The Mole. 6.022 x 10 23



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Transcription:

The Mole 6.022 x 10 23

Background: atomic masses Look at the atomic masses on the periodic table. What do these represent? E.g. the atomic mass of Carbon is 12.01 (atomic # is 6) We know there are 6 protons and 6 neutrons in C-12 Protons and neutrons have roughly the same mass of about 1.66 x10-24 grams Set this mass equal to 1 amu (atomic mass units). Carbon-12 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.

The Mole Scientist set out to develop a basic unit of measurement to convert from atomic mass to grams. Used Carbon-12 to set the standard. Scientists set one mole = the number of atoms of C-12 in 12 grams of C-12. Experiments show that there are 6.022 x 10 23 carbon atoms in 12 grams of Carbon-12. 6.022 x 10 23 things is a mole of any thing! Also known as Avogadro s number

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 10 23 bills x sec/$1,000,000,000 bills = 6.02 x 10 14 payments = 6.02 x 10 14 seconds 6.02 x 10 14 seconds / 60 = 1.003 x 10 13 minutes 1.003 x 10 13 minutes / 60 = 1.672 x 10 11 hours 1.672 x 10 11 hours / 24 = 6.968 x 10 9 days 6.968 x 10 9 days / 365.25 = 1.908 x 10 7 years A: It would take 19 million years

Molar mass The mass of one mole of something is called its molar mass. Since one atom of C-12 = 12 amu and one mole of C-12 = 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

Molar mass What are the following element molar masses? S =32.06 g/mol Ag = 107.9 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 = 44.00 g/mol C x 1 = 12.01 x 1 = O x 2 = 16.00 x 2 = 12.01g/mol 32.00g/mol 44.00 g/mol

Molar Mass Calculations Determine the mass in grams of: 1. 2.0 mol of Au 2. 4.37 mol of Zn Determine the number of moles of: 1. 254 g of Cu 2. 12 g of Na 2.0 mol Au x 196.9 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

Cu 3 (BO 3 ) 2 Molar mass =308.27 g/mol Cu x 3 = 63.55 x 3 = 190.65 g/mol B x 2 = 10.81 x 2 = 21.62 g/mol O x 6 = 16.00 x 6 = 96.00 g/mol 308.27 g/mol Calculate molar masses (to 2 decimal places) CaCl 2 110.98 g/mol (Ca x 1, Cl x 2) (NH 4 ) 2 CO 3 96.11 g/mol (N x 2, H x 8, C x 1, O x 3) O 2 32.00 g/mol (O x 2) C 6 H 12 O 6 180.18 g/mol (C x 6, H x 12, O x 6)

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 10 23 in each) No, sugar has more (45:3 ratio)

Converting between grams and moles If we are given the # of grams of a compound we can determine the # of moles, & vise-versa 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

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 36.46 9.1 mol (n) 0.25 H 2 SO 4 98.08 53.15 0.5419 NaCl Cu 58.44 63.55 207 1.27 3.55 0.0200 Equation g= g/mol x mol mol= g x mol/g g= g/mol x mol mol= g x mol/g

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)

% 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 = (12.01 + 2.02 + 16.00)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%

Pathway to figure out empirical formula