Chemical
Equation
7
• We will learn in this section to calculate
quantities of reactants and products in a
chemical reaction.
• Need a balanced chemical equation for the
reaction of interest.
• Keep in mind that the coefficients
represent the number of moles of each
substance in the equation.
Molar Interpretation of a Chemical
Equation
• Suppose we wished to determine the number of
moles of NH3 we could obtain from 4.8 mol H2.
N 2 (g)  3H 2 (g)  2 NH 3 (g )
Because the coefficients in the balanced
equation represent mole-to-mole ratios, the
calculation is simple.
2 mol NH 3
4.8 mol H 2 
 3.2 mol NH 3
3 mol H 2
Limiting Reagent:
Reactant used up first in
the reaction.
2NO + O2
2NO2
NO is the limiting reagent
O2 is the excess reagent
60
Limiting Reagent
• Zinc metal reacts with hydrochloric acid by
the following reaction.
Zn(s)  2 HCl(aq)  ZnCl 2 (aq)  H 2 (g )
If 0.30 mol Zn is added to hydrochloric acid
containing 0.52 mol HCl, how many moles of
H2 are produced?
61
Limiting Reagent
Take each reactant in turn and ask how much product
would be obtained if each were totally consumed. The
reactant that gives the smaller amount is the limiting
reagent.
1 mol H 2
0.30 mol Zn 
 0.30 mol H 2
1 mol Zn
1 mol H 2
0.52 mol HCl 
 0.26 mol H 2
2 mol HCl
Since HCl is the limiting reagent, the amount of H2
produced must be 0.26 mol.
62
Reaction Yield
Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted.
Actual Yield is the amount of product actually obtained
from a reaction.
% Yield =
Actual Yield
x 100
Theoretical Yield
63
Theoretical and Percent Yield
• The theoretical yield of product is the maximum
amount of product that can be obtained from
given amounts of reactants.
The percentage yield is the actual yield
(experimentally determined) expressed as a
percentage of the theoretical yield (calculated).
actual yield
%Yield 
 100%
theoretical yield
64
Theoretical and Percent Yield
• To illustrate the calculation of percentage yield,
recall that the theoretical yield of H2 in the
previous example was 0.26 mol (or 0.52 g) H2.
If the actual yield of the reaction had been 0.22 g
H2, then
0.22 g H 2
%Yield 
 100%  42%
0.52 g H 2
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Summary
• Atomic mass unit (amu)
• Avagadro’s number
• Percent composition by mass of a compound,
Empirical formula, molecular formula
• Chemical Equations
• Stoichiometry and limiting reagents
• Theoretical yield, actual yied, and percent
yield
66
Homework
• Practice all the examples worked out in the
book
• 3.2,3.5,3.9,3.13,3.14,3.16,3.22,3.24,3.25,3.26,
3.38,3.40,3.48,3.50,3.52,3.59,3.60,3.66,3.67,
3.68,3.89
67