Topic 3: Stoichiometry

Stoichiometry is the arithmetic of chemistry: it links the symbols in a balanced equation to the actual masses, moles and volumes of substances that react. Once you can count particles using the mole, almost every quantitative chemistry problem becomes a sequence of clear, repeatable steps. This topic gives you the tools to predict how much product a reaction yields and how pure a sample really is.

Formulae and balanced chemical equations

A chemical formula shows the ratio of atoms in a substance, for example H2O means two hydrogen atoms joined to one oxygen atom. A balanced equation must obey the law of conservation of mass: atoms are never created or destroyed during a reaction, so the number of each type of atom must be identical on both sides. To balance, you place large numbers (coefficients) in front of formulae but never change the small subscripts inside a formula. For example, the burning of hydrogen starts as H2 + O2 -> H2O, which is not balanced because there are two oxygen atoms on the left but only one on the right. Adjusting the coefficients gives 2H2 + O2 -> 2H2O, where both sides now have four hydrogen atoms and two oxygen atoms. State symbols (s) solid, (l) liquid, (g) gas and (aq) aqueous can be added to show the physical state of each species.

Relative atomic mass and relative formula mass

The relative atomic mass (Ar) of an element is the average mass of its atoms compared with one twelfth of the mass of a carbon-12 atom; it is the weighted average that accounts for an element's isotopes. The relative formula mass (Mr), sometimes called relative molecular mass for simple molecules, is found by adding together the Ar values of every atom shown in the formula. For water, H2O, the Mr is (2 x 1) + 16 = 18. For calcium carbonate, CaCO3, the Mr is 40 + 12 + (3 x 16) = 100. When a formula contains brackets, multiply everything inside the bracket by the number outside it first: for magnesium hydroxide, Mg(OH)2, the Mr is 24 + 2 x (16 + 1) = 58. These mass values have no units because they are ratios.

The mole and the Avogadro constant

A mole is simply a counting unit, much like a dozen means twelve, except that one mole contains 6.02 x 10^23 particles. This huge number is the Avogadro constant, and it is chosen so that one mole of any substance has a mass in grams equal to its Ar or Mr. The central relationship is: number of moles = mass in grams divided by molar mass. For example, the number of moles in 36 g of water is 36 divided by 18, which equals 2 mol. Rearranging the same equation, the mass of 0.5 mol of calcium carbonate (Mr 100) is 0.5 x 100 = 50 g. Because one mole always contains the same number of particles, moles let chemists convert measurable masses into countable quantities of atoms or molecules.

Reacting-mass calculations

A balanced equation tells you the mole ratio in which substances react, and this ratio is the key to working out unknown masses. The reliable method has three steps: convert the known mass to moles, use the equation ratio to find moles of the wanted substance, then convert those moles back to mass. Worked example: what mass of magnesium oxide forms when 12 g of magnesium burns completely? The equation is 2Mg + O2 -> 2MgO. Moles of Mg = 12 divided by 24 = 0.5 mol. The ratio of Mg to MgO is 2:2, that is 1:1, so 0.5 mol of MgO forms. Mass of MgO = 0.5 x (24 + 16) = 0.5 x 40 = 20 g. Always start by checking the equation is balanced, or every later step will be wrong.

Concentration of solutions

The concentration of a solution describes how much solute is dissolved in a given volume of solution. It can be expressed in grams per cubic decimetre (g/dm3) or, more usefully for calculations, in moles per cubic decimetre (mol/dm3). Remember that 1 dm3 equals 1000 cm3. The key equation is: concentration in mol/dm3 = number of moles divided by volume in dm3. Worked example: if 0.2 mol of sodium chloride is dissolved to make 500 cm3 of solution, first convert the volume to 0.5 dm3, then concentration = 0.2 divided by 0.5 = 0.4 mol/dm3. To convert between the two units, multiply concentration in mol/dm3 by the Mr of the solute to obtain g/dm3. Concentration calculations are essential for titration work, where reacting volumes of solutions are measured precisely.

Percentage yield and percentage purity

In real laboratories the amount of product collected is usually less than the maximum predicted by the equation, because some product is lost during transfer, some reactions do not go to completion, and side reactions may occur. Percentage yield = (actual yield divided by theoretical yield) x 100. Worked example: if a reaction should theoretically produce 20 g of product but only 16 g is obtained, the percentage yield is (16 divided by 20) x 100 = 80 percent. Percentage purity measures how much of a sample is the desired substance: percentage purity = (mass of pure substance divided by total mass of sample) x 100. For instance, a 50 g sample containing 45 g of pure compound has a purity of (45 divided by 50) x 100 = 90 percent.

Empirical and molecular formulae

The empirical formula is the simplest whole-number ratio of atoms in a compound, while the molecular formula gives the actual number of atoms in one molecule. To find an empirical formula from masses or percentages: divide each element's mass or percentage by its Ar to get moles, then divide all the answers by the smallest of those values to obtain the simplest ratio. Worked example: a compound contains 40 percent carbon, 6.7 percent hydrogen and 53.3 percent oxygen. Dividing by Ar gives 40/12 = 3.33, 6.7/1 = 6.7 and 53.3/16 = 3.33. Dividing each by the smallest (3.33) gives 1 : 2 : 1, so the empirical formula is CH2O. If the relative molecular mass is known to be 60, divide 60 by the empirical mass of 30 to get 2, so the molecular formula is C2H4O2.

Gas volume calculations

Equal volumes of all gases at the same temperature and pressure contain the same number of molecules, a result known as Avogadro's law. At room temperature and pressure (r.t.p., roughly 20 degrees C and 1 atmosphere) one mole of any gas occupies a molar volume of 24 dm3, which is the same as 24000 cm3. The working equation is: number of moles of gas = volume in dm3 divided by 24. Worked example: the volume of 0.25 mol of carbon dioxide at r.t.p. is 0.25 x 24 = 6 dm3. Because the mole ratio in a balanced equation also applies to gas volumes, you can read volume ratios straight from the coefficients. For 2H2 + O2 -> 2H2O, 40 cm3 of hydrogen reacts with exactly 20 cm3 of oxygen, matching the 2:1 coefficient ratio.