# Percent Composition

Document Sample

```					                   The Mole!
When you look at the periodic table, the
average atomic mass below each element
represents the average mass of one atom in
amu. (Review: what is an amu?) This is an
incredibly small mass and not really reasonable
to work with in a lab setting. SO, if we take the
average atomic mass and treat it as a quantity
of grams, we call this quantity a mole.
Example:
1 atom of carbon has the average mass of 12.01
amu.
1 mole of carbon has the mass of 12.01 grams.
Question: How many grams in one mole of lead?
How many grams in 4.5 moles of lead?
A few more details about the mole:
***A guy named Avogadro calculated how many
atoms would be in one mole of an element and
determined that it would equal 6.022 x 1023
atoms of that element. In this sense, we use
the term mole in the same way we use the term
"1 dozen". A dozen always equals 12 of
something; a mole always represents 6.022 x
1023 of something.
***The mole is also used to describe quantities
of compounds and molecules. The "molar mass"
(or grams in a mole) of any substance can be
determined by adding up the individual masses
of the atoms in the compound.
For example:
Water, H2O, contains two hydrogen atoms and
one oxygen atom.
H = 1.01 gram per mole
H = 1.01 gram per mole
O= 16.00 gram per mole
H2O = 18.02 grams per mole
Question:
How many grams would be in 1 mole of ammonia,
NH3?
How many grams would be in 1 mole of
Mg(NO3)2?
This quantity (grams in 1 mole) is called:
molar mass or molecular mass or molecular weight

Percent Composition

Percent composition is a way to describe what a
substance is made of by listing the percent of
each element in the compound.
Percent = part/whole x 100

To determine the percent composition of one or
all elements in a compound:
1. Determine the molar mass of the compound
by adding up the individual masses of all the
atoms in the compound.
2. Divide the total individual mass of each
element by the overall molar mass and multiply
by 100.
Example:
H3PO4
3 H = 3(1.01)   = 3.03
P = 30.97       = 30.97
4 O = 4 (16.00) = 64.00
98.00
H     3.03/98.00 x 100 =         %H
P    30.97/98.00 x 100 =         %P
O 64.00/98.00 x 100 =            %O

Practice: what is the percent composition of
glucose, C6H12O6?
Empirical Formula

So, we have gone from the chemical formula to
the percent composition. Now, we learn to go
from percent composition back to the chemical
formula, in a way. From these percents, we can
determine the empirical formula, which is the
simplest whole number ratio of atoms in a
molecule. The empirical formula may or may not
be the same as the molecular formula which is
the actual ratio of atoms in a molecule.
Example:
Molecular formula of glucose: C6H12O6
Empirical formula of glucose: CH2O

Determining empirical formula:
1. Treat the percentages as grams.
2. Convert the grams to moles.
3. Divide each mole quantity by the smallest
mole quantity to achieve an initial ratio.
4. If this ratio is extremely close to whole
numbers, STOP. This is your ratio for your
empirical formula.
5. If this ratio is not close to whole numbers,
multiply by a whole number to achieve a whole
number ratio. (For example, if the ratio is 1:1.5,
you will need to multiply by 2 to get to 2:3.)

Example: A compound is made of 23.29% Mg,
30.72% S, and 45.99% O. What is its empirical
formula?

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 7 posted: 2/13/2012 language: pages: 6
How are you planning on using Docstoc?