Percent Comp
Percentage composition
Indicates the relative amount of each element present in a compound.
Calculating percentage composition
Step 1 : Calculate molar mass
Step 2 : Divide the subtotal for each element’s mass by the molar mass.
Step 3: Multiply by 100 to convert to a percentage.
Example 1
1. Calculate the molar mass of water, H2O.
Step 1 – H- 2(1.01)= 2.02 O – 1(16.0)= 16.0 ________ 18.02
Step 2 2.02/18.02 X 100 = 11.21%
16.0 /18.02 X 100 = 88.79 %
Water is composed of 11.21 % hydrogen and 88.79 % oxygen.
Example 2
Calculate the percentage composition of sucrose , C12H22O11.
Example 3
Find the percentage composition of hydrogen in sulfuric acid.
Empirical Formula
From percentage to formula
Empirical Formula
The empirical formula gives the ratio of the
number of atoms of each element in a
compound.Compound Formula Empirical Formula
Hydrogen peroxide H2O2 OH
Benzene C6H6 CH
Ethylene C2H4 CH2
Propane C3H8 C3H8
Calculating Empirical
Pretend that you have a 100 gram sample of the compound.
That is, change the % to grams. Convert the grams to mols for each element. Write the number of mols as a subscript in a
chemical formula. Divide each number by the least number. Multiply the result to get rid of any fractions.
EXAMPLE 1
A compound was analyzed to be 82.67% carbon
and 17.33% hydrogen by mass. What is the
empirical formula for the compound?
Assume 100 g of sample, then 82.67 g are C and
17.33 g are H
EXAMPLE 1
Convert masses to moles:
82.67 g C x mole/12.011 g = 6.88 moles C
17.33 g H x mole/1.008 g = 17.19 mole H
Find relative # of moles (divide by smallest number)
EXAMPLE 1
Convert moles to ratios:
6.88/6.88 = 1 C
17.19/6.88 = 2.50 H
Or 2 carbons for every 5 hydrogens
C2H5
Example 2
Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.
Assume 100 g so38.67 g C x 1mol C = 3.220 mole C
12.01 gC 16.22 g H x 1mol H = 16.09 mole H
1.01 gH45.11 g N x 1mol N = 3.219 mole N
14.01 gN
Example 2
3.220 mole C 16.09 mole H 3.219 mole N
•C3.22H16.09N3.219
If we divide all of these by the smallest one It will give us the empirical formula
Example 2
The ratio is 3.220 mol C = 1 mol C 3.219 molN 1 mol N
The ratio is 16.09 mol H = 5 mol H 3.219 molN 1 mol N
C1H5N1 is the empirical formula
Molecular Formula
Is the actual FormulaOnly applies to covalent compoundsFind the Mass of the Empirical Formula Divide the actual molar mass by the mass of
one mole of the empirical formula.Caffeine has a molar mass of 194 g. what is
its molecular formula?
Empirical Formula C4H5N2O1
Caffeine Example
Find the mass of the empirical formula
Empirical Formula C4H5N2O1
C= 4 X 12.01
H = 5 X 1.01
N = 2 X 14.01
O = 1 X 15.99
Total Mass = 97g
Caffeine Example
Find x if massformulaempirical
massmolarx
•194 g
•97 g= 2
C4H5N2O1
C8H10N4O2.
2 X
Example 2
A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula?
Example 2
71.65 Cl
24.27C
4.07 H.
g
mol
5.35
1
g
mol
12
1
g
mol
1
1
= 2.0mol
= 2.0mol
= 4.0mol
Example 2
Cl2C2H4
Its molar mass is known (from gas density)
is known to be 98.96 g. What is its molecular
formula?
We divide by lowest (2mol )
•Cl1C1H2
would give an empirical wt of 48.5g/mol
Its molar mass is known (from gas density)
is known to be 98.96 g. What is its molecular
formula?
•
would give an empirical wt of 48.5g/mol
massformulaempirical
massmolarx
g
g
5.48
96.98= 2=
2 X Cl1C1H2
= Cl2C2H4