1. Moles!

2. Whatcha talkn’ about a mole? First…a dozen. 12 cookies = 12 bagels = 12 cans = 12 footballs = 12 dance moves = As scientist we need to determine something to quantify atoms, molecules, particles, or something very, very small 1 dozen

3. As scientist we don't work with dozen....we work with something called a mole! Back to original question: "What the heck is a mole?!!" Same concept as a dozen but larger....much larger. If I want to have a mole of cookies…. 6.022 x 10 23 times!! I mole of cookies = 6.022 x 10 23 cookies

4. What about a mole of dance moves? 6.022 x 10 23 times!! I mole of dance moves= 6.022 x 10 23 dance moves

5. Mole in Chemistry? Obviously in chemistry you don’t have cookies and dance moves….we have Atoms Molecules Particles Formula units (ionic compounds) Electrons Abbreviation for mole is mol

6. Moles! 1 mole of atoms = 1 mole of molecules = 1 mole of particles = 1 mole of formula units = 1 mole of electrons = 6.022 x 10 23 atoms 6.022 x 10 23 molecules 6.022 x 10 23 particles 6.022 x 10 23 formula units 6.022 x 10 23 electrons 6.022 x 10 23 is called Avogadro number (named after Amadeo Avogadro)

7. So what does this mean? Converting….DA!!! Yeah! 1. How many moles are in 4.56 x 10 24 atoms of hydrogen? 4.56 x 10 24 atoms 6.022 x 10 23 atoms 1 mole = 7.57 moles 2. How many formula units are in 8.92 moles of NaCl? 8.92 moles6.022 x 10 23 formula units 1 mole = 5.37 x 10 24 formula units

8. More moley practice… 1. How many atoms of oxygen are in 5.00 moles of carbon dioxide (CO 2 )? 5.00 mol CO 2 1 mole of CO 2 6.022 x 10 23 molecules CO 2 1 molecule CO 2 2 atoms O = 6.02 x 10 24 atoms of O

9. Why moles, not atoms? 1 atom of Al = 4.48 x 10 -23 grams Using moles to work with easier numbers 1 mole of Al = 26.98 grams Which is called molar mass ◦ Mass (in grams) of a mole Guess what….another conversion Yeah!! Boom! No doubt!

10. Molar Mass 1. Determine the molar mass of Sucrose (C 12 H 22 O 11 ) C – 12(12.01) H – 22(1.008) O – 11(15.999) 2. Determine the molar mass of Ca(NO 3 ) 2 Ca – 1(40.078) N – 2(14.0067) O – 6(15.999) = 342.29 g/mol = 164.09 g/mol

11. Converting Moles to grams or grams to moles All conversions come from molar mass…aka periodic table 1. How many grams of aluminum are in 4.89 moles of Al? 2. How many moles of CaCl 2 are in 13.5 g of CaCl 2 ? 4.89 mol of Al mol of Al grams of Al26.98 1 = 132 g of Al 13.5 g of CaCl 2 g of CaCl 2 mol of CaCl 2 110.984 1 = 0.122 mol of CaCl 2 Molar mass of Al Molar mass of CaCl 2

12. Practice 1. How many moles are in 25.67 grams of nickel? 0.4374 mol of Ni 2. How many grams are in 0.234 moles of lead(II) iodide (PbI 2 )? 108 grams of PbI 2

13. More Practice 1. How many atoms are in 14.0 grams of manganese? 1.53 x 10 23 atoms of Mn 2. How many molecules are in 46.7 grams of nitrogen gas? 1.00 x 10 24 molecules of N 2 3. How many oxygen atoms are in 3.45 grams of CO 2 ? 9.44 x 10 22 atoms of O

14. Molarity What are these bottles we keep working with numbers and a big M? Well, Molarity is how scientist describe concentration of a solution Molarity = how many moles per 1 liter of solution

15. Molarity What does this mean? For example: 2.00 M HCl 2.00 moles of HCl per 1 liter OR 2.00 moles of HCl = 1 L That’s right another conversion factor!! Yeah!

