2. The Mole

3. Calculating Formula/Molar Mass Calculate the molar mass of carbon dioxide, CO 2. 12.01 g + 2(16.00 g) = 44.01 g  One mole of CO 2 (6.02 x 10 23 molecules) has a mass of 44.01 grams

4. Mole Relationships Mole Atoms or molecule s Atoms or molecule s Mass 6.02 X 10 23 Atoms 1 Mole 6.02 X 10 23 Atoms 1 Mole Molar Mass (g) 1 Mole

5. How would you convert, Atoms  Moles Moles  Atoms Moles  Mass (g) Mass (g)  Moles Mass (g)  Atoms Atoms  Mass (g) atoms mole 6.02 x 10 23 1 moles atoms mole 6.02 x 10 23 1 moles= = atoms moles mole 1 g g= g g 1 moles= g g 1 mole atoms6.02 x 10 23 = atoms 6.02 x 10 23 1 mole g g=

6. Calculations with Moles: Converting moles to grams How many grams of lithium are in 3.50 moles of lithium?grams 3.50 mol Li = g Li 1 mol Li 7 g Li 24.5

7. Calculations with Moles: Converting grams to moles How many moles of lithium are in 18.2 grams of lithium?moles 18.2 g Li = mol Li 7 g Li 1 mol Li 2.60

8. Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 3.50 moles of lithium?atoms 3.50 mol = atoms 1 mol 6.02 x 10 23 atoms 2.11 x 10 24

9. Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 18.2 g ofatoms lithium? 18.2 g Li = atoms Li 1 mol Li 6.02 x 10 23 atoms Li 1.57 x 10 24 7 g Li1 mol Li (18.2)(6.02 x 10 23 )/6.94

10. 1 mol NaOH 40.00 g NaOH How many moles is 5.69 g of NaOH?moles Na = 22.99 g/mol O = 16.00 g/mol H = 1.01 g/ mol 40.00 g/mol 5.69 g NaOH = 0.142 mol NaOH

11. How many grams are in 9.45 mol of dinitrogen trioxide? N 2 O 3 = 2(14.01) + 3(16.00) = 76.02 g/mol 9.45 mol N 2 O 3 76.02 g N 2 O 3 1 mol N 2 O 3 = 718.2 =718 g N 2 O 3

12. Find the number of sodium ions, Na+, in 3.00 mol of Na 2 CO 3 3.00 mol Na 2 CO 3 2 mol Na+ 6.022 x 10 23 ions 1mol Na 2 CO 3 1 mol Na+ = 3.61 x 10 24 Na+ ions

13. Find the number of sodium ions, Na+, in 3.00 mol of Na 4 P 2 O 7 3.00 mol Na 4 P 2 O 7 4 mol Na+ 6.022 x 10 23 ions 1mol Na 4 P 2 O 7 1 mol Na+ = 7.23 x 10 24 Na+ ions

14. Percent Composition, Empirical and Molecular Formulas Courtesy www.lab-initio.com

15. 7.3: Calculating Percent Composition of a Compound Like all percent problems: Part whole Find the mass of each component, then divide by the total mass (assume one mole). x 100

16. What is the % composition of water. –Find MM: 2(1.01) + 16.00 = 18.02 g/mol Double check: % ’ s should add up to 100.  100 = Percent Composition %H = 2.02 g H 18.02 g H 2 O  100 = %O = 16.00 g O 18.02 g H 2 O 11.2 % H 88.8 % O

17. What is the percent carbon in C 5 H 8 NO 4 (MSG monosodium glutamate), a compound used to flavor foods and tenderize meats? 1.Find mass of C 1.5 x 12.0 g = 60.0 g C 2.Find mass of MSG 1.5(12.0) + 8(1.0) +14.0 +4(16.0)=146.0 g 3.Mass of element/ mass of cmpd x 100 1.60.0 g C/146.0 g x100 = 41.1% C Practice Problem

18. %Fe = 28 g 36 g  100 = 78% Fe %O = 8.0 g 36 g  100 = 22% O Find the percentage composition of a sample that is 28 g Fe and 8.0 g O. –Determine total mass: Percent Composition 28 g + 8.0 g = 36 g

19. Calculating Percentage Composition Calculate the percentage composition of magnesium carbonate, MgCO 3. Formula mass of magnesium carbonate: 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g 100.00

20. How many grams of copper are in a 38.0-gram sample of Cu 2 S? –Multiply mass of cmpd with % element (38.0 g Cu 2 S)x(0.79852) = 30.3 g Cu Using Percent as a Conversion Factor Cu 2 S is 79.8% Cu (Sample 7-10 p 227)

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

22. EF vs MF Empirical Formula (EF) Def ’ n: lowest whole # ratio H 2 O HO CH CH 2 O Molecular Formula (MF) Def ’ n: actual ratio H 2 O H 2 O 2 C 6 H 6 C 6 H 12 O 6 Water peroxide benzene sugar

23. Calculating Empirical Formula It is not just the ratio of atoms, it is also the ratio of moles of atoms. In 1 mole of CO 2 there is 1 mole of carbon and 2 moles of oxygen.

24. Steps for Calculating the Empirical Formula We get a ratio from the percent composition. Assume you have a 100 g sample. 1.% = g 2.Convert grams to moles (÷ atomic mass) 3.Find lowest whole number ratio. (÷ by smallest # of moles) (if this step gives a decimal, multiply by 2, 3, or 4 to get whole #’s) 4. Use ratio to write the EF

25. Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Part 1 Calculate the number of moles Assume 100 g 38.67 g C x 1mol C = 3.220 mole C 12.01 g C 16.22 g H x 1mol H = 16.09 mole H 1.01 g H 45.11 g N x 1mol N = 3.219 mole N 14.01 g N

26. Part 2 calculate the formula! 3.220 mol C 16.09 mol H 3.219 mol N 3.219 = = = 1 mol C 5 mol H 1 mol N Divide by the smallest number of moles CH 5 N

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

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

29. Try this! Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid? 1. Convert to grams, calculate # of moles

30. 2. Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

31. 3. Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x 2 352 Empirical formula: C3H5O2C3H5O2

32. Molecular Formula “True Formula” - the actual number of atoms in a compound CH 3 C2H6C2H6 empirical formula molecular formula ?

33. Steps to calculating Molecular Formula 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empirical mass. 4. Multiply each subscript by the answer from step 3.

34. Molecular Formula The empirical formula for ethylene is CH 2. Find the molecular formula if the molecular mass is 28.1 g/mol? 28.1 g/mol 14.03 g/mol = 2.00 empirical mass = 14.03 g/mol (CH 2 ) 2  C 2 H 4 1. Find the empirical formula and its mass 2. Divide the molecular mass by the empirical mass. 3. Multiply each subscript by the answer from step 3.

35. Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C 3 H 5 O 2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

36. Finding the Molecular Formula 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g 2. Divide the molecular mass by the empirical mass. 3. Multiply each subscript by the answer from step 3. (C 3 H 5 O 2 ) x 2 = C 6 H 10 O 4

37. Example A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) to be 98.96 g. What is its molecular formula? –EF: CClH 2 –MF: C 2 Cl 2 H 4