Transcript Figure 1.01a: (a.)The surface of a single grain of table salt.

Chapter 3:
STOICHIOMETRY
Stoichiometry - The study of quantities of
materials consumed and produced in
chemical reactions.
Atomic Masses
• IUPAC define: The system of atomic masses is based on
12C as the standard, 12C has 12 atomic mass units (amu).
• (Mass 13C)/Mass 12C = 1.0836129
• Mass 13C = (12 amu )(1.0836129) = 13.003355
• Elements occur in nature as mixtures of isotopes
• Carbon =
•
•
98.89% 12C
1.11% 13C
<0.01% 14C
• Carbon atomic mass =
0.9889 x 12
amu
.0111 x 13.0034 amu
~
12.01
• Average atomic mass = Atomic mass
amu
Figure 3.1: (left) A scientist injecting a sample into
a mass spectrometer. (right) Schematic diagram of
a mass spectrometer.
zV = (½)mv2; v = (2zV/m)0.5
Figure 3.2: (a) Neon gas glowing in a discharge tube.
(b) "peaks" and (c) a bar graph.
20Ne
= 90.92%, 21Ne = 0.257%, 22Ne = 8.82%
Figure 3.3: Mass spectrum of natural copper.
(69.09
amu
atoms)(62.93 atom
)+
amu
(30.91 atoms) (64.93 atom )
= 6355 amu
Average mass =
6355 amu
100 atom
= 63.55 amu/atom
The Mole
•The number equals to the number of carbon
atoms in exactly 12 grams of pure 12C.
•1 mole of anything = 6.022  1023 units of that
thing
Avogadro’s number equals
6.022  1023 units
Figure 3.4: Proceeding clockwise from the top
samples containing one mole each of copper,
aluminum, iron, sulfur, iodine, and (in the center)
mercury.
The mass of 1mole of an element = atomic mass (g)
12C:
(6.022 x
12C:
1023
12amu
atoms) (
1atom
6.022 x 1023 amu = 1 g
) = 12 g
Ex 3.2 : The mass in grams of a sample of Am
containing 6 atoms.
12C:
amu
6 atoms x 243
= 1.46 x 103 amu
atom
∵