16. Other examples 1. 0.500 M of Pb(NO 3 ) 2 0.500 moles of Pb(NO 3 ) 2 = 1 L 2. 1.56 M of NaCl 1.56 moles of NaCl = 1 L

17. Using molarity…All about the DA 1. A 250. mL container has a solution of NaCl with a molarity of 1.54 M. How many moles of NaCl is present? 2. How many grams of NaCl would that contain? 250. mL 1000 mL 1 L L mol1.54 1 0.385 mol of NaCl = mol g58.443 1 = 22.5 g of NaCl Molar mass of NaCl

18. Why finding mass? You need the mass… - how much you need to measure in the lab (because you can’t measure moles on a scale) - make accurate concentrations of your solutions

19. Practice 1. How many moles are present in 500.0 mL of 0.800 M solution of HCl? 2. How many grams are present in 250. mL of 0.500 M solution of Pb(NO 3 ) 2 ? 500. 0 mL 1000 mL 1 L0.800 mol 1 L = 0.400 mol of HCl 250. 0 mL 1000 mL 1 L0.500 mol 1 L 331.21 g Pb(NO 3 ) 2 1 mol = 41.4 g of Pb(NO 3 ) 2 Molar mass of Pb(NO 3 ) 2

20. More Practice… 1. How many grams are needed to make 4.00 L of 0.250 M solution of NaCl? 2. How many grams are needed to make 2.50 L of 0.500 M solution of Na 2 CO 3 ? = 58.4 g of NaCl = 132 g of Na 2 CO 3

21. More Practice… Complete the table: FormulaGrams Dissolved Moles Dissolved Volume of Solution Molarity NaCl4.00 grams?1.50 L? KOH?2.56 x 10 -3 1.80 L? Moles NaCl = 0.0684 mol Molarity of NaCl = 0.0456 M Grams KOH = 0.144 g Molarity of KOH= 0.00142 M

22. Making Solutions 1. Obtain correct volumetric flask 2. Determine the correct number of grams needed to measure 3. Add the massed out compound to flask 4. Rinse weigh boat with DI into flask 5. Add DI to etched line on flask

23. Example 1. How many grams are needed to make 250 mL of 0.250 M solution of CuSO 4 ? 250. 0 mL 1000 mL 1 L0.250 mol 1 L 159.608 g CuSO 4 1 mol = 9.98 g of CuSO 4 Molar mass of CuSO 4 -Mass out 9.98 grams of copper(II) sulfate -Hot dog weigh boat and use DI water to add CuSO 4 to 250 mL volumetric flask -Add DI water to 250 mL mark -Add stir bar and stir till all dissolves -Add solution to appropriate containers

24. Percent Mass and Determining Compound Formulas Find percent of each element in the compound Define Molecular and Empirical Formulas

25. Mass Percent What percent of each element in a compound? Example: Find the percent of each element in ethanol (C 2 H 5 OH). C – (2)(12.01) = 24.02 H – (6)(1.008) = 6.048 O – (1)(15.999) = 15.999 46.07 g/mol % C = 24.02 g/mol 46.07 g/mol x 100 =52.14% of C

26. Mass Percent What percent of each element in a compound? Example: Find the percent of each element in ethanol (C 2 H 5 OH). C – (2)(12.01) = 24.02 H – (6)(1.008) = 6.048 O – (1)(15.999) = 15.999 46.07 g/mol % H = 6.048 g/mol 46.07 g/mol x 100 =13.13% of H

27. Mass Percent What percent of each element in a compound? Example: Find the percent of each element in ethanol (C 2 H 5 OH). C – (2)(12.01) = 24.02 H – (6)(1.008) = 6.048 O – (1)(15.999) = 15.999 46.07 g/mol % O = 15.999 g/mol 46.07 g/mol x 100 =37.73% of O

28. Mass Percent Practice 1. Find the percent of each element in CuSO 4. 2. Find the percent of each element in C 2 H 2. Cu = (1)63.546 = 63.546 S = (1)32.065 = 32.065 O = (4)15.999 = 63.996 159.607 g/mol %Cu = 39.81% %S = 20.09% %O = 40.10% H = (2)1.008 = 2.016 C = (2)12.01 = 24.02 26.04 g/mol %H = 7.743% %C = 92.26%

