cum
top of page
Search

# CSEC Chemistry: Mole Concept

Relative Atomic Mass of Elements

Before we get to talking about moles, we need to cover relative masses.

The relative atomic mass of an element is the ratio of the average mass of one atom of a specific element compared to 1/12 the mass of one atom of carbon-12. Relative atomic mass is dimensionless, that is, it has no units.

Carbon-12 is assigned a mass of exactly 12.00 atomic mass units.

For example, to find the relative atomic mass of hydrogen, we divide the mass of 1 hydrogen atom by 1/12 the mass of 1 atom of carbon:

(1.67 Ã— 10^-27 kg)/ (1/12)(2.00 Ã— 10^-26) = 1

The relative atomic mass of hydrogen is 1.

Relative Molecular Mass of Compounds

The relative molecular mass is the average mass of one molecule or formula unit of the compound compared with 1/12 the mass of one atom of carbon-12.

Finding the relative molecular mass is fairly simple:

1. Write the chemical formula of the compound

2. Identify the atoms or ions present

3. Multiply the relative atomic mass of each atom or ion by the number of like atoms present in the compound

Â

Moles

The term mole is similar to the word "dozen," or "pair," because it represents a specific, fixed number of objects. In the same way that dozen refers to 12 items, or pair refers to 2, mole constitutes a fixed number of particles.

The mole is the amount of substance which contains as many particles (whether ions, atoms or molecules) as there are atoms in 12 grams of carbon-12.

(You will find that lots of measurements in Chemistry are based around the Carbon-12 isotope).

To show you the number of atoms present in 12 grams of carbon-12, which is the number or particles in a mole, we can perform a calculation:

• The mass of one carbon-12 atom is: 1.99 Ã— 10^-26 kg, or 1.99 Ã— 10^-23 g

• Therefore, the number of carbon atoms in 12 grams of carbon-12 will be:

12 g/1.99 Ã— 10^-23 g = 6.023 Ã— 10^23

This number, when rounded off, is approximately 6.0 Ã— 10^23, known as Avogadro's Number, the number of particles in one mole of a substance.

This is the chemical equivalent as knowing that a dozen is twelve- one mole corresponds to the same number of particles in every chemical substance.

Now that you know how many particles are in a mole of any substance (Avogadro's Number), you can find the molar mass of a chemical substance.

The molar mass is the mass in grams of 1 mole of a specific substance, and you find it by multiplying the mass of one atom of the element by Avogadro's Number.

Using hydrogen as an example again:

Molar mass of hydrogen = (1.67 Ã— 10^-24 g) Ã— (6.0 Ã— 10^23) = 1 g/mol

You will realize that the molar mass of hydrogen is the same value as the relative atomic mass, however, relative atomic mass has no units, while molar mass is measured in g/mol (or gmol^-1).

Note: There is a difference between one mole of atoms and one mole of molecules, especially in molecular elements like chlorine, nitrogen and hydrogen gas (which exist as diatomic molecules).

1 mole of atoms is 6 Ã— 10^23 atoms.

1 mole of molecules is 6 Ã— 10^23 molecules.

Conversions Involving the Mole

At this point, you know about moles, how they relate to Avogadro's number and the number of particles in a mole, and relative atomic masses. CSEC requires you to know how to perform certain calculations and conversions with the information given to you.

Converting Mass to Moles

number of moles= (given mass in grams of element or compound)/(molar mass of element or compound)

Simply divide the mass of the substance mentioned in the question by the molar mass of the substance.

For example: What is the number of moles of sodium chloride formula units present in 1.17g of sodium chloride?

The mass of 1 mole of NaCl = 23+35.5= 58.5 g/mol

Hence, 58.5 g of NaCl contains 1 mole of NaCl formula units.

1.17g NaCl will contain (1.17/58.5) = 0.02 mol of NaCl formula units

Converting Moles to Mass

mass of a certain number of moles= molar mass Ã— number of moles

For example: What is the mass of 0.25 moles of nitric acid?

Molar mass of nitric acid= [1+14+(3Ã—16)]= 63 g/mol

Hence, the mass of 0.25 moles of nitric acid= 0.25 Ã— 63 g/mol= 15.75 g

Converting Moles to Number of Particles

number of particles= Avogadro's Number(L) Ã— number of moles

Example: How many particles are in 3.5 moles of Lithium?

