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Unit 4 – The Mole

Unit 4 – The Mole. Honors Chemistry Part 1. WHAT IS A MOLE?. 602214199000000000000000 6.02 x 10 23. Mole Facts. 6.02 X 10 23  Pennies: Would make at least 7 stacks that would reach the moon. 6.02 X 10 23  Watermelon Seeds: Would be found inside a melon slightly larger than the moon.

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Unit 4 – The Mole

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  1. Unit 4 – The Mole Honors Chemistry Part 1

  2. WHAT IS A MOLE? • 602214199000000000000000 • 6.02 x 1023

  3. Mole Facts • 6.02 X 1023  Pennies: Would make at least 7 stacks that would reach the moon. • 6.02 X 1023 Watermelon Seeds: Would be found inside a melon slightly larger than the moon. • 6.02 X 1023  Blood Cells: Would be more than the total number of blood cells found in every human on earth. • 1 Liter bottle of Water contains 55.5 moles H20

  4. Definition of Mole • The amount of atoms in 12.0 grams of Carbon 12 (6.02 x 1023 atoms known as Avogadro’s number). • A sample of any element with a mass equal to that element's atomic weight (in grams) will contain precisely one mole of atoms (6.02 x 1023 atoms).

  5. Molecular Mass • The sum of the masses of all the atoms in a molecule of a substance CaCO3 1 atom of Ca = 40.08 amu 1 atom of C = 12.00 amu 3 atoms of O = 48.00 amu 100.08 amu } Add these

  6. Formula Mass (Molar Mass) • The mass of 1 mole (in grams) • Equal to average atomic mass but the unit is grams 1 mole of C atoms = 12.01 g 1 mole of Na atoms = 22.99 g 1 mole of Cu atoms = 63.55 g

  7. Example Problem Find the mass of 1 mole of KAl(SO4)2● 12H2O 1 K = 39.10 1 Al = 26.98 2 (SO4) 2(32.06 + ((16.00 x 4))=192.12 12 H2O 12(2.02 + 16.00) =216.24 Mass of 1 mole = 474.44 g/mol

  8. Try These: • Find the molecular mass for these : • HNO3 • CO2 • Find the molar mass for these compounds: • C6H10O5 • H2SO4

  9. The Mole • 1 mole of gas always contains 6.02 x 1023 molecules of that gas • 1 mole Cl2 gas = 6.02 x 1023 molecules of Cl2 • 1 mole NO2 gas = 6.02 x 1023molecules of NO2 • 1 mole CO gas = 6.02 x 1023 molecules of CO • 1 mole CO2 gas = 6.02 x 1023molecules of CO2

  10. The Mole • Also applies to other particles! (not only molecules in a gas) • 1 mole C = 6.02 x 1023 C atoms • 1 mole H2O = 6.02 x 1023H2O molecules • 1 mole NaCl = 6.02 x 1023NaCl formula units • 1 mole of Na+ = 6.02 x 1023 Na+ ions • 1 mole of Cl- = 6.02 x 1023Cl– ions

  11. Avogadro’s Number • We can use Avogadro’s # as a conversion factor: 1 mole 6.02 x 1023 particles Or 6.02 x 1023 particles 1 mole • Note that a particle could be an atom, molecule, formula unit, or ion !

  12. Example Problems • How many molecules are in 3.5 moles of H2O? • How many moles are present in 4.65 molecules of NO2?

  13. Mass and Mole Relationship • 1 mole of any substance = the molar mass of that substance (in grams) • Find the number of moles present in 56.7 g of HNO3. 56.7 g HNO3 1 mole HNO3 63.01 g HNO3

  14. Example Problems • Find the number of grams present in 4.5 moles of C6H10O5. • Find the number of moles present in 12.31 g of H2SO4. • How many molecules are in 4.5 grams of NaCl?

  15. Gas Volumes and Molar Mass • Avogadro’s Law • Equal volumes of gases under the same conditions of temperature and pressure contain equal numbers of molecules • 1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 liters • Standard temperature: 0ºC or 273K • Standard pressure: 1 atm or 101.325 kPa

  16. Gas Volumes and Molar Mass • 32.00 g O2 = 1 mole = 22.4 L • 2.02 g H2 = 1 mole = 22.4 L • 44.01 g CO2 = 1 mole = 22.4 L

  17. ExampleProblems • How many liters are present in 5.9 moles of O2? • How many liters are present in 3.67 moles of CO2? • How many atoms of O are present in 78.1 g of O2?

