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
Molecular mass – the sum of the average atomic masses of all the atoms in a molecule

***Remember***
A molecule is a neutral group of atoms that are held together by covalent bonds!!!

The Mole and Molar Mass

Molar Mass


The atomic molar mass is the mass of one mole of atoms. This is the value found on your Periodic Table.
 All values are rounded to three decimal places.
The term molar mass is a general term referring to the mass of one mole of a compound. Also called molecular mass , formula mass , molar mass .
 Add the molar mass together for each atom bonded together


The molar mass (mm) of a compound is determined the same way, except now you add up all the atomic masses for the molecule (or compound)

Ex. Molar mass of CaCl
2

Avg. Atomic mass of Calcium = 40.08g

Avg. Atomic mass of Chlorine = 35.45g

Molar Mass of calcium chloride =
40.08 g/mol Ca + (2 X 35.45) g/mol Cl
 110.98 g/mol CaCl
2
20
Ca
40.08
17
Cl
35.45

The Mole

Atoms, ions and molecules are so small it is impractical to refer to 1, 2 or 3 of these particles.
Mole: 6.02 x 10 23 items.
This is the number of atoms required to produce the atomic mass of an element when that mass is expressed in grams.
Example:
1 atom Mg = 24.31 amu
(4.04 x 10 -23 g)
6.02 x 10 23 atoms Mg = 24.31 g


Avogadro
’ s Number (symbol N ) is the number of atoms in 12.01 grams of carbon.

Its numerical value is 6.02 × 10 23 .

Therefore, a 12.01 g sample of carbon contains 6.02 × 10 23 carbon atoms.

The Mole

Reading Balanced Equations
2 Mg (s) + O
2
(g) 
2 MgO (s)




2 Mg atoms + 1 O
2 molecule produces 2 MgO formula units.
12.04 x 10 23 Mg atoms + 6.02 x 10 23 O
2
10 23 MgO formula units.
molecules produces 12.04 x
2 mol Mg atoms + 1 mol O
2 formula units.
molecules produces 2 mol MgO
The terms atoms , molecules and formula units are usually omitted:
Two moles magnesium react with one mole oxygen to produce two moles magnesium oxide.

The Mole
Other Examples:
Cu (s) + 2 AgNO
3
(aq)

2 Ag (s) + Cu(NO
3
)
2
(aq)
One mole of copper metal reacts with two moles of silver nitrate solution to produce two moles of silver metal and one mole of copper(II) nitrate solution .


The volume occupied by one mole of softballs would be about the size of the Earth.

One mole of Olympic shot put balls has about the same mass as the
Earth.


What is the molar mass of KCl?

What is the molar mass of
Ba(NO
3
)
2
?


4
2

Calculations Involving Mass and
Moles

Mass to Moles Calculations: ma ss
# of m oles = mola r mass m and n =
M
OR
# of m ol = ? g ×
1 mo l x g
Calculated number of moles
=
Given number of grams x
1
Calculated Molar Mass

Calculations Involving Mass and
Moles
Example: 1000. g NaCl is how many moles?
1 Na 1 x 22.99
= 22.99
1 Cl1 x 35.45
= 35.45
58.44 g/mol n m 10 00.
g
= =
M 58.44 g/mol
= 17.11
mo l NaCl
1 mol
OR 1 000 . g ×
58.44 g
= 17.11
mol NaCl

Atoms or
Molecules
Divide by 6.02 X 10 23
Multiply by 6.02 X 10 23
Moles
Divide by atomic/molar mass from periodic table
Multiply by atomic/molar mass from periodic table
Mass
(grams)

Grams molar mass Avogadro ’ s number
Moles particles

Calculations Involving Mass and
Moles

Moles-to-Mass Calculations:
# of grams = moles x molar mass m = n M
OR x g
# of gra m s = ? mol ×
1 mol
Calculated number of grams =
Given number of moles x
Calculated molar mass

Calculations Involving Mass and
Moles
Example: 0.21 mol NaCl is how many grams?
1 Na 1 x 22.99
= 22.99
1 Cl1 x 35.45
= 35.45
58.44 g/mol m = n M = 0.21 mol x 58.44 g/mol = 12 g
NaCl
OR 0 .21 m ol ×
58.44 g
= 1 2 g NaCl
1 mo l

Percent Composition

Law of Definite Proportions:


Elements in a compound combine in a constant proportion by mass.