6.022 x 1023 amu = 1 g
1.46 x 103 amu = 2.42 x 10-21 g
Pure aluminum.
Aluminum alloys are used for many highquality bicycle components, such as this
chain wheel.
Ex 3.3 Determining Moles of Atoms
The number of moles of atoms and the number of
atoms in a 10.0g Al sample.
1 m ol Al
(10.0 g Al) x
= 0.371 mol Al atoms
26.98 g  Al
23
6
.
022
x
10
 atom s = 2.33 x 1023 atoms
0.371 mol Al x
1 m ol Al
Molar Mass
• A substance’s molar mass (molecular weight)
is the mass in grams of one mole of the
compound.
• CH4 = 16.04 grams per mole
Ex 3.6 Calculating Moles Mass
(a) Calculate the molar mass of C10H6O3 of juglone
(b) The number of moles of 1.56 x 10-2g juglone sample.
10 C:
10 x 12.01 g
6H :
6 x 1.008 g
3O :
3 x 16.000 g
Mass of 1 mole juglone
(b)
1.56 x 10-2g x
1  m ol
174.1  g
=
=
=
=
120.1 g
6.048 g
48.00 g
174.1 g
= 8.96 x 10-5 moles
Percent Composition
• Mass percent of an element:
mass of element in compound
mass % 
 100%
mass of compound
• For iron in iron (III) oxide, (Fe2O3)
55.85 x 2
55 .85 x 2  16 .00 x3
x 100% = 70.07 %
Determining the Formula of a compound
Figure 3.5: A schematic diagram of the combustion device
used to analyze substances for carbon and hydrogen.
A 0.1156 sample, containing C, H, and N only, reacts with excess O,
0.1638 g CO2 and 0.1676 H2O are collected. Determining the formula
of the compound.
• 0.1638 g CO2 x
»
• 0.1676 g H2O x
»
Then,
• 38.67 g C x
12.01 g  C
= 0.04470 g C,
44.01 g  CO2
= 38.87 % C = 38.67 g C/100g
2.016 g  H
18.02  g  H 2O = 0.01875 g H,
1 mol C
12.01 g  C
= 16.22 % H = 16.22 g H,
= 45.11% N = 45.11 g N
= 3.220 mol C,
= 16.09 mol H,
= 3.219 mol N
C:
3.220
3.220
= 1.00 = 1
H:
16.09
3.220
N:
3.219
3.220 = 1.000 = 1
= 4.997 = 5
⇒CH5N (empirical formula)
To specify the molecular formula, we must know the molecular mass.
Determining the Formula of a compound
• molecular formula = (empirical formula)n
[n = integer]
• molecular formula = C6H6 = (CH)6
• empirical formula = CH
Ex. 3.11 Determine the empirical and molecular formula for a
compound that gives the following percentages upon analysis (in mass
percents):
71.65% Cl, 24.27% C, 4.07% H.
The molar mass is known to be 98.96 g/mol
1  m ol Cl
34.45  g  Cl
= 2.021 mol Cl,
• 24.27 g C x
1 m ol C
12.01 g  C
= 2.021 mol C
• 4.07 g H x
1  m ol H
1.008 g  H
• 71. 65 g Cl x
⇒
= 4.04 mol H,
ClCH2 (empirical formula)
Empirical formula mass = 49.48 g/mol
98.96  g / m ol
Molar m ass
=
=
49.48  g / m ol
Em pirical form ular m ass
Molecular formula = (ClCH2)2 = Cl2C2H4
2
Figure 3.7: The two forms of dichloroethane.
Figure 3.8:
The structure
of P4O10.
Computer-generated molecule of caffeine.
Empirical Formula Determination
• 1. Base calculation on 100 grams of
compound.
• 2. Determine moles of each element in 100
grams of compound.
• 3. Divide each value of moles by the
smallest of the values.
• 4. Multiply each number by an integer to
obtain all whole numbers.
Molecular Formula Determination
Method I
1.
2.
3.
4.
Obtain the empirical formula.
Calculate the empirical formula mass.
Molar m ass
Calculate the ratio (n) = Em pirical form ular m ass
Molecular formula = (empirical formula)n
Method II
1.
2.
3.
Using the mass percentage and molar mass to determine the
mass of each element per mole of compound.
Determine the # of moles of each element /mol compound.
The integers of # of moles of each element are the subscript in
the molecular formula.
Chemical Equations
Chemical change involves a
reorganization of the atoms in one or
more substances.
Chemical Equation
• A representation of a chemical reaction:
•
• CH4 + O2
reactants

CO2 + H2O
products
Molecular rep of a rxn
Molecular rep of a rxn
Hydrochloric
acid reacts
with solid
sodium
hydrogen
carbonate to
produce
gaseous
carbon
dioxide.
Chemical Equation
• C2H5OH + 3O2  2CO2 + 3H2O
• The equation is balanced.
• 1 mole of ethanol reacts with 3 moles of
oxygen
• to produce
• 2 moles of carbon dioxide and 3 moles of
water
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3a–33
Decomposition of ammonium dichromate.
Decomposition of ammonium dichromate.
(cont'd)
Calculating Masses of Reactants and
Products
•
•
•
•
•
1. Balance the equation.
2. Convert mass to moles.
3. Set up mole ratios.
4. Use mole ratios to calculate moles of
desired substituent.
5. Convert moles to grams, if necessary.
Figure 3.9: Three different stoichiometric
mixtures of methane and water, which react
one-to-one.
Figure 3.10: A mixture of CH4 and H2O
molecules.
Figure 3.11: Methane and water have
reacted to form products according to the
equation CH4 + H2O
3H2 + CO.
Figure 3.12: Hydrogen and nitrogen react
to form ammonia according to the equation
N2 + 3H2
2NH3.
Limiting Reactant
The limiting reactant is the reactant
that is consumed first, limiting the
amounts of products formed.
Solving a Stoichiometry Problem
•
•
•
•
•
1. Balance the equation.
2. Convert masses to moles.
3. Determine which reactant is limiting.
4. Use moles of limiting reactant and mole
ratios to find moles of desired product.
5. Convert from moles to grams.
Solving
stoichiometry
problems
flow chart.