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Understanding Mole, Avogadro’s Number, and Mass in Chemistry

Learn the importance of composition in chemistry, counting atoms by mass, molar mass, and Avogadro’s number. Practice with examples and explore real-life applications of chemical formulas.

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Understanding Mole, Avogadro’s Number, and Mass in Chemistry

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  1. Mole, Avogadro’s Number Molar Mass Mass percent composition Empirical Formula

  2. Why is Knowledge of Composition Important? • Everything in nature is either chemically or physically combined with other substances • Some Applications: • the amount of sodium in NaCl for diet • the amount of iron in iron ore for steel production • the amount of carbon in fossil fuel in terms of green house effect

  3. Counting Pennies by Weight • What if a person doesn’t have bills but pounds of pennies when he wants to buy a $10 pizza? • Seinfeld: Kramer’s attempt to buy calzone with pennies https://www.youtube.com/watch?v=ywidjw9oQVw • http://www.youtube.com/watch?v=kMimygVTgbU • Assuming each penny weighs exactly the same (2.500 g), we can count pennies by weight the total weight of pennies…

  4. Counting Pennies by Weight Kramer brought 2,500. g (ca. 5.5 lbs) of new pennies to the Italian restaurant. Do you think he can buy three calzones ($10.0 total)? Each new penny weighs 2.500 g.

  5. Counting Coins by Weight • What if Kramer bought a different coin? • Would the mass of single dime be 2.500 g? • Would there be $10.00 in 2,500. g dimes? • How would this affect the weight-dollar conversion factors?

  6. Mass of atoms • Atoms of the same element have on average the same mass. Mass of atom in amu(atomic mass unit): 1 amu = 1.66054×10-27 kg • Mass of a C-12 atom = 12 (exact) amu • Mass of a gold atom on average = 197.0 amu • Do I have to memorize these numbers?!

  7. Counting atoms by mass • Similar to how we count the number of pennies, we can count the number of atoms using their mass • The number of atoms in a sample often is an astronomical number, so we use a big unit to account for the number of atoms. Just like 1 dozen = 12 items

  8. “Chemical Dozen” – the Mole • The number of particles in 1 mole: Avogadro’s Number = _______________ units • 1 mole of any atomic element has 6.022 x 1023atoms • 1 mole of oxygen gas has 6.022 x 1023 O2molecules • 1 mole of NaCl solid has 6.022 x 1023 Na+ ions and 6.022 x 1023 Cl- ions

  9. Avogadro’s Number as Conversion Factor • Given number of Moles  number of Units • Given number of Units  Mole ´ = # Mole #Units

  10. Information Given: 0.041 moles of Ag atoms Find: ? number of Ag atoms Example:A silver ring contains 0.041 moles of silver. How many silver atoms silver are in the ring? 1 mole Ag atoms = 6.022 x 1023 Ag atoms 2.5 x 1022 Ag atoms (2 sig. figs)

  11. Information Given: 1.234 x 1020 water molecules Find: ? moles of water molecules Example:How many moles of water are in 1.234 x 1020 water molecules? 1 mole water = 6.022 x 1023 water molecules 2.049 x 10-4 moles water (4 sig. figs)

  12. Mole, Atomic Mass, Mass • The mass of exactly one mole of any element equals the Atomic Mass in grams. Example: • Exactly 1 mole He = _______ g • Exactly 1 mole Li = _______ g • If given the mass of a particular number of atoms, we can determine the number of atoms in any mass of the element!

  13. The Mass of one mole substance depends on the Element

  14. Molar Mass • The mass of one mole of atoms is called the Molar mass. Unit as g/mole or g/mol. • The molar massof an element, in grams, is numerically equal to the element’s atomic mass. Example: • Molar mass of Sodium = ___________

  15. Molar Mass as Conversion Factor • Example: molar mass He = 4.003 g So 1 mole He = 4.003 gram He • Given number of Moles  Mass Mass (gram) = _________________ • Given Mass  Mole Mole = ___________________

  16. Information Given: 57.8 g S Find: ? moles S Conv. Fact.: 1 mole S = 32.06 g Example:Calculate the number of moles of sulfur in 57.8 g of sulfur 1.80 moles S (3 sig. figs)

  17. Information Given: 3.34 x 10-3 moles He Find: ? Grams of He Conv. Fact.: 1 mole He = 4.00 g Example:Calculate the mass of 3.34 x 10-3 mole of helium gas. 0.0133 g He

