1. …Chemical Composition

2. Why it’s important? Everything is either chemically or physically combined with other things. Cookie=physically combined The atom is chemically combined. To know the amount of a material in a sample, you need to know what fraction of the sample it is

3. Counting Nails by the Pound I want to buy a certain number of nails for a project, but the hardware store sells nails by the pound. How do I know how many nails I am buying when I buy a pound of nails? Analogy ◦ How many atoms in a given mass of an element?

4. EXAMPlE A hardware store customer buys 2.60 pounds of nails. A dozen of the nails has a mass of 0.150 pounds. How many nails did the customer buy? Solution map:

5. Counting Nails by the Pound … The customer bought 2.60 lbs of nails and received 208 nails. He counted the nails by weighing them!

6. Counting Nails by the Pound What if he bought a different size nail? ◦ Would the mass of a dozen be 0.150 lbs? ◦ Would there be 208 nails in 2.60 lbs? ◦ How would this effect the conversion factors?

7. Atomic Mass May be better described as an average atomic mass … Carbon has two common isotopes … C-12 … 99.89% (.9989)(12) = 11.8668 C-13 … 1.11% (.0111)(13) = 0.1443 12.011

8. Atomic Mass Actual masses of elements … ◦Hydrogen = 1.66 X 10 -24 g ◦Carbon = 1.99 X 10 -23 g ◦Oxygen = 2.66 X 10 -23 g Atomic Mass Scale (amu) ◦Hydrogen = 1.0 amu ◦Carbon = 12.0 amu ◦Oxygen = 16.0 amu

9. Problem There is no easy way to measure the mass of one atom Solution … ◦Change amu to grams ◦Determine how many atoms are in 12.011 g of Carbon ◦Give this number a name

10. Counting by MOLES. The number of atoms is 6.022 x 10 23. MOLE THIS NUMBER IS CALLED A MOLE ◦ 1 mole = 6.022 x 10 23 things ◦ OR ◦602,000,000,000,000,000,000,000.

11. One Mole 1 mole of marbles would cover the earth 50 miles deep 1 mole of grains of sand would cover the US 1 cm deep Could you count to Avogadro’s number?

12. Moles & Masssss The mass of one mole is called molar mass!

13. Mole and Mass Relationships

14. Molar Mass of Compounds Formula Mass = 1 molecule of H 2 O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu *since 1 mole of H 2 O contains 2 moles of H and 1 mole of O Molar Mass = 1 mole H 2 O = 2(1.01 g H) + 16.00 g O = 18.02 g

15. Moles to Molecules If 3 moles of oxygen gas are contained in a sealed glass jar; how many molecules are present? 3 moles O 2 x 6.02 x 10 23 molecules O 2 = 1.806 x 10 24 O 2 Molecules 1 mole O 2

16. Grams to Moles Calculate the number of moles of each element in the following samples: 24.02 g C x 1 mole C = 2 mole C 12.01 g C 5.00 kg LiOH x 1000 g LiOH x1 mole LiOH = 1 kg LiOH 23.94 g LiOH = 209 mole LiOH

17. Chemical Formulas  Conversion Factors 1 loaf of bread =12 pieces 1 minute = 60 seconds 1 H 2 O molecule  2 H atoms  1 O atom

18. Converting Molecules to Grams Given 2.00 x 10 25 molecules of O 2. How many grams of O 2 is present? Solution: 2.00 x 10 25 molec O 2 x 1 mole O 2 x 32.0 g O 2 = 6.02 x 10 23 molec of O 2 1 mole O 2 = 1060g O 2

19. Converting Grams to Molecules Given 2.00 grams of O 2. How many molecules of O 2 are present? Solution: 2.00 g O 2 x 1 mole O 2 x 6.02 x 10 23 molec of O 2 = 32.0 g O 2 1 mole O 2 = 3.76 x 10 22 molec O 2

20. Solution Concentrations

21. 21 Solution Concentration Descriptions dilute solutions have low solute concentrations concentrated solutions have high solute concentrations _________________________________ Saturated solutions have the maximum amount of solute in them Unsaturated solutions have less than the maximum amount of solute in them

22. 22 Concentrations – Quantitative Descriptions of Solutions Solutions have variable composition To describe a solution accurately, you need to describe the components and their relative amounts Concentration = amount of solute in a given amount of solution ◦Occasionally amount of solvent

23. 23 Mass Percent parts of solute in every 100 parts solution ◦if a solution is 0.9% by mass, then there are 0.9 grams of solute in every 100 grams of solution  or 10 kg solute in every 100 kg solution since masses are additive, the mass of the solution is the sum of the masses of solute and solvent

24. 24 Example: Calculate the mass percent of a solution containing 27.5 g of ethanol (C 2 H 6 O) and 175 mL of H 2 O.

