1 / 36

The Mole

The Mole. CA Standards. The Mole. 1 dozen =. 12. 1 gross =. 144. 1 ream =. 500. 1 mole =. 6.02 x 10 23. There are exactly 12 grams of carbon-12 in one mole of carbon-12. Calculating Formula/Molar Mass. Calculate the formula mass of carbon dioxide, CO 2. 12.01 g + 2(16.00 g) =.

colin
Download Presentation

The Mole

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. The Mole

  2. CA Standards

  3. The Mole 1 dozen = 12 1 gross = 144 1 ream = 500 1 mole = 6.02 x 1023 There are exactly 12 grams of carbon-12 in one mole of carbon-12.

  4. Calculating Formula/Molar Mass Calculate the formula mass of carbon dioxide, CO2. 12.01 g + 2(16.00 g) = 44.01 g  One mole of CO2 (6.02 x 1023 molecules) has a mass of 44.01 grams

  5. Mole Relationships 1 Mole 6.02 X 1023 Atoms Atoms or molecules 1 Mole Molar Mass (g) Mole Mass 1 Mole Molar Mass (g) 6.02 X 1023 Atoms 1 Mole

  6. How would you convert, Atoms  Moles Moles  Atoms atoms mole moles 1 6.02 x 1023 atoms = moles atoms = 6.02 x 1023 atoms 1 mole Moles  Mass (g) Mass (g)  Moles moles g mole g 1 = g = moles g mole 1 Atoms  Mass (g) Mass (g)  Atoms atoms g 1 mole 1 mole 6.02 x 1023 atoms g = g atoms = 1 mole 1 mole 6.02 x 1023 atoms g

  7. Calculations with Moles:Converting moles to grams How many grams of lithium are in 3.50 moles of lithium? 3.50 mol Li 6.94 g Li = g Li 45.1 1 mol Li

  8. Calculations with Moles:Converting grams to moles How many moles of lithium are in 18.2 grams of lithium? 18.2 g Li 1 mol Li 2.62 = mol Li 6.94 g Li

  9. Calculations with Moles:Using Avogadro’s Number How many atoms of lithium are in 3.50 moles of lithium? 6.02 x 1023 atoms 3.50 mol = atoms 2.07 x 1024 1 mol

  10. Calculations with Moles:Using Avogadro’s Number How many atoms of lithium are in 18.2 g of lithium? 18.2 g Li 6.022 x 1023 atoms Li 1 mol Li 6.94 g Li 1 mol Li (18.2)(6.022 x 1023)/6.94 1.58 x 1024 = atoms Li

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

  12. How many grams are in 9.45 mol of dinitrogen trioxide? N2O3 = 2(14.01) + 3(16.00) = 76.02 g/mol 9.45 mol N2O3 76.02 g N2O3 1 mol N2O3 = 718.2 =718 g N2O3

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

  14. 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

  15. 2.02 g H 18.02 g H2O 16.00 g O 18.02 g H2O Percent Composition • 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 = %H = 11.2 % H 88.8 % O %O =  100 =

  16. Practice Problem What is the percent carbon in C5H8NO4 (MSG monosodium glutamate), a compound used to flavor foods and tenderize meats? • Find mass of C • 5 x 12.0 g = 60.0 g C • Find mass of MSG • 5(12.0) + 8(1.0) +14.0 +4(16.0)=146.0 g • Mass of element/ mass of cmpd x 100 • 60.0 g C/146.0 g x100 = 41.1% C

  17. 28 g 36 g 8.0 g 36 g Percent Composition • Find the percentage composition of a sample that is 28 g Fe and 8.0 g O. • Determine total mass: 28 g + 8.0 g = 36 g  100 = 78% Fe %Fe =  100 = 22% O %O =

  18. Calculating Percentage Composition Calculate the percentage composition of magnesium carbonate, MgCO3. Formula mass of magnesium carbonate: 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g 100.00

  19. Using Percent as a Conversion Factor • How many grams of copper are in a 38.0-gram sample of Cu2S? • Multiply mass of cmpd with % element Cu2S is 79.8% Cu (Sample 7-10 p 227) (38.0 g Cu2S)x(0.79852) = 30.3 g Cu

  20. Empirical Formula (EF) Def’n: lowest whole # ratio H2O HO CH CH2O Molecular Formula (MF) Def’n: actual ratio H2O H2O2 C6H6 C6H12O6 EF vs MF Water peroxide benzene sugar

  21. CH3 B. Empirical Formula • Smallest whole number ratio of atoms in a compound C2H6 reduce subscripts

  22. Calculating Empirical Formula • It is not just the ratio of atoms, it is also the ratio of moles of atoms. • In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen.

  23. Steps for Calculating the Empirical Formula We get a ratio from the percent composition. Assume you have a 100 g sample. • % = g • Convert grams to moles (÷ atomic mass) • Find lowest whole number ratio. (÷ bysmallest # 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

  24. 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

  25. Part 2 calculate the formula! Divide by the smallest number of moles 3.220 mol C 16.09 mol H 3.219 mol N = 1 mol C 3.219 = 5 mol H CH5N 3.219 = 1 mol N 3.219

  26. 1.85 mol 1.85 mol 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 = 1 N 74.1 g 1 mol 16.00 g = 4.63 mol O = 2.5 O

  27. N2O5 Empirical Formula N1O2.5 Need to make the subscripts whole numbers  multiply by 2

  28. 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

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

  30. 3. Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x 2 x 2 x 2 3 5 2 Empirical formula: C3H5O2

  31. C2H6 Molecular Formula • “True Formula” - the actual number of atoms in a compound CH3 empirical formula ? molecular formula

  32. 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.

  33. 28.1 g/mol 14.03 g/mol Molecular Formula • The empirical formula for ethylene is CH2. Find the molecular formula if the molecular mass is 28.1 g/mol? 1. Find the empirical formula and its mass empirical mass = 14.03 g/mol 2. Divide the molecular mass by the empirical mass. = 2.00 3. Multiply each subscript by the answer from step 3. (CH2)2  C2H4

  34. Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C3H5O2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

  35. 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. (C3H5O2) x 2 = C6H10O4

  36. 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: CClH2 • MF: C2Cl2H4

More Related