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