2.1.3  Amount of Substance
The Mole
In Chemistry, reactions are often thought of in terms of the number of atoms. However, there are over 600 sextillion atoms in just 12 grams of carbon. An alternative to using these large numbers is to say ‘the number of atoms in 12 grams of carbon’, which we give a placeholder, the letter N, which is equal to this number over 600 sextillion.
Exactly what this value is, is the number of elemental entities in exactly n grams of an isotope with relative isotopic mass n. The most accurate approximation of this number was found by dividing the total charge of N electrons by the charge of an individual electrons. Finally, the number was named after an Italian scientist, Amedeo Avogadro. Thus this number is called Avogadro’s Constant.
$$ N_A=6.02\times{10^{23}} $$
A mole is defined as:
The amount of substance containing exactly $6.02214076×10^{23}$ elementary entities,
and so one mole of a substance has a mass in grams equal to its relative atomic, molecular or formula mass. The units for moles are $mol$.
Given this relationship between atomic mass, molar mass and the number of moles, an equation can be formed:
$$ \text{Molar mass}=\text{Number of Moles}\times\text{Relative atomic mass} $$
and, using appropriate units, where molar mass is in $g$ and number of moles is in $mol$, relative atomic mass must be in $g\space mol^{1}$.
It is also known that one mole of any gas at a certain temperature occupies a specific volume (varies depending on temperature), and so another equation can be used to calculate number of moles in a given volume, provided the molar gas volume is known.
$$ \text{Volume}=V_{m}\times\text{Number of moles} $$
Where $V_m$ is a constant specific to the temperature (e.g. 24.0 at room temperature, 298K). Again, using known units from volume and moles, the units of $V_m$ must be $dm^3mol^{1}$.
Determination of Formulae
Atoms of different elements can form molecules and compounds (molecules involve only covalent bonds), and each element in a compound contributes a certain percentage of its mass. If these proportions are known, the empirical formula can be determined.
The empirical formula is the simplest whole number ratio of atoms of each element present in a compounds, and can be deduced from the ratio of the moles of each element in the compound.
For example, if a compound has 43.7% of its mass from phosphorous and 56.3% from oxygen, then (assuming the sample is 100g), there are:
 $\frac{43.7}{31.0}$ = 1.41 moles of phosphorous
 $\frac{56.3}{16.0}$ = 3.53 moles of oxygen
in the compound. This has a ratio of 1.41 : 3.52, which when simplified, gives a ratio of approximately 2:5. Therefore the empirical formula of this compound is $P_2O_5$.
This formula has relative formula mass of 142 g mol$\small^{1}$, but if the actual compound had mass 284, we would need to multiply the number of moles by $\frac{284}{142}$= 2, getting us $P_4O_{10}$ (while an impossible compound, it’s use here is for explanation purposes), which matches the experimental data. This formula is the molecular formula  the number and type of atoms of each element in a molecule. Sometimes, the empirical and molecular formulae can be the same.
Water of Crystallisation
A salt in its crystalline form is very likely not pure. While crystallisation is said to produce a pure, dry salt, it does in fact produce a crystal that contains water. This is known as the water of crystallisation, and a salt that contains it is described as a hydrated salt. A salt that does not is instead described as anhydrous.
With this, the formula of a salt can in fact be expanded to include a set number of water molecules per salt compound. Using copper sulphate as an example, it is written as CuSO$\small_{4}$.5H$\small_{2}$O, but the number of water molecules may be different depending on the salt.
A hydrated salt can be made anhydrous by heating the ‘dry’ crystal until there is no further change in mass. This fact can be utilised to calculate the number of water molecules in the hydrated salt.
To do this, take a sample of the salt, measure its mass, and then heat. Remove from the heat and weigh regularly to see if the mass is constant and thus all water has been evaporated. Then measure the now anhydrous sample’s mass again. Calculate the difference in mass. This is the mass of water that was stored in the crystal.
The process that occurs here is known as thermal decomposition, and its equation is:
$$ \text{CuSO$_4$.5H$_2$O} _{\text{(s)}} \leftrightharpoons \text{CuSO$_4$} _\text{(s)} + \text{5H$_2$O} _\text{(g)} $$
Example Question:
13.2 g of a sample of zinc sulphate, ZnSO$\small_{4}$.xH$\small_{2}$O, was strongly heated until no further change in mass was recorded. Given that the final mass of the heated salt was 7.4g, calculate the value of x.
Solution:
First, calculate the number of moles of the zinc sulphate by dividing the mass by the A$_r$:
$\frac{7.4}{65.4\space+\space32.1\space+\space4\times16.0}$ = 0.0458 mol of ZnSO$\small_{4}$
Second, calculate the difference in mass:
13.2  7.4 = 5.8g of water
Third, divide the mass by the relative molecular mass of water:
$\frac{5.8}{18.0}$ = 0.322 mol of water
Finally, divide the number of moles of water by the number of moles of ZnSO$\small_{4}$ to find the value of x:
$\frac{0.322}{0.0458}$ = 7.03056… moles of water per mole of ZnSO$\small_{4}$, according to experimental data. This means that the value of x is essentially 7.
Therefore, ZnSO$\small_{4}$.7H$\small_{2}$O is the full equation of the hydrated salt.
Concentration
When using fluids instead of solids as reactants, concentration is used instead. Concentration is the amount of substance per unit volume of solution. It can be calculated using the equation:
$$ \text{Concentration}=\frac{\text{Amount of Substance}}{\text{Volume}} $$
where Amount of Substance may be expressed in either grams or moles, and volume is in $dm^3$. Therefore the accepted units for concentration are either $g\space dm^{3}$ or $mol\space dm^{3}$. This applies for either liquid or gas.
Using this equation and the one above involving mass, relative mass and moles; a lot can be determined about a reaction. For instance, if you know that exactly 26.55 cm$\small^{3}$ of 0.200 mol dm$\small^{3}$ NaOH neutralises 25.0 cm$\small^{3}$ of an unknown concentration of HCl, you can work out the concentration:

