International Mole Day 2017

Hi there,

Welcome to International Mole Day wherever you are in the world!!

Today is the 23rd October 2017 and it is International Mole Day.

That’s 10.23.17.

On International Mole Day we celebrate all that is the Mole, that’s not the furry, little critters who bury under your lawn but the massive number sometimes called Avogadro's Number with the symbol L.

It is 6.0223 ×10²³ particles or any thing else for that matter that's a really big number of things.

602,000,000,000,000,000,000,000 things actually.

So why not have a day to celebrate this vast quantity of stuff.

This number of atoms of carbon weigh 12.00000g if the carbon atom is a carbon-12 isotope.

This number of grams is the atomic mass of carbon and its the standard atomic mass against which all other masses of atoms are measured.

So a mole of magnesium atoms weigh 24.000g.

A mole of helium atoms weigh 4.000g.

There are four posts I've put up to show how you can use the mole in your chemistry.

You can find them here, and here, and here, and here.

Look them over on this blog.

To celebrate today here are some images:

and don't forget Mole Time which is 6.02pm on 10.23.17!!

The Mole (4) Using the mole to determine equation stoichiometry

Ok so what is a chemical equation?

If a chemical formula reveals the mole ratio of elements in  compound then a chemical equation reveals the ratio of reactants to each other and to the products of a chemical change.

This mole ratio is called the equation stoichiometry.

You always wanted know what that word meant, didn't you!!

Here's an example, a simple one:

2H2O2     =      2H2O   +     O2

You can see here that the mole ratio or stoichiometry is given by the large numbers in front of the chemical formulas.

That's 2 moles hydrogen peroxide decomposing to form 2 moles water and a mole of oxygen gas.

Each large number represents the number of moles (or the amount) of that compound in the balanced symbol equation for the chemical change.

(This blog post describes how to balance combustion equations)

Question is how can anyone measure an equation stoichiometry?

Here is one approach using the mass of a precipitate formed in a chemical reaction.

You can find this example here

But we need a simpler example to illustrate the idea.

Suppose we react a known mass of magnesium with excess hydrochloric acid and measure the volume of hydrogen evolved.

Here's the apparatus you could use.

Then we have a way of comparing the mass of magnesium used with the volume of hydrogen formed.

Here are a typical set of results that could be obtained:

Magnesium (g)       Hydrogen (cm3)

0.24                         240
0.12                         120
0.06                          60
0.10                         100

 If these mass results are converted into amounts in moles then this is what we find:

Amt Magnesium (mol)   Amt Hydrogen (mol)
0.01                                 0.01
0.005                               0.005
0.0025                             0.0025
0.00417                           0.00417

The mole ratio of magnesium to hydrogen in this chemical change is 1:1

We can begin to construct the chemical equation for the reaction between magnesium and hydrogen like this:

Mg    +    x HCl    =      XMgCl2     +     H2

It follows that x is then 2 and X  is 1 and the completed chemical equation for this reaction is

Mg    +     2HCl     =   MgCl2       +     H2

Calculations both at GCSE and Advanced level often require you to calculate masses of reacting substances or products formed.

These questions often rely on you using the molar quantities in the chemical equation for the reaction

Here is a typical example for you:

What mass of carbon dioxide is formed from the complete thermal decomposition of 25g of sodium hydrogen carbonate  NaHCO3?


Write the equation:                            2NaHCO3     =      Na2CO3    +     H2O     +     CO2

Put molar quantities to the symbols:   168g                      106g                 18g               44g                  
(you need to calculate the molar
mass (Mr) of each compound)

From the question. identify the           168g                                                                    44g
compounds involved

Work out what 1g would give               1g                                                                      44/168

Scale up by the quantity in                    25g                                                                  (25*44)/168
the question

                                                                                  Answer 6.55g carbon dioxide
Some example problems:

Zinc chloride can also be prepared in the laboratory by the reaction between zinc and hydrogen chloride gas.
Zn + 2HCl =  ZnCl2 + H2
An impure sample of zinc powder with a mass of 5.68 g was reacted with
hydrogen chloride gas until the reaction was complete. The zinc chloride produced had a mass of 10.7 g.
Calculate the percentage purity of the zinc metal. Give your answer to 3 significant figures.

