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

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Unit V: The Mole Concept – Atomic Mass, Avogrados Hypothesis, and the Mole (pg , 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 6.02x10 23  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 6.02x10 23  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 =  6.02x10 23 Mg Atoms  1 mole Mg(OH) 2 =  6.02x10 23 Mg(OH) 2 molecules  1 mole O 2 =  6.02x10 23 O 2 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/ ¥  = 6.02 x 10^14 payments = 6.02 x 10^14 seconds  6.02 x 10^14 seconds/60 = x 10^13 minutes  x 10^13 minutes/60 = x 10^11 hours  x 10^11 hours/24 = x 10^9 days  x 19^9 days/ = x 10^7 years  It would take 19 million years!

How gases combine  Early chemist John Dalton ( ) 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 H 2 reacted with 88.9g O 2  Interpretation O 2 is 8 times heavier (look at PT)  46.7g of N 2 reacted with 53.3g O 2  42.9g C reacted with 57.1g O 2  No real pattern

Joseph Gay-Lussac  Combined gas  1L of H 2 reacts with 1L Cl 2  2L of HCl  1L of N 2 reacts with 3L H 2  2L of NH 3  2L of CO reacts with 1L O 2  2L of CO 2  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 N 2 reacts with 3L H 2  2L of NH 3  Lets say each volume contains 1 molecule, we could then say:  1 molecule of N 2 reacts with 3 molecules of H 2 to form 2 molecules of NH 3  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 H 2 reacts with 1L Cl 2  2L of HCl  Reactants: 2 hydrogen atoms, 2 Cl atoms  Products: 2 hydrogen atoms, 2 Cl atoms 2L of CO reacts with 1L O 2  2L of CO 2  Reactants: 2 carbon atoms, 4 oxygen atoms  Products: 2 carbon atoms, 4 oxygen atoms  If 2L of H 2 reacts with 1L of O 2, how many litres of H 2 O would be produced?  4 H, 2 O = 2H 2 O = 2L H 2 O Do exercises 2-5 on p. 78

Who can explain this? AAvogadro ’ s Hypothesis E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  H 2, O 2, N 2, F 2, Cl 2, Br 2, I 2, At 2  you must remember these  Special elements  Sometimes Phosphorus is P  Sometimes P 4  Sometimes Sulphur is S  Sometimes S 8  Assume the rest of the elements are monatomic

Finding the Molar Mass of Compounds H2OH2O = 2(1.0) = 18.0 g/mol  Ca(NO 3 ) 2 = (14.0) + 6(16.0) = 164.1g/mol  Ammonium phosphate  (NH 4 ) 3 PO 4 = 3(14.0) +12(1.0) (16.0) = g/mol HMWK: p80 #6-7