  18. Percent Composition • Finding what percent of the total weight of a compound is made up of a particular element • Formula for calculating % composition: Total mass of the element in the compound Total formula mass X 100

  19. Example Problem: • Calculate the % composition of calcium in Ca(OH)2.

  20. Example Problems • Find the percentage composition of a compound that contains 1.45 g of carbon, 4.23 g of sulfur, and 1.00 g of hydrogen in a 6.68 g sample. • 21.7% carbon • 63.3% sulfur • 15.0% hydrogen

  21. Example Problem • A sample of an unknown compound with a mass of 5.00 grams is made up of 75% carbon and 25% hydrogen. What is the mass of each element? • 3.75 g of carbon • 1.25 g of hydrogen

  22. Formulas • Empirical Formula - expresses the smallest whole number ratio of atoms present • E.g. CH2O • Ionic formulas are always empirical formulas • Molecular Formula - states the actual number of each kind of atom found in one moleculeof the compound. • E.g. C6H12O6

  23. Empirical Formula • Determine mass in grams of each element • Calculate the number of molesof each • Divide each by the smallest number of moles to obtain the simplest whole number ratio • If whole numbers are not obtainedin step 3, multiply all by the smallest number that will give whole numbers

  24. Empirical Formula Remember this: • Percent to mass • Mass to mole • Divide by small • Multiply ‘till whole

  25. Example Problem • Given that a compound is composed of 60.0% Mg and 40.0% O, find the empirical formula.

  26. Example Problem • A compound is found to contain 68.5% carbon, 8.63% hydrogen, and 22.8% oxygen. The molecular weight of this compound is known to be approximately 140.00 g/mol. Find the empirical and molecular formulas.

  27. Hydrates • Ionic compounds • Water is bonded to the crystal structure • Ex: CuSO3 • 7H2O • The percentage of water in a hydrate can easily be calculated using the formula: % Water = Mass of water x 100 Mass of hydrate

  28. Example Problem • What is the percentage of water in CuSO3 • 7H2O? • A 3.5 g sample of a hydrate is heated and only 1.7 g of the anhydrous salt remain. What is the percentage of water?

  29. Law of Definite Proportions • Formulas give the numbers of atoms or moles of each element • Always a whole number ratio • 1 molecule NO2 : 2 atoms of O for every 1 atom of N • 1 mole of NO2 : 2 moles of O atoms to every 1 mole of N atoms

  30. Law of Multiple Proportions • When any two elements, A and B, combine to form more than one compound, the different masses of B that unite with a fixed mass of A have a small whole-number ratio • Example: • In H2O, the proportion of H:O = 2:16 or 1:8 • In H2O2, H:O is 2:32 or 1:16

  31. How Do We Determine Concentration? • Molarity • Molality

  32. How do we make solutions? M1 = m1/V1 rearrange to M1V1 = m1 M2 = m2/V2 rearrange to M2V2 = m2 If m1=m2 then, M1V1 = M2V2

  33. M1V1 = M2V2 • M1 = concentration of the first solutionV1 = volume of the first solutionM2 = concentration of the second solutionV2 = volume of the second solution • Let's consider a sample problem: • You have 1 L of a 0.125 M aqueous solution of table sugar.  You want to dilute the solution to 0.05 M.  What do you do?

  34. Dilution • To solve the problem, you simply plug in the three numbers you know: • (0.125 M) (1 L) = (0.05 M) V2 • 2.5 L = V2 • Using the equation, you determine that the volume of the diluted solution should be 2.5 L.  • So we simply add enough water to the first solution so that the solution's volume becomes 2.5 L.

  35. What is Saturation? • A solution is saturated if it contains as much solute as can possibly be dissolved under the existing conditions of temperature and pressure • Unsaturated: Has less than maximum amount of solute that can be dissolved • Supersaturated: Contains more than maximum (How can this happen?)

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