Chemistry I | © Ray Lovegrove |

Chemical Calculations

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 (
A_{r})**

The relative atomic mass
or relative isotopic mass (*A _{r}*) 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

**Relative Molecular Mass (
M_{r})**

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:

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 + O_{2} ==> CO_{2}

M_{r} of C is 12, while the Mr of O_{2} 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 O_{2}. But if we measure
them with Amount of substance, we can see that 1 mole of C react with exactly 1 mole of O_{2}.
In more complicated reactions, the usage of "amount of substance" is even more helpful.

**The Mole**

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
quantity
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
(6.023
×10
^{2
3}) for
particles.
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 (N_{A}) and is
equal to 6.022 x 10^{23} 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
manageable.

**The Avogadro Constant (
L)**

Avogadro was an Italian physicist
who lived
in the 19^{th} 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, *N _{A}*.

**Moles of Atoms**

To find the number of atoms in a known number of moles,
multiply the moles by 6.022 x 10^{23}. For example, 2 moles of Iron atoms contains
2× (6.022 ×10 ^{23})=1.2044 x 10^{24} Iron atoms.

To find the moles of atoms, divide the number of atoms by 6.022 x 10^{23}.
For example, if we have 3.011×10^{23} Iron atoms,

the moles of iron atoms=3.011×10^{23} ÷ 6.022 x 10^{23}= 0.5 mol .

**Mole of
Molecules**

1 mole of molecules contains 6.022 x 10^{23} molecules. For example,
1 mole of oxygen molecules (O_{2}) contains 6.022 x 10^{23} oxygen molecules
(O_{2}).

To find the moles of molecules, divide the number of molecules by 6.022 x 10^{23}.
So if we have 6.022 x 10^{22} chlorine molecules (Cl_{2}),
the moles of Cl_{2} molecules = (6.022 x 10^{22}) ÷ (6.022 x 10^{23}) =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 10^{23}.
For example, ˝ mole of carbon dioxide molecules (CO_{2}) contains
˝ x 6.022 x 10^{23} =3.011 x 10^{23} carbon dioxide molecules (CO_{2}).

**Other Examples**

Chlorine gas Cl
2, has
a
relative molecular
mass of 71 because it is made up of two Cl atoms, each of _{r}
35.5. A mole of chlorine gas has a mass of 71g, and
contains two moles of chlorine atoms, each of mass
35.5g.

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
the
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’.

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.

Example
The mass of one mole of
magnesium chloride MgCl
m = n × M |

**Moles to
Masses**

A quantity expressed in moles can
be
converted to grams by multiplying by the Relative Formula Mass.

Example 2.5 moles of magnesium
chloride: = 2.5 × 95 = 237.5 g. |

**Masses to
Moles**

The process can be reversed by dividing by the molar mass.

Example Express 10g of
magnesium chloride (MgCl = 10/95 = 0.105 mole. |

**Chemical Formula **

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
C_{2}H_{2} and C_{6}H_{6}, respectively.

Example 1.20 g of magnesium reacts with excess
chlorine to form 4.75g of magnesium chloride. Calculate the empirical formula
of magnesium chloride.
Moles : 0.05 (1.2/24) and 0.10
(3.55/35.5) Ratio is 1 : 2 Therefore the
empirical formula is MgCl |

**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*
M _{r}* must be known. The molecular formula
is always a
simple multiple of the empirical formula.

**Percentage
Composition**

If a chemical formula is known, the
percentage composition by mass of each element present may be calculated.

Example
Calculate the percentage of aluminium
present in aluminium oxide whose formula is
Al %
Al = (2 ×
M Al_{r}_{2}O_{3})
´ 100 = 54/102
×
100 = 52.9 |

**Reacting
Quantities**

If the chemical equation for a
reaction is
known, the quantities taking part in that reaction can be
calculated.

Example
Calculate the fraction of a mole of sodium
chloride formed when 0.2 mole of sodium carbonate reacts with excess
hydrochloric
acid.
1 mole sodium carbonate gives 2 moles of sodium
chloride. Therefore, |

**Calculation of Reacting
Masses**

The above method can be extended
to the
calculation of reacting masses:

Example
Calculate the mass of sodium chloride
produced when 20g of sodium carbonate reacts with excess hydrochloric
acid.
Na
M = 58.5_{r}20 g. sodium carbonate = 20/106 mole =
0.189
mole 2
×
0.189
= 0.378 mole = 0.378
×
58.5 = 22.1 g. NaCl |

**Gas
Volumes**

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 H_{2}O 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.

**Concentration of
Solutions**

A molar solution (a 1 mol/dm^{3
}
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/dm^{3} 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.

^{3}
solution

1 mole material dissolved in
0.25dm
3
water gives a 4 mol/dm^{3}
solution

0.5 mole material dissolved in 1dm
3
water gives a 0.5 mol/dm^{3}
solution

0.1 mole material dissolved in 1dm
3
water gives a 0.1 mol/dm^{3}
solution

^{3}
solution of hydrogen chloride in water

20dm
3
of a 0.1
mol/dm^{3} solution of
hydrogen chloride in water

1dm
3
of a 2
mol/dm^{3}
solution of hydrogen chloride in water, etc.

**Conclusion
**

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