  18. Practice: #atoms, moles, molar mass • The largest uncut diamond (pure carbon) weighs 755 carat (1 carat = 0.2 grams exactly). How many moles of carbon atoms are in this diamond? • What is mass for 1.2 x 10-3 mole gold? • (optioinal) 70% of the mass of the Sun is hydrogen. The mass of the Sun is 1.99 x 1030 kg. How many moles of hydrogen atoms are in the Sun? 12.6 mol C 0.24 g mol Au 1.38 x 1033 mol H atom

  19. Molar Mass and Avogadro’s Number • Since 1 mole contains Avogadro’s number of units whilst has a mass of molar mass • there is direct connection/conversion factor between Avogadro’s number and molar mass. Example: • 6.022 x 1023Al atoms = _______ g Al

  20. Information Given: 16.2 g Al Find: ? atoms Al 1 mol Al = 26.98 g = 6.022 x 1023 Al atoms Example:How many aluminum atoms are in an aluminum can with a mass of 16.2 g? g Al mole Al atoms Al 3.62 x 1023 atoms Al

  21. Given: 1.34 x 1020 Ag Find: ? grams Ag 1 mol Ag = 107.87 g Ag = 6.022 x 1023 Ag atoms SM: atoms mol g Example:What is the mass of 1.34 x 1020 silver atoms? 0.0240 g Ag

  22. Practice: Mass vs. Molec/Atom • Diamond is made of pure carbon. The largest uncut diamond weighs 755 carat (1 carat = 0.2 grams exactly). How many carbon atoms are in this diamond? • What is the mass of one gold atom in grams? 7.57 x 1024 C atoms 3.2078 x 10-22 grams

  23. Molar Mass of Compounds Molar Mass: Mass of molecules per exactly one mole of this molecule. Unit = g/mol Example: 1 mole of H2O = 2 moles H + 1 mole O Molar Mass of H2O = 2(1.008 g/mol H) + 16.000 g/mol O = 18.016 g/mol 1 mole Al(NO3)3 = 1 mole Al + 3 mole N + 9 mole O Molar Mass of Al(NO3)3= 1(26.98) + 3(14.01) + 9(16.00) = 213.01 g/mol

  24. Practice: Find the Molar Mass • Sulfuric acid • Aluminum sulfate • Ammonium chlorite • Hydrobromic acid • Nickel(II) phosphate

  25. Mass of Compound ↔ Moles of Compound

  26. Information Given: 1.75 mol H2O Find: ? g H2O C F: 1 mole H2O = 18.02 g H2O Example:Calculate the mass (in grams) of 1.75 mol of water 31.5 g H2O

  27. Information Given: 454 g CO2 Find: ? mol CO2 C F: 1 mole CO2 = 44.01 g CO2 Example:Calculate the mole of CO2 in 454 g dry ice (solid carbon dioxide). 10.3 mol CO2

  28. Mass of Compound ↔ Number of molecules

  29. Information Given: 4.8 x 1024 molec NO2 Find: ? g NO2 1 mole NO246.01 g NO2 Example:Find the mass of 4.8 x 1024 NO2 molecules • 3.7 x 102 g

  30. Information Given: 1.00 mL H2O Find: #H2O molecules 1 mole H2O = 18.02 g H2O Example:How many water molecules in 1.00 mL DI water? Density of water = ______ g/mL • 3.34 x 1022H2O molecules

  31. Practice • Online: http://www.sciencegeek.net/Chemistry/taters/Unit4GramMoleVolume.htm • How many moles of water in 10.00 mL pure water if the density of water is 1.00 g/mL? • Determine the mass of 2.0 × 103 mole NaCl.

