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## Lecture Notes:

• Atomic Mass Units
• The mass numbers we have talked about relate to atomic mass units (amu)
• An amu is based on a relative scale of masses which assigns the isotope Carbon-12 a mass of exactly 12 amu
• Hydrogen-1 has a mass of 1 amu or 1/12 of a carbon atom; Copper-63 has a mass of 62.94 amu
• The masses listed in the periodic table are tabulated in amu; what you see is actually an average atomic mass
• An average atomic mass can be calculated if you know the percentage abundance of the isotopes of an element and each isotope’s atomic mass
• Moles and Mass
• Chemists do not deal with individual atoms and molecules because nobody has tweezers that small
• One Carbon-12 atom has a mass of 1.99 x 10-23 kg; a mole of Carbon-12 atoms on the other hand has a handy mass of 0.012 kg (12 g)
• The mole is an extraordinarily large number: 6.022 137 x 1023 (OK, without using scientific notation it is
602 213 700 000 000 000 000 000...woulnd’t you rather use the scientific notation?)
• Just how big is this number? Well, if you counted to a million once every second for the age of the universe from its beginning until now, you would have just managed to count up to one mole!
• You can have a mole of any entity or object: there are 6.022 x 1023 doughnuts in a mole of doughnuts; there are 6.022 x 1023 electrons in a mole of electrons; there are 6.022 x 1023 rabbits in a mole of rabbits
• Besides being impressively large, the mole has an incredible utility: you can use it to get the same number of different atoms or molecules regardless of their individual atomic/molecular masses (trust me, this is important)
• If you express Avogadro’s number in SI units it would be 6.022 137 x 1023 1/mol; the unit is in the denominator because it means that there are that many entities per mole
• In answer to your question (if you haven’t already asked) the word mole comes from German: Molekül; the German abbreviation is ‘mol’ and so is the English abbreviation; we add an ‘e’ so that everybody says it correctly
• Calculations
• Average Atomic Mass
Multiply the mass of each isotope times its percentage abundance and add the products together.
For Nitrogen: nitrogen-14’s mass is 14.003074, nitrogen-15’s mass is 15.000108
(14.003074)(0.9963) + (15.000108)(0.0037) = 14.007
• Number of Entities
Given a number of moles you can calculate how many items you have.
If you have 2 moles of oxygen molecules (O2) you can find out how many oxygen molecules you have by multiplying by Avogadro’s number (N = 6.022 x 1023). You have 12.044 x 1023 O2 molecules.
Given a number of entities you can calculate how many moles you have by dividing by Avogadro’s number.
If you have 3.011 x 1023 pieces of chalk, you have 0.5 moles of pieces of chalk.
You can use the factor-label method to solve problems of this type.
• Molar mass of an element
The molar mass of an element is easy. It is just its average atomic mass expressed in g instead of amu. For example, the mass of one mole of oxygen atoms is 15.9994 g/mol. Note that the unit shows that for 1 mole there are approx. 16 g. Be careful, because the mass of one mole of oxygen gas is not the same...
• Molar mass of a compound
To find the mass of one mole of a compound you just use the molar masses of the elements it is composed of, multiplied by how many atoms of that element are in the compound.
How to calculate the molar mass of Al(NO3)3
(1 x 26.98 g/mol) + (3 x 14.007 g/mol) + (9 x 16.00 g/mol) = 213.00 g/mol
213.00 grams is the mass of one mole of aluminum nitrate and contains 6.022 x 1023 entities of Al(NO3)3
• Moles from a Mass
To find the number of moles given a mass of a substance you need to know its chemical formula. From the formula you can find its molar mass. From there this calculation is just a matter of setting up a conversion.
You have 32.0 g of nitrogen gas in a bottle. Nitrogen gas has the formula N2 and has a molar mass of 2 x 14.00674 g/mol or 28.0135 g/mol.
```          1 mol
32.0 g x -------    =  1.14 mol N2
28.0135 g
```
• Percent Composition
There are three steps to finding the percent composition of a compound from its formula:
1. figure out the molar mass from the formula
2. figure out the grams each atom contributes by multiplying the atomic weight by the subscript
3. divide the answer for each atom by the molar mass of the whole compound and multiply the product by 100 to get a percentage.
Example: CO (carbon monoxide) has a molar mass of 28.01 g/mol. 12.01 g/mol is from carbon, and 16.00 g/mol is from oxygen. 100%(12.01 g/mol ÷ 28.01 g/mol) = 42.88% carbon. 100%(16.00 g/mol ÷ 28.01 g/mol) = 57.12%
Note that more than one compound can have the same empirical formula: C6H12O6, CH2O, CH3COOH all have the empirical formula CH2O
• Empirical Formulas
To find the empirical formula of a compound using its percent composition, just start by assuming you have 100 g of the substance. Divide each percent as if it were grams by the molar mass of each element. Find the simplest whole number ratio of your answers and use them as subscripts in a formula.