For example, 1 g hydrogen combines with 8 g oxygen to produce 9 g water.
If atoms of an element have a specific mass, it must be true that atoms combine in a specific ratio.
Commonly expressed as percent composition mass pe rce nt X = mass of element X mass o f compo und
× 100

Percent Composition
Example (using experimental data):
0.250 g of magnesium are burned to produce
0.415 g of magnesium oxide. Determine the mass percent of magnesium in the compound.
0.250 g magnesium
0.415 g magn e si um oxi de
× 1 0 0 = 6 0.2
% Mg

Percent Composition
Example (using molar mass calculations):
What is the percent by mass of each of the elements in magnesium sulfate, MgSO
4
?
1 Mg 1 x 24.31
= 24.31
1 S 1 x 32.07
= 32.07
4 O 4 x 16.00 = 64.00
120.38 g/mol
24 .31
120.
38
× 10 0 = 2 0.19
% Mg
3 2.
07
120.
38
× 1 0 0 = 2 6.6
4% S
6 4.
00
120.
38
× 1 0 0 = 5 3.1
6% O

4. Combustion analysis of 0.500 g of an unknown hydrocarbon yielded 1.541 g of
CO
2 and 0.710 g H
2
O.
a) What is the percent by mass of carbon in the compound?
12.01 g C
1.541 g CO ×
2
44.01 g CO
2
= 0.420
5 g C
0.4205 g C
0.500 g hydroc arb o n
× 100 = 8 4.1
% C

The Empirical Formula is the simplest integer ratio of the elements in a compound.
For example: C4H8 is NOT an empirical formula. CH2 IS.


Assume you have 100 g of the substance (makes the math easier because everything is a straight percent).

Consider the amounts you are given as being in units of grams.

Convert the grams to moles for each element.

Find the smallest whole number ratio of moles for each element.


EXAMPLE:

Find the empirical formula for a compound consisting of 63% Mn and 37% O


Assuming 100 g of the compound, there would be 63 g Mn and 37 g O Look up the number of grams per mole for each element using the Periodic Table. There are 54.94 grams in each mole of manganese and 16.00 grams in a mole of oxygen. 63 g Mn × (1 mol
Mn)/(54.94 g Mn) = 1.1 mol Mn 37 g O × (1 mol
O)/(16.00 g O) = 2.3 mol O


Find the smallest whole number ratio by dividing the number of moles of each element by the number of moles for the element present in the smallest molar amount. In this case there is less Mn than O, so divide by the number of moles of Mn: 1.1 mol Mn/1.1 = 1 mol
Mn 2.3 mol O/1.1 = 2.1 mol O The best ratio is Mn:O of
1:2 and the formula is MnO2 The empirical formula is
MnO2


Molecular Formula is the exact ratio of the elements in the compound.

You can find the Molecular Formula of a compound by using the empirical formula and the molar mass.

The mass of the molecular formula is called the molecular mass (in the same way that the mass of the empirical formula is called the formula mass)


1) Calculate the mass of the empirical formula and divide the molar mass of the compound by the mass of the empirical formula in order to find the ratio between the molecular formula and the empirical formula.

2)Then multiply all the atoms by this ratio to find the molecular formula!


EXAMPLE:

What is the molecular formula of a compound that has a molecular weight of 240g? A 5.00g sample contains 2.00g of Carbon, 0.34g of Hydrogen and
2.69g of Oxygen.