# World of Chemistry Chapter 6

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```					World of Chemistry Chapter 6—Chemical Composition
Section 6.5 Percent Composition of Compounds
It’s useful and easy to determine the percentage of each element
in a compound…

Percent* composition            = mass of the elment in 1 mole of compound
of Compound                                 molar mass of the compound
*multiply the result by 100 to obtain the percentage

Section 6.6 Formulas of Compounds
 Empirical Formula: The formula of a compound that expresses
the smallest whole-number ratio of the atoms present (a.k.a.
simplest formula)
 Molecular Formula: The actual formula of a compound that gives
the composition of the molecule.
 Example: Glucose is C6H12O6 this is the molecular formula
because each molecule of glucose contains 6 carbon atoms, 12
hydrogen atoms and 6 oxygen atoms. Notice that each subscript
is divisible by 6…the empirical formula for glucose is
CH2O…whenever possible it is best to use the molecular formula.

Section 6.7 Calculation of Empirical Formulas
Steps for Determining the Empirical Formula
Of a Compound
1. Determine the mass (in grams) of each element present
2. Convert mass to moles using the periodic table
3. Divide the number of moles of each element by the smallest
number of moles to convert the smallest number to 1. These
numbers are the subscripts.
4. If one or more of the numbers from step 3 are decimals multiply
all the numbers by the smallest number that will convert all of
them to whole numbers. These numbers are the subscripts.
Section 6.8 Calculation of Molecular Formulas
Now that we know how to calculate the empirical formula we can
determine the molecular formula if we know the molar mass of the
compound.
I.        Molecular Formula = (empirical formula) x n*
*where n=a small whole number
A. Through the magic of algebra we know that
n=         molar mass
empirical formula mass
B. Once you have determined n multiply the subscripts in the
empirical formula by n to obtain the molecular formula
Example Problem: self-check 6.11 p.185
A compound used as an additive for gasoline to help prevent engine knock shows the following
percentage composition:
71.65% Cl                24.27% C                  4.07% H
The molar mass is known to be 98.96g. Determine the empirical formula and the molecular formula for this
compound.

First, to figure out the empirical formula we determine the moles of each element. We need to know how many
grams of each element we have:

(98.96g of compound) x (71.65% of Cl) = 98.96 x .7165 =   70.9g of Chlorine
x (24.27% of C) = 98.96 x .2427 =    24g of Carbon
x (4.07% of H) = 98.96 x .0407 =      4g of Hydrogen

Second determine how many moles of each element:

70.9g Cl x          1     mol Cl = 2 moles of Chlorine
35.45 g Cl

24g C x       1 mol C = 2 moles of Carbon
12 g C

4g H x        1    mol H = 4 moles of Hydrogen
1g H
Third, write the empirical formula: Cl2C2H4 reduced by 2 is ClCH2

Finally, to write the molecular formula we divide

n=      Molar mass
Empirical mass

Determine the empirical mass: 2 moles H =      2   g
1 mole of C = 12     g
1 mole of Cl= + 35.45 g
47.45 g

n =     98.96 9 (molar mass)   =   2.02 or about 2
47.45 g (empirical mass)

Remember: molecular Formula = (empirical formula) x n

So    ClCH2 x 2 = Cl2C2H4

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