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Section 3.4—Counting Molecules

Section 3.4—Counting Molecules. So the number of molecules affects pressure of an airbag…how do we “ count ” molecules?. What is a Mole? Ted Ed video. http://ed.ted.com/lessons/daniel-dulek-how-big-is-a-mole-not-the-animal-the-other-one. What is a mole?. Mole – metric unit for counting.

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Section 3.4—Counting Molecules

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  1. Section 3.4—Counting Molecules So the number of molecules affects pressure of an airbag…how do we “count” molecules?

  2. What is a Mole? Ted Ed video • http://ed.ted.com/lessons/daniel-dulek-how-big-is-a-mole-not-the-animal-the-other-one

  3. What is a mole? Mole– metric unit for counting We use it just like we use the terms dozen and ream! The only acceptable abbreviation for “mole” is “mol”…not “m”!!

  4. What is a counting unit? You’re already familiar with one counting unit…a “dozen” A dozen = 12 “Dozen” 12 A dozen doughnuts 12 doughnuts A dozen books 12 books A dozen cars 12 cars

  5. What can’t we count atoms in “dozens”? Atoms and molecules are extremely small We use the MOLE to count particles

  6. A mole = 6.02  1023 particles (called Avogadro’s number) 6.02  1023 = 602,000,000,000,000,000,000,000 “mole” 6.02  1023 1 mole of doughnuts 6.02  1023 doughnuts 1 mole of atoms 6.02  1023 atoms 1 mole of molecules 6.02  1023 molecules This number was named after Amadeo Avogadro. He did not calculate it!

  7. FUNNY!

  8. Representative Particles • Remember, matter is broken down into either SUBSTANCES or mixtures • Substances are broken down into either ELEMENTS or COMPOUNDS

  9. Example: Particles & MolesUse the conversion factor (1 mol = 6.02 x 1023) particles to convert Example 1: How many molecules of water are in 1.25 moles?

  10. Example: Molecules & Moles Example 1: How many molecules of water are in 1.25 moles? 1 mol = 6.021023 molecules 1.25 mol H2O Molecules H2O 6.02  1023 = _______ molecules H2O 7.531023 1 mol H2O

  11. Let’s Practice #2 Example: How many moles are equal to 2.8 × 1022 formula units of KBr?

  12. Let’s Practice #2 Example: How many moles are equal to 2.8 × 1022 formula units KBr? 1 mol = 6.021023 formula units 1 mole 2.8 × 1022formula units = _______ moles 0.047 6.02  1023 Formula units

  13. Let’s Practice #3 Example: How many atoms are equal to 3.56 moles of Fe?

  14. Let’s Practice #3 Example: How many atoms are equal to 3.56 moles of Fe? 1 mol = 6.021023 molecules 6.02 x 10 23 atoms 3.56 moles Fe = _______ atoms 2.14 x 1024 1 moles

  15. Molar Mass Molar Mass– The mass for one mole of an atom or molecule. Other terms commonly used for the same meaning: Molecular Weight Molecular Mass Formula Weight Formula Mass

  16. Molar Mass for Elements The average atomic mass = grams for 1 mole Average atomic mass is found on the periodic table Element Mass 1 mole of carbon atoms (C) 12.01 g 1 mole of oxygen atoms (O2) 16.00g x 2 = 32.00 g O2 1 mole of hydrogen atoms (H2) 1.01g x 2 = 2.02 g H2 Unit for molar mass: g/mole or g/mol

  17. Molar Mass for Compounds The molar mass for a molecule = the sum of the molar masses of all the atoms

  18. Calculating a Molecule’s Mass To find the molar mass of a molecule: 1 Count the number of each type of atom 2 Find the molar mass of each atom on the periodic table 3 Multiply the # of atoms by the molar mass for each atom 4 Find the sum of all the masses

  19. Example: Molar Mass Example: Find the molar mass for CaBr2

  20. Example: Molar Mass 1 Count the number of each type of atom Example: Find the molar mass for CaBr2 Ca 1 Br 2

  21. Example: Molar Mass 2 Find the molar mass of each atom on the periodic table Example: Find the molar mass for CaBr2 Ca 1 40.08 g/mole Br 2 79.90 g/mole

