1. The Mole

2. What is a mole? Well, yes, but we’re not discussing biology or dermatology now. We want the CHEMIST’S mole.

3. Relating Mass to Numbers of Atoms Recall that we defined the amu in terms of the carbon-12 atom’s mass. Now we want to relate this to the number of atoms that corresponds to this mass. Chemists devised a counting unit called the MOLE. You use counting units all the time— dozen, pair—to mean a specific number. We are going to introduce the MOLE, the SI unit for amount of substance.

4. The Mole A mole is the amount of a substance that contains as many particles as there are atoms in exactly 12 g of carbon-12. A mole is 6.022 x 10 23 units of a substance. That’s 602 000 000 000 000 000 000 000 water molecules or 602 000 000 000 000 000 000 000 atoms of copper. This number is commonly known as Avogadro’s number, named for Amedeo Avogadro, the 19 th Century Italian chemist.

5. Molar Mass The MOLAR MASS of a PURE substance is the mass of one mole (6.022 x 10 23 units) of that substance. The mass in grams of one mole of an element is equal to the atomic weight of that element. What is the molar mass of carbon? What is the atomic weight of carbon? The molar mass is 12.01 grams. 1 mol C = 12.01 g C or 12.01 g/mol C So 12.01 g C contains 1 mole of atoms.

6. But what about molecules? Well, one mole of water (H 2 O) has 6.022 x 10 23 molecules in it, because the unit here is a ‘molecule’, not an atom. To find the molar mass of water, we need the atomic masses of the elements that make up water: hydrogen and oxygen. 1 mol H = 1.0079 g H = 1.01 g H 1 mol O = 16.00 g O

7. Molar Mass of Water In each mole of water, there are two moles of H 2 mol H x 1.01 g H = 2.02 g H 1 mol H and one mole of O 1 mol O x 16.00 g O = 16.00 g O 1 mol O Therefore the molar mass of H 2 O is the sum = 2.02 g + 16.00g = 18.02 g or 18.02 g/mol H 2 O

8. Molar Masses of Diatomic Molecules What is the molar mass of O 2 ? What is the atomic weight of oxygen? 16.00 g The subscript tells us we have two atoms of oxygen in each molecule, so the molar mass is 2 mol O x 16.00 g O = 32.00 g O 1 mol O so 1 mol O 2 = 32.00 g O 2 or 32.00 g/mol O 2

9. What is the molar mass of carbon dioxide (CO 2 )? Start with the molar mass of each element: 1 mol C = 12.01 g C 1 mol O = 16.00 g O From the compound’s formula, we see that we need 1 mole of carbon and 2 moles of oxygen: 1 mol C x 12.01 g C = 12.01 g C 1 mol C 2 mol O x 16.00 g O = 32.00 g O 1 mol O So the molar mass of CO 2 = (12.01 g + 32.00g) = 44.01 g CO 2 or 44.01 g/mol CO 2

10. Moles to Gram Conversions What is the mass in grams of 3.50 mol of the element copper, Cu? Given: 3.50 mol Cu Unknown: Mass of Cu in grams PLAN: We need to convert the amount of Cu in moles into mass of Cu in grams. We multiply the amount of moles by the molar mass (g/mol) to find the amount in grams.

11. This is just the molar mass!

12. Grams to Moles Conversions This is just the inverse of the molar mass!

13. Moles to Grams Conversions

14. Moles to Number of Atoms

15. Conversions Mass of element or molecule in grams Amount of element or molecule in moles Number of atoms of element or molecule.

16. How to Calculate Percent Mass Let’s use carbon dioxide as an example. First, we need to know the molar mass of CO 2 : 1 mol C x12.01g/mol C+ 2 mol O x16.00g/mol O= 44.01 g/mol CO 2 Second, now find the percentage of each constituent. Oxygen: Carbon: NOTE: The total percent should equal 100%

17. Determining Chemical Formulae When a new substance is discovered or synthesized, it is analyzed quantitatively to reveal its percentage compositions. From these data, the empirical formula is then determined. An empirical formula consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole-number mole ratio of the different atoms in the compound.

18. Empirical Formulae For an IONIC COMPOUND, the formula unit is usually the compound’s empirical formula. For a MOLECULAR COMPOUND, unfortunately, the empirical formula doesn’t necessarily show the actual numbers of atoms present in each molecule. For instance, the gas diborane has an empirical formula of BH 3 but the molecular formula is B 2 H 6.

19. Calculation of Empirical Formulae

21. Example 1: Quantitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula of this compound. Plan: Convert the percentage composition into mass composition (based on a 100 g sample), then convert into moles, and then find the smallest whole-number mole ratio of atoms. For a 100 g sample, we would have: 32.38 g Na, 22.65 g S, and 44.99 g O.

24. Calculation of Empirical Formulae Example 2: Analysis of a 10.150 g sample of a compound known to contain only phosphorus and oxygen gives a phosphorus content of 4.433 g. What is the empirical formula of this compound? Plan: Use the masses of the sample and the phosphorus content to find the mass of the oxygen content. Then convert these masses to moles,, and then find the smallest whole-number mole ratio of atoms.

27. Molecular Formulae We have just practiced finding empirical formulas that contain the smallest possible whole-number ratios of atoms. The molecular formula is the actual formula of a molecular compound. The empirical formula may or may not be a correct molecular formula. We used diborane earlier as an example of this (empirical formula of BH 3 but the molecular formula is B 2 H 6 ). Ethene, C 2 H 4, and cyclopropane, C 3 H 6, have the same atomic ratio (1C:2H) yet they are very different substances.

28. Relationship between Empirical and Molecular Formulas

29. Calculation of Molecular Formula