unit v: the mole concept
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Unit V: The Mole Concept. 5.1-5.2 – Atomic Mass, Avogrados Hypothesis, and the Mole (pg. 77-85, Hebden ). Today’s Objectives. Explain the significance of the mole, including: Recognize the significance of relative atomic mass, with reference to the periodic table - PowerPoint PPT PresentationTRANSCRIPT
Unit V: The Mole Concept
5.1-5.2 – Atomic Mass, Avogrados Hypothesis, and the Mole(pg. 77-85, Hebden)
Today’s Objectives Explain the significance of the mole, including:
Recognize the significance of relative atomic mass, with reference to the periodic table
Identify the mole as the unit for counting atoms, molecules, or ions
Perform calculations involving the mole, including: Determine the molar mass of an element or
compound
The Mole Question: how long would it take to spend a
mole of 1 Yuan coins if they were being spent at a rate of 1 billion coins per second?
What is a mole? Atoms are REALLY small! We can’t work with individual atoms or amu’s
(atomic mass units) in the lab Why? Because we can’t see things that small
The Mole Instead, we work with samples large enough for
us to see and weigh on a balance using units of grams
This creates a problem…. A pile of atoms big enough for us to see contains
billions of atoms! Billions of atoms are hard to keep track of in
calculations So, chemists made up a new unit:
THE MOLE
The Mole Just as a dozen eggs equals 12 eggs, a mole =
602,000,000,000,000,000,000,000 or 6.02x1023
It is equal to that number no matter what kind of particles you’re talking about
It could represent marbles, pencils, or chicken feet
Usually, the mole deals with atoms and molecules
The mole, whose abbreviation is mol, is the SI base unit for measuring amount of a pure substance
The Mole The mole, as a unit, is only used to count very
small items It represents a number of items, so, we can know
exactly how many items are in 1 mole The experimentally determined number a
mole is called is Avogrado’s Number, or 6.02x1023
The term representative particle refers to the species present in a substance: Atoms (most often) Molecules Formula units (ions)
Pop Quiz 1 dozen Mg atoms =
12 Mg atoms 1 mole Mg atoms =
6.02x1023 Mg Atoms 1 mole Mg(OH)2 =
6.02x1023 Mg(OH)2 molecules 1 mole O2 =
6.02x1023 O2 molecules
How big is a Mole? 1 Mole of soft drink cans is enough
to cover the surface of the earth to a depth of over 320 km
If you had Avogrado’s number of unpopped popcorn kernels, and spread them across China, the country would be covered in popcorn to a depth of over 15 km
If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole
Mollionaire Back to that question: How long would it take to
spend a mole of 1 Yuan coins if they were being spent at a rate of 1 billion per second?
Answer: ¥ 6.02 x 10^23/ ¥1 000 000 000 = 6.02 x 10^14 payments = 6.02 x 10^14 seconds 6.02 x 10^14 seconds/60 = 1.003 x 10^13 minutes 1.003 x 10^13 minutes/60 = 1.672 x 10^11 hours 1.672 x 10^11 hours/24 = 6.968 x 10^9 days 6.968 x 19^9 days/365.25 = 1.908 x 10^7 years It would take 19 million years!
How gases combine Early chemist John Dalton (1766-1844) wondered how
much of a given element would bond (react) with a given amount of another element
He did not assign an absolute mass for individual atoms of any given element, but rather assigned an arbitrary (relative) mass to each element
He assumed that hydrogen was the lightest and assigned hydrogen a unit mass of 1
Through experimentation, he determined that C was 6 times heavier than oxygen, so he assigned C a mass of 6
Oxygen was found to have a mass 16 times heavier than hydrogen, so he assigned O a mass of 16
Using this same process, he was able to determine the relative masses of all of the elements
John Dalton’s Experiment Looked at masses of gases
11.1g H2 reacted with 88.9g O2 Interpretation O2 is 8 times heavier (look at PT)
46.7g of N2 reacted with 53.3g O2 42.9g C reacted with 57.1g O2
No real pattern
Joseph Gay-Lussac Combined gas
1L of H2 reacts with 1L Cl2 2L of HCl 1L of N2 reacts with 3L H2 2L of NH3 2L of CO reacts with 1L O2 2L of CO2
Concluded that gases combine in simple volume ratios
But why aren’t the volumes of the reactants and products equal?
Avogrado’s Hypothesis Equal volumes of any gas at standard temperature and
pressure contain the same number of molecules Example:
1L of N2 reacts with 3L H2 2L of NH3 Lets say each volume contains 1 molecule, we could then
say: 1 molecule of N2 reacts with 3 molecules of H2 to form 2
molecules of NH3 Lets count the atoms to prove this: Reactants: 2 nitrogens, 6 hydrogens Products: 2 nitrogens, 6 hydrogens
Mass is always conserved in a chemical reaction, volume is not always conserved in a chemical reaction
Avogrado’s Hypothesis Let’s look at the other 2 examples (again
assuming each volume of gas contains 1 molecule):
1L of H2 reacts with 1L Cl2 2L of HCl Reactants: 2 hydrogen atoms, 2 Cl atoms Products: 2 hydrogen atoms, 2 Cl atoms
2L of CO reacts with 1L O2 2L of CO2 Reactants: 2 carbon atoms, 4 oxygen atoms Products: 2 carbon atoms, 4 oxygen atoms
If 2L of H2 reacts with 1L of O2, how many litres of H2O would be produced? 4 H, 2 O = 2H2O = 2L H2O
Do exercises 2-5 on p. 78
Who can explain this? Avogadro’s Hypothesis
Equal volumes of any gas at standard temperature and pressure contain the same number of molecules
This Explains the simple volume ratio for gases
Atomic Mass
The mass of 1 mole of atoms of an element. The mass of one mole of “C” atoms is 12.0g The mass of one mole of “Ca” atoms is 40.1g
Molar Mass (Molecular Mass)The mass of 1 mole of molecules of an element or compound
Diatomic Elements Some elements are naturally diatomic.
Remember the “gens” Hydrogen, nitrogen, oxygen, halogens
H2, O2, N2, F2, Cl2, Br2, I2, At2 you must remember these
Special elements Sometimes Phosphorus is P
Sometimes P4
Sometimes Sulphur is S Sometimes S8
Assume the rest of the elements are monatomic
Finding the Molar Mass of Compounds H2O = 2(1.0) + 16.0 = 18.0 g/mol Ca(NO3)2
= 40.1 + 2(14.0) + 6(16.0) = 164.1g/mol Ammonium phosphate
(NH4)3PO4
= 3(14.0) +12(1.0) + 31.0 + 4(16.0)= 149.0 g/mol
HMWK: p80 #6-7