1. Empirical and Molecular Formulas

2. Chemistry Joke Q: What happens to rock that has been heated to 6.02 X 10 23 degrees? A: It becomes molten!

3. Definitions Empirical formula – the lowest whole-number ratio of the atoms of the elements in a compound. Empirical formula – the lowest whole-number ratio of the atoms of the elements in a compound. Molecular formula – chemical formula that shows the actual number of atoms in a compound. Molecular formula – chemical formula that shows the actual number of atoms in a compound.

4. Which is it: empirical, molecular, or possibly both?? NO N2O4N2O4 C 6 H 12 O 6 Ba(OH) 2 C4H8C4H8

5. Empirical Formulas The simplest whole-number ratio of atoms in a compound. H2O2H2O2 H2OH2OH2OH2O C 6 H 12 O 6 CH 2 O CaCl 2 HO All are divisible by 6. Correct ionic formulas are always empirical. Both are divisible by 2. Already simplified

6. An unknown compound contains: 0.0806 grams of C 0.01353 grams of H 0.1074 grams of O What is the empirical formula of the compound? 1.Grams  mole 2.Divide by smallest 3.Round ‘til whole Finding an Empirical Formula

7. 0.0806 g C0.01353 g H0.1074 g O 0.0806 g C 12.01 g C 1 mol C 0.01353 g H 1.01 g H 1 mol H0.1074 g O 16.00g O 1 mol O 0.0068 mol C0.01340 mol H0.0067 mol O Step 1: Change grams to moles

8. 0.0068 mol C0.01340 mol H0.0067 mol O Step 2: Divide by smallest mole value Step 3: Round to the whole 0.0067 1.0149 mol C2 mol H1 mol O 1 mol C2 mol H1 mol O

9. 1 mol C2 mol H1 mol O CH 2 O This is the empirical formula.

10. Let’s try another – Let’s try another – This time, using This time, using percent composition!!

11. An unknown compound is known to have the following composition: 11.11% hydrogen 88.89% oxygen What is the empirical formula of the compound? In these problems we will choose to use a 100 gram sample. So, in this sample we have 11.11 grams H and 88.89 grams O.

12. 11.11 g H1 mol H 1.01 g H = 11.00 mol H 88.89 g O1 mol O 16.00 g O = 5.56 mol O Step 1: Change grams to moles.

13. 11.00 mol H 5.56 mol O 5.56 2 1 Step 2: Divide by smallest mole value. Step 3: Round to a whole number.

14. The ratio of moles of hydrogen to moles of oxygen is 2:1 H2OH2O

15. Going one step further… Finding molecular formulas from empirical formulas. Remember: The empirical formula is the simplest whole number ratio; whereas, the molecular formula shows the actual number of each atom. Step 1: Find molar mass of empirical formula. Step 2: Divide the molar mass of the compound by the molar mass of the empirical formula. Step 3: Multiply the empirical formula by that number.

16. CH 2 O is the empirical formula. The molar mass of the actual compound is 180 g/mol. What is the molecular formula? Practice Problem

17. Step 1: Find molar mass of empirical formula. 180 g / 30 g = 6 6 (CH 2 O) (1 x 12.01) + (2 x 1.01) + (1 x 16.00) = 30. 03 g/mol Step 2: Divide the molar mass of the compound by the molar mass of the empirical formula. Step 3: Multiply the empirical formula by that number. = C 6 H 12 O 6 The Molecular Formula

18. You try one… Empirical formula – CH 4 Empirical formula – CH 4 The actual molecular mass is 64g. The actual molecular mass is 64g. Find the molar mass of empirical formula  C + 4(H) = 16g Divide molecular mass by empirical mass  64g / 16g = 4 Multiply the empirical formula by that number  4 (CH 4 ) C 4 H 16  molecular formula

19. One more thing to mention… Hydrates vs. Anhydrates Hydrates vs. Anhydrates Hydrates – salts with water molecules attached Hydrates – salts with water molecules attached BaCl 2 2 H 2 O BaCl 2 2 H 2 O Name the salt, then use a prefix + word “hydrate”—Barium Chloride Dihydrate Name the salt, then use a prefix + word “hydrate”—Barium Chloride Dihydrate Anhydrates – the salt without its water Anhydrates – the salt without its water BaCl 2 BaCl 2

20. Finding the Formulas of Hydrates --Similar to Finding an Empirical Formula 1. Find the mass of the water by subtraction. 1. Find the mass of the water by subtraction. The mass of the hydrate and the anhydrate will be given. The mass of the hydrate and the anhydrate will be given. 2. Change g to mol 2. Change g to mol Mass of water to mol Mass of water to mol Mass of salt to mol Mass of salt to mol 3. Divide by the smallest mole value 3. Divide by the smallest mole value 4. Round to a whole # if necessary. 4. Round to a whole # if necessary.

21. Finding the Formulas of Hydrates BaCl 2 X H 2 O Hydrate mass (mass with water) – 24.40 g Anhydrate mass (mass w/o water) - 20.80 g Mass H 2 O = 24.40 g – 20.80 g = 3.60 g H 2 O For the formula we want the ratio of moles of water to moles of anhydrate. Step 1: Find the mass of the water.

22. 20.80 g BaCl 2 208 g BaCl 2 1 mol = 0.1 mol BaCl 2 3.60 g H 2 O 18 g H 2 O 1 mol = 0.2 mol H 2 O Step 2: Change grams to moles Grams of Salt Grams of Water

23. 0.1 mol BaCl 2 0.2 mol H 2 O 0.1 1 2 BaCl 2 2 H 2 O Barium chloride dihydrate Step 3: Divide by the smallest mole value. Step 4: Round to a whole # if necessary.

24. Calculating the percent water in a hydrate is similar to calculating percent composition: Calculating Percent Water in a Hydrate % H 2 O = mass of the water x 100 mass of the hydrate

25. In the lab, a five gram sample of hydrous copper (II) nitrate is heated. If 3.9 grams of the anhydrous salt remains, what is the percent water in the hydrate? Example Problem Mass of water: 5 g – 3.9 g = 1.1 g H 2 O (1.1 g H 2 O / 5 g total) x 100 = 22 % H 2 O Mass of hydrate: 5 g

26. Chemistry Joke Q: What is a dog lover’s favorite part of chemistry? A: The “Lab” Work!