|Chemistry I||© Ray Lovegrove|
Chemistry is a subject
highly dependent on measurement and number - for this reason we can not progress
far without coming up against a wide variety of chemical equations. Whether you view
the prospect of these calculations with joy or horror probably depends on whether you are
a keen mathematician or not. However, the best way to become skilled at chemical
calculations is with constant practice. If you do find some of these ideas tough going,
be sure to revisit this chapter often.
Relative Atomic Mass ( Ar)
The relative atomic mass
or relative isotopic mass (Ar) is the relative mass of the isotope, scaled
with carbon-12 as exactly 12. The actual mass of a atom is extremely small, for example, the mass
of a hydrogen atom is 1.674x10-27Kg and of a carbon atom is 1.993×10-26Kg.
So for chemical calculations, very accurate values of masses are not usually required and we use relative
atomic masses which have been rounded off to a convenient number. It is scaled that the mass of a carbon
atom is 12 (Carbon has different isotopes within which the C-12 was selected as the standard). These are
the masses given on the periodic table in your data sheets. Note that the
Relative Molecular Mass ( Mr)
The relative molecular mass of a substance is simply found by adding together all the relative masses of the constituent. For example, in lead nitrate, Pb(NO 3) 2, there are one lead, two nitrogen and six oxygen atoms. Its relative formula mass is therefore:
+ (2 × 14) + (6
× 16) = 331
We treat the relative masses of ions in a similar way.
The term ‘relative molecular mass’ is used for all covalent substances;
for ionic substances the term ‘relative
formula mass’ is often used. Note that the Mr of
specified compounds is given in atomic mass units (amu).
Amount of Substance
The amount of substance, n, of a sample or system is a physical quantity which is proportional to the number of elementary entities present. "Elementary entities" may be atoms, molecules, ions, electrons, or particles. Just as length or time, The amount of substance is one of the seven physical quantities in the SI base units. The SI unit for amount of substance is the mole. Why use amount of substance instead of mass or volume to tell how much of a substance there is? This is because in chemical reactions, the reagents react molecule-to-molecule, ion-to-ion, etc. For example, in the reaction below,
C + O2 ==> CO2
Mr of C is 12, while the Mr of O2 is 32. So in this reaction, if we measure the substance with "mass", we can see that 1 gram of C does not react exactly with 1 gram of O2. But if we measure them with Amount of substance, we can see that 1 mole of C react with exactly 1 mole of O2. In more complicated reactions, the usage of "amount of substance" is even more helpful.
We shall use the concept of ‘the mole’ quite frequently in chemistry, so it is
important to get this next section clear before going on to the calculations.
Basically, the mole is a quantity that chemists use to
prevent them having to use very large numbers. We use
words all the time; for instance, we buy eggs in "dozens" (12) and paper in
"reams" (500) . But if we count atoms or molecules, it would be impossible because
they are so small that even only 0.001gram of substance may contain millions of atoms
or molecules. so it should be easy to use moles
1 mole contains the same number of particles as there are in 12g of carbon-12 atoms
by definition.This number is called Avogadro's number or Avogadro's constant (NA) and is
equal to 6.022 x 1023 particles. Despite being a much larger number than a dozen
or a ream, moles serve the same purpose they are there to make dealing with quantities more
The Avogadro Constant ( L)
Avogadro was an Italian physicist who lived in the 19th century. He worked on the idea that a mole of any species always contains the same number of particles. Later it was found that it contained 6.023 ×10 23 particles and we now call this value the Avogadro number, NA.
Moles of Atoms
To find the number of atoms in a known number of moles,
multiply the moles by 6.022 x 1023. For example, 2 moles of Iron atoms contains
2× (6.022 ×10 23)=1.2044 x 1024 Iron atoms.
To find the moles of atoms, divide the number of atoms by 6.022 x 1023. For example, if we have 3.011×1023 Iron atoms,
the moles of iron atoms=3.011×1023 ÷ 6.022 x 1023= 0.5 mol .
Mole of Molecules
1 mole of molecules contains 6.022 x 1023 molecules. For example, 1 mole of oxygen molecules (O2) contains 6.022 x 1023 oxygen molecules (O2).
