The Mole x 10 23


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1 The Mole x 10 23
2 Background: atomic masses Look at the atomic masses on the periodic table. What do these represent? E.g. the atomic mass of Carbon is (atomic # is 6) We know there are 6 protons and 6 neutrons in C12 Protons and neutrons have roughly the same mass of about 1.66 x1024 grams Set this mass equal to 1 amu (atomic mass units). Carbon12 thus has a mass of 12 amu. The atomic mass shown on the Periodic Table is a weighted average of masses of all isotopes of an element But more useful to associate atomic mass with a mass in grams.
3 The Mole Scientist set out to develop a basic unit of measurement to convert from atomic mass to grams. Used Carbon12 to set the standard. Scientists set one mole = the number of atoms of C12 in 12 grams of C12. Experiments show that there are x carbon atoms in 12 grams of Carbon x things is a mole of any thing! Also known as Avogadro s number
4 Who wants to be a Mollionaire? Q: how long would it take to spend a mole of $1 bills if they were being spent at a rate of one billion per second? A: $ 6.02 x bills x sec/$1,000,000,000 bills = 6.02 x payments = 6.02 x seconds 6.02 x seconds / 60 = x minutes x minutes / 60 = x hours x hours / 24 = x 10 9 days x 10 9 days / = x 10 7 years A: It would take 19 million years
5 Molar mass The mass of one mole of something is called its molar mass. Since one atom of C12 = 12 amu and one mole of C12 = 12 grams, we can use atomic masses to directly convert from amu/atom to grams/mol. For an element, molar mass = Atomic Mass from Periodic Table, but use g/mole rather than amu Example: Lithium s atomic mass = 6.94 amu Thus, 1 mole of Li = 6.94 g Li This is expressed a molar mass of 6.94 g/mol Sometimes referred to as gram formula mass
6 Molar mass What are the following element molar masses? S =32.06 g/mol Ag = g/mol For a compound, molar mass = sum of molar masses of each element times the number of atoms of that element in the compound. CO 2 = g/mol C x 1 = x 1 = O x 2 = x 2 = 12.01g/mol 32.00g/mol g/mol
7 Molar Mass Calculations Determine the mass in grams of: mol of Au mol of Zn Determine the number of moles of: g of Cu g of Na 2.0 mol Au x g/mol = 3.9 x 10 2 g 4.37 mol Zn x 65.4 g/mol = 2.85 x 10 2 g 254g Cu x mol/63.5g = 4.0 mol Cu 12g Na x mol/23.0g = 0.52 mol Na
8 Cu 3 (BO 3 ) 2 Molar mass = g/mol Cu x 3 = x 3 = g/mol B x 2 = x 2 = g/mol O x 6 = x 6 = g/mol g/mol Calculate molar masses (to 2 decimal places) CaCl g/mol (Ca x 1, Cl x 2) (NH 4 ) 2 CO g/mol (N x 2, H x 8, C x 1, O x 3) O g/mol (O x 2) C 6 H 12 O g/mol (C x 6, H x 12, O x 6)
9 Comparing sugar (C 12 H 22 O 11 ) & H 2 O Same volume? mass? # of moles? # of molecules? # of atoms? 1 gram each No, they have different densities. Yes, that s what grams are! No, they have dif. molar masses No, they have dif. molar masses No 1 mol each No, molecules have dif. sizes. No, molecules have dif. masses Yes. Yes (6.02 x in each) No, sugar has more (45:3 ratio)
10 Converting between grams and moles If we are given the # of grams of a compound we can determine the # of moles, & viseversa In order to convert from one to the other you must first calculate the molar mass (g/mol) Then use dimensional analysis to convert: moles to grams: mol x g/mol = g grams to moles: g x mol/g = mol This can be represented in an equation triangle mol g x g/mol
11 Converting between grams and moles First: Determine the compound s molar mass (g/mol) using the Periodic Table. mol g x g/mol Formula g/mol g HCl mol (n) 0.25 H 2 SO NaCl Cu Equation g= g/mol x mol mol= g x mol/g g= g/mol x mol mol= g x mol/g
12 Empirical and molecular formula Consider NaCl (ionic) vs. H 2 O 2 (covalent) Na Cl Na Cl Cl Na Cl Na Chemical formulas are either simplest (a.k.a. empirical ) or molecular (all bonded atoms). Ionic compounds are always expressed as the simplest ratio of the ions (formula units like NaCl or Li 2 O). Thus ionic formulas are always empirical. Covalent compounds can be shown as either molecular formulas (e.g. H 2 O 2 ) or empirical (e.g. HO)
13 % composition Percent Composition: Identifies the elements present in a compound as a mass percent of the total compound mass. The mass percent is obtained by dividing the mass of each element by the total mass of a compound and converting to percentage. Example problem: CH 2 O 1 mole of CH 2 O = 1 mole C, 2 moles H and 1 mole O Total mass = ( )g/mol = 30.03g/mol Percent composition: %C = 12.01/30.03 x 100% = 39.99% %H = 2.02/30.03 x 100% = 6.73% %O = 16.00/30.03 x 100% = 53.28%
14 Pathway to figure out empirical formula
15
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