Advanced Higher Chemistry Cfe Unit 1 Inorganic Chemistry Lesson 1: Avogadro's Constant Learning Intention

Advanced Higher Chemistry CfE Unit 1 Inorganic Chemistry

Lesson 1: Avogadro’s Constant

Learning Intention -Learn to use Avogadro’s constant in basic calculations involving moles.

Background Avogadro’s constant was part of the old Higher course. However, an in depth knowledge of Avogadro’s constant is now no longer required. The purpose of this lesson is to highlight that Avogadro’s constant is just a numerical value which is applied in many calculations.

An Italian scientist named Amedeo Avogadro (1776 -1856) suggested that: equal volumes of all gases have the same number of molecules.

Some time later, the idea of the mole was introduced from his work. Amedeo Avogadro’s work was so important that a constant was named after him:

Avogadro’s constant (L) 6.02 x 10 23 formula units

602000000000000000000000 formula units

Example 1

12.0 g of carbon (1 mole) contains 6.02 x 1023 atoms of carbon.

This is one carbon atom, so if you had one mole of carbon (12.0g) you would have 6.02 x 1023 atoms of carbon.

C This is just one carbon atom, try to imagine 23 6.02 x 10 atoms of carbon!

Advanced Higher Chemistry CfE Unit 1 Inorganic Chemistry

Example 2 23 18.0 g of water (1 mole) contains 6.02 x 10 molecules of water.

This is one molecule of water, so if you had one mole of water (18.0g) you would have 6.02 x 1023 molecules of water.

also 1 mole of H O: 2

6.02 x 1023 atoms of oxygen.

2 x 6.02 x 1023 atoms of hydrogen.

Example 3 58.5 g of Na+Cl-/salt (1 mole) contains:

6.02 x 10 23 Cl- ions

6.02 x 10 23 Na+ ions

Also 1 mole of Na+Cl- contains: 6.02 x 1023 Na+ Cl- formula units

Worked Example 1 Calculate the mass of 3.01 x 1023 atoms of sodium, Na. (1 mole) 6.02 x 1023 Na atoms  23g 3.01 x 1023 Na atoms  (23/6.02 x 1023) x 3.01 x 1023 3.01 x 1023 Na atoms  11.5g

Advanced Higher Chemistry CfE Unit 1 Inorganic Chemistry

Worked Example 2

How many atoms are present in 15g of ethane, C2H6?

23 (1 mole) 30g C2H6  8 x 6.02 x 10 atoms

23 15g C2H6  (8 x 6.02 x 10 /30) x 15

24 15g C2H6  2.408 x 10 atoms

Worked Example 3

How many molecules are present in 8.8g of carbon dioxide, CO2?

23 (1 mole) 44g CO2  6.02 x 10 molecules

23 8.8g CO2  (6.02 x 10 /44) x 8.8

23 15g C2H6  1.204 x 10 molecules

Additional Resources

-Watch the short clip on Youtube: Crash Course Chemistry #6 Stoichiometry (5.00 min-7.30min). -Answer the Question from “Sheet 1.1 Avogadro” and check the answers when you have completed them. -If there are any questions regarding this lesson or the questions from sheet 1.1, then please leave a post on Microsoft Teams.

Advanced Higher Chemistry CfE Unit 1 Inorganic Chemistry

1.1 Avogadro

1. What is the mass of: (a) 6.02 x 1023 Na atoms (b) 1.204 x 1024 methane molecules (c) 3.01 x 1023 glucose molecules

2. Calculate the number of moles present in 1.806 x 1024 atoms of protactinium

3. A glass of water contains 72g of water. Calculate: (a) How many moles of water there are in the glass. (b) How many water molecules there are in the glass.

4. How many of each of the following are present in 4.4g of propane? (a) molecules of propane (b) atoms of carbon (c) atoms of hydrogen

5. Calculate the number of atoms present in: (a) 4.6g of sodium (b) 1.25 moles of oxygen

6. Calculate the mass of: (a) 1 atom of magnesium (b) 1000 atoms of carbon (c) 3.01 x 1022 zinc atoms (d) 6.02 x 1020 sulfur atoms