Introduction to the Mole
How many things are in a...
dozen?
pair?
couple?
quartet?
century?
Chemists also have a word which represents a
certain number. A mole contains 6.02 x 1023 (602 000 000 000 000 000 000 000)
particles and this number is called Avogadro’s number (NA) in honour
of Amedeo Avogadro (1776-1856).
The mole concept is important because we
can’t measure out individual particles since they are microscopic (no scale
could weigh out one particle), but we can measure a mole of particles since that amount would be macroscopic (we are now looking at masses which are measurable).
For example, a measurable amount of water,
like 1 teaspoon full, contains approximately 2 x 1023
molecules.
I did a few fun calculations hoping to impart how astronomically large is the value of a mole:
I did a few fun calculations hoping to impart how astronomically large is the value of a mole:
- A mole of marbles would cover the Earth’s surface to a depth of 5 km.
- If one mole of pennies were distributed to every Canadian citizen (around 37 million people), each person would have approximately 190 trillion dollars
- If a home computer counts 100 million #s per second, it would take 191 million years to count to one mole.
- 6.02 X 1023 watermelon seeds would be found inside a melon slightly larger than the moon.
- 6.02 X 1023 doughnut holes would cover the earth and be 8 km deep.
- 6.02 X 1023 pennies would make at least 7 stacks that would reach the moon.
- 6.02 X 1023 grains of sand would be more than all of the sand on Miami Beach.
- 6.02 X 1023 blood cells would be more than all the blood cells found in every human on earth.
WOAH!! That's Crazy!
Determining the Mass of Atoms – Average
Atomic Mass
The atom consists of protons and neutrons (which are approximately equal in mass), as well as electrons (which have insignificant mass).
The atom’s mass is essentially equal to the mass of its nucleons (protons and neutrons). However, since an atom is so tiny, its mass cannot be directly measured. So, the C-12 atom was assigned as the standard and atoms are measured in unified atomic mass units (u), which is equal to 1/12 of the mass of a C-12 atom (1u = 1.66054 × 10-24 g).
An element can have several isotopes, each with the same number of protons and electrons, but differing numbers of neutrons.
For instance, any sample of lithium will contain 92.58% Li-7
(7.015 u) and 7.42% Li-6 (6.015 u); these percentages are known as the isotopic abundances (short-formed "iso abund" below) of lithium.
Since any sample of lithium consists of this
mixture of isotopes, we can calculate the average
atomic mass.
m = (mass of isotope 1)(iso abund/100) + (mass of isotope 2)(iso abund/100) + ...
m = 7.015(0.9258) + 6.015(0.0742)
m =
6.494 487 u + 0.446 313 u
m = 6.940 8 u
m = 6.941 u
Need extra help figuring this out? Check out this video about isotopes and average atomic mass.
Homework: Practice
Problems, p. 19 # 1-5
Question 1: How do you round the average atomic mass questions?
Since there are two types of calculations (multiplication and addition) involved in these questions, we must do two types of rounding (first, to the least number of sig digs, then to the least number of decimal places). However, although it is fine to think of how each step in a calculation would be rounded (and thus, how many sig digs or decimal places that number would have as it goes into future calculation steps), don't do any actual rounding until the very last step in the calculation. Rounding at every step in a multi-step calculation can actually round error into a value (Boo! Hiss!).
Question 2: Help! #5 is wacky.
Here you go...