To find the moles of molecules, divide the number of molecules by 6.022 x 1023. So if we have 6.022 x 1022 chlorine molecules (Cl2), the moles of Cl2 molecules = (6.022 x 1022) ÷ (6.022 x 1023) =0.1 mol
But if we know the mass of the substance, we can also use relative formula mass to calculate the moles of atoms or molecules. For example, if we have 20 gram of solid sodium hydroxide(NaOH), the moles of NaOH is 20 g ÷ 40g/mol= 0.5 mol.
To find the number of molecules, multiply the moles of molecules by 6.023 x 1023. For example, ˝ mole of carbon dioxide molecules (CO2) contains ˝ x 6.022 x 1023 =3.011 x 1023 carbon dioxide molecules (CO2).
Chlorine gas Cl
mass of 71 because it is made up of two Cl atoms, each of
A mole of lead nitrate Pb(NO 3) 2 , contains a mole of lead atoms, two moles of nitrogen atoms and six moles of oxygen atoms.
Defining a Mole
For calculation purposes, a mole is
formula mass - the sum of masses of all the atoms involved - expressed in
grams. From Avogadro’s work, the mole can be defined as
‘an Avogadro’s number of particles’. This may look like two different
one is given in gram units, the other as a number of particles.
We can now start to think about calculations using our understanding of what moles are.
Calculation of Mole Quantities
One mole is equal to the Relative Formula Mass expressed in grams.
m = n × M r = 1 mol × [24 + ( 2 × 35.5)] = 95 g.
Moles to Masses
A quantity expressed in moles can be converted to grams by multiplying by the Relative Formula Mass.
= 2.5 × 95 = 237.5 g.
Masses to Moles
The process can be reversed by dividing by the molar mass.
Express 10g of magnesium chloride (MgCl2) in moles.
= 10/95 = 0.105 mole.
The Chemical Formula is a formula that expresses the constitution of a substance by the symbols of elements. For molecular compounds the chemical formula is also known as the molecular formula, while for ionic compounds and other non-molecular substances, it is known as the empirical formula. Chemical formula also includes structure formula. These are all important and frequently used in chemical calculations.
The empirical formula of a chemical compound is a simple expression of the relative numbers of each type of atom in it. It expresses the simplest whole number mole ratio of the atoms of different elements present. Many compounds do not exist in the form of molecule, so an empirical formula will be useful. For example, NaCl means in the crystal of sodium chloride, the ratio of atoms of Na and Cl is 1:1. But NaCl is not a molecule. Different organic compounds often have the same empirical formula. Such as the empirical formulae of ethyne and benzene are both CH, but their molecular formula are C2H2 and C6H6, respectively.
1.20 g of magnesium reacts with excess chlorine to form 4.75g of magnesium chloride. Calculate the empirical formula of magnesium chloride.
masses are:    1.20 (Mg)
3.55 g (Cl)
Moles : 0.05 (1.2/24) and 0.10 (3.55/35.5)
Ratio is 1 : 2
Therefore the empirical formula is MgCl2.
The Molecular Formula
The Molecular Formula specifies the actual number of atoms of each element in a molecule. This is the true formula of a molecular compound And we should contrast it to the above definition of empirical formula. To find this formula, both the empirical formula and the Mr must be known. The molecular formula is always a simple multiple of the empirical formula.
The empirical formula of hydrogen peroxide is HO and its Mr 34. Calculate the molecular formula.
Therefore, the Mr is 17n.
17n = 34
n = 2
Molecular formula = (HO)2 or H2O2.
If a chemical formula is known, the percentage composition by mass of each element present may be calculated.
Calculate the percentage of aluminium present in aluminium oxide whose formula is Al2O3.
% Al = (2 × Ar Al / Mr Al2O3) ´ 100
= 54/102 × 100
If the chemical equation for a reaction is known, the quantities taking part in that reaction can be calculated.
Calculate the fraction of a mole of sodium chloride formed when 0.2 mole of sodium carbonate reacts with excess hydrochloric acid.
Na2CO3(s)+ 2HCl(aq) ==> 2NaCl(aq) + CO2(g) + H2O(l)
1 mole sodium carbonate gives 2 moles of sodium chloride.