25. 25 Using Concentrations as Conversion Factors concentrations show the relationship between the amount of solute and the amount of solvent ◦12% by mass sugar(aq) means 12 g sugar  100 g solution The concentration can then be used to convert the amount of solute into the amount of solution, or visa versa

26. 26 Example: A soft drink contains 11.5% sucrose (C 12 H 22 O 11 ) by mass. What volume of soft drink in milliliters contains 85.2 g of sucrose? (assume the density is 1.00 g/mL)

27. 27 Preparing a Solution Preparing a Solution need to know amount of solution and concentration of solution calculate the mass of solute needed ◦start with amount of solution ◦use concentration as a conversion factor ◦5% by mass solute 5 g solute  100 g solution Example - How would you prepare 250.0 g of 5.00% by mass glucose solution (normal glucose)? dissolve 12.5 g of glucose in enough water to total 250 g

28. 28 Solution Concentration Molarity moles of solute per 1 liter of solution used because it describes how many molecules of solute in each liter of solution If a sugar solution concentration is 2.0 M, 1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar molarity = moles of solute liters of solution

29. 29 Preparing a 1.00 M NaCl Solution Weigh out 1 mole (58.45 g) of NaCl and add it to a 1.00 L volumetric flask. Step 1 Step 2 Add water to dissolve the NaCl, then add water to the mark. Step 3 Swirl to Mix

30. 30 Example: Calculate the molarity of a solution made by putting 15.5 g of NaCl into a beaker and adding water to make 1.50 L of NaCl solution.

31. 31 Example: How many liters of a 0.114 M NaOH solution contains 1.24 mol of NaOH?

32. 32 How would you prepare 250 mL of 0.20 M NaCl? 0.250 L x 0.20 moles NaCl 1 L x 58.44 g 1 mole NaCl = 2.9 g NaCl Dissolve 2.9 g of NaCl in enough water to total 250 mL Sample - Molar Solution Preparation

33. 33 Molarity and Dissociation When strong electrolytes dissolve, all the solute particles dissociate into ions By knowing the formula of the compound and the molarity of the solution, it is easy to determine the molarity of the dissociated ions simply multiply the salt concentration by the number of ions

34. 34 Molarity & Dissociation NaCl(aq) = Na + (aq) + Cl - (aq) 1 “molecule”= 1 ion + 1 ion 100 “molecules”= 100 ions + 100 ions 1 mole “molecules”= 1 mole ions + 1 mole ions 1 M NaCl “molecules”= 1 M Na + ions + 1 M Cl - ions 0.25 M NaCl= 0.25 M Na + + 0.25 M Cl -

35. 35 Molarity & Dissociation CaCl 2 (aq) = Ca 2+ (aq) + 2 Cl - (aq) 1 “molecule”= 1 ion + 2 ion 100 “molecules”= 100 ions + 200 ions 1 mole “molecules”= 1 mole ions + 2 mole ions 1 M CaCl 2 = 1 M Ca 2+ ions + 2 M Cl - ions 0.25 M CaCl 2 = 0.25 M Ca 2+ + 0.50 M Cl -

36. 36 Find the molarity of all ions in the given solutions of strong electrolytes 0.25 M MgBr 2 (aq) 0.33 M Na 2 CO 3 (aq) 0.0750 M Fe 2 (SO 4 ) 3 (aq)

37. 37 Find the molarity of all ions in the given solutions of strong electrolytes MgBr 2 (aq) → Mg 2+ (aq) + 2 Br - (aq) 0.25 M0.25 M0.50 M Na 2 CO 3 (aq) → 2 Na + (aq) + CO 3 2- (aq) 0.33 M 0.66 M 0.33 M Fe 2 (SO 4 ) 3 (aq) → 2 Fe 3+ (aq) + 3 SO 4 2- (aq) 0.0750 M 0.150 M 0.225 M

38. Percent Composition Percentage of each element in a compound can be determined from 1.the formula of the compound 2.the experimental mass analysis of the compound

39. Problem Calculate the percent composition of Oxygen in Carbon Dioxide. 1. Write the formula of the molecule ◦CO 2 2. Calculate the molar mass of the molecule ◦(12.01g + (2 x 16.0g)) = 44.01g  THE WHOLE 3. Calculate the total mass of the desired element(total number of atoms of the desired element x the molar mass of the element) in the formula ◦ (2x 16.0g) =31.98g  THE PART 4. Divide the total mass of the desired element by the total mass of the molecule ◦31.98g / 44.01g =.7267 THE PART/ THE WHOLE 5. Multiply by 100% ◦.7267 x 100% = 72.67%

40. Mass Percent as a Conversion Factor The mass percent tells you the mass of a constituent element in 100 g of the compound ◦the fact that NaCl is 39% Na by mass means that 100 g of NaCl contains 39 g Na This can be used as a conversion factor ◦100 g NaCl  39 g Na