Chemistry 161
Lecture 10: Composition of Earth, The Mole, Intro to Writing Balanced
Chemical Equations
Sections 3.1, 3.2, 3.3 in Gilbert, Kirss, Foster, and Davies 2nd edition
10/22/08
Learning Objectives:
1. Use basic physical properties to appreciate the composition of Earth’s
early atmosphere
2. Define the atomic mass unit
3. Understand the basis for the relative mass scale
4. Know the value of Avogadro’s number
5. Appreciate the convenience of employing Avogadro’s number to define
the mole
6. Define Molar mass, formula mass (weight), and molecular mass
(weight)
7. Relate the mole to the gram molecular weight and determine the
amount of a substance present or required for a particular purpose
3.1: Composition of the Earth
 Light elements swept away by solar wind – No H, no water
Question: So where did H come from?
HW 3 Question (2 points): Using available information, accept or refute the
hypothesis that comets and asteroids brought us most of our water.
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 A question that immediately comes to mind is why is there not an equal
or greater abundance of H2O on Mars?
Oxygen trapped as oxides. Some examples are CO2, SO2, SO3, and NO2
On the corrosiveness of the early atmosphere – advanced Lewis structures
and an introduction to covalent reaction mechanisms
3.2 & 3.3: The Mole & Writing Balanced Chemical Equations
Question: What is the central law governing a balanced (non-nuclear)
chemical reaction? Consider the acid-base reaction between sodium
hydroxide and hydrochloric acid
NaOH + HCl  H2O + NaCl
Answer:
Atomic Mass
Question: Only
12
C has an integral value. Why is this so?
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Answer: We require a mass standard – since what are we really trying to do
is determine the proportions in which atoms, molecules, and ions react
 The mole is defined by international agreement as the quantity of
substance that contains the same number of particles as the number
of C atoms in exactly 12 g 12C
 The atomic mass unit is also a relative scale, defined such that 1 C
atom – 6 p, 6 n, 6 e- - is exactly 12 amu
 The connector is Avogadro’s number – 6.022 x 1023 (that is, it allows
us a conversion between g and amu from it) and is given the symbol NA
o NA = 6.022 x 1023mol-1
HW 3 Question (1 point): What is an amu in grams?
Question: How can it be that 12C = 12 amu? 12C contains 6 protons and 6
neutrons, and the proton has a mass of 1.00728, the neutron a mass of
1.00867.
Answer:
 Bear in mind the atomic weight (mass) reflects the weighted
contribution of the isotopes of that element
o An (average) atom of B = 10.81 amu, a mole of B = 10.81 g
 Our definition of the mole allows us to translate molecular and
formula mass into molar mass
o The molar mass is just that – the mass required for 1 mole of
that compound
o Molecular weight (mass) refers to covalent species
o Formula weight (mass) refers to ionic species
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 As an example, let’s say a large protein has a molecular mass of
100,000 amu1. How many molecules are in 1 mg?
 Given H2O has a molecular mass of 18 amu, how many molecules are in
1 mg (1 L) water?
Before moving on, let’s employ the concept of molar mass and mass balance
to solve fundamental questions concerning chemical reactions
Acetic acid (CH3CO2H) will undergo a complete reaction with the strong base
NaOH as follows:
CH3CO2H + NaOH  CH3CO2-Na+ + H2O
Question: What mass of NaOH pellets would be required to completely
convert 6.0 g acetic acid to sodium acetate?
Question: What mass of sodium acetate would result from the complete
reaction of 6.0 g acetic acid?
Question: If 6.0 g acetic acid completely reacts with 3.0 g sodium
hydroxide, how much product will be formed?
1
It would be common to refer to this as a 100 kDa protein; a Dalton = 1 amu
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