People who have a zinc deficiency can take hydrated zinc sulfate (ZnSO4.xH2O) as a dietary supplement.
A student heated 4.38 g of hydrated zinc sulfate and obtained 2.46 g of anhydrous zinc sulfate.

Use these data to calculate the value of the integer x in ZnSO4.xH2O Show your working. 

The Mole (3) Using the mole to determine the simplest (empirical) formulae
The Mole (1) Relative Atomic Mass and the Mass Spectrometer
Mole (2) Amount of substance and Molar Mass

The Mole (3) Using the mole to determine the simplest formula of a compound

What is a chemical formula?

A compound's chemical formula gives the mole ratio of the elements in that compound.

Water has the formula  H2O

The formula tells us that there are two moles of hydrogen atoms for every mole of oxygen atoms in a mole of water molecules (18g/mol)

Or take sodium chloride  formula NaCl

The formula tells us that for every mole of sodium ions Na+  there is also one mole of chloride ions Cl- in a mole of sodium chloride (58.5g/mol)

These are the simplest mole ratios of elements in the compound.

The simplest mole ratio of elements in a  compound is called an empirical formula.

But how do we calculate these and other formulae like them?

School experiments to determine simple formulae tends to focus on binary or two element compounds.

The methods are usually to either take away an element from the compound and find the masses of each element in the sample by difference.

Or the other method is to add an element to another and again by weighing determine the masses of the two elements in a given mass of the compound and hence from that data find the formula.

In the first method a common approach is to remove oxygen from copper oxide using natural gas.

In the second it is to burn magnesium in air and add oxygen to the metal to create magnesium oxide.

If you have done any kind of useful practical chemistry you will have carried out at least one of these experiments in a laboratory yourself.

In the UK they can be found in most GCSE and A level courses.

Here is a link to heating magnesium in air if you have never done this practically.

Here is link to the experiment to remove copper from copper oxide to find the formula of copper oxide

Let's have a look at some results from a typical experiment and see how to use the mole to arrive at the empirical formula for the compound.

Let's begin with finding the formula of magnesium oxide.

Here's the data ( it's not "real" data but imagined to serve the purpose of the calculation!! if ever you see results like these in a real experiment you ought to be suitably suspicious!!):

Mass of magnesium(g)   Mass of magnesium oxide(g) Mass of oxygen(g)
0.0                                   0.0                                           0.0
0.6                                   1.0                                           0.4
1.2                                   2.0                                           0.8
1.8                                   3.0                                           1.2
2.4                                   4.0                                           1.6

There are several things we can see from these results.

First if we were to plot a graph of mass of oxygen against mass of magnesium we'd have  straight line.

The mass of oxygen added is directly proportional to the mass of magnesium burned.

This illustrates the law of constant composition for a compound i.e. the mole radio of the elements in the compound never changes no matter how many moles of the compound you look at and wherever it is found in the Universe (though having never been to Mars or Jupiter i'm taking the last bit of that definition on trust).

We can calculate the amount of each substance used.

We use the equation from Mole(2)

So if n=m/M then amt magnesium = mass of magnesium(g)/molar mass of magnesium (g/mole)

Therefore   Amt magnesium =   0.6g /24 g/mol   =    0.025 mol

The answers are below but you ought to just try a few calculations from memory to check you can actually come up with the same results as here.

Amt magnesium(mol)  Amt magnesium oxide(mol)  Amt oxygen(mol)
0.0                                 0.0                                          0.0
0.025                             0.025                                      0.025
0.050                             0.050                                      0.050
0.075                             0.075                                      0.075
0.10                               0.10                                        0.10

What do you notice about the numbers of moles?

In every example they are the same !!

Why is that?

