# FINDING THE PERCENT COMPOSITION. 1. Find the molar mass by hpx14343

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FINDING THE PERCENT COMPOSITION.
1. Find the molar mass of each element in the molecule or compound. If there is
more than 1 atom in the molecule, multiply the molar mass by that number.
2. Find the molar mass of the molecule or compound.
3. Divide the total mass of each element by the mass of the molecule or compound
and multiply by 100.
4. Check by adding the percents. It should equal 100.

EXAMPLE: % composition of CO2
Step 1 C 1 x 12.01 = 12.01
O 2 x 16.0 = 32.00

Step 2                     44.01

Step 3     C       12.01
44.01 x100      = 27.3%

O       32.00
44.01 x 100     = 72.7%
Step 4                             100%

FINDING THE EMPIRICAL FORMULA.
1. If given % composition, assume 100-gram sample and change % to grams.
2. Find moles by dividing the given mass by the molar mass of the element.
3. Divide all moles by the smallest number of moles.
4. If the quotients are whole numbers then you have the empirical formula.
5. If the quotients are 0.5, 0.333, or 0.667 then you must multiply all quotients
by a whole number to make the decimal into a whole number.

EXAMPLE: A compound is 38.77% Cl and 61.32% O.
What is the empirical formula?
Step 1 & 2    Cl 38.77 g x 1mol/35.45g    = 1.09 mol
O 61.32 g x 1mol/16.00g     = 3.83 mol

Step 3             Cl      1.09/1.09 = 1
O       3.83/1.09 = 3.5

Step 5           Cl     1x2 =2
O      3.5 x 2 = 7
Empirical Formula = Cl2O7
FINDING THE MOLECULAR FORMULA.
1. If you are not given the empirical formula then find it as directed in the steps
above.
2. Find the molar mass of the empirical formula.
3. The molar mass of the molecular formula will be given. Divide the empirical
formula’s molar mass into the molecular formula’s molar mass. This will
equal a whole number.
4. Multiply all atoms’ subscripts by the whole number from step 3. This is the
molecular formula.

EXAMPLE:
Step 1 Empirical formula = Cl2O7
Step 2 Empirical formula’s molar mass = 182.9 g/mol
Step 3 Molecular formula’s molar mass = 548.7 g/mol
548.7/182.9 = 3
Step 4 Cl2O7 x 3 = Cl6O21

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