Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2015

Lecture Notes 8: 4 February

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Stoichiometry, cont.

Determination of Atomic Mass

We want to be able to figure out the atomic mass of a sample with a particular isotopic composition.

Example: Cu occurs as an isotopic mixture of 69.09% 63Cu (mass = 62.93 amu) and 30.91% 65Cu (64.93 amu). What is the atomic mass of copper in this sample?

Assume the sample consists of 1 atom for convenience, then

(0.6909 atoms)(62.93 amu/atom) + (0.3091 atoms)(64.93 amu/atoms) =

43.478 amu + 20.070 amu = 63.558 amu for 1 atom

= 63. 558 amu/atom

How about sig figs? 1 is a count, therefore exact. The two multiplications each have 4 sig figs so the calculations each have 4 sig figs (note I keep one extra, that is 5 sig figs, in the calculations to avoid rounding errors.) . For the addition we use the add/subt. rule and look at decimal place, for our four sig figs the hundredth's place is then the sig fig (again, during calculation its best to keep one extra sig fig to avoid rounding errors). The final answer then has 4 sig figs:

= 63. 56 amu/atom

An example of the reverse problem can be found on the posted Final, number II.3.

 

Most samples of matter consist of combinations of atoms or ions to give compounds characterized by molecules or formulae. We are thus interested in molecular masses, formula masses etc.

Examples:

0.25 mole x (39.10 g/mole + 79.90 g/mole) = 29.75 g = 3.0 x 101g

so weight = 0.25 mole x (6 [12.01]g/mole + 12 x [1.008 g/mole] + 6 [16.00] g/mole) = 45.04 g = 45 g

so MW = (12 [12.01]g/mole + 22 x [1.008 g/mole] + 11 [16.00 ] g/mole) = 342.30 g/mole

and moles = (23.5 g)/(342.30 g/mole) = 6.865 x 10-2mole = 6.87 x 10-2mole.

 

First need FW = (39.10 g/mole + 79.90 g/mole) = 119.00 g/mole

then find moles = (1.00 g)/(119.00 g/mole) = 8.403 x10-3mole

then multiply by Avogadro's Number: (8.403 x10-3mole) x (6.022 x 1023 atoms/mole) = 5.061 x1021 KBr "units"

But - there are two atoms/formula, So 1.01 x1022 atoms

(Of course there are no atoms actually present, just ions. However, all the parts for the atoms are there, so we still ask the question in terms of atoms.)

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© R A Paselk

Last modified 6 February 2015