1. Chapter 7.3

2.  How do we use these?  These indicate which of the elements make up a substance.  These also indicate the number of ions or atoms that make up a given substance. C 6 H 12 O 6

3.  Atomic mass is the weighted average of the masses of the isotopes of that element.  This reflects the mass and the relative abundance of the isotopes as they occur in nature.

4.  A mole is a unit of measurement of number of atoms of an element ◦ 602,000,000,000,000,000,000,000 atoms are in one mole of an element. ◦ 1 mole = 6.02 x 10 23 particles ◦ This is called Avogadro’s number  The atomic mass of an element is the mass of one mole of that element in grams g / mol.  Using moles allows us to work with formulas in a measurable way – using grams.

5. “Atomic masses and chemical formulas”  Take out your periodic tables  Complete this assignment We will review this assignment in five minutes

6.  One mole of an element is equal to its atomic mass in grams Equals 1 mole of Nitrogen 6.02 x 10 23 atoms of Nitrogen 14.007 g of Nitrogen

7.  Moles are a way to make working with teeny tiny particles manageable.  Have you ever purchased a dozen eggs? Moles are not that much different. These are both ways of counting a number of items

8. The mole is how we can measure something … …very small (atoms of an element) …in a manner that we can see (mass in grams). For Bromine, the mass of one mole is 79.9 grams. This is also the mass of 6.02 x 10 23 particles of Bromine … AND one mole of Bromine is 6.02 x 10 23 particles of Bromine

9. This activity will help you to see the similarities between the mole (which is unfamiliar) and the dozen (which is quite familiar).

10.  Now that we know how to work with single elements’ atomic masses, how can we apply this to more complicated substances?  If we can determine the mass of mole of a single element in grams, we can determine the mass of a mole of a compound in grams

11.  As you know, a mole reflects 6.02 x 10 23 representative particles.  What is a representative particle?  A representative particle is one piece of the substance …  Depending on what you are talking about, it might be ◦ An atom of a single element ◦ A molecule of a substance ◦ An ion ◦ Anything!

12.  The molar mass of a substance is the mass of one mole of a substance in grams.  To determine it, you must know the chemical formula for the substance, and the atomic masses for each of the elements in it.

13. MgCl 2  Each particle of this compound contains: ◦ 1 atom of Magnesium ◦ 2 atoms of chlorine  The molar mass of MgCl 2 = ◦ 1 x atomic mass of Mg plus ◦ 2 x atomic mass of Cl= ◦ 1x 24.305 g Mg / mole ◦ +2x35.453 g Cl / mole = 95.211 g/mol MgCl 2

14. Remember, the sum of the atomic masses times the number of atoms of each kind of element is equal to the mass of one mole of the substance. Examples: Na = 22.990 g/mol + Cl = 35.453 g/mol Therefore, NaCl has a molar mass of 58.4743 g/mol Nitric acid is HNO 3. Its molar mass is H = 1.0079 x 1 = 1.0079 g/mol N = 14.007 x 1 = 14.007 g/mol O = 15.9997 x 3 = 47.9991 g/mol Total = 63.014 g/mol HNO 3

15.  The subscripts in the formula tell you how many of each atom to include in your calculations  Ca 3 N 2 will have three Ca atoms and 2 N atoms  If there are parenthesis in a formula, remember that the number outside the parenthesis acts as a multiplier for everything within the parenthesis.  Mg(OH) 2 particles contain 1 Mg, 2 O and 2 H atoms per particle

16.  There are certain compounds called hydrates which are salts that bind water molecules in their crystal structure.  For example, consider ZnSO 4 * 7 H 2 O Zinc sulfate heptahydrate  When determining the molar mass of a hydrate, you add the molar masses of that number of water molecules to the salt portion of the formula

17. ZnSO 4 * 7 H 2 O Molar mass of zinc sulfate + 7 x molar mass of water ZnSO 4 molar mass = 1 Zn (65.39 g/mol) + 1 S (32.06 g/mol) + 4 O (15.999 g/mol) = 161.4476 g/mol zinc sulfate 2H + 1 O = 18.0148 g/mol water x 7 (hepta) = 126.1036 g/mol H 2 O Total molar mass is  161.4476 g/mol zinc sulfate plus 126.1036 g/ 7 mol H 2 O  287.5512 g/mol ZnSO 4 * 7 H 2 O

