1. IIIIII II. Formula Calculations Ch. 10 – The Mole

2. Hydrates n Some compounds contain H 2 O in their structure. These compounds are called hydrates. n This is different from (aq) because the H 2 O is part of the molecule (not just surrounding it). n The H 2 O can usually be removed if heated.

3. Hydrates n A dot separates water in the formula  CuSO 4 5H 2 O n When naming a Greek prefix indicates the # of H 2 O groups.  copper(II) sulfate pentahydrate

4. Naming Hydrates n Na 2 SO 4 10H 2 O n NiSO 4 6H 2 O n CoCl 2 6H 2 O n MgSO 4 7H 2 O n Sodium sulfate decahydrate n Nickel (II) sulfate hexahydrate n Cobalt (II) chloride hexahydrate n Magnesium sulfate heptahydrate

5. Writing Formulas of Hydrates n sodium carbonate monohydrate n barium chloride dihydrate n Tin (IV) chloride pentahydrate n Barium hydroxide octahydrate n Na 2 CO 3 H 2 O n BaCl 2 2H 2 O n SnCl 4 5H 2 O n Ba(OH) 2 8H 2 O

6. What is Percent Composition?

7. A. Percent Composition n the percentage by mass of each element in a compound

8. %Fe = 28 g 36 g  100 = 78% Fe %O = 8.0 g 36 g  100 = 22% O n Find the percent composition of a sample that is 28 g Fe and 8.0 g O. A. Percent Composition Known: n Mass of Fe = 28 g n Mass of O = 8.0 g n Total Mass = 28 + 8.0 g = 36 g Unknown: n % Fe = ? n % O = ? Check: 78% + 22% = 100%

9.  100 = A. Percent Composition n Find the % composition of Cu 2 S. n Use this formula when finding % composition from a chemical formula: %Cu = 127.10 g Cu 159.17 g Cu 2 S  100 = %S = 32.07 g S 159.17 g Cu 2 S 79.85% Cu 20.15% S Known: n Mass of Cu in 1 mol Cu 2 S = 2(63.55g) = 127.10 g Cu n Mass of S in 1 mol Cu 2 S = 32.07 g S n Molar Mass = 127.10 g + 32.07 g = 159.17 g/mol Unknown: n % Cu = ? n % S = ?

10. n How many grams of copper are in a 38.0-gram sample of Cu 2 S? n Use answer from last question as a conversion factor. Cu 2 S is 79.85% Cu = A. Percent Composition 38.0 g Cu 2 S 79.85 g Cu 100 g Cu 2 S = 30.3 g Cu 79.85 g Cu 100 g Cu 2 S

11. n Find the percent composition of Cu 2 SO 4. A. Percent Composition Known: n Mass of Cu in 1 mol Cu 2 SO 4 = 2(63.55 g) = 127.10 g Cu n Mass of S in 1 mol Cu 2 SO 4 = 32.07 g S n Mass of 0 in 1 mol Cu 2 SO 4 = 4(16.00 g) = 64.00 g O n Molar Mass = 127.10 g + 32.07 g + 64.00 g = 223.17 g/mol Unknown: n % Cu = ? n % S = ? n % O = ?

12. n Find the percent composition of Cu 2 SO 4. A. Percent Composition %Cu = 127.10 g 223.17 g  100 = 56.95% Cu %S = 32.07 g 223.17 g  100 = 14.37% S Check: 56.95% + 14.37% + 28.68% = 100% %O = 64.00 g 223.17 g  100 = 28.68% O

13.  100 = %H 2 O = 36.04 g/mol 147.02 g/mol 24.51% H 2 O n Find the mass percentage of water in calcium chloride dihydrate, CaCl 2 2H 2 O. A. Percent Composition Known: n Mass of H 2 O in 1 mol compound = 2(2(1.01g) + 16.00 g) = 36.04 g H 2 O n Molar Mass = 40.08 g + 2(35.45 g) + 36.04 g = 147.02 g Unknown: n % H 2 O = ?

14. B. Empirical Formula C2H6C2H6 CH 3 reduce subscripts n Smallest whole number ratio of atoms in a compound

15. B. Empirical Formula 1. Find mass (or %) of each element. 2. Find moles of each element. 3. Divide moles by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

16. B. Empirical Formula n Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9 g N 1 mol N 14.01 g N = 1.85 mol N 74.1 g O 1 mol O 16.00 g O = 4.63 mol O 1.85 mol = 1 N = 2.5 O N 1.85 O 4.63

17. B. Empirical Formula N 1 O 2.5 Need to make the subscripts whole numbers  multiply by 2 N2O5N2O5

18. B. Empirical Formula n Find the empirical formula for a sample of 94.1% O and 5.9% H. 94.1 g O 1 mol O 16.00 g O = 5.88 mol O 5.9 g H 1 mol H 1.01 g H = 5.84 mol H 5.84 mol = 1 O = 1 H EF = OH

19. C. Molecular Formula n “True Formula” - the actual number of atoms in a compound, either the same as or a whole-number multiple of the empirical formula CH 3 C2H6C2H6 empirical formula molecular formula ?

20. C. Molecular Formula 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molar mass by the empirical formula mass. 4. Multiply each subscript in your EF by the answer from step 3.

21. C. Molecular Formula n The empirical formula for ethylene is CH 2. Find the molecular formula if the molar mass is 28.1 g/mol? 28.1 g/mol 14.03 g/mol = 2.00 empirical formula mass = 14.03 g/mol (EF) n = (CH 2 ) 2  C 2 H 4 n =

22. D. Put it all together!! n 1,6-diaminohexane is 62.1% C, 13.8% H, and 24.1% N. What is the empirical formula? If the molar mass is 116.21 g/mol, what is the molecular formula? 62.1 g C 1 mol C 12.01 g C = 5.17 mol C 13.8 g H 1 mol H 1.01 g H = 13.7 mol H = 3 C = 8 H 24.1 g N = 1 N 1 mol N 14.01 g N = 1.72 mol N 1.72 mol EF = C 3 H 8 N

23. C3H8NC3H8N 116.21 g/mol 58.12 g/mol = 2.00 empirical formula mass = 58.12 g/mol ( C 3 H 8 N ) 2  n = C 6 H 16 N 2 D. Put it all together!!