Unit V: The Mole Concept

1. Unit V: The Mole Concept 5.1-5.2 – Atomic Mass, Avogrados Hypothesis, and the Mole (pg. 77-85, Hebden)

2. 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

3. 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?

4. 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

5. 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

6. 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

7. 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)

8. Pop Quiz • 1 dozen Mg atoms = • 12 Mg atoms • 1 mole Mg atoms = • 6.02x1023Mg Atoms • 1 mole Mg(OH)2 = • 6.02x1023 Mg(OH)2 molecules • 1 mole O2 = • 6.02x1023 O2 molecules

9. 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

10. 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!

11. 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

12. 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

13. 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?

14. 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

15. 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

16. Avogadro’s Hypothesis Equal volumes of any gas at standard temperature and pressure contain the same number of molecules Who can explain this? This Explains the simple volume ratio for gases

17. 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 Atomic Mass

18. Molar Mass (Molecular Mass) • The mass of 1 mole of molecules of an element or compound

19. 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

21. 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