1 mole - St John Brebeuf

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Transcript 1 mole - St John Brebeuf

Unit V: The Mole Concept
Lesson 1
Chemical Calculations
Atoms and molecules are extremely small.
If they are so small and so light, how can we weigh them?
We weigh large numbers of them.
Avogadro took 1.00 g of the smallest atom (H) and determined how
many H atoms there are in 1.00 g of H.
He found that:
1.00 g H
= 6.02 x 1023 atoms = 1.00 mole
This is called Avogadro’s number
Your new best friend…
Avogadro’s number = 6.02 x 1023
1 dozen donuts = 12 donuts
1 century
1 millennium
1.00 mole
= 100 years
= 1000 years
= 6.02 x 1023 particles
The mole is a large number of particles
Particle
Atom
elements
Cu
6.02 x 1023 atoms
1 mole
Molecule
compounds
CH4
6.02 x 1023 molecules
1 mole
The Mole & Avogadro’s number
1. Convert 2.5 x 1025 atoms of Carbon to moles
2.5 x 1025 atoms x
1 mole
6.02 x 1023 atoms
=
42 moles C
2. Convert 16.3 moles CO2 to molecules
16.3 moles
x
6.02 x 1023 molecules
1mole
= 9.81 x 1024 molecules
3. Convert 8.9 x 1024 molecules CO2 to moles
4. Convert 28 moles NaCl to molecules
Avogadro’s Hypothesis
Avogadro’s Hypothesis: Equal volumes of different gases, at the
same temperature and pressure, contain the same number of particles.
V.1 AVOGADRO’S HYPOTHESIS
• Avogadro’s hypothesis: Equal volumes of different gases contain the
same number of particles (at the same temperature and pressure).
• If 1L of gas A reacts with 1L of gas B, then the formula for the
compound is AB
• If 2 L of gas A reacts with 1L of gas B, then the formula for the
compound is A2B
• If 2 L of gas A reacts with 3L of gas B, then the formula for the
compound is __________________
Avogadro’s hypothesis
• In other words, however many litres of a
certain gas you have, it will be represented
as a subscript for that particular gas.
• Only under constant temperature &
pressure.
Try…
• If 2 L of an unknown gas X contains 4 x
1023 molecules at a certain temperature and
pressure. How many molecules are present
in 4 L of oxygen gas at the same
temperature and pressure?
Homework
• Questions: p. 78 #2-5