Let's look at the mole ratio of elements in magnesium oxide:

                               magnesium                       oxygen
mass (g)                     1.2                                      0.8
molar mass (g/mol)   24                                       16
amount (mol)            0.05                                    0.05
simplest
mole ratio                  1.0                    :                 1.0
empirical
formula                      Mg                                      O

You'll see here the table method I have used to set out the calculation

Try some examples for yourself and see how easy this stuff is.

Then try some of these examples below.

If you want the answers you'll have to send in an email comment for them

Happy calculating

Example empirical formula calculations:

A compound of carbon, hydrogen and oxygen contains 40.0% carbon, 6.6% hydrogen and 53.4% oxygen. Calculate its empirical formula.

Determine the formula of a mineral with the following mass composition: Na = 12.1%, Al = 14.2%, Si = 22.1%, O = 42.1% and H2O = 9.48%.

A 10.00g sample of a compound contains 3.91g pf carbon , 0.87g of hydrogen and the rest is oxygen. Calculate the empirical formula of this compound.

The next examples calculate the formula of hydrated salts i.e x in this type of formula MO.x H2O

A sample of a hydrated compound was analysed and found to contain 2.10g of cobalt, 1.14g of sulphur, 2.28g of oxygen and 4.5g of water.  Calculate its empirical formula.

Mole (1)
Mole (2)
The Mole (4) Using the mole to determine equation stoichiometry
Mole (5)

The Mole (1) Relative Atomic Mass (Ar) and the Mass Spectrometer

Well, here goes, let's see what we can all make of this inscrutable, challenging and vital topic.

The topic of the Mole starts a bit further back with the attempt to determine atomic mass and molecular mass.

Incredibly difficult tasks need inventive ways to solve them.

Solving the problem of measuring atomic mass taxed the minds of chemists throughout the 19th century.

In principle the way into the problem came with the dawning of the idea of comparative mass or relative mass.

In other words, how do you solve the problem of measuring atomic mass?

Answer: you don't, instead you compare atoms against one another to see which is the more massive.

Take a standard atom and determine that its mass is the standard mass which we'll denote as 1 and see how many times heavier other atoms are compared to this particular atom.

If it happens that you choose the standard mass as the mass of the lightest atom then it works.

And for some years in the late 19th century and early 20th century that's what chemists did.

They used hydrogen (but you'd guessed that already I think) as the standard mass.

Chemists developed intricate machines to measure the relative mass of other elements relative to hydrogen's mass.

Couple the measurement of relative atomic mass with Mendeleev's brilliant piece of imaginative thinking that gave us the Periodic Table and in the late 19th century you had a rough set up.

Adding precision was the real challenge.

It was left to a Brit, Aston by name who around 1920 invented the first mass spectrometer to measure relative atomic mass.

His design remains the basic design students are asked to study on most A level and college courses today.

It's like this:

The process of determining the relative mass works like this:

Vaporisation:  Mass spectrometers these days are used to determine the fragmentation pattern of the breakdown of an organic molecule in the machine.

So molecules are usually injected in the machine as liquids which are then vaporised.

Ionisation: Atoms or molecules are then bombarded by fast moving electrons which ionise the atoms or break up (fragment) and ionise the organic molecules (M).

So     M      +       e-(fast)       =         M+         +      2e- (slow)

 (Other particles/fragments M- and M• also form together with M+)

Acceleration: As these ionised fragments (M+, M- and M•) enter an electric field, the positively charged fragments only are accelerated into a magnetic field.

All other fragments (M- and M•) are evacuated from the machine.

Deflection:  Positively charged fragments (M+) enter a magnetic field and are deflected according to their mass the lighter particles more than the heavier ones.

Detection: A detector picks up the incredibly small changes in charge as a small current at different mass values ( the machine is calibrated with a substance of known mass ) and converts these into a fragmentation pattern print out.

Here is a mnemonic to help remember this detail VIADD
Vaporisation
Ionisation
Acceleration
Deflection
Detection

Use it if it helps.

