Download 06.1 - Chemical formulas and composition stoichiometry

Chemical formulas and
composition stoichiometry
Stoichiometry
The word “stoichiometry” is derived from the Greek
stoicheion, which means “element,” and metron, which means
“measure.”
Stoichiometry describes the quantitative relationships among elements in
compounds (composition stoichiometry) and among substances as they
undergo chemical changes (reaction stoichiometry).
Main division of chemical compounds: organic vs inorganic
Many of the molecules found in nature are organic compounds.
• Organic compounds contain C-C or C-H bonds or both, often in combination
with nitrogen, oxygen, sulfur, and other elements.
• All of the other compounds are inorganic compounds.
Name
Formula
Name
Formula
Name
Formula
water
H2O
sulfur dioxide
SO2
butane
C4H10
hydrogen peroxide
H2O2
sulfur trioxide
SO3
pentane
C5H12
hydrogen chloride*
HCl
carbon monoxide
CO
benzene
C6H6
sulfuric acid
H2SO4
carbon dioxide
CO2
methanol (methyl alcohol)
CH3OH
nitric acid
HNO3
methane
CH4
ethanol (ethyl alcohol)
CH3CH2OH
acetic acid
CH3COOH
ethane
C2H6
acetone
CH3COCH3
ammonia
NH3
propane
C3H8
diethyl ether (ether)
CH3CH2OCH2CH3
Chemical formulas
The chemical formula gives the number of atoms of each type in the molecule.
But this formula does not express the order in which the atoms in the
molecules are bonded together.
For instance, the structural formula of propane is C3H8 and shows that the molecule is composed of 3
carbons and 8 hydrogens.
The structural formula shows the order in which atoms are connected. The
lines connecting atomic symbols represent chemical bonds between atoms
The structural formula of propane shows that the three C atoms are linked in a chain, with three H
atoms bonded to each of the end C atoms and two H atoms bonded to the center C.
H
H
H
H
C
C
C
H
H
H
H
Chemists sometimes write out the formula in a way to better convey the connectivity information,
e.g., CH3CH2CH3, which is a longer representation of the propane chemical formula.
Chemical formulas
Each molecule of acetic acid, CH3COOH, contains 2 carbon atoms, 4 hydrogen
atoms, and 2 oxygens.
Writing it as CH3COOH (instead of C2H4O2) includes useful bonding and
structural information.
H
H
C
H
OH
C
O
Law of Definite Proportions
A compound can be decomposed by chemical means into simpler substances,
always in the same ratio by mass.
E.g.
Water originates from the reaction between two H and one O. This leads to the ratio 1:8 which
is the ratio of the masses of H and O in the molecule of H2O.
AWH = 1,0079 amu = 1
AWO = 15,9994 amu = 16
One oxygen is sixteen times the mass of a H. Hence, in the molecule of water
H2
O
2 amu
16 amu
1
:
H2O
FW = 18
8
Water is 11.1% hydrogen and 88.9% oxygen by mass.
Law of Definite Proportions
Carbon dioxide (CO2) is 27.3% carbon and 72.7% oxygen by mass
Calcium oxide (CaO) is 71.5% calcium and 28.5% oxygen by mass
Calcium carbonate (CaCO3) is 40.1% calcium, 12.0% carbon, 47.9% oxygen by mass
Observations such as these on many pure compounds (as CaCO3)
led Joseph-Louis Proust in1799 to the statement of the Law of
Definite Proportions (also known as the Law of Constant
Composition):
Different samples of a compound always contain the same elements in the same
proportion by mass;
this corresponds to atoms of these elements combined in fixed numerical ratios.
Percent
composition
and
chemical
formulas
2-7 Percent Composition and Formulas of Compounds
If the formula
of aiscompound
is known,
its chemical
composition
be expressed as
as
mass)
If the chemical
formula
known, its
chemical
composition
can can
be expressed
ht (mass)
the mass percent
of each elementininthe
the compound (percent
composition). For example,
the mass
C H A P percent
T E R 2 • Cof
H Eeach
M I C A Lelement
F O R M U L A S A N D Ccompound
O M P O S I T I O N (percent
S T O I C H I O M composition).
ETRY
he
%.E.g.
one carbon dioxide molecule, CO2, contains one C atom and two O atoms. Percentage is
the part divided by the whole times 100% (or simply parts per hundred), so we can represent the percent composition of carbon dioxide as follows:
2-7
Percent
Composition
and
Formulas
of
Compounds
One CO2, contains one C atom and two O atoms.
