Reacting Mass Calculations.

When dealing with reacting masses we use a type of code well known to chemists. In everyday life we deal with familiar units such as weight and height. The Chemist however has different ways of measuring chemicals. There are four main measures, and they are:

MASS:

The basic unit of any chemical is the ATOM. Atoms have a mass, and the more atoms present the greater the mass will be. However the masses of all the atoms of a particular element are not all the same due to the fact that isotopes exist for numerous elements. These are defined as atoms of the same elements with contain the same number of protons but different numbers of protons. For example the element has two different types of atoms having masses 35 and 37. These masses result due to the following scenarios:

Cl 35

Cl 37

Most other elements also have isotopes. As a result, chemists take the average mass of the atoms of the element.

In the case of the element chlorine, the average mass turns out to be 35.5, not 36. This is because the Cl 35 isotope is much more abundant than the other. Therefore we have to ask ourselves how we get these average masses.

Well the answer is, from the Periodic table, where there are so called relative atomic masses or ram for short.

Below are some very important points outlining RAM:

They have no units for they are in fact just ratios

They compare the average mass of an atom of the element with the mass of an atom of carbon (carbon 12)

They are found as the larger of the two numbers in the periodic table as shown below:

Therefore one can see that compared to carbon 12 magnesium atoms are twice as heavy. In the same way throughout the periodic table we can make numerous comparisons.

How does relative atomic mass (RAM) relate to how chemicals actually exist?

The way that all chemicals exist is given by their formulae. E.g. The element oxygen does not exist as millions of separate oxygen atoms. Instead its atoms pair up to form molecules and as a result oxygen is given the formula O₂. This therefore means that oxygen particles have an average mass of twice the relative atomic mass of the element oxygen.

=2 x 16.0 = 32.0

This is referred to as the formula mass or molecular mass or molecular of oxygen.

Here are some examples of how other elements actually exist:

In view of this, you can see the importance of being able to write chemical formulae properly.

Don’t panic, however, help is at hand!!

You will be provided with a Data Sheet in all three of your tests, which provides much of what is needed.  Let us have a look at it and see how it may be used

THE ION TABLES THAT YOU WILL BE GIVEN.

Cations:

Anions:

If you look closely at these tables, you will notice that some of the ions are missing.  There is an excuse for this, for they are obtainable from the Periodic Table, which you are supplied with. Some of the missing ions are:

We see that:

Group I elements form + ions

Group II elements form 2+ ions

Aluminium in  group III forms a 3+ ion

We also see that:

Group VII elements form - ions

Group VI elements form 2- ions

NOTE: hat the negative ions shown, have a name ending with ide (e.g. Chloride, oxide and sulphide)

So far so good, but we are still missing important information.  We do not have the formulae of covalent substances.  You need to remember them. They are as follows:

Hydrogen (H₂)

Oxygen (0₂)

Nitrogen (N₂)

Chloride (Cl₂)

Bromine (Br₂)

Iodine (I₂)

Water (H₂0)

Carbon Dioxide (CO₂)

Carbon Monoxide (CO)

Sulphur Dioxide (SO₂)

Sulphur Trioxide (SO₃)

Hydrogen Sulphide (H₂S)

Hydrogen Peroxide (H₂0₂)

Nitrogen Dioxide (NO₂)

Nitrogen Monoxide (NO)

Amongst these ions are what are called Compound Ions.  These contain more than one element (e.g. CO₃^2- has both carbon and oxygen).  When you take more than one of such an ion, to balance the formula, you must place brackets around it.  (E.g. (NH⁴)₂CO₃ )

Perhaps you could now try to write formulae for the following:

(a)     Aluminium Oxide

(b)     Lead Sulphide

(c)     Calcium Hydrogencarbonate

Molecular Mass Of Elements and Compounds

(1)                 Start by writing the formula. E.g.

Chlorine Cl₂

Nitrogen N₂

Water H₂0

Sodium Chloride NaCl

Calcium Nitrate Ca(NO₃)₂

(2)           Break up the formula to add up the masses of everything present.

                (eg. For calcium nitrate above) Ca(NOO)(NOO)

1              x              Ca           =              1              x              40            =              40

2              x              N             =              2              x              14            =              28

6              x              O             =              6              x              16            =              96

                                                                                                                Total       164

You now have the molecular mass of the chemicals (164 for calcium nitrate.)

Perhaps you could now try to work out some molecular masses for yourself.  Below are some questions with the answers put in brackets.

(1)                 Calcium Carbonate (100)

(2)                 Lead Sulphate       (303)

(3)                 Copper (II) Chloride (134.5)

(4)                 Calcium Hydrogencarbonate (162)

2 NUMBERS OF PARTICLES

In chemistry we use a mole to describe the specific number of particles.  This is an extremely large number known as Avogadros number (6x10²³).  However since the particles of chemicals (atoms, molecules and ions) are so small, this usually turns out to be nothing more than would fit comfortably onto a watch glass.

How would you measure out the a mole of chemical?