  22. Example: Molar Mass 3 Multiple the # of atoms  molar mass for each atom Example: Find the molar mass for CaBr2  Ca 1 40.08 g/mole 40.08 g/mole =  159.80 g/mole Br 2 79.90 g/mole =

  23. Example: Molar Mass 4 Find the sum of all the masses Example: Find the molar mass for CaBr2  Ca 1 40.08 g/mole 40.08 g/mole =  + 159.80 g/mole Br 2 79.91 g/mole = 199.88 g/mole 1 mole of CaBr2 =199.90 g

  24. Example 2: If you see a Parentheses in the Formula Be sure to distribute the subscript outside the parenthesis to each element inside the parenthesis. Example: Find the molar mass for Sr(NO3)2

  25. Example 2: Molar Mass & Parenthesis Be sure to distribute the subscript outside the parenthesis to each element inside the parenthesis. Example: Find the molar mass for Sr(NO3)2 Sr  1 87.62 g/mole 87.62 g/mole =  N 2 14.01 g/mole 28.02 g/mole =  + 96.00 g/mole O 6 16.00 g/mole = 211.64 g/mole 1 mole of Sr(NO3)2 =211.64 g

  26. Let’s Practice #3 Example: Find the molar mass for Al(OH)3

  27. Let’s Practice #2 Be sure to distribute the subscript outside the parenthesis to each element inside the parenthesis. Example: Find the molar mass for Al(OH)3 Al  1 26.98 g/mole 26.98 g/mole =  O 3 16.00 g/mole 48.00 g/mole =  + 3.03 g/mole H 3 1.01 g/mole = 78.01 g/mole 1 mole of Al(OH)3 =78.01 g

  28. Using Molar Mass in Conversions

  29. Example: Moles to Grams Example: How many grams are in 1.25 moles of water?

  30. H 2  1.01 g/mole 2.02 g/mole =  + 16.00 g/mole O 1 16.00 g/mole = 18.02 g/mole Example: Moles to Grams When converting between grams and moles, the molar mass is needed Example: How many grams are in 1.25 moles of water? 1 mole H2O molecules = 18.02 g 1.25 mol H2O g H2O 18.02 = _______ g H2O 22.5 1 mol H2O

  31. Example: Grams to Moles Example: How many moles are in 25.5 g NaCl?

  32. Na 1  22.99 g/mole 22.99 g/mole =  + 35.45 g/mole Cl 1 35.45 g/mole = 58.44 g/mole Example: Grams to Moles Example: How many moles are in 25.5 g NaCl 1 moles NaCl molecules = 58.44 g 25.5 g NaCl mol NaCl 1 58.44 g NaCl = ____ moles NaCl .436

  33. Example: Grams to Molecules Example: How many formula units are in 25.5 g NaCl?

  34. Na 1  22.99 g/mole 22.99 g/mole =  + 35.45 g/mole Cl 1 35.45 g/mole = 58.44 g/mole Example: Grams to Moles Example: How many formula units are in 25.5 g NaCl 1 moles NaCl formula units = 58.44 g 25.5 g NaCl mol NaCl 1 6.02 x 1023 FU’s 58.44 g NaCl 1 mol NaCl = ____ FU’s NaCl 2.63 x 1023

  35. Percent Composition • Defined as the percent by mass of each element in a compound • Steps to Finding Percent Composition • Add up the mass of each element within the compound to get the mass of the compound. • Divide each element’s mass by the mass of the compound. • Multiply by 100 % composition= mass of element x 100 mass of compound

  36. Percent Composition by Mass of Air

  37. Example:Calculate the % composition of each element in calcium carbonate. CaCO3 Molar mass = 100.09 g % C = 12.01/100.09 x 100 = 12.00 % %Ca = 40.08/100.09 x 100 = 40.04% %O = 48.00/100.09 x 100 = 47.96%

  38. Example: What is the % of each element in a compound that contains 29.00g Ag and 4.30g S only? Total mass of compound = 33.30 g % Ag = 29.00/33.30 x 100 = 87.09 % %S = 4.30/33.30 x 100 = 12.9%

  39. Hydrates • A HYDRATE is an ionic compound with water trapped • in its crystal. • Examples are: • CuSO4 5H2O MgSO4 7 H2O CoCl2 H2O • Heating a hydrate removes the water and leaves • behind just the salt which is called the anhydrate.