If the formula
of C
adivided
compound
is known,
chemical
can be expressed
Percentage
is the
part
by the
whole
times
100%,composition
so12.0
we amu
can represent
the as
mass
of
AW of its
C
%C
5 percent of each
3 100%
5 in the
3 100%
5
3 100%For
5 example,
27.3% C
s)percent
the
mass
element
(percent
composition).
composition
dioxide
follows:
mass of CO 2carbon
MWasofcompound
CO 2
44.0
amu
one carbon dioxide molecule, CO2, contains one C atom and two O atoms. Percentage is
)
2 ( 16.0
amu
massdivided
of O by the whole
2 times
3 AW100%
of O (or simply parts
the
part
per
hundred),
so we
can rep%O 5
3 100% 5
3 100% 5
3 100%
5 72.7%
O
mass
of
CO
MW
of
CO
44.0
amu
2
resent the percent
composition of carbon2 dioxide as follows:
One mole of CO (44.0 g) contains one mole of C atoms (12.0 g) and two moles of O atoms
2
mass of C
AW of C
12.0 amu
therefore
masses5in the preceding
calculation.
These
%C 5 (32.0 g). We could
3 100%
5 have used these
3 100%
3 100%
5 27.3%
C
mass
of
CO
MW
of
CO
44.0
amu
2
numbers are 2the same as the ones usedionly
the units are different. In Example 2-12 we
base our calculation
on AW
one mole
( 16.0 amu )
masswill
of O
23
of Orather than one 2molecule.
%O 5
3 100% 5
3 100% 5
3 100% 5 72.7% O
mass
of
CO
MW
of
CO
44.0
amu
2 2 (44.0 g) contains one2 mole of C atoms (12.0 g) and two moles of O
One mole of CO
atoms (32.0 g).
One mole of CO2 (44.0 g) contains one mole of C atoms (12.0 g) and two moles of O atoms
(32.0 g). We could therefore have used these masses in the preceding calculation. These
numbers are the same as the ones usedionly the units are different. In Example 2-12 we
EXAMPLE 2-12 Percent Composition
will base our calculation on one mole rather than one molecule.
Calculate the percent composition by mass of HNO3.
Percent composition and chemical formulas
Sample exercise
• Calculate the percent composition by mass of HNO3
We first calculate the mass of one mole (molar mass). Then we express the mass of each element as a
percent of the total.
The molar mass of HNO3 is:
Number of Mol of Atoms x Mass of One Mol of Atoms = Mass Due to Element
1x H =1 mol
x 1,0 g
=1,0 g of H
1x N =1 mol
x 14,0 g
= 14,0 g of N
3x O = 3 mol
x 16,0 g
= 48,0 g of O
Mass of 1 mol of HNO
Now, its percent composition is:
•% of H = (1.0 g /63,0 g) 100% = 1,6 %
•% of N = (14.0 g /63,0 g) 100% = 22,2 %
•% of O = (48,0 g /63,0 g) 100% = 76,2 %
Remember
The amount of substance that contains the mass in grams numerically equal to its formula
weight in amu contains one mole of the substance. This is the molar mass and is
numerically equal to the formula weight and has the units grams/mole (g/mol).
Percent composition and chemical formulas
Practice exercise
• Calculate the percent composition by mass of each of the following
compounds
Dopamine: C8H11NO2
Vitamin E: C29H50O2
Vanillin: C8H8O3
Derivation of Formulas from Composition
The molecular formula indicates the numbers of atoms in a
molecule.
The empirical formula is the smallest whole-number ratio of atoms
present.
E.g. The empirical and molecular formulas for water are both H2O; for hydrogen
peroxide, the empirical formula is HO, and the molecular formula is H2O2
Molecular
formula
Simplest
formula
C2H4O2
C6H12O6
H3PO4
H2O
H2O2
CH2O
CH2O
H3PO4
H2O
HO
Derivation of Formulas from Composition
One of the first steps in characterizing a new compound is the determination of its
percent composition.
A qualitative analysis is performed to determine which elements
are present in the compound.
A quantitative analysis is performed to determine the amount of
each element.
Once the percent composition of a compound is known, the empirical formula can be
determined
CHAPTER 2 • CHEMICAL FORMULAS AND COMPOSITION STOICHIOMETRY
Derivation
of Formulas from Composition
m
obtain
mol x Fw
Sample
exercise
Step
2: Now(Empirical
we know thatFormula)
100.0 g of the compound contains 1.56 mol of S atoms and
several
3.12 mol of O atoms. We obtain a whole-number ratio between these numbers
ch • Compounds containing S and O are serious air pollutants and they represent the major
ber, and cause
all the mass.
llest
ate
the ratioofofthis
atoms
in the simplest
formula.
of acidthat
rain.gives
A sample
compound
contains
50.1% sulfur and 49.9% oxygen by
What is the empirical formula of1.56
the compound?
1.56
5 1.00 S
SO2
1.Let’s consider 100.0 g of compound, which contains 50.1 g of S and
49.9 g of O. We calculate the
3.12
number of moles of atoms of each.
