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Ch 6 Student Notes

Ch 6 Student Notes - 1 Chapter 6 Chemical Calculations...

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1 Chapter 6 Chemical Calculations Submicroscopic Macroscopic

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2 Chapter Outline 1. Formula Masses (Ch 6.1) 2. Percent Composition (supplemental material) 3. The Mole & Avogadro’s Number (Ch 6.2) 4. Molar Mass (Ch 6.3) 5. Chemical Formula & Mole Calculations (Ch 6.4 & 6.5) 6. Empirical & Molecular Formula (supplemental material & Lab Exp 8) 7. Chemical Equations & Stoichiometric Calculations (Ch 6.6, 6.7, 6.8) 8. Heat of Reaction (supplemental material) 9. Percent Yield (supplemental material) 10. Limiting Reagent (supplemental material)
3 1. Formula Masses (Ch 6.1) Molecular mass/weight – sum of atomic masses of the atoms in a molecule. Example: ________ M.W. = 46.0 amu C: 2 x 12.0 amu = 24.0 amu H: 6 x 1.0 amu = 6.0 amu O: 1 x 16.0 amu = 16.0 amu 46.0 amu

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4 Formula mass/weight – sum of atomic masses of atoms in an ionic substance (formula unit, NOT a molecule). Example: Ammonium sulfide _____________ F.W. = 68.1 amu N: 2 x 14.0 amu = 28.0 amu H: 8 x 1.0 amu = 8.0 amu S: 1 x 32.1 amu = 32.1 amu 68.1 amu Due to its offensive smell, ammonium sulfide it is the active ingredient in a variety of foul pranks, including the common stink bomb.
5 2. Percent Composition (supplemental material) % Composition = # of g of each element in 100 g of a compound = mass of element x 100 total mass Example: Ammonium sulfide (NH 4 ) 2 S F.W. = 68.1 amu % N = 2 x 14.0 amu N x 100 = 41.1 11 % N 68.1 amu (NH 4 ) 2 S % H = 8 x 1.0 amu H x 100 = 11.7 47 % H 68.1 amu (NH 4 ) 2 S % S = 32.1 amu S x 100 = 47.1 36 % S 68.1 amu (NH 4 ) 2 S

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6 3. The Mole & Avogadro’s Number (Ch 6.2) 1 dozen = 12 objects 1 ream = 500 sheets 1 mole = _________ objects 6.02 21415 x 10 23 Avogadro’s Number
7 Sample problem: How many He atoms are in 2.55 moles of He ? information x _________ = information given factor sought 2.55 mole He x 6.02 x 10 23 He atoms = __________ He atoms 1 mole He

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8 4. Molar Mass (Ch 6.3) 1 mole 12 C = exactly 12 g 12 C demo sample Molar mass of 12 C = __________ We now have three conversion factors to relate moles, number of atoms, and mass of Carbon: 6.02 x 10 23 C atoms 12 g C 1 mole 1 mole 6.02 x 10 23 C atoms 12 g C
9 Figure 6.5 1 mole of S, Zn, C, Mg, Pb, Si, Cu, Hg (start counterclockwise from yellow S and Hg in center) 12 C is our standard and when we compare it to other elements, we find that there are Avogadro’s number of atoms of any element in a sample whose mass in grams is numerically equal to its atomic weight. Mg = 24.31 amu 1 mole Mg = 24.31 g Mg = 6.02 x 10 23 atoms Mg Pb = ________ amu 1 mol Pb = _______ g Pb = 6.02 x 10 23 atoms Pb

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10 Now we can do the same for ionic compounds as well as for molecules because the molar mass is the mass (in grams) of a substance that is numerically equal to the substance’s formula mass. Ammonium sulfide (NH 4 ) 2 S formula mass = 68.1 amu 68.1 g (NH 4 ) 2 S = 1 mole (molar mass or formula weight, F.W. : ____________) 68.1 g (NH 4 ) 2 S = 6.02 x 10 23 formula units of (NH 4 ) 2 S Carbon dioxide CO 2 formula mass = 44.01 amu 44.01 g CO 2 = 1 mole (molar mass or molecular weight, M.W. : _____________) 44.01 g CO 2 = 6.02 x 10 23 molecules of CO 2
11 Sample Problem: If 7.50 moles of ammonia, NH 3 , are required for a certain experiment, what mass of ammonia is needed?

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Ch 6 Student Notes - 1 Chapter 6 Chemical Calculations...

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