Calculation of Reacting Masses
The above method can be extended to the calculation of reacting masses:
Calculate the mass of sodium chloride produced when 20g of sodium carbonate reacts with excess hydrochloric acid.
Na2CO3(s) + 2HCl(aq) ==> 2NaCl(aq) + CO2(g) + H2O(l)
Mr = 106 Mr = 58.5
20 g. sodium carbonate = 20/106 mole = 0.189 mole
0.189 mole sodium carbonate gives 2 × 0.189 mole NaCl
2 × 0.189 = 0.378 mole = 0.378 × 58.5 = 22.1 g. NaCl
We have learned the relation between the volume, density and mass of substance (density=mass/volume). So now in chemical calculations, if we know the amount of a substance, its relative formula mass and its density, then we can calculate its volume. For solid and liquid, the volumes of 1 mole of substance are different. But Avogadro noted that the volume occupied by one mole of various gases under the same conditions of temperature and pressure was always similar. The reason is that the volume of a substance depends on three factors, which are the number of the particles, the volume of the particles and the distance between the particles. The distances between gas molecures are much bigger than that between solid or liquid molecures. For example, the volume of 1 mol of H2O is 18mL at 273K, but the volume can be expanded 1700 times when the temperature becomes 373K. So the gas volume depends more on the distance rather than on the volume of each molecure. When under the same conditions, the distances in between are always similar for different gas. Hence 2g of hydrogen and 44g of carbon dioxide always occupy the same volume under the same conditions.
This volume is referred to as the molar volume of a gas. It is very dependent on temperature and pressure, but under standard conditions of temperature and pressure (known as S.T.P.) it is taken as 22.4dm 3. S.T.P. is defined as 273K temperature (that is 0 °C, but should always be given using the absolute scale in calculations) and 1 atmosphere (atm) pressure.
Consider the reaction which occurs when calcium carbonate is dissolved in hydrochloric acid to release carbon dioxide:
CaCO3(s)+ 2HCl(aq) ==> CaCl(s) + CO2(g) + H2O(g)
From the equation, we can write:
40 + 12 + 3 (
×16) calcium carbonate
100g calcium carbonate produces 44g of carbon dioxide
100g calcium carbonate produces 22.4dm 3 of carbon dioxide at S.T.P.
Concentration of Solutions
A molar solution (a 1 mol/dm3 solution) is obtained by dissolving 1 mole of material in 1dm 3 water (that is 1000cm 3 water). Molarity is a concentration term, so a 1 mol/dm3 solution can also be obtained by dissolving:
1 mole material in 1dm 3 water
0.5 mole material in 0.5dm 3 water
2 mole material in 2dm 3 water, etc.
Note that the abbreviation for ‘mole’ is ‘mol’ but this should only be used when writing units and not in sentences. Also you will find some textbooks use ‘M’ instead of ‘mol/dm3’ – ‘M’ is in fact obsolete and should no longer be used. Those of you who are mathematicians will know that mol/dm3’ can be written as ‘mol dm-3’ it’s just another way of stating the same thing. If you are not a mathematician, just be aware that both forms are in common use and mean exactly the same.
Solutions of other molarities can be obtained as follows:
1 mole material dissolved in 0.5dm 3 water gives a 2 mol/dm3 solution
1 mole material dissolved in 0.25dm 3 water gives a 4 mol/dm3 solution
0.5 mole material dissolved in 1dm 3 water gives a 0.5 mol/dm3 solution
0.1 mole material dissolved in 1dm 3 water gives a 0.1 mol/dm3 solution
If we refer back to the earlier part of the chapter, we can see that we need 2 mole hydrogen chloride to dissolve 1 mole calcium. Hence we could use:
20dm 3 of a 0.1 mol/dm3 solution of hydrogen chloride in water
1dm 3 of a 2 mol/dm3 solution of hydrogen chloride in water, etc.
In calculations such as these, the method is always to calculate the molar mass needed, and then convert it into a concentration of solution.
We will be using these chemical calculations in later chapters so
please return to the examples as often as you need to. Remember
- practise makes perfect! Conclusion
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