5 2.00 O
1.56
50,1Exercise
g
49,9 g
You
should
now
work
54.
mol of S =
mol of O =
= 1,56 mol of S
= 3,12 mol of O
32,1 g mol-1
16 g mol-1
2.We obtain a whole-number ratio between these numbers that gives the ratio of atoms in the
empirical formula
1,56
= 1,00 S
1,56
2
3.12
The solution for Example 2-13 can be set up in tabular form.
= 2,00 O
1,56
nt”
mass of
his
n could
of Moles
mn
wholems. But
ame
SO
Element
Relative
Mass of
Element
S
50.1
O
49.9
Relative Number
of Atoms
(divide mass by AW)
Divide by
Smallest Number
50.1
5 1.56
32.1
49.9
5 3.12
16.0
1.56
5 1.00 S
1.56
3.12
5 2.00 O
1.56
Smallest WholeNumber Ratio
of Atoms
SO2
Derivation of Formulas from Composition
m
mol x Fw
Sample exercise (Empirical Formula)
g of an ionic compound is found to contain 6.072 g of Na, 8.474 g of S, and 6.336 g
• 20.882
The solution for Example 2-13 can be set up in tabular form.
of O. What is its empirical formula?
t”
ass of
s
could
Moles
hole. But
me
atio is
Element
Relative
Mass of
Element
Relative Number
of Atoms
(divide mass by AW)
Divide by
Smallest Number
Smallest WholeNumber Ratio
of Atoms
0,264
6,072
50.1g
1.56
=
0,264
mol
= 1,00
1,00x2 = 2,00
5 1.56
5 1.00
S
-1
1.56
0,264
23,0 32.1
g mol
SO2
49.9
3.12
0,2645 2.00 O
8,474 g 5 3.12
49.9 g
SO
8,474
1,00x2 = 2,00
=
0,264
mol
= 1,00
16.0 -1
1.56
0,264
32,0 g mol
6,336 g
0,396
O
6,336
g
1,50x2
= 3,00as the
=-1 0,396 way
mol to solve simplest-formula
This tabular format provides a convenient
problems,
= 1,50
16,0 g mol
0,264
next example illustrates.
Na
S
6,072
50.1 g
Empirical formula =
Na 2 S 2 O 3
ormula
N.B. The ratio of atoms in the empirical formula must be a whole-number ratio (by definition). To convert
ratio 1:1:1.5
to ato
wholenumber
each
number
in and
the 6.336
ratio was
which gave the
nicthe
compound
is found
contain
6.072 gratio,
of Na,
8.474
g of S,
g of multiplied
O. What isbyits2,simplest
empirical formula Na2S2O3
Derivation of Formulas from Composition
Practice exercise
• The hormone norepinephrine is released in the human body during stress
and increases the body’s metabolic rate. Like many biochemical compounds,
norepinephrine is composed of carbon, hydrogen, oxygen, and nitrogen.
The percent composition of this hormone is 56.8% C, 6.56% H, 28.4% O,
and 8.28% N. What is the empirical formula of norepinephrine?
Derivation of Formulas from Composition
Sample exercise (Percent composition)
•
A 0.1647 g sample of hydrocarbon is burned in a C-H combustion train to produce 0.4931 g of CO2
and 0.2691 g of H2O. Determine the masses of C and H in the sample and the percentages of these
elements in this hydrocarbon.
1. With a proportion, we use the observed masses to determine the masses of C and H in the original sample.
There is one mole of carbons, (aw=12.01), in each mole of CO2, 44.01 g; there are two moles of hydrogen atoms,
(aw=2.016), in each mole of H2O, 18.02 g:
mass of
element
mass of the
compound
=
AW
FW
g of C = 0.4931 g CO2 x12.01 g/mol C = 0.1346 g C
44.01 g/mol CO2
g of H = 0.2691 g H2O x 2.016 g/mol H = 0.0301 g H
18.02 g/mol H2O
2. Then we calculate the percentages by mass of each element:
% C = 0.1346 g C x 100% = 81.72% C
0.1647 g sample
% H = 0.0301 g C x 100% = 18.28%H
0.1647 g sample
Remember: Hydrocarbons are organic compounds composed entirely of hydrogen and carbon.
Molecular Formulas from Empirical Formulas
To determine the molecular formula of a substance, both its empirical formula and its
formula weight must be known.
The molecular formula is a multiple of the empirical formula.
E.g.
Butane, C4H10.
The empirical formula for butane is C2H5, but the molecular formula contains twice as many atoms;
that is, 2x(C2H5)=C4H10.
Benzene, C6H6.
The empirical formula for benzene is CH. the molecular formula has six times as many atoms;
6x(CH)=C6H6.
The molecular formula for a compound is either the same as, or an integer multiple of, the
(empirical) formula.
so
n=
molecular weight
empirical formula weigh
A practical example: Combustion Analysis
One technique for determining empirical formulas in the laboratory is combustion
analysis, commonly used for organic compounds.
