Welcome to Accelerated Study Notes

Message of the day (that never changes)

All systems are normal right now.

If you do get locked out for whatever reason, you can always email us at admin@acceleratedstudynotes.com to recover your password, or follow the instructions on the screen.

Registering a new account

If for any reason you can not see the captcha (try to refresh page using Ctrl + Shift + r or Ctrl + Shift + F5), please email us at admin@acceleratedstudynotes.com with your desired username and email. We will create the account for you and a random password will be emailed to you.

Captcha Error

If you do not see the captcha, please either refresh or navigate to http://www.acceleratedstudynotes.com/wp-login.php

Member Login

Lost your password?

Not a member yet? Sign Up!

Blue Captcha Image
Refresh



Connect with:

IGCSE Coordinated Science: The Mole Concept

By (Administrator)

Click here to go back to Table of Contents

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 1023, 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 1023 magnesium atoms. 1 mole of calcium (40g) also has 6.02 x 1023 Calcium
atoms. This is the same for any element. The nucleon number of a substance is equivalent to one mole and has 6.02 x1023 atoms.

2 Use the molar gas volume, taken as 24 dm3  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 dm3.

 

3 Calculate stoichiometric reacting masses and reacting volumes of solutions; solution concentrations will be expressed in mol/dm3.

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 24dm3.

Therefore, 0.5 mole will occupy 0.5 x 24 dm3 = 12 dm3.

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

Mr of H2O = (1+1) + 16 = 18

1 mole = 18 g

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

1 mole = 24dm3, as we learned just now, so as 36g of water is 2  moles, 2 moles = 24dm3 x 2 = 48 dm3.

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 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 : 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 dm3, 48 dm3 = 24 dm3 * x ; x= 2

So 48 dm3 = 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 dm3, as 1 mole = 24dm3, 4 moles = 24 x 4 = 96 dm3

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?

N2 + O2 → 2NO

Now, let’s write down the coefficients:

N2 = 1

O2 = 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 dm3,

Volume = 0.286 moles x 24  dm3 = 6.864 dm3

————————————————————————————-

The concentration of a substance is the amount of solute dissolved in 1 dm3 of solution.

The formula to finding concentration is:

  • Concentration = [Amount of constituent (mol) / Volume of the solution (dm3)]

We measure concentration using the units : mol/dm3

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

  • Amount of constituent (mol) = Concentration (mol/dm3) x Volume (dm3)

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

:Mass of Solute= Number of Moles x Molecular Mass

Photo Credithttps://en.wikipedia.org/wiki/File:Kochendes_wasser02.jpg

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

13 Responses to IGCSE Coordinated Science: The Mole Concept

  1. patrick on February 19, 2013 at 7:08 PM

    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?

  2. patrick on February 19, 2013 at 7:06 PM

    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?

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

  4. Ju on January 2, 2013 at 5:30 PM

    “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?

  5. Taariq on October 28, 2012 at 9:20 PM

    we have to learn limiting reagent also

    • Andrew Chan on November 4, 2012 at 8:03 PM

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

  6. Taariq on October 28, 2012 at 8:22 PM

    where is percentage yield?

    • Andrew Chan on November 4, 2012 at 8:01 PM

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

  7. james mokaya on October 11, 2012 at 2:58 PM

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

    • Andrew Chan on November 4, 2012 at 7:54 PM

      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

    • patrick on March 16, 2013 at 1:08 PM

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

      • Acceleratedstudynotes on March 16, 2013 at 1:20 PM

        40 g is the mass of one mole of calcium.

  8. [...] Click here to go to the next lesson: 4.2 The Mole Concept Print PDF Share this:FacebookTwitterDiggPrintTags: Construct and use symbolic equations with state symbols, Construct and use word equations., Deduce the formula of a simple compound from a model or a diagrammatic representation, Deduce the formula of a simple compound from the relative numbers of atoms present, Determine the formula of an ionic compound from the charges on the ions present, including ionic equations, Use the symbols of the elements to write the formulae of simple compounds. [...]