Conservation of Mass & Moles
The Law of Conservation of Mass
In any chemical reaction, no atoms are created or destroyed — they are simply rearranged to form new substances. This means that the total mass of the reactants always equals the total mass of the products. This principle is known as the law of conservation of mass.
If you carry out a reaction in a sealed (closed) container — for example, a precipitation reaction in a stoppered flask — the mass before and after the reaction will be identical. This is because nothing can enter or leave the system.
In an open container, mass may appear to change. If a gas is produced (e.g. \(CO_2\) from a carbonate reacting with acid), the gas escapes and the mass decreases. If a gas is taken in (e.g. when magnesium burns in air and gains oxygen), the mass of the solid product increases. In both cases, mass is actually conserved — but some of it has entered or left the container.
Relative Formula Mass
The relative formula mass (\(M_r\)) of a compound is found by adding up the relative atomic masses (\(A_r\)) of every atom in its formula. For example, for water (\(H_2O\)):
$$M_r = (2 \times 1) + 16 = 18$$
For calcium carbonate (\(CaCO_3\)):
$$M_r = 40 + 12 + (3 \times 16) = 100$$
You can calculate the percentage by mass of an element in a compound using:
$$\text{percentage by mass} = \frac{A_r \times \text{number of atoms of element}}{M_r} \times 100$$
For example, the percentage by mass of oxygen in water (\(H_2O\), \(M_r = 18\)):
$$\frac{16}{18} \times 100 = 88.9\%$$
Balancing Equations and Reacting Masses
A balanced chemical equation shows the same number of atoms of each element on both sides. This reflects the law of conservation of mass. You can use a balanced equation to calculate the mass of reactants or products.
Calculate the mass of carbon dioxide produced when 10 g of calcium carbonate reacts with excess hydrochloric acid:
$$CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$$
Step 1: Find the \(M_r\) of \(CaCO_3 = 100\) and \(CO_2 = 44\).
Step 2: From the equation, 1 mole of \(CaCO_3\) produces 1 mole of \(CO_2\), so 100 g of \(CaCO_3\) produces 44 g of \(CO_2\).
Step 3: Scale down: \(\frac{10}{100} \times 44 = 4.4\) g of \(CO_2\).
To calculate reacting masses: (1) write the balanced equation, (2) find the \(M_r\) of the substances involved, (3) use the mole ratio from the equation to scale up or down from the known mass.
The Mole
The mole is a unit used to measure the amount of a substance. One mole of any substance contains exactly \(6.02 \times 10^{23}\) particles (atoms, molecules, or ions). This number is called the Avogadro constant.
The mass of one mole of a substance in grams is numerically equal to its relative formula mass. For example, one mole of carbon (\(A_r = 12\)) has a mass of 12 g. One mole of water (\(M_r = 18\)) has a mass of 18 g.
The key equation linking mass and moles is:
$$\text{moles} = \frac{\text{mass (g)}}{M_r}$$
One mole of any substance contains \(6.02 \times 10^{23}\) particles (the Avogadro constant). To convert between mass and moles, use: moles = mass \(\div\) \(M_r\).
The concentration of a solution tells you how much solute is dissolved in a given volume. It is calculated using:
$$\text{concentration (g/dm}^3\text{)} = \frac{\text{mass of solute (g)}}{\text{volume of solution (dm}^3\text{)}}$$
For example, if 5 g of sodium chloride is dissolved in 250 cm\(^3\) of water (= 0.25 dm\(^3\)):
$$\text{concentration} = \frac{5}{0.25} = 20 \text{ g/dm}^3$$
Empirical and Molecular Formulae
The empirical formula shows the simplest whole-number ratio of atoms of each element in a compound. It can be calculated from reacting masses or percentage composition. The molecular formula shows the actual number of atoms in one molecule and can be deduced from the empirical formula and the relative molecular mass.
For example, ethane has the molecular formula \(C_2H_6\). Its empirical formula is \(CH_3\) (dividing by the common factor of 2).
You can determine the empirical formula of magnesium oxide by heating a known mass of magnesium ribbon in a crucible. Weigh the crucible before and after heating. The increase in mass equals the mass of oxygen that reacted. Divide each mass by its \(A_r\) to find the mole ratio, then simplify to get the empirical formula (MgO).
Limiting Reactants
In most reactions, one reactant is used up before the other. The reactant that is completely used up is called the limiting reactant — it controls the amount of product formed. The other reactant is said to be in excess. Adding more of the excess reactant will not produce any more product.