The answer is to use the chemicals molecular mass, as shown earlier, and then take the mass of the chemical in grams.  You then have a mole of the chemical.  You have 6x10²³ formula particles.  Here are two illustrations:

1 mole of CO₂:                      6x10²³ CO₂ particles

O.2 mole of NaCl:                                0.2 x 6 x 10²³ NaCl particles

                                                                1.2 x 10²³ NaCl particles

3. VOLUME (WHERE THE CHEMICAL IS A GAS.)

As you can well imagine, it would be difficult for chemists to use mass when they want to measure out a particular amount of a gaseous chemical.  In this case volume is an easier approach.  Fortunately,

A Mole of any gas:

(1)                 At room temperature and pressure occupies 24dm³

[RTP = 25 ^oC and 1 atm]

(2)           At standard temperature and pressure occupies 22.4dm³

                [STP = 0^oC and 1 atm]

Here are some illustrations to show what arises from this information.

[at RTP]

1 mole of Oxygen:

(1)                 Occupies 24dm³

(2)                 Has 6x10²³ Oxygen Molecules (O₂)

(3)                 Has 12x10²³ Oxygen Atoms (O)

(4)                 Weighs 2x16= 32g

[at STP]

1 mole of Carbon Dioxide:

(1)                 Occupies 22.4dm³

(2)                 Has 6x10²³ Carbon Dioxide molecules (CO₂)

(3)                 Has 6x10²³ Carbon atoms

(4)                 Has 12x10²³= 1.2x10²⁴ O atoms

(5)                 Has 3x6x10²³ = 18x10²³ = 1.8x10²⁴ atoms in total

4 CONCENTRATION (WHERE A CHEMICAL IS DISSOLVED IN WATER)

When you make a solution two different measures become important.

(1)           How much water was used:                    VOLUME (cm³ or dm³)

(2)           How much chemical was used:                          MASS OR MOLES

You might say for example that you put:

                10g in 100cm³ water

Or            2.5g in 1 dm³ water

Chemists however, prefer to talk in terms of how many moles are dissolved in a particular volume of water.  Indeed they prefer to talk about how many moles you would have if you had 1dm³ (1000cm³) of the solution.

Here are some illustrations to show what this means.

1 Mole of Sodium Chloride (NaCl)

(1)                 Weighs (23 + 35.5) = 58.5g

(2)                 58.5g NaCl dissolved in 1000cm³ water would have the concentration 1.0mol dm^-3.

(3)                 5.85g NaCl dissolved in 100cm³ water would have the concentration 0.1mol dm^-3.

(4)                 5.85g NaCl dissolved in 100cm³ water would have the concentration 0.1mol in 100cm³                                                                          =1mol in 1000cm3

=1mol dm^-3

(5)                 If you take less water the solution becomes more concentrated, and vice versa.

This completes our look at the four different measures used by chemists.  You can be sure that you will meet examination questions that will assess your knowledge of these.  In some cases the questions will be as easy as the examples I have given.  However, you should expect to get some questions that involve several of these calculations all tied up together.  Once again, you should not panic, or worse still, give up!  Instead, begin by identifying the separate steps that will be needed, and then carry them out in the correct sequence.

Some students like to have the assistance of FORMULAE when it comes to calculations.  The diagram below shows all the possible steps that you may be required to carryout, together with formulae for carrying out the steps.  Note that each step can go in either direction.

Concentration:                                                                                          moles x 1000

         Volume

THE MOLE

Number of Particles                                                                                                                                                                             Mass=  mole x molecular mass

1mol= 6x10²³

Volume of Gas

1 mol at RTP = 24dm³

1 mol at STP = 22.4dm³

                What you can see from this diagram is that the mole measure is the most important, or most used by chemists.  You cannot, for instance work out the mass that a particular volume of gas would be, unless you first take the step of changing the volume into moles.  Having changed it into moles (e.g. 2.4dm³ of oxygen at RTP = 0.1 mole) you then change it from moles into mass (e.g. 0.1 mole of oxygen = 0.1 x 32 = 3.2g Oxygen).

CALCULATIONS FROM EQUATIONS

Most often, the balanced equation is given.  Where it is not provided, you must ensure that the equation you write is fully balanced.

Such equations are generally done in two easy stages.

(a)     Make an equivalence between what you are asked and what you are given.

E.g.                         2A           +              3B           =              C             +              4D

Let's say the question asks how much D would be produced when chemical B reacts fully with an excess of chemical A.

Here B will clearly run out first, and so its amount will determine how much C and D are produced.  So make an equivalence between B and D.

                                                                i.e.           3 mol. B = 4 mol. D

(b)     Put in the value you have been given ( in mole units) and cross multiply.

E.g.         3 mol. B = 4 mol. B

                1.5 mol. B               =              ?

Now try this technique with the following questions.

Qu. 1

In the reaction

                CuO        +              2HCl       =              CuCl₂      +              H₂O

7.95g copper oxide was completely reacted in an excess of Hydrochloric acid. Calculate the mass of copper chloride salt formed.

Qu. 2

In the reaction between sodium and an excess of an aqueous solution of zinc chloride, 5.75g of the metal was used.  Calculate the mass of zinc metal formed.

Qu. 3

In the reaction

                N₂            +              3H₂          =              2NH₃

28g of nitrogen was reacted with 8g of hydrogen.  Calculate the volume of ammonia produced, as measured at RTP. [ RTP = 24dm³]

   Garreth Mc Bride: Chemistry Department: Loreto College, Coleraine.