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**4.2 The Mole Concept**

**1 Define the mole in terms of a specific number of particles called Avogadro’s constant.**

*Mole *is a quantity used to express the amount of substance as there are in the number of carbon atoms in 12 grams of Carbon-12.

Let’s say we have the element Calcium. It’s Nucleon Number (Protons + Neutrons) is 40. So 40g of Calcium will contain the same number of atoms as 12g of Carbon-12. We call this 1 mole of Calcium atoms.

1 mole of atoms has a certain quantity. It’s something like 6.02 x 10^{23}, atoms, ions, molecules, whatever is in the substance. That’s like 602 000 000 000 000 000 000 000… I don’t have to have aced GCSE Mathematics to know that that’s one big number. Luckily, you

won’t have to do calculations with this ridiculously large number.

To sum this up, 1 mole of magnesium (24g) has 6.02 x 10^{23} magnesium atoms. 1 mole of calcium (40g) also has 6.02 x 10^{23} Calcium

atoms. This is the same for any element. The nucleon number of a substance is equivalent to one mole and has 6.02 x10^{23} atoms.

**2 Use the molar gas volume, taken as 24 dm ^{3} at room temperature and pressure.**

If you get 4 flasks, with different elements, with standard conditions applied, and you weigh the four flasks with 4 distinct elements, you see something magical. Each flask has exactly 1 mole of each gas!

We call this **Avagadro’s Law.**

The conditions to get this magical effect (1 mole of each gas) are:

- Same Volume
- Same temperature
- Same pressure

At room temperature (298K) and 1atm pressure, the volume is 24 dm^{3}.

**3 Calculate stoichiometric reacting masses and reacting volumes of solutions; solution concentrations will be expressed in mol/dm ^{3}.**

I’m going to split this into parts. Part 1 is calculating the volume of a gas:

This is simply a mathematical application of Avagadro’s Law.

**Example Question: **What is the volume occupied in 0.5 moles of a gas? (Standard Conditions Apply)

We know that 1 mole of a substance occupies 24dm^{3}.

Therefore, 0.5 mole will occupy 0.5 x 24 dm^{3} = 12 dm^{3}.

**Example Question: **Find the volume occupied by 36 g of Water in Standard Conditions?

M_{r} of H_{2}O = (1+1) + 16 = 18

1 mole = 18 g

36 g will therefore be (36/18) = 2 moles

1 mole = 24dm^{3}, as we learned just now, so as 36g of water is 2 moles, 2 moles = 24dm^{3} x 2 = 48 dm^{3}.

We can also calculate the gas volume from an equation. To do this, we can simply find the ratio of coefficients subscripts and use the ratio to make the appropriate calculations. This may sound kind of fuzzy right now, but I’ll do an example to illustrate this:

**What volume of oxygen will react with 48 dm**^{3}of hydrogen under standard conditions?

To approach this question, we should first write an equation expressing this reaction.

2H_{2} + O_{2} –> 2H_{2}O

Now, there are three substances here. But we don’t need to make use of all of them! Let’s look carefully in the question. We only need to do calculations with the substance “oxygen” and “hydrogen”.

The coefficient subscript is the number in front of the substance.

For the reaction above:

2H_{2} + 1O_{2} –> 2H_{2}O

The underlined values are the coefficient subscripts of the substance. Now the ratio is simply expressed as:

2: 1 : 2

Basically, what this this is saying is that 2 moles of hydrogen will react with 1 mole of oxygen to produce 2 moles of water!

Since 1 mole = 24 dm^{3}, 48 dm^{3} = 24 dm^{3} * x ; x= 2

So 48 dm^{3} = 2 moles, that’s just 1st grade math.

Since the ratio of hydrogen and oxygen is 2:1, two moles of 2 will give you 4 moles of hydrogen.

And if you want to express the moles in dm^{3}, as 1 mole = 24dm^{3}, 4 moles = 24 x 4 = 96 dm^{3}

If you get the concept of comparing coefficients and doing some simple multiplication problems, you are on your way to acing any problem given to you in the exam.

Let’s do one more

**The reaction between nitrogen and oxygen forms nitrogen oxide. What volume of nitrogen oxide is formed when 2 g of Nitrogen is allowed to burn?**

When we solve this problem, let’s first ask ourselves, what is it that we want to find?

We want to find the volume of nitrogen oxide formed when 2g of nitrogen is burned.

Let’s first write down the reaction between nitrogen and oxide?

N_{2} + O_{2} → 2NO

Now, let’s write down the coefficients:

N_{2} = 1

O_{2} = 1

NO = 2

**Ratios between coefficients:**

1:1:2

We want to find the volume of nitrogen oxide formed.

We know that the Nucleon Number of Nitrogen = 14 g

2g of nitrogen = (2/14) mole = (1/7) mole

We know that 1 mole of nitrogen will react with 2 moles of nitrogen oxide

Since the mole reaction between nitrogen and nitrogen oxide is 1:1, (1/7) mole of nitrogen will react with (1/7)*2 = 2/7 mole of nitrogen oxide.

2/7 mole = 0.286 moles, and as 1 mole = 24 dm^{3},

Volume = 0.286 moles x 24 dm^{3} = 6.864 dm^{3}

————————————————————————————-

The concentration of a substance is the amount of solute dissolved in 1 dm^{3} of solution.

The formula to finding concentration is:

**Concentration = [Amount of constituent (mol) / Volume of the solution (dm**^{3})]

We measure concentration using the units : **mol/dm ^{3}**

** **To find the amount of constituent or solute in a solution, we can rearrange the equation to get:

**Amount of constituent (mol) = Concentration (mol/dm**^{3}) x Volume (dm^{3})

To convert the moles to grams, simply use the traditional formula:

:Mass of Solute= Number of Moles x Molecular Mass

**Photo Credit**: https://en.wikipedia.org/wiki/File:Kochendes_wasser02.jpg

Click here to move on to the next unit: 5.1 Electricity and Chemistry

Sorry a typo on my part:: In the N2/2NO example above…N2 has a mass of 28? Your moles number for N2 should be given from 2/28 not 2/14?

In the N2/NO example above…N2 has a mass of 28? Your moles number for N2 should be given from 2.28 not 2/14?

[...] Click here to go to the next lesson: 4.2 The Mole Concept [...]

“What volume of oxygen will react with 48 dm3 of hydrogen under standard conditions?

To approach this question, we should first write an equation expressing this reaction.

2H2 + O2 –> 2H2O

Now, there are three substances here. But we don’t need to make use of all of them! Let’s look carefully in the question. We only need to do calculations with the substance “oxygen” and “hydrogen”.

The coefficient subscript is the number in front of the substance.

For the reaction above:

2H2 + 1O2 –> 2H2O

The underlined values are the coefficient subscripts of the substance. Now the ratio is simply expressed as:

2: 1 : 1″

When you were explaining this question, is the ratio not 2:1:2 instead of 2:1:1?

we have to learn limiting reagent also

Taariq, this will be added as soon as possible, thank you for the reminder!

where is percentage yield?

Taariq, thank you for reminding us, we will include that as soon as possible.

your nucleon number for calcium is not correct. it is supposed to be 40g

James, thank you for pointing this out, the relative atomic mass, also know as the mass number and nucleon number, for Calcium is indeed 40.078g

i believe its 40 NO UNITS, its a number of particles, protons and neutrons…..no masss is assigned to this quantity.

40 g is the mass of one mole of calcium.

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