# of particles= 3.5 Ã— 6.0Ã—10^23 = 2.1 Ã— 10^24

Converting Number of Particles to Moles

number of moles= number of particles/Avogadro's Number(L)

Example: How many moles of silicon would contain 3.72 Ã— 10^24 particles?

# of moles of silicon= (3.72 Ã— 10^24)/(6.0 Ã— 10^23)= 6.2 mol

Converting Mass to Number of Particles

# of particles= Avogadro's Number Ã— (mass in grams)/(molar mass)

Example: How many molecules of carbon dioxide are present in 440 g of the compound?

molar mass of Carbon dioxide= 12 + (2 Ã— 16) = 44 g/mol

# of particles of carbon dioxide= 6.0 Ã— 10^23 Ã— (440 g)/(44 g/mol)

= 6.0 Ã— 10^24 molecules of carbon dioxide

Converting Number of Particles to Mass

mass= (number of particles/Avogadro's Number) Ã— molar mass

Example: What is the mass of 3 Ã— 10^21 molecules of nitric acid?

mass of nitric acid= (3 Ã— 10^21/6.0 Ã— 10^23) Ã— 63 g/mol = 3.15 Ã— 10^-1 grams

Empirical and Molecular Formulae

Mole concept can also be used to calculate the formulae of compounds. The exact number of atoms or ions in one molecule or formula unit of a compound is the molecular formula of the compound. The simplest whole number ratio of atoms or ions in a compound is the empirical formula of the compound.

For example, the molecular formula for tartaric acid is C4H6O6, while the empirical formula would be the simplest form, that is, C2H3O3.

The molecular and empirical formula of a compound do not have to be different, for example water's molecular formula is H2O, and its empirical formula is also H2O.

Different compounds may also have the same empirical formula, for example, butene (molecular formula C4H8) and propene (C3H6) would have the same empirical formula, CH2.

It is possible to calculate the empirical formula of a compound based on the percentage composition of the substance. The percentage composition by mass of a compound is the percentage by mass of each element in 1 mole of the compound.

Finding the Empirical Formula

1. Calculate the percentage composition by mass of each element: percentage by mass of an element= (total mass of the particular element/mass of the compound) Ã— 100

2. Find the mass of each element in 100 g of the compound (which is the same as the percentage value expressed as grams).

3. Calculate the number of moles by dividing the mass by the molar mass of the element.

4. Calculate the relative number of moles of each element by dividing by the smallest number of moles.

5. The values given correspond to the ratio of the elements in the compound.

For example:

Find the empirical formula of the compound with the percentage composition by mass of Na= 43.4%, C= 11.3%, and O= 45.3%.

1. The percentage composition of mass is given: Na= 43.4%, C= 11.3%, and O= 45.3%.

2. The mass of each element in 100 grams of the compound is: Na= 43.4 g, C= 11.3 g, and O= 45.3 g.

3. Find the number of moles by dividing by the molar mass: Na= 43.4 g/23 gmol^-1= 1.89 mol, C= 11.3 g/12 gmol^-1= 0.94 mol, and O= 45.3 g/16 gmol^-1= 2.83 mol.

4. Divide by the smallest number of moles (0.94 moles): Na=1.89 mol/0.94 mol= 2.0, C= 0.94 mol/0.94 mol= 1.0, O= 2.83 mol/0.94 mol= 3.0

5. Therefore, the empirical formula of the compound is Na2CO3.

Determining the Molecular Formula

Once the empirical formula is determined, we can determine the mass of 1 mol of empirical formula units, and therefore the number of empirical formula units in 1 mol of the compound.

number of empirical formula units= mass of 1 mol of a compound/mass of 1 empirical formula unit

Continuing with the example, if the molar mass of sodium carbonate (empirical formula Na2CO3) is 106 gmol^-1, then what is the molecular formula?

We first find the molar mass of 1 empirical formula unit by adding the masses of each element in the compound:

mass of compound= (2 Ã— 23)+12+(3 Ã— 16)= 106 g

Then, dividing we get:

106 g/106 g = 1 empirical unit/mol

You multiply the empirical formula by 1, giving us Na2CO3.