18.  Now that we know the amount of mass contributed by each individual element in a formula, we can determine the percentage of that element by mass  Mass of element in one mole x 100 = % element in compound molar mass of compound

19. The percentage composition of each element in a compound can be determined using only the correct formula and the atomic masses. Example: Sodium chloride or NaCl Na = 22.99 % Na = 22.99 x 100 = 39.3%Na Cl = 35.453 58.443 58.443g/mol % Cl = 35.453 x 100 = 60.7% Cl 58.443 Notice that the total of the percentages is always equal or very close to 100%.

20.  Percent of phosphorus in phosphate (PO 4 )  In each phosphate, there are 1 P and 4 O atoms  1 P x 30.97376 g/mol = 30.97376 g/mol  4 O x 15.9994 g/mol = 63.9976 g/mol  94.97136 g/mol PO 4  30.97376 g P *100 = 32.6% P  94.97136 g PO 4

21. What is the percent of water in ZnSO 4 * 7 H 2 O? From our previous calculations, the t otal molar mass is  161.4476 g/mol zinc sulfate plus +126.1036 g/ 7 mol H 2 O  287.5512 g/mol ZnSO 4 * 7 H 2 O  126.1036 g/ 7 mol H 2 O x 100 = 43.9% water  287.5512 g/mol ZnSO 4 * 7 H 2 O

22.  We use dimensional analysis to work with moles, grams and particles.  Atomic masses, one mole, and Avogadro’s number can be used as conversion factors to convert between moles, grams and particles of an element

23. Equals 1 mole of Nitrogen 6.02 x 10 23 atoms of Nitrogen 14.007 g of Nitrogen

24.  How many grams are there in 5.40 moles of Nitrogen?  Converting from MOLES to GRAMS The conversion factor you will use is: 1 mole N = 14.007 g N  5.40 moles N x 14.007 g N = 1 mole 75.6 g N

25.  How many atoms are there in 40.6 g Nitrogen?  Converting from GRAMS to ATOMS  The conversion factor that you will use is 14.007 g Nitrogen = 6.02 x 10 23 atoms  40.6 g N x 6.02 x 10 23 atoms = 14.007 g N 1.74 x 10 24 atoms of N

26.  How many moles are there in 2.3 x 10 23 atoms of Nitrogen?  Converting from ATOMS to MOLES  The conversion factor that you will use is: 1 mole Nitrogen = 6.02 x 10 23 atoms  2.3 x 10 23 atoms x 1 mole = 6.02 x 10 23 atoms 0.38 moles N

27.  What is an empirical formula?  The empirical formula represents the lowest whole number ratio between elements in a compound.  For example, peroxide has 1 H for every 2 O in the compound

28.  BUT … the empirical formula may not match the way that the actual molecules form.  For that, we have the molecular formula, which represents the whole number ratio of atoms in a molecule.  Sometimes these do match!  Ex: For water, H 2 O, the empirical and molecular formulas are the same.

29.  Let’s work through a problem in which you need to determine the empirical and molecular formulas for a substance.  Peroxide is a substance made from hydrogen and oxygen atoms.  We will determine the empirical and molecular formulas for peroxide:

30.  A 6.8 g sample of a compound contains 4.0 g H and 6.4 g O. It has a molar mass of 34.0146 g/mol.  Step 1. Determine the number of moles of each of the elements in the formula.

31.  From our previous calculation, we determine that there is a one-to-one ratio between H and O in this substance.  0.4 moles H = 0.4 moles O = 1:1 ratio  But this does not match the molar mass!  HO = 1.0079 g/mol H + 15.9994 g/mol O = 17.0073 g/mol HO

32. But ….. 17.0073 g/mol X 2 does equal 34.0146 g/mol.  So, the correct molecular formula for peroxide is H 2 O 2.

33.  We will work through some problems on this topic.  And, this is the last topic in this chapter, so we will move into chemical reactions next!