2. Unit III Chemical Composition (i.e. The Mole)

3. Atomic Masses Atomic masses use Carbon 12 ( 12 C) as the standard Calculated with the aide of a mass spectrometer

4. This process is also known as gas chromatography As gas chromatograph counts the number of particles present in a given sample Produces a graph like the one above With this information percentages can be calculated and formulas can be determined

5. Atomic mass is the average of all the naturally occurring isotopes of that element Carbon = 12.011 IsotopeSymbolComposition of the nucleus % in nature Carbon- 12 12 C6 protons 6 neutrons 98.89% Carbon- 13 13 C6 protons 7 neutrons 1.11% Carbon- 14 14 C6 protons 8 neutrons <0.01%

6. The Mole Abbrev. mol 1 mol = # C atoms in 12 g of pure 12 C Avogadro’s number Equal to 6.022 x 10 23 atoms in 1 mol C Named in honor of the Italian chemist Amadeo Avogradro (1776-1855) I didn’t discover it. Its just named after me!

7. 1 mol = 6.022 x 10 23 atoms = molar mass (g) = 22.4 L

9. To covert amongst mol, atoms, grams, & liters, use the following equivalencies: 1 mol = 6.022 x 10 23 = molar mass = 22.4 L atoms (grams) Or use ½ of the following chart:

10. Keys to Use: 1. Only use 1 side of the chart 2. Proceed from one shaded box to another shaded box 3. When following the arrows, perform the indicated function 4. When going against the arrows, perform the opposite function

11. Molar Mass The molar mass is determined by summing the masses of the component atoms Example: What is the molar mass of MgCO 3 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g

12. Molar Calculations Example 1: How many grams of lithium are in 3.50 mol of lithium? 3.50 mol Li → g Li (3.50 6.94) / 1 = 24.29 (3 SF’s in original problem) → Round 24.3 → SSN → 2.43 x 10 1 Don’t forget the units → g Li 3.50 mol Li 1 mol Li 6.94 g Li = 2.43 x 10 1 g Li

13. Example 2: How many mol are in 98.2 g of NaCl? 98.2 g NaCl → mol NaCl (98.2 1) / 58.44 = 1.680355921 (3 SF’s in original problem) → Round 1.68 → SSN → 1.68 x 10 0 Don’t forget the units → mol NaCl 98.2 g NaCl 58.44 g NaCl 1 mol NaCl = 1.68 x 10 0 mol NaCl

14. Example 3: How many atoms are in 411 g of calcium phosphate? 411 g Ca 3 (PO 4 ) 2 → atoms Ca 3 (PO 4 ) 2 (411 6.022 x 10 23 ) / 310.18 = 7.9793 x 10 23 (3 SF’s in original problem) → Round 7.98 x 10 23 → SSN → 7.98 x 10 23 Don’t forget the units → atoms Ca 3 (PO 4 ) 2 411 g Ca 3 (PO 4 ) 2 310.18 g Ca 3 (PO 4 ) 2 6.022 x 10 23 atoms Ca 3 (PO 4 ) 2 = 7.98 x 10 23 atoms Ca 3 (PO 4 ) 2

15. Example 4: How many liters of dioxygen heptioide gas does 7.5 x10 24 atoms occupy? 7.5 x 10 24 atoms O 2 I 7 → L O 2 I 7 (7.5 x 10 24 22.4) / 6.022 x 10 23 = 278.977084 (2 SF’s in original problem) → Round 280 → SSN → 2.8 x 10 2 Don’t forget the units → L I 2 O 7 7.5 x 10 24 atoms O 2 I 7 6.022 x 10 23 atoms O 2 I 7 22.4 L I 2 O 7 = 2.8 x 10 2 L O 2 I 7

16. Percent Composition of Compounds Mass Percent = (Part / Whole)  100% Calculate the percent composition of magnesium carbonate (MgCO 3 ) Molar Mass = 84.32 g 24.31 g + 12.01 g + 3 (16.00) Mg = (24.31 / 84.32)  100 = 28.83 % C = (12.01 / 84.32)  100 = 14.24 % O = (48.00 / 84.32)  100 = 56.93 % 100 %

17. Determining the Formula of a Compound There are 2 basic formulas: 1. Empirical Simple – Can not be simplified Examples: H 2 0, MgCl 2 2. Molecular Complex – Can be simplified Examples: H 8 O 4, Mg 3 Cl 6 Molecular Formula = (Empirical Formula ) n Example: (H 2 0) 4 = H 8 O 4 n = molar mass / Empirical Formula Mass

18. Note: % are grams (if based on 100) Adipic acid contains 49.32% C, 6.85% H, & 43.84% O by mass. What is the E.F.? C: (49.32/12.01)= 4.10 / 2.74 = 1.50  2 = 3 H: (6.85 /1.01) = 6.78 / 2.74 = 2.47  2 = 5 O: (43.84 /16.00)= 2.74 / 2.74 = 1  2 = 2 Answer: C 3 H 5 O 2 Moles Divide by smallest # of mol Mole Ratio X by integer to obtain a whole # # of atoms Empirical Formula Determination

19. Empirical Formula Determination Sample Problems Sample Problem 1:.6884 g of lead combined with.2356 g of Cl to form a binary compound. Calculate the empirical formula. Sample Problem 2: A compound’s percent by mass is as follows: Copper = 33.88%, Nitrogen = 14.94%, and Oxygen = 51.18%. Determine the empirical formula of the compound.

20. Problem: The E.F. for adipic acid is C 3 H 5 O 2 The molar mass is 146 g/ mol Steps: 1. Determine the E.F. mass of C 3 H 5 O 2 (3  12.01) + (5  1.01) + (2  16.00) = 73.08 g 2. Find the n factor n = molar mass / E.F. mass n = 146 / 73.08 = 2 3. M.F. = (E.F.) n = (C 3 H 5 O 2 ) 2 = C 6 H 10 O 4 Molecular Formula Determination

21. Molecular Formula Determination Sample Problems Sample Problem 1: Calculate the M.F. of a compound with the E.F. CH 2 O and a molar mass of 150 g/mol. Sample Problem 2: A gas is composed of 71.75% Cl, 24.27% C, and 4.07% H (by mass) & its molar mass is 98.96 g/mol. What is the M.F.?

22. Formulas of Hydrates A hydrate is a compound that has a specific number of water molecules bound to its atoms Written with each formula unit following a dot Examples: 1. Calcium chloride dihydrate → CaCl 2 2 H 2 O 2. Magnesium sulfate heptahydrate→ MgSO 4 7 H 2 O Samples: 1. Sodium cabonate decahydrate (molar mass ?) 2. Barium hydroxide octahydrate (molar mass ?)