  32. Chemical Formulas as Conversion Factors • 1 spider  8 legs + 1 head + 1 abdomen • 1 chair  4 legs + 1 seat + 1 back • 1 H2O molecule  2 H atoms + 1 O atom

  33. Mole Relationships inChemical Formulas Given: • #moles of the compound • Chemical formula of the compound #moles of a constituent element within the compound

  34. Chemical formula: __________ Molar mass = 389.56 g/mol *Example:How many oxygen atoms are in 100. g of iron(II) phosphate. For exactly 1 mole compound, there are ___ mole of oxygen • 0.2567 mol_________ • 2.054 mol O • 1.24 x 1024 O atoms

  35. Mass Percent Composition Mass% represents the mass of constituent element in grams per exactly 100 g of compound: Example: mass percent of nitrogen in Al(NO3)3

  36. Example: Find mass% • Find Mass% of Nitrogen in Ammonium phosphate. Formula: Molar mass = 181.17 g/mol For exactly 1 mole compound, there are ___ mole of Nitrogen Exactly 1 mole compound weighs ______ g ____ moles nitrogen weighs 42.03 g Mass% of Nitrogen = 42.03 g/181.17 g x 100% = 23.20%

  37. Example: Use mass% to find mass of constituent element • Given the mass% of nitrogen in Ammonium phosphate (23.20%), find the mass of nitrogen in 454 g of ammonium phosphate. Meaning of mass%: A conversion factor  exact 100 g ammonium phosphate = 23.20 g N 454 g _______ x Mass Nitrogen = 105 g

  38. Empirical Formula Chemical Formula  Mass% of each element Mass% of each element  Chemical Formula? Empirical Formula: The simplest, whole-number ratio of atoms in a molecule • It shows the mole ratio in the compound • It can be determined from percent composition or combining masses

  39. Hydrogen Peroxide (in store) Molecular Formula = H2O2 Empirical Formula = HO Benzene (in gasoline) Molecular Formula = C6H6 Empirical Formula = CH Glucose (essential energy in body) Molecular Formula = C6H12O6 Empirical Formula = CH2O Empirical Formula vs. Molecular Formula

  40. Finding an Empirical Formula • mass% of each element  grams of each element • skip if already grams • grams  moles of each element • use atomic mass of each element • pseudoformula w/ moles as subscripts • divide all by smallest number of moles • multiply all mole ratios by number to make all whole numbers • if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3, etc. • skip if already whole numbers

  41. MMA mass A (g) moles A MMB mass B (g) moles B How to Determine Empirical Formulas. A • From the masses of constituents moles A moles B

  42. mol Fe, O mol ratio empirical formula Empirical Formula: Mass composition • One 1.205-g sample of iron-oxygen compound contains 0.872 g iron. Find the empirical formula of this compound. Solution Map: g Fe, O Atomic mass (g/mol): O 16.00, Fe 55.85 mass of O = 1.205 – 0.872 = 0.333 g 0.0156 mol Fe 0.0208 mol O

  43. Find: 0.0156 mol Fe, 0.0208 mol O Example: One 1.205-g sample of iron-oxygen compound contains 0.872 g iron. Find the empirical formula of this compound. (Contd.)

  44. Practice: A 1.144-g sample of manganese ore contains 0.320 g oxygen. Find its empirical formula.

  45. mol C, H, O mol ratio empirical formula Example: Empirical formula from mass%Find the empirical formula of aspirin with the given mass percent composition: 60.00% C, 4.48% H, 35.53% O g C, H, O

  46. Assuming 100 (exact) grams of compound aspirin, mass of each element in grams equals percentage. Mass C = 60.00 g Mass H = 4.48 g Mass O = 35.52 g Given: 60.00% C, 4.48% H, 35.52% O 1 mol C = 12.01 g; 1 mol H = 1.008 g; 1 mol O = 16.00 g Example:Find the empirical formula of aspirin with the given mass percent composition. C4.996H4.44O2.221 C9H8O4

  47. Note: Empirical formula comes from REDUCED Molecular formula The “Molar mass” based on Empirical formula is multiplied by an integer n should give the Molecular Formula From Empirical Formula to Molecular Formula

  48. Example: Find Empirical Formula and Molecular Formula • Vitamic C has the following mass percent composition: C 40.91%, H 4.58%, O 54.41%. Find the empirical formula. • Ans: C3H4O3 • The molar mass of vitamin C is 176.1 g/mol. Find the molecular formula of vitamin C. • Ans: solve n = 2, so molecular formula C6H8O6

  49. Same Empirical Formula, Different Molecular Formula!

  50. Answer key: Molar Mass (all in g/mol) • Sulfuric acid: 98.09 g/mol • Aluminum sulfate: 342.17 g/mol • Ammonium chlorite: 181.70 g/mol • Hydrobromic acid: 80.91 g/mol • Nickel(II) phosphate: 366.01 g/mol

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