Form the full balanced equation for the reaction that is occurring:
$$ \text{NaOH} _\text{(aq)} + \text{HCl} _\text{(aq)} \rightarrow \text{NaCl} _\text{(aq)} + \text{H$_2$O} _\text{(l)} $$

Then calculate the number of moles of NaOH being used by multiplying the concentration by the volume, making sure to convert from cm$\small^{3}$ to dm$\small^{3}$:
0.200 x 0.02655 = 0.00531 mol of NaOH

Then compare the stoichiometry of the reaction, and deduce the number of moles of HCl that would react with 0.00531 moles of NaOH. In this case, the ratio is 1:1, and so 0.00531 moles of HCl react also.

This means that in the solution of HCl, there are 0.00531 moles of it per 0.025 dm$\small^{3}$, so we can calculate the concentration:
$\frac{0.00531}{0.025}$ = 0.2124 mol dm$\small^{3}$ of HCl.
Another example will be something along the lines of ‘Calculate the mass of pure, dry NaOH required to produce 250cm$\small^{3}$ of 0.200 mol dm$\small^{3}$ NaOH solution:’

Calculate the number of moles in that volume of that concentration:
0.250 x 0.200 = 0.0500 mol of pure, dry NaOH

Calculate the relative formula mass of NaOH:
23.0(Na) + 16.0 (O) + 1.0(H) = 40.0 g mol$\small^{1}$

Multiply the relative mass by the number of moles to get the mass of NaOH required:
0.05 x 40.0 = 2.0 g of NaOH
Ideal Gas Equation
An equation linking pressure, volume and temperature of an amount of ideal gas in a closed system is known as the ideal gas equation, and is:
$$ pV=nRT $$
where:
 ‘p’ is the pressure of the system measured in Pa (Pascals).
 ‘V’ is the volume in m$\small^{3}$.
 ‘n’ is the number of moles of gas (mol).
 ‘T’ is the temperature in K (Kelvin).
 ‘R’ is a constant, known as the gas constant. Its value and units are 8.314 J mol$\small^{1}$ K$\small^{1}$.
Percentage Yield and Atom Economy
Two separate numerical measurements as to the efficiency and practicability of reactions are percentage yield and atom economy.
Percentage yield is an experimental value, specifically the percentage that the actual yield is of the theoretical yield. The actual yield is determined by measuring the mass or volume of the product and converting into moles; whereas the theoretical yield is determined by comparing the stoichiometry of the reactant(s) and product. The equation to calculate it is then:
$$ \text{Percentage yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times100 $$
Atom economy is instead an entirely theoretical value, specifically the percentage by mass of all the products that is of useful product. For example, in the production of cyclohexane, C$\small_{6}$H$\small_{12}$ from hexane, C$\small_{6}$H$\small_{14}$, a byproduct of hydrogen gas, H$\small_{2}$, is produced. The total relative mass of the products is 86 g mol$\small^{1}$, whereas the useful product cyclohexane has 84 g mol$\small^{1}$. The equation to calculate atom economy is:
$$ \text{Atom economy}=\frac{\text{relative mass of useful product(s)}}{\text{total relative mass of all products}}\times100 $$
and so this reaction has 97.7% atom economy.
High atom economies and/or percentage yields are preferred as they reduce the cost of reactants over time, reduce the production of a dangerous byproduct, reduces the cost of disposing of waste products and reduces the cost of separating the desired product from a mixture containing byproducts or unreacted reactants.