The standard atomic mass is no longer hydrogen but the isotope of Carbon: Carbon -12.

Carbon-12 is assigned a Relative Isotopic Mass of 12.00000 amu (atomic mass units)

Relative Atomic Mass is the weighted isotopic average mass for an element relative to 1/12th the mass of the Carbon-12 isotope.

So let's use the Chlorine mass spectrum (that's for atomic chlorine: Cl, not molecular chlorine Cl2) to work out its RAM:

Here is the print out showing the two isotopes of atomic chlorine:

Look at the peaks: Chlorine-35 is three times higher than chlorine-37.

Chlorine-35 is three times more abundant on earth than chlorine-37

Therefore, 25% of chlorine atoms are mass 37 and  75% are of mass 35.

Therefore, the RAM of chlorine is

And this number, 35.5, is what you'll find on your common copy of the Periodic Table.

You'll find more precise calculations on the web if you go looking, like this one below:

Just to finish you'll probably have realised by now that Relative Isotopic Mass refers to the relative mass of one particular isotope of an element relative to the Carbon-12 isotope.

Relative Formula Mass is the relative mass of a compound's formula relative to the mass of the Carbon-12 isotope.

You just have to add up the RAM's of the right number of elements to calculate an RFM.

Like NaCl or KH2PO4 or (NH4)2SO4 

You should satisfy for yourself that you an calculate these values for these RAM's correctly and then try: (NH4)2Fe(SO4)2.6H2O which is the formula of Mohr's Salt the hexagonal crystals shown below:

Similarly with Relative Molecular Mass (Mr)when the compound is an actual discrete group of atoms like H2SO4, just add up the relative atomic masses of the right number of element atoms.

You can check the answers to these problems from the web quite easily.

Next we'll look at how you can use RAM data in  calculations involving formulae and equations.

Here is a typical problem on this topic:

1The mass spectrum of the isotopes of element Germanium is shown in the diagram below                                   Use data from the diagram to calculate the relative atomic mass of Germanium. Give your answer to one decimal place. ............................................................................................................................................ ............................................................................................................................................ ............................................................................................................................................ ............................................................................................................................................ ............................................................................................................................................ Define the term relative atomic mass .............................................................................................................................................

Identify the ion responsible for the peak at 72 ............................................................................................................................................

Identify which one of the isotopes of Germanium is deflected the most in the magnetic field of a mass spectrometer. Give a reason for your answer.

Isotope ................................................................................................................................ 
Reason ...............................................................................................................................

In a mass spectrometer, the relative abundance of each isotope is proportional to the current generated by that isotope at the detector.
Explain how this current is generated.
............................................................................................................................................
............................................................................................................................................
............................................................................................................................................
............................................................................................................................................

Ge and Zn are different elements.
Explain why the chemical properties of 70Ge and 70Zn are different.
............................................................................................................................................
...........................................................................................................................................

You can get answers to these questions if you add a comment and email contact

The pages of the Mole topic are listed below.

Click on any one to go there (When they are written and uploaded!!!)

Mole (2) Amount of substance and Molar Mass
The Mole (3) Using the mole to determine the simplest (empirical) formulae
The Mole (4) Using the mole to determine equation stoichiometry
Mole (5)
Mole (6)

The Mole (2) Amount of Substance and Molar Mass

So how can we relate the mass of elements and compounds to the relative mass of their particles and the number of those particles?

Let's suppose you have an opaque bottle which your friend has weighed for you and put in the bottle a number of 2p coins.

How can you find out how many coins are in the jar without opening it?

What would you need to know about the jar and its contents?

You'd need the mass of a typical 2p coin and the mass of the empty jar.

Here's what you'd do:

First you'd weigh the jar and coins and then subtract the mass of the jar to find the mass of the coins themselves.

If you then divide the mass of the coins on their own by the mass of one 2p coin you will have the number of coins in the jar.

Simples!!

