Transcript Section 3.1 Counting by Weighing

Chapter 3
Stoichiometry
Chapter 3
Chemical Stoichiometry
• Stoichiometry – The study of quantities of materials
consumed and produced in chemical reactions.
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Section 3.1
Counting by Weighing
•
•
Need average mass of the object.
Objects behave as though they were all
identical.
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Section 3.1
Counting by Weighing
Exercise
A pile of marbles weigh 394.80 g. 10 marbles
weigh 37.60 g. How many marbles are in the
pile?
Avg. Mass of 1 Marble =
37.60 g
= 3.76 g / marble
10 marbles
394.80 g = 105 marbles
3.76 g
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Section 3.2
Atomic Masses
Counting
by Weighing
•
•
Elements occur in nature as mixtures of
isotopes.
Carbon = 98.89% 12C
1.11% 13C
< 0.01% 14C
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Section 3.2
Atomic Masses
Counting
by Weighing
Average Atomic Mass for Carbon
98.89% of 12 amu + 1.11% of 13.0034 amu =
exact number
(0.9889)(12 amu) + (0.0111)(13.0034 amu) =
12.01 amu
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Section 3.2
Atomic Masses
Counting
by Weighing
Average Atomic Mass for Carbon
•
•
Even though natural carbon does not
contain a single atom with mass 12.01, for
stoichiometric purposes, we can consider
carbon to be composed of only one type of
atom with a mass of 12.01.
This enables us to count atoms of natural
carbon by weighing a sample of carbon.
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Section 3.2
Atomic Masses
Counting
by Weighing
Schematic Diagram of a Mass Spectrometer
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Section 3.2
Atomic Masses
Counting
by Weighing
Exercise
An element consists of 62.60% of an isotope
with mass 186.956 amu and 37.40% of an
isotope with mass 184.953 amu.
• Calculate the average atomic mass and
identify the element.
186.2 amu
Rhenium (Re)
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Section 3.3
The Mole by Weighing
Counting
•
•
•
The number equal to the number of carbon
atoms in exactly 12 grams of pure 12C.
1 mole of anything = 6.022 x 1023 units of that
thing (Avogadro’s number).
1 mole C = 6.022 x 1023 C atoms = 12.01 g C
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Section 3.3
The Mole by Weighing
Counting
Concept Check
Calculate the number of iron atoms in a 4.48
mole sample of iron.
2.70×1024 Fe atoms
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Section 3.4
Molar Mass
•
Mass in grams of one mole of the substance:
Molar Mass of N = 14.01 g/mol
Molar Mass of H2O = 18.02 g/mol
(2 × 1.008 g) + 16.00 g
Molar Mass of Ba(NO3)2 = 261.35 g/mol
137.33 g + (2 × 14.01 g) + (6 × 16.00 g)
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Section 3.4
Molar Mass
Concept Check
Which of the following is closest to the average
mass of one atom of copper?
a)
b)
c)
d)
e)
63.55 g
52.00 g
58.93 g
65.38 g
1.055 x 10-22 g
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Section 3.4
Molar Mass
Concept Check
Calculate the number of copper atoms in a
63.55 g sample of copper.
6.022×1023 Cu atoms
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Section 3.4
Molar Mass
Concept Check
Which of the following 100.0 g samples
contains the greatest number of atoms?
a) Magnesium
b) Zinc
c) Silver
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Section 3.4
Molar Mass
Exercise
Rank the following according to number of
atoms (greatest to least):
a) 107.9 g of silver
b) 70.0 g of zinc
c) 21.0 g of magnesium
b)
a)
c)
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Section 3.4
Molar Mass
Exercise
Consider separate 100.0 gram samples of
each of the following:
H2O, N2O, C3H6O2, CO2
 Rank them from greatest to least number
of oxygen atoms.
H2O, CO2, C3H6O2, N2O
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Section 3.5
Learning to Solve Problems
Conceptual Problem Solving
•
Where are we going?

•
How do we get there?

•
Read the problem and decide on the final
goal.
Work backwards from the final goal to decide
where to start.
Reality check.

Does my answer make sense? Is it
reasonable?
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Section 3.6
Percent Composition of Compounds
•
Mass percent of an element:
mass of element in compound
mass % =
× 100%
mass of compound
•
For iron in iron(III) oxide, (Fe2O3):
2( 55.85 g)
111.70 g
mass % Fe =
=
× 100% = 69.94%
2( 55.85 g)+ 3( 16.00 g) 159.70 g
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Section 3.6
Percent Composition of Compounds
Exercise
Consider separate 100.0 gram samples of
each of the following:
H2O, N2O, C3H6O2, CO2
 Rank them from highest to lowest percent
oxygen by mass.
H2O, CO2, C3H6O2, N2O
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Section 3.7
Determining the Formula of a Compound
Formulas
•
Empirical formula = CH
 Simplest whole-number ratio
• Molecular formula = (empirical formula)n
[n = integer]
• Molecular formula = C6H6 = (CH)6
 Actual formula of the compound
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Section 3.7
Determining the Formula of a Compound
Analyzing for Carbon and Hydrogen
•
Device used to determine the mass
percent of each element in a compound.
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Section 3.7
Determining the Formula of a Compound
Exercise
The composition of adipic acid is 49.3% C,
6.9% H, and 43.8% O (by mass). The molar
mass of the compound is about 146 g/mol.
 What is the empirical formula?
C3H5O2
 What is the molecular formula?
C6H10O4
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