Percent Composition of Compounds
Assume 1 mole of the compound
Percent = |
mass of the element |
x100 |
molar mass of compound |
Find the mass percents for the elements present in H2SO4.
molar mass of H2SO4 = 2(1) + 32 + 4(16) = 98 g
mass of H = 2 mol H x 1.008g/mol = 2.016 g
mass of S = 1 mol S x 32.00g /mol = 32.00 g
mass of O = 4 mol O x 16.00g/mol = 64.00 g
mass % H = 2.016g/98g x 100 = 2.06%
mass % S = 32.00g/98g x 100 = 32.6%
mass % O = 64.00g/98g x 100 = 65.3%
Calculate the mass % of the elements present in glucose, C6H12O6.
Determining Empirical Formulas
When chemists have an unknown substance, they can perform experiments to determine the mass of each element in a certain amount of the compound. From this information they can determine the empirical formula of the substance.
In order to determine the empirical formula, first convert the mass of each element to moles. Determine the simplest mole ratio for the elements. The empirical formula will have the same ratio of atoms.
Example- Determine the empirical formula for a compound that contains 3.238 g sodium, 2.265 g sulfur, and 4.499 g oxygen.
mol Na = 3.238 g x 1 mol/22.99g = 0.1408 mol
mol S = 2.265 g x 1 mol/32.00g = 0.07078 mol
mol O = 4.499 g x 1mol/16.00g = 0.2812 mol
Next find the simplest ratio of elements by dividing the smallest number into all of the numbers. This will tell us the ratio of the elements in the formula.
mol Na = 0.1408/0.07078 =2.0 (1.99)
mol S = 0.0708/0.07078 = 1.0
mol O = 0.2812/0.07078 = 4.0 (3.97)
These whole numbers represent the number of each kind of atom in the formula , Na2SO4.
Example - When 1.916 g of titanium is heated in oxygen, an oxide sample weighing 3.196 g results. Calculate the empirical formula of the compound.
g O = 3.196 g - 1.916 g = 1.280 g
mol Ti = 1.916 g x 1mol/47.88 g = 0.0400 mol
mol O = 1.280 g x 1mol/16.00g = 0.0800 mol
Simplest whole number ratio is
mol Ti = 0.0400/0.0400 = 1
mol O = 0.0800/0.0400 = 2
The empirical formula is TiO2.
Sometimes the numbers determined by the above process are not close to integers. If this happens, you must find a small integer to multiply all of the ratios by to convert them to whole numbers.
Example - Analysis of a compound containing aluminum and oxygen shows
mol Al = 0.99
mol O = 1.49
These ratios should be multiplied by 2
mol Al = 1.98
mol O = 2.98
So the formula is Al_{2}O_{3}.
Determining An Empirical Formula by Combustion Analysis.
This method is frequently used for compounds that contain carbon and hydrogen and possibly oxygen. In this method a sample of the compound is combusted.
C_{x}H_{y}O_{z} + O_{2} à CO_{2} + H_{2}O (unbalanced equation)
All of the carbon in the original sample is present as CO_{2} and all of the hydrogen in the original sample is present as H_{2}O after combustion.
Knowing the mass of carbon dioxide and water formed allows the calculation of moles and mass of carbon and hydrogen in the original sample.
Subtracting the mass of carbon and the mass of hydrogen from the mass of the original sample allows the determination of the mass (and moles) of oxygen in the sample.
Determining a Chemical Formula from Combustion Analysis Data
Formula of the type C_{x}H_{y}O_{z}.
Step 1: Convert grams of CO_{2} to moles of C. This is the subscript x.
Convert grams of H_{2}O to moles of H. This is y.
Step 2: Convert moles of C to grams of C.
Convert moles of H to grams of H.
Add these masses to get grams of C + H.
Step 3: Determine the mass of O.
mass of sample - mass C+H.
Determine moles of O.
Step 4: Find simplest ratios of moles.
Example - A 0.1000 g sample of a compound containing carbon, hydrogen, and oxygen is combusted in oxygen gas to form 0.1953 g of carbon dioxide and 0.1000 g of water. What is the empirical formula of the compound?
Empirical vs Molecular Formula
Molecular formula show what atoms and how many of each kind are present in a molecule.
Empirical formula is the simplest formula containing whole number ratios for a substance.
Compound |
Empirical formula |
Molecular formula |
Molar mass (g) |
formaldehyde |
CH2O |
CH2O |
30.03 |
acetic acid |
CH2O |
C2H4O2 |
60.06 |
glucose |
CH2O |
C6H12O6 |
180.18 |
Molecular formulas are integer multiples of empirical formulas.
Integer = Multiplier |
molecular molar mass empirical formula molar mass |
= x |
where x is an integer (1,2,3, ...)
Example - Determine the molecular formula of glucose which has a molar mass of 180. g. The empirical formula of glucose is CH_{2}O.
x = |
180. g 30.0 g |
= 6 |
In order to determine the molecular formula, you need to know the molar mass of the compound. Then divide the molar mass by the molar mass of the empirical formula. This will give the integer x. Multiply the subscripts of the empirical formula by this integer to determine the molecular formula.
The molecular formula of glucose is
C_{1x6}H_{2x6}O_{1x6} = C_{6}H_{12}O_{6}
Determining Empirical Formulas from Percent Composition of a Compound
Propanol contains 60.0% carbon, 13.4% hydrogen, and 26.6% oxygen. Determine the empirical formula of propanol.
60.0% C means there are 60.0 g C in every
100 g of the compound
13.4 % H mean there are 13.4 g H in every
100 g of the compound
26.6% O means there are 26.6 g O in every
100 g of the compound
Find the number of mole of each element.
60.0g x 1mol/12.0g = 5.0 mol C
13.4g x 1mol/1.01g = 13.3 mol H
26.6g x 1mol/16.0g = 1.66 mol O
Find the integer ratios
5.0 mol/ 1.66 = 3.0 mol C
13.3 mol/1.66 = 8.0 mol H
1.66 mol/1.66 = 1.0 mol O
The empirical formula is
C3H8O
Putting it all together.....
What is the molecular formula of para-dichlorobenzene, an ingredient in mothballs, that has a molar mass of 147.0 g and a percent composition of 49.0% C, 48.2% Cl, and 2.74% H.
Answer = C6Cl2H4
Flow Chart For Formula Problems