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Lecture Notes on Atomic Weights and Molar Masses for Chemistry 101, Study notes of Chemistry

Dr. Joy Heising's Fall 2001 lecture notes for Chemistry 101 cover the topics of atomic weights, amu units, grams, moles, formula weights, and the relationship between grams and moles. The notes include examples of calculating the number of atoms in a mole of a substance, the molar mass of a substance, and the percent composition of a substance.

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CHEM 101 LECTURE NOTES Fall 2001 Dr. Joy Heising S 572-580
Tuesday, September 4, 2001 Chapter 2
How much does a single atom weigh?
Different elements weigh different amounts related
to what makes them unique.
What units do we use to define the weight of an atom?
amu units of atomic weight. (atomic mass unit)
1 amu 1/12 the mass of 1 “carbon-12” atom.
But in the lab, scientists use grams it’s more practical.
When scientists do experiments with substances (elements,
molecules, ionic compounds), they usually
- Use the chemical formula to plan (ratio of atoms, NaCl)
- Measure the amount needed in grams
How can we relate grams to the chemical formula?
1.0 g C – 5.0 x 1022 C atoms
1.0 g Ba – 4.4 x 1021 Ba atoms
Inconvenient to work with such large numbers all the time.
Chemists use another unit which works like a ‘dozen’:
Mole – 6.022 x 1023 particles
(atoms, molecules or formula units)
Avogadro’s number
Why do we use such a weird number?
Because it gives us an ‘easy’ relationship between
amus and grams.
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Tuesday, September 4, 2001 Chapter 2

How much does a single atom weigh?

Different elements weigh different amounts – related

to what makes them unique.

What units do we use to define the weight of an atom?

amu – units of atomic weight. (atomic mass unit)

1 amu – 1/12 the mass of 1 “carbon-12” atom.

But in the lab, scientists use grams – it’s more practical.

When scientists do experiments with substances (elements,

molecules, ionic compounds), they usually

- Use the chemical formula to plan (ratio of atoms, NaCl)

- Measure the amount needed in grams

How can we relate grams to the chemical formula?

1.0 g C – 5.0 x 10

22

C atoms

1.0 g Ba – 4.4 x 10

21

Ba atoms

Inconvenient to work with such large numbers all the time.

Chemists use another unit which works like a ‘dozen’:

Mole – 6.022 x 10

23

particles

(atoms, molecules or formula units)

Avogadro’s number

Why do we use such a weird number?

Because it gives us an ‘easy’ relationship between

amus and grams.

Tuesday, September 4, 2001 Chapter 2

**one mole of atoms of an element has a mass in grams

numerically equal to the atomic weight of the element**

1 H atom 1 amu

1 mol H 1 g

1 C atom 12 amu

1 mol C 12 g

The ratio of atoms in the simplest formula for a compound

is the same as the ratio of moles of atoms of the elements in

a sample of the compound

1 H 2 O molecule

2 H atoms

1 O atom

1 mol H 2 O

2 mol H

1 mol O

How much does H 2 O weigh?

Formula weights – the sum of the atomic weights of the

elements in the formula, each taken the number of times the

element occurs.

water H 2 O 2 (1.01) + 1(16.00) = 18.02 amu

Chlorine Cl 2 2(35.45) = 70.90 amu

Calcium nitrate Ca(NO 3 ) 2 nitrate - 14.00 + 3(16.00) = 62.00 amu

1(40.08) + 2(62.00) = 164.08 amu

Tuesday, September 4, 2001 Chapter 2

Percent composition – the % of each element in the

compound (by mass)

ex. calculate the percent composition of C and H in C 3 H 8.

Molar mass of C 3 H 8 = C 3 (12.01 g/mol) = 36.03 (g/mol) H 8 (1.008 g/mol) = 8.064 (g/mol) 44.09 (g/mol)

% carbon: = 36.03 (g/mol) x 100 = 81.71% 44.09 (g/mol)

% hydrogen = 8.064 (g/mol) x 100 = 18.29% 44.09 (g/mol)

empirical formula – the smallest, whole number ratio of

elements present

NaCl H 2 O H 2 O 2 - HO

Molecular formula – actual number of atoms present in a

molecule

Tuesday, September 4, 2001 Chapter 2

If we have the % composition of a substance, we can

determine the empirical formula

Ex. 24.74% K 24.74 g K

34.76% Mn 34.76 g Mn

40.50% O 40.50 g O

100.00% 100 g substance

24.74 g K x 1 mol K = 0.6327 mol K 39.10 g

34.76 g Mn x 1 mol Mn = 0.6327 mol Mn 54.94 g

40.50 g O x 1 mol O = 2.531 mol O 16.00 g

divide through by smallest # moles to get the ratio:

0.6327 mol K/0.6327 = 1 0.6327 mol Mn/0.6327 = 1 2.532 mol O/0.6327 = 4

empirical formula is KMnO 4

but we still don’t know the molecular formula unless we

also have the molar mass of the substance

ex. in previous example, if molar mass = 158 g/mol

empirical formula = molecular formula

if a compound has the empirical formula HO

and its molar mass = 34.0 g/mol

HO = 1.0 + 16.0 = 17.0 g/mol

34/17 = 2

2(HO) = H 2 O 2 molecular formula