And you are saying: why isn't the mole like that?!!

Well, if you were a bank cashier you would use the coin method all the time to count the coins in a bag.

And that's the point I think, the Mole is really about using the numbers in simple problem calculations so that we get savvy at solving them.

Let's get to the mole and see what can do with it.

First the mole is a number of particles.

We have names for numbers of things: a dozen eggs (12), a score of pencils (20), a ream of paper (500) so why not a mole of particles (6.02214129(27)×10²³ ) or more briefly 6.022×10²³ 

The number is called the Avogadro number. (Symbol: L)

You can see how its calculated if you divide the standard RAM of Carbon 12 by the mass of an atom of carbon 12 (1.99252×10⁻²³ g) that gives you the number of atoms in the 12g of carbon.

Try it for yourself and see that it approximates to the Avogadro number.

Actually, the Relative Atomic Mass of any substance measured in grams contains a Mole of particles.

This mass per mole of particles is called the Molar Mass (symbol M) with units g/mol or g.mol⁻¹ 

You can probably figure out that therefore the molar mass of carbon-12 is 12.0g.mol⁻¹

And if technetium has a RAM of 99.0 its molar mass is 99.0g.mol⁻¹

And if molar mass applies to all substances then lets see how it works out for ethanol C2H5OH molecules.

C2H5OH

    There are two carbon atoms (2*12.0)
There are six hydrogen atoms  (6*1.0)
              and one oxygen atom  (1*16.0)
                   Total molar mass is 46.0g.mol⁻¹

It is important you watch for or specify the kind of particle that's being referred to

So for example:

1.0 mol of Br  i.e. 1 mole of bromine atoms     RAM  80     Molar mass 80g.mol⁻¹

1.0 mol of Br2 i.e. 1 mole of bromine molecules RAM  80*2 =160  Molar mass 160g.mol⁻¹

So what would happen if we had say 1.75g of carbon and wanted to know how many moles of atoms of carbon-12 there were in that sample?

Using our analogy with the coins experiment we said we would find the number of coins as follows:

number of coins   =    mass of all coins /mass of one coin

so to find the number of moles of a substance (its Amount)  we divide the mass of that substance by the mass of one mole

In words:   number of moles (n)   =   mass of substance (m)
                                                     mass of one mole of the substance (M)


You are probably familiar with this simple little equation     n    =   m/M
                                                                                                     
It comes in other shapes and sizes which you can find on the web:

These I picked because they happen to be correct unlike several others which use the RAM for the mass of one mole.

RAM has the same integer value but it doesn't carry a unit and so doesn't belong in this equation.

Thing to do now is to get to using this relationship in real problems from books and exam papers so that you are thoroughly saturated in this mathematical chemistry.

There are no short cuts just damned hard work required to really get this.

Some of us will need to do much more than others to train our brains.

Here is a collection of typical exam problems:

Typical problem and its solution:

A sample of magnesium weighs 128g What amount of magnesium is this?

Substitute in to n=m/M

m=128g and M = 24g.mol⁻¹

thus  n =  128g/24g.mol⁻¹   =   5.33mol



Some further problems:

Calculate the mass of    a) 2.5mol H2   b) 0.75mol copper sulfate  CuSO4   c) 0.66 mol ethene C2H4

Calculate the amount of particles in a) 21.0g  Chlorine molecules Cl2  b) 156g  Helium atoms He

Calculate the amount of carbon dioxide in 1.5g and what mass of ethanol contains the same number of particles?

Calculate  the number of particles in a) 15g of iron, b) 12g of zinc, c) 14.4g of phosphorus P4

How many ions are present in a) 0.3mol of potassium hydroxide, b) 20g of copper sulphate

If you need answers you will have to add a comment (below) and email for me to get back to you.

Some related fun mole stuff:

The Mole (1) Relative Atomic Mass and the Mass Spectrometer

The Mole (3) Using the mole to determine the simplest (empirical) formulae

The Mole (4) Using the mole to determine equation stoichiometry