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(I have now prefaced the slides for my talk on Sept. 5 with the solution to the atomic mass problem from Sept. 3. The lecture on moles follows those slides.).
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(I have now prefaced the slides for my talk on Sept. 5 with the solution to the atomic mass problem from Sept. 3. The lecture on moles follows those slides.)
Problem: Given the exact masses, 35Cl = 34.969mu and 37Cl = 36.966 u, and the average atomic mass of chlorine = 35.453 u, find the abundances of the two isotopes.
Problem: Given the exact masses, 35Cl = 34.969 u and 37Cl = 36.966 u, and the average atomic mass of chlorine = 35.453 u, find the abundances of the two isotopes. Ave. atomic mass = a1m1 + a2m2 + . . .
Problem: Given the exact masses, 35Cl = 34.969 u and 37Cl = 36.966 u, and the average atomic mass of chlorine = 35.453 u, find the abundances of the two isotopes. Ave. atomic mass = a1m1 + a2m2 + . . . 35.453 u = a1(34.969 u) + a2(36.966 u)
Problem: Given the exact masses, 35Cl = 34.969 u and 37Cl = 36.966 u, and the average atomic mass of chlorine = 35.453 u, find the abundances of the two isotopes. Ave. atomic mass = a1m1 + a2m2 + . . . 35.453 u = a1(34.969 u) + a2(36.966 u) Also: a1 + a2 = 1
Problem: Given the exact masses, 35Cl = 34.969 u and 37Cl = 36.966 u, and the average atomic mass of chlorine = 35.453 u, find the abundances of the two isotopes. Ave. atomic mass = a1m1 + a2m2 + . . . 35.453 u = a1(34.969 u) + a2(36.966 u) Also: a1 + a2 = 1 So 35.453 u = (1 – a2)(34.969 u) + a2(36.966 u)
So 35.453 u = (1 – a2)(34.969 u) + a2(36.966 u) 35.453 u = 34.969 u – 34.969 u(a2) + 36.966 u(a2) 35.453 u = 34.969 u – 34.969 u(a2) + 36.966 u(a2) 35.453 u = 34.969 u + 1.997 u(a2) a2 0.2424 and a1 = 1 – 0.2424 = .7576
How to Count Atoms(when they are really really small) Review: 1 12C atom = 12.0000 u = 1.9926x 10-23 g (from mass spectroscopy experiments) So
Goal: a number that is capable of expressing numbers of atoms in convenient terms. H and O react in simple numbers of atoms to form water (H2O): 2 H : 1 O There is not a simple relation between the masses that react. Yet – masses are what we can readily measure in a laboratory. To get a quantity capable of expressing numbers of atoms, Define: 1 mole = number of atoms in 12.0000 g of 12C.
We now have enough information to calculate the number of atoms in one mole of carbon: But this is also the number of atoms in one atomic mass, expressed in grams, of any element. This important number is known as Avogadro’s Number, NA.
Both a mole and a dozen express quantities of things by a collective number. A dozen eggs weighs more than a dozen ping-pong balls because each individual unit is heavier.
16.0 amu 12.0 amu O C 192.0 amu in carton 144.0 amu in carton (A dozen oxygen atoms weigh more than a dozen carbon atoms) Take the same number of atoms as at the left. The total mass will be 16.0 grams Take enough cartons to have 6.02 x 1023 atoms The total mass will be 12.0 gram Moral: a mole of oxygen weighs more than a mole of carbon because each individual oxygen atom weighs more.
So now, atomic masses have two interpretations: 1 atom of C weighs 12.0000 u 1 mole of C weighs 12.0000 g Some conversion factors: 1 mole of C = 12.00 g 1 mole of C = 6.02 x 1023 atoms C
Now we can calculate the mass of any atom: 1 atom of U = 238 u
Now we can calculate the mass of any atom: 1 atom of U = 238 u Find the number of carbon atoms in 3.0 g C.
A mole road-map Atomic mass NA atoms Moles grams
This can be applied to compounds, which have a molecular mass (in amu) or a molar mass (in grams). For instance: In one molecule of H2O, there are 2 H atoms = 2 x 1 amu = 2 u 1 O atom = 1 x 16 amu = 16 u 18 u 1 mole of H2O has 1 mole of O atoms and 2 moles of H atoms. 1 mole of water weighs 18 g and has 6.02 x 1023molecules
For Ca(OH)2: 1 Ca = 40.1 u 2 O = 32.0 u 2 H = 2.0 u 74.1 u 1 mole of Ca(OH)2 has a mass of 74.1 g Ca(OH)2 is an ionic compound so we can’t talk about molecules. Sometimes the simplest formula – which indicates only the ratio of the ions – is referred to as a formula unit.