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Unit III Chemical Composition (i.e. The Mole )

Unit III Chemical Composition (i.e. The Mole ). Atomic Masses. Atomic masses use Carbon 12 ( 12 C ) as the standard Calculated with the aide of a mass spectrometer. This process is also known as gas chromatography

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Unit III Chemical Composition (i.e. The Mole )

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  1. Unit IIIChemical Composition(i.e. The Mole)

  2. Atomic Masses • Atomic masses use Carbon 12 (12C) as the standard • Calculated with the aide of a mass spectrometer

  3. This process is also known as gas chromatography • As gas chromatographcounts the number of particles present in a given sample • Produces a graph like the one above • With this information percentages can be calculated and formulas can be determined

  4. Atomic mass is the average of all the naturally occurring isotopes of that element • Carbon = 12.011

  5. The Mole • Abbrev. mol • 1 mol = # C atoms in 12 g of pure 12C • Avogadro’s number • Equal to 6.022 x 1023atoms in 1 mol C • Named in honor of the Italian chemist Amadeo Avogradro (1776-1855) I didn’t discover it. Its just named after me!

  6. 1 mol = 6.022 x 1023 atoms = molar mass (g) = 22.4 L

  7. To covert amongst mol, atoms, grams, & liters, use the following equivalencies: • 1 mol = 6.022 x 1023 = molar mass = 22.4 L atoms (grams) • Or use ½ of the following chart:

  8. Keys to Use: 1. Only use 1 side of the chart 2. Proceed from one shaded box to another shaded box 3. When following the arrows, perform the indicated function 4. When going against the arrows, perform the opposite function

  9. Molar Mass • The molar mass is determined by summing the masses of the component atoms • Example: What is the molar mass of MgCO3 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g

  10. Molar Calculations • Example 1: How many grams of lithium are in 3.50 mol of lithium? • 3.50 mol Li→ g Li • (3.50 • 6.94) / 1 = 24.29 • (3 SF’s in original problem) → Round • 24.3 → SSN → 2.43 x 101 • Don’t forget the units → g Li 3.50 mol Li 6.94 g Li 2.43 x 101 g Li = 1 mol Li

  11. Example 2: How many mol are in 98.2 g of NaCl? • 98.2 g NaCl→ mol NaCl • (98.2 • 1) / 58.44 = 1.680355921 • (3 SF’s in original problem) → Round • 1.68 → SSN → 1.68 x 100 • Don’t forget the units → mol NaCl 98.2 g NaCl 1 mol NaCl 1.68 x 100 mol NaCl = 58.44 g NaCl

  12. Example 3: How many atoms are in 411 g of calcium phosphate? • 411 g Ca3(PO4)2→ atoms Ca3(PO4)2 • (411 • 6.022 x 1023) / 310.18 = 7.9793 x 1023 • (3 SF’s in original problem) → Round • 7.98 x 1023 → SSN → 7.98 x 1023 • Don’t forget the units → atoms Ca3(PO4)2 6.022 x 1023atoms Ca3(PO4)2 411 g Ca3(PO4)2 7.98 x 1023atoms Ca3(PO4)2 = 310.18 g Ca3(PO4)2

  13. Example 4: How many liters of dioxygen heptioide gas does 7.5 x1024 atoms occupy? • 7.5 x 1024 atoms O2I7→ L O2I7 • (7.5 x 1024 • 22.4) / 6.022 x 1023 = 278.977084 • (2 SF’s in original problem) → Round • 280 → SSN → 2.8 x 102 • Don’t forget the units → L I2O7 7.5 x 1024atoms O2I7 22.4 L I2O7 2.8 x 102 L O2I7 = 6.022 x 1023atoms O2I7

  14. Percent Composition of Compounds • Mass Percent = (Part / Whole)  100% • Calculate the percent composition of magnesium carbonate (MgCO3) • Molar Mass = 84.32 g • 24.31 g + 12.01 g + 3 (16.00) Mg = (24.31 / 84.32)  100 = 28.83 % C = (12.01 / 84.32)  100 = 14.24 % O = (48.00 / 84.32)  100 = 56.93 % 100 %

  15. Determining the Formula of a Compound • There are 2 basic formulas: 1.Empirical • Simple – Can not be simplified • Examples: H20, MgCl2 2.Molecular • Complex – Can be simplified • Examples: H8O4, Mg3Cl6 • Molecular Formula = (Empirical Formula )n • Example: (H20)4 = H8O4 • n = molar mass / Empirical Formula Mass

  16. Note: % are grams (if based on 100) • Adipic acid contains 49.32% C, 6.85% H, & 43.84% O by mass. What is the E.F.? • C: (49.32/12.01)= 4.10 / 2.74 = 1.50  2 = 3 • H: (6.85 /1.01) = 6.78 / 2.74 = 2.47  2 = 5 • O: (43.84 /16.00)= 2.74 / 2.74 = 1  2 = 2 • Answer: C3H5O2 Empirical Formula Determination # of atoms Divide by smallest # of mol Moles Mole Ratio X by integer to obtain a whole #

  17. Empirical Formula DeterminationSample Problems • Sample Problem 1: • .6884 g of lead combined with .2356 g of Cl to form a binary compound. Calculate the empirical formula. • Sample Problem 2: • A compound’s percent by mass is as follows: Copper = 33.88%,Nitrogen = 14.94%, and Oxygen = 51.18%.Determine the empirical formula of the compound.

  18. Problem: • The E.F. for adipic acid is C3H5O2 • The molar mass is 146 g/ mol • Steps: 1. Determine the E.F. mass of C3H5O2 • (3  12.01) + (5  1.01) + (2  16.00) = 73.08 g 2. Find the n factor • n = molar mass / E.F. mass • n = 146 / 73.08 = 2 3. M.F. = (E.F.)n = (C3H5O2)2 = C6H10O4 Molecular Formula Determination

  19. Molecular Formula DeterminationSample Problems • Sample Problem 1: • Calculate the M.F. of a compound with the E.F. CH2O and a molar mass of 150 g/mol. • Sample Problem 2: • A gas is composed of 71.75% Cl, 24.27% C, and 4.07% H (by mass) & its molar mass is 98.96 g/mol. What is the M.F.?

  20. Formulas of Hydrates • A hydrate is a compound that has a specific number of water moleculesbound to its atoms • Written with each formula unit following a dot • Examples: 1. Calcium chloride dihydrate → CaCl2 • 2 H2O 2. Magnesium sulfateheptahydrate→ MgSO4 • 7 H2O • Samples: 1. Sodium cabonate decahydrate (molar mass ?) 2. Barium hydroxide octahydrate (molar mass ?)

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