29. Empirical and Molecular Formulas Look at C 2 H 2 (Acetylene gas) Acetylene + O 2 + pumpkin = awesome! What is the ratio between hydrogen and carbon atoms? What about the mole ratio between hydrogen and carbon? 2 atoms of H 2 atoms of C = 1 H : 1 C Same: 1 H : 1 C Why? 2 atoms H 6.022 x 10 23 atoms 1 mol 3.32 x 10 -24 moles H 2 atoms O 6.022 x 10 23 atoms 1 mol 3.32 x 10 -24 moles O

30. What does this tell us? A chemical compound’s formula can be determined by finding the moles of each element. Yeah!! Moles! I Moles This formula is called Empirical (strikes back) formulaEmpirical

31. The empirical formula only tells you the ratio between elements… For example: C 2 H 2 and C 6 H 6 both have the same empirical formula… CH To find the correct formula (molecular formula)…you need the molar mass of the compound to be given. Example: Given molar mass of 78.1 g/mol and empirical formula of CH… Molar mass of CH = 13.02 g/mol (Periodic table) It takes 6 times of 13.02 to get 78.1… therefore the empirical formula (CH) gets multiplied by 6 and thus C 6 H 6

32. Molecular FormulaEmpirical Formula The exact formula for the compound. Ex. C 4 H 10 or P 4 O 8 The reduced formula for the compound. Ex. C 2 H 5 or PO 2 The empirical can be the molecular just as MgCl 2 or Fe 2 O 3

33. Determine Empirical Formula 1. Must find moles of each element (molar mass) 2. Find ratios between each element (large over smallest) Problem: A compound is made up of sulfur and oxygen. 25.0 grams each of sulfur and oxygen are present. Find the empirical formula. 25.0 g S 1 mol of S 32.06 g S = 0.780 mol of S 25.0 g O 1 mol of O 15.999 g O 1.56 mol of O = 1.56 mol of O 0.780 mol of S = 2 O : 1 S Formula: SO 2

34. Determine Molecular Formula 1. Must find molar mass of empirical formula (periodic table) 2. Find ratio between given molecular molar mass and empirical molar mass (large over small) Problem: A compound is made up of sulfur and oxygen. 25.0 grams each of sulfur and oxygen are present. The molar mass of the compound is 128.12 g/mol. Determine the molecular formula. Empirical Formula: SO 2 Molar Mass of SO 2 = 64.06 g/mol 128.12 g/mol 64.06 g/mol = 2 : 1 Molecular Formula: S 2 O 4

35. Determine Empirical/Molecular Formula 1. Must find moles of each element (molar mass) 2. Find ratios between each element (large over smallest) to get empirical formula 3. Find molar mass of the empirical formula 4. Find the ratio between molar mass 5. Multiply the ratio to each element in the empirical formula

36. Example 1 An unknown compound contains both hydrogen and carbon has a molecular molar mass of 26.04 g/mol. The compound contains 1.85 grams of hydrogen and 22.15 grams of carbon. Determine the empirical and molecular formulas. 1.85 g H 1 mol of H 1.008 g H = 1.84 mol of H 22.15 g C 1 mol of C 12.01 g C 1.84 mol of H = 1.84 mol of C 1.84 mol of C = 1 H : 1 C Emp Formula: CH Molar mass of CH = 13.02 g/mol 26.04 g/mol 13.02 g/mol = 2:1 Mol. Formula: C 2 H 2

37. Example 2 An unknown compound contains both sulfur and oxygen has a molecular molar mass of 160.1 g/mol. The compound contains 40.0% sulfur and 60.0% oxygen Determine the empirical and molecular formulas. 40.0 g S 1 mol of S 32.06 g S = 1.25 mol of S 60.0 g O 1 mol of O 15.999 g O 3.75 mol of O = 3.75 mol of O 1.25 mol of S = 3 O : 1 S Emp. Formula: SO 3 Molar mass of SO 3 = 80.06 g/mol *****Math Alert!!***** Attention: This is a math ALERT!!! Treat percentage just like grams….it doesn’t matter how much sample you have!! Therefore if you have %...make it grams! 160.1 g/mol = 2:1 Mol. Formula: S 2 O 6 80.06 g/mol