Lesson 2: Quantitative Analysis of Compounds

Part c: Percent Composition

Part a: Molar Mass of Compounds
Part b: Grams-Moles-Particles Relationship
Part c: Percent Composition
Part d: Empirical and Molecular Formulae


 

Composition of Compounds

A central question in Chemistry is …

What is the substance in this sample of matter? That is, what is its name and chemical formula?
 
This is a question about composition. A chemical formula is one of a variety of ways to describe the composition of a compound. The formula indicates the types of atoms (elements) and the number of atoms of each type that is present in a molecule or formula unit. When you see the formula H2O, then you know that there are two elements (hydrogen and oxygen). And you know there are two atoms of hydrogen and one atom of oxygen in a molecule. A chemical formula indicates the composition of a compound in terms of atoms.
 
We learned in Lesson 1c that the mole is a means of supersizing 1 atom up to a measurable number of atoms – 6.022x1023 atoms. While a single atom is too small to count or to mass, a mole of atoms has a measurable mass. And if you can mass a sample, you can use Avogadro’s number to determine the number of atoms. So, a second interpretation of a formula is to interpret the subscripts as indicating the relative number of moles of atoms of each type of element. For H2O, there are two moles of hydrogen and one mole of oxygen in every mole of the compound.
 

 

 
A third means of thinking about composition is the idea of mass composition. Hydrogen has a mass of 1.0079 grams per 1 mole. And oxygen has a mass of 15.9994 grams per 1 mole. If the relative number of moles of hydrogen and oxygen in water are known, then it makes sense that the relative mass of hydrogen and oxygen would also be known. One mole of water would have a mass of 18.0152 grams. This mass of water is composed of 2.0158 grams of hydrogen and 15.9994 grams of oxygen. This is mass composition data and a third means of describing the composition of water.
 
 
 

A Percent Composition Calculation

The most common means of expressing the mass composition is by use of percentages. A common problem would be: Determine the percentage by mass of each element in a sample of compound X. As an example, let’s consider the task of calculating the percent composition of the two elements in H2O. As mentioned above, the molar mass of water is 18.0152 g/mol. This accounts for the two moles of H (2.0158 g) and the 1 mole of O (15.9994 g). The percentages of these two elements are determined by dividing the mass of the element by the total mass (18.0152) and multiplying by 100.

% H = (2.0158) / (18.0152) • 100 = 11.19 % Hydrogen
 
% O = (15.9994) / (18.0152) • 100 = 88.81 % Oxygen
 
You will note that the two percentages add up to 100%.
 
 

 


 

 

Calculating Percent Composition – a Procedure

The following procedure will prove useful in performing percent composition calculations:

  1. Determine the chemical formula (if not given).
  2. Assuming 1 mole of the compound, use molar mass values and subscripts to determine the mass of each element.
  3. Determine the molar mass of the compound – the grams per 1 mole.
  4. For each element, divide the mass of the element (step 2) by the mass of the compound (step 3) and multiply by 100. This is the percent by mass of that element.
  5. As a check for accuracy, make sure all percentages add to 100%. If they do not add to 100% (or within 0.1% of 100%), then recheck your calculations.
 
The following examples illustrate the use of this procedure in calculating percent composition of the elements in a compound.
 
 
Example 1
Determine the percent composition of copper(II) nitrate.
 
Step 1: Using a polyatomic ion list, we can determine the chemical formula to be Cu(NO3)2.
 
Step 2: Assuming 1 mole of Cu(NO3)2, the masses of the three elements are:
 
Mass Cu = 1 mol•(63.546 g/mol) = 63.546 g Cu
 
Mass N = 2 mol•(14.0067 g/mol) = 28.0134 g N
 
Mass O = 6 mol•(15.9994 g/mol) = 95.9964 g O
 
Step 3: The molar mass of Cu(NO3)2 is ...
 
Molar Mass = 63.546 + 28.0134 + 95.9964 = 187.5558 g/mol
 
Step 4: The percent by mass of each element can now be calculated:
 
% Cu = (63.546 g) / (187.5558 g) • 100 = 33.88 % Cu
 
% N = (28.0134 g) / (187.5558 g) • 100 = 14.94 % N
 
% O = (95.9964 g) / (187.5558 g) • 100 = 51.18 % O
 
Step 5: As a check, the three percentages are added to see if they add to 100%.
 
Check: 33.88% + 14.94% + 51.18% = 100.00%Checkbox Checked with solid fill
 
 
 
About Significant Digits: It’s worth noting that the percentage values were rounded to the second decimal place. This was an arbitrary decision. Given the precision of the molar mass values that were used in the solution, the percentage values could be expressed to the third decimal place. Your instructor would be the final authority as to how many decimal places to include in your answer.
 
 

 
 
Example 2
Determine the percent composition of ethanol, C2H5OH.
 
Step 1: The chemical formula to be C2H5OH. Note that there are six H atoms.
 
Step 2: Assuming 1 mole of C2H5OH, the masses of the three elements are:
 
Mass C = 2 mol•(12.0107 g/mol) = 24.0214 g C
 
Mass H = 6 mol•(1.0079 g/mol) = 6.0474 g H
 
Mass O = 1 mol•(15.9994 g/mol) = 15.9994 g O
 
Step 3: The molar mass of C2H5OH is ...
 
Molar Mass = 24.0214 + 6.0474 + 15.9994 = 46.0682 g/mol
 
Step 4: The percent by mass of each element can now be calculated:
 
% C = (24.0214 g) / (46.0682 g) • 100 = 52.14 % C
 
% H = (6.0474 g) / (46.0682 g) • 100 = 13.13 % H
 
% O = (15.9994 g) / (46.0682 g) • 100 = 34.73 % O
 
Step 5: As a check, the three percentages are added to see if they add to 100%.
 
Check: 52.14% + 13.13% + 34.73% = 100.00%Checkbox Checked with solid fill
 
 
As we always say, Chemistry is not a great spectator sport. You won’t master a skill from the sidelines. Take some time to insure you got this by doing some practice problems. The Before You Leave section provides some ideas.
 
 
 
 

Before You Leave

 
 

 

Check Your Understanding

Use the following questions to assess your understanding. Tap the Check Answer buttons when ready.
 
1. Identify the formula and calculate the percent composition of the elements in dinitrogen tetroxide. (Review Formulas for Molecular Compounds.)

Check Formula
 

Check Percent Calculations



 
2. Identify the formula and calculate the percent composition of the elements in copper(II) acetate. (Review: Formulas for Ionic Compounds || Polyatomic Ion List)
 
Check Formula
 

Check Percent Calculations



 
3. Identify the formula and calculate the percent composition of the elements in lead(II) phosphate. (Review: Formulas for Ionic Compounds || Polyatomic Ion List)

Check Formula
 

Check Percent Calculations



 

Next Part of this Lesson: Empirical and Molecular Formulae

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