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# Moles!

## Moles!

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##### Presentation Transcript

1. Moles!

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

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 1023 times!! I mole of cookies = 6.022 x 1023 cookies

4. What about a mole of dance moves? 6.022 x 1023 times!! I mole of dance moves= 6.022 x 1023 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 1023 atoms 6.022 x 1023molecules 6.022 x 1023particles 6.022 x 1023formula units 6.022 x 1023electrons 6.022 x 1023 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 1024 atoms of hydrogen? 1 mole 4.56 x 1024 atoms = 7.57 moles 6.022 x 1023 atoms 2. How many formula units are in 8.92 moles of NaCl? 8.92 moles 6.022 x 1023 formula units 1 mole = 5.37 x 1024 formula units

8. More moley practice… 1. How many atoms of oxygen are in 5.00 moles of carbon dioxide (CO2)? 2 atoms O 5.00 mol CO2 6.022 x 1023 molecules CO2 1 mole of CO2 1 molecule CO2 = 6.02 x 1024 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 (C12H22O11) C – 12(12.01) H – 22(1.008) O – 11(15.999) 2. Determine the molar mass of Ca(NO3)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 • How many grams of aluminum are in 4.89 moles of Al? • How many moles of CaCl2 are in 13.5 g of CaCl2? Molar mass of Al 4.89 mol of Al 26.98 grams of Al = 132 g of Al 1 mol of Al Molar mass of CaCl2 1 mol of CaCl2 13.5 g of CaCl2 = 0.122 mol of CaCl2 110.984 g of CaCl2

12. Practice • How many moles are in 25.67 grams of nickel? 0.4374 mol of Ni • How many grams are in 0.234 moles of lead(II) iodide (PbI2)? 108 grams of PbI2

13. More Practice • How many atoms are in 14.0 grams of manganese? 1.53 x 1023 atoms of Mn • How many molecules are in 46.7 grams of nitrogen gas? 1.00 x 1024 molecules of N2 • How many oxygen atoms are in 3.45 grams of CO2? 9.44 x 1022 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 • 0.500 M of Pb(NO3)2 0.500 moles of Pb(NO3)2 = 1 L • 1.56 M of NaCl 1.56 moles of NaCl = 1 L

17. Using molarity…All about the DA • A 250. mL container has a solution of NaCl with a molarity of 1.54 M. How many moles of NaCl is present? • How many grams of NaCl would that contain? 1 L 1.54 mol 250. mL 0.385 mol of NaCl = 1000 mL 1 L Molar mass of NaCl 58.443 g = 22.5 g of NaCl 1 mol

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 • How many moles are present in 500.0 mL of 0.800 M solution of HCl? • How many grams are present in 250. mL of 0.500 M solution of Pb(NO3)2? 500. 0 mL 1 L 0.800 mol = 0.400 mol of HCl 1 L 1000 mL Molar mass of Pb(NO3)2 331.21 g Pb(NO3)2 250. 0 mL 1 L 0.500 mol 1 L 1 mol 1000 mL = 41.4 g of Pb(NO3)2

20. More Practice… • How many grams are needed to make 4.00 L of 0.250 M solution of NaCl? • How many grams are needed to make 2.50 L of 0.500 M solution of Na2CO3? = 58.4 g of NaCl = 132 g of Na2CO3

21. More Practice… Complete the table: Moles NaCl = 0.0684 mol Grams KOH = 0.144 g Molarity of NaCl = 0.0456 M Molarity of KOH= 0.00142 M

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

23. Example • How many grams are needed to make 250 mL of 0.250 M solution of CuSO4? Molar mass of CuSO4 159.608 g CuSO4 250. 0 mL 1 L 0.250 mol 1 L 1 mol 1000 mL = 9.98 g of CuSO4 • Mass out 9.98 grams of copper(II) sulfate • Hot dog weigh boat and use DI water to add CuSO4 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 (C2H5OH). C – (2)(12.01) = 24.02 H – (6)(1.008) = 6.048 O – (1)(15.999) = 15.999 46.07 g/mol 24.02 g/mol % C = x 100 = 52.14% of C 46.07 g/mol

26. Mass Percent • What percent of each element in a compound? Example: Find the percent of each element in ethanol (C2H5OH). C – (2)(12.01) = 24.02 H – (6)(1.008) = 6.048 O – (1)(15.999) = 15.999 46.07 g/mol 6.048 g/mol % H = x 100 = 13.13% of H 46.07 g/mol

27. Mass Percent • What percent of each element in a compound? Example: Find the percent of each element in ethanol (C2H5OH). C – (2)(12.01) = 24.02 H – (6)(1.008) = 6.048 O – (1)(15.999) = 15.999 46.07 g/mol 15.999 g/mol % O = x 100 = 37.73% of O 46.07 g/mol

28. Mass Percent Practice • Find the percent of each element in CuSO4. • Find the percent of each element in C2H2. Cu = (1)63.546 = 63.546 S = (1)32.065 = 32.065 O = (4)15.999 = 63.996 %Cu = 39.81% %S = 20.09% %O = 40.10% 159.607 g/mol H = (2)1.008 = 2.016 C = (2)12.01 = 24.02 %H = 7.743% %C = 92.26% 26.04 g/mol

29. Empirical and Molecular Formulas Look at C2H2 (Acetylene gas) Acetylene + O2 + 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? 1 mol 2 atoms O 1 mol 2 atoms H 6.022 x 1023 atoms 6.022 x 1023 atoms 3.32 x 10-24 moles O 3.32 x 10-24 moles H

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) formula

31. The empirical formula only tells you the ratio between elements… For example: C2H2 and C6H6 both have the same empiricalformula… CH To find the correct formula (molecularformula)…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 C6H6

32. Molecular Formula Empirical Formula The exact formula for the compound. Ex. C4H10 or P4O8 The reduced formula for the compound. Ex. C2H5 or PO2 The empirical can be the molecular just as MgCl2 or Fe2O3

33. Determine Empirical Formula • Must find moles of each element (molar mass) • 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. 1 mol of S 25.0 g S = 0.780 mol of S 32.06 g S 1.56 mol of O = 2 O : 1 S 1 mol of O 0.780 mol of S 25.0 g O = 1.56 mol of O 15.999 g O Formula: SO2

34. Determine Molecular Formula • Must find molar mass of empirical formula (periodic table) • 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: SO2 128.12 g/mol = 2 : 1 Molar Mass of SO2 = 64.06 g/mol 64.06 g/mol Molecular Formula: S2O4

35. Determine Empirical/Molecular Formula • Must find moles of each element (molar mass) • Find ratios between each element (large over smallest) to get empirical formula • Find molar mass of the empirical formula • Find the ratio between molar mass • 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 mol of H 1.85 g H = 1.84 mol of H 1.008 g H 1.84 mol of H = 1 H : 1 C 1 mol of C 1.84 mol of C 22.15 g C = 1.84 mol of C 12.01 g C Emp Formula: CH Molar mass of CH = 13.02 g/mol 26.04 g/mol Mol. Formula: C2H2 = 2:1 13.02 g/mol

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. *****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! 1 mol of S 40.0 g S = 1.25 mol of S 32.06 g S 3.75 mol of O = 3 O : 1 S 1 mol of O 1.25 mol of S 60.0 g O = 3.75 mol of O 15.999 g O Emp. Formula: SO3 Molar mass of SO3 = 80.06 g/mol 160.1 g/mol Mol. Formula: S2O6 = 2:1 80.06 g/mol