When a compound containing carbon and hydrogen is combusted, the carbon is converted to CO2 and
the hydrogen is converted to H2O.
The amounts of CO2 and H2O produced are determined by measuring the mass increase in the CO2
SECTION 3.5 Empirical Formulas from Analyses
95
and H2O absorbers.
From the masses of CO2 and H2O we can calculate the number of moles of C and H in the original
sample and thereby the empirical formula.
mulas in the laboratory is
ounds containing princi-
Sample combusted,
producing CO2 and H2O
H2O and CO2 are trapped
in separate absorbers
d hydrogen is completely
Sample
n in ! FIGURE 3.14, the O2
n is converted to H2O.
H2O absorber CO2 absorber
H2O produced are deterFurnace
CO2 and H2O absorbers.
︎Apparatus
for combustion
analysis
ate the number of moles Mass
gained
by each absorber
corresponds to mass of
the empirical formula. If CO2 or H2O produced
If acan
third
is present in the compound, its mass can be determined by subtracting the measured
mass
beelement
determined
" FIGURE 3.14 Apparatus for
masses
C and H
from the original sample mass.
H from
the of
original
samcombustion analysis.
Law of Multiple Proportions
Two (or more) elements may form more than one compound (e.g. CO and CO2, NO and NO2,
H2O and H2O2). The Law of Multiple Proportions states that
“When two elements, A and B, combine and form more than one compound,
the ratio of the masses of element B, in each of the compounds, can be
expressed by small whole numbers”
E.g.: SO2 and SO3 provide an example.
In SO2, two moles of oxygen combine with one mole of sulfur atoms
In SO3, three moles of oxygen combine with one mole of sulfur atoms.
Thus the ratio of oxygen atoms in the two compounds compared to a given number of sulfur
atoms is 2:3.
Many similar examples, such as CO and CO2 (1:2 oxygen ratio) and H2O and H2O2 (1:2 oxygen
ratio), are known.
H2
O
2 amu
16 amu
1
:
8
H2O
FW = 18
Law of Multiple Proportions
Sample exercise (Empirical Formula)
• What is the ratio of the numbers of oxygen atoms that are combined with a given number
of nitrogen atoms in the compounds N2O3 and NO?
1.To compare the number of oxygen atoms, we must have equal numbers of nitrogen atoms.
Because NO has half as many nitrogen atoms in its formula relative to N2O3, we must multiply it by
a factor of 2 to compare the two elements on the basis of an equal number of nitrogen atoms.
Once we show the number of atoms of each element present, we can cancel out the equal amounts
of nitrogen atoms, leaving the ratio of oxygen atoms.
Oxygen
ratio
=
N2O3
2N 3O
=
=
2 (NO)
2N 2O
3O
2O
=
3
2
The oxygen ratio in the two compounds N2O3 and NO is 3:2
Law of Multiple Proportions
Practice exercise
• Show that the compounds water, H2O, and hydrogen peroxide, H2O2, obey
the Law of Multiple Proportions.
•
Nitric oxide, NO, is produced in internal combustion engines. When NO
comes in contact with air, it is quickly converted into nitrogen dioxide,
NO2, a very poisonous, corrosive gas.
1. What mass of O is combined with 3.00 g of N in (a) NO and (b) NO2?
2. Show that NO and NO2 obey the Law of Multiple Proportions.
Other Interpretations of Chemical Formulas
Sample exercise (Empirical Formula)
• Once we master the mole concept and the meaning of chemical formulas, we can use them
in many ways. For example, what mass of chromium (Cr) is contained in 35.8 g of
(NH4)2Cr2O7?
1.The formula tells us that each mole of (NH4)2Cr2O7 contains two moles of Cr atoms, so we first
find the number of moles of (NH4)2Cr2O7. We know that the molar mass is 252.0 g/mol.
m
mol x Fw
moles of
(NH4)2Cr2O7
=
35.8 g
=
252.0 g/mol
0.142 mols
2.Each mol of (NH4)2Cr2O7 contains 2 moles of Cr, so we convert the number of moles of
(NH4)2Cr2O7 into the number of moles of Cr atoms it contains, using the proportion
moles
of Cr
=
0.142 mol (NH4)2Cr2O7 x
2 mol Cr
=
1 mol (NH4)2Cr2O7
0.284 mols
3.We then use the atomic weight of Cr to convert the number of moles of chromium atoms to the
mass of chromium.
g of Cr
= 0.284 mol Cr x 51.99 g/mol Cr=
14.76 g Cr
Other Interpretations of Chemical Formulas
Practice exercise
• Mercury occurs as a sulfide ore called cinnabar, HgS.
How many grams of mercury are contained in 578 g of pure HgS?