  40. Example: What is the % water in the hydrate, CuCl2 2H2O Molar mass of hydrate = 170.48 g % water = 36.04/170.48 x 100 = 21.14%

  41. http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/flashfiles/stoichiometry/empirical.htmlhttp://group.chem.iastate.edu/Greenbowe/sections/projectfolder/flashfiles/stoichiometry/empirical.html Heating of A Hydrate Animation Calculating the experimental % composition of water in a hydrate.

  42. Empirical Formula • A chemical formula showing the simplest whole number ratio of moles of elements (subscipts) • Hydrogen Peroxide has an actual formula (molecular formula) of H2O2 but an • empirical formula of HO

  43. How to Calculate Empirical Formula • RHYME: Percent to Mass • Mass to Mole • Divide by Small • Multiply til Whole • Assume 100 grams of the sample of compound. Switch the percent sign to grams • Convert each element’s mass into moles. • Divide each element’s mole amount by the smallest mole amount in the entire problem. The answer is the subscript of the element within the compound. • OPTIONAL: If mole ratio is not within .1 of a whole number, multiply each amount by the smallest whole number that will produce either a whole number itself or a number within .1 of a whole number.

  44. Example: What is the empirical formula for 40.05% S and 59.95% O? • Switch the percent sign to grams & convert each element’s mass into moles • 40.05 g S / 32.01g = 1.250 mol S • 59.95 g O / 16.00 g = 3.747 mol O • Divide each element’s mole amount by the smallest mole amount in the entire problem. • 1.250 mol S = 1 3.747 mol O = 2.99 = 3 • 1.250 mole 1.250 mol S1O3 SO3

  45. Example: What is the empirical formula for 43.64% P and 56.36% O? • Switch the percent sign to grams & convert each element’s mass into moles • 43.64 g P / 30.97g = 1.409 mol S • 56.36 g O / 16.00 g = 3.522 mol O • Divide each element’s mole amount by the smallest mole amount in the entire problem. • 1.409 mol S = 1 3.522 mol O = 2.49 ≠ 3 • 1.409 mole 1.409 mol • If mole ratio is not within .1 of a whole number, multiply each amount by the smallest whole number that will produce either a whole number itself or a number within .1 of a whole number. • 1 x 2 = 2 2.49 x 2 = 4.998 = 5 P2O5

  46. Molecular Formula • Is the ACTUAL, true formula of the compound. • They are usually multiples of their empirical formula • N2O4 is the molecular formula; the empirical formula is NO2 • Notice that the molecular formula is 2 times larger than the empirical formula

  47. Molecular Formula

  48. How to Calculate Molecular Formula 1. You need to find the empirical formula and calculate its molar mass. Call this empirical formula mass EFM. 2. Find the mass of the actual formula which will most likely be given to you in grams. Call this molecular formula mass MFM. 3. Divide the MFM by the EFM to get a factor. 4. Multiply the factor by the empirical formula to get the MOLECULAR FORMULA

  49. Example:What is the molecular formula of a compound whose empirical formula is CH4N and the molecular mass is 60.12 g/mol? 1. Empirical Formula Mass (EFM) = 12.01 + 4.04 + 14.01 = 30.06 g 2. Molecular Formula Mass (MFM) = 60.12 g 3. 60.12 / 30.06 = 2 4. 2(CH4N) = C2H8N2

  50. Section 3.5—Gas Behavior How does the behavior of gases affect airbags? What is PRESSURE? Force of gas particles running into a surface.

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