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W3 REACTION STOICHIOMETRY
Amounts of Chemicals
Atoms and molecules are so small that we cannot directly measure the number in a sample or the number that change in a reaction. Instead, we use measurements such as the mass or volume to measure the amount. The volume can be converted to a mass using the density. The number of atoms or molecules present is related to the mass or volume through the atomic or molecular mass respectively.
For both cooks and chemists, it is most convenient to measure mass in grams and volume in litres (or millilitres).
A cook buying a large number of eggs will ask for them by the dozen, because it is less fiddly than dealing with the number of individual eggs. The chemist has the same problem with molecules, except that the number of molecules we deal with is huge. We count atoms and molecules not by the dozen but by the **mole ***. For the cook,
1 dozen eggs corresponds to 12 eggs. 2 dozen eggs corresponds to 24 eggs. 3 dozen eggplants corresponds to 36 eggplants. 2 dozen hydrogen atoms corresponds to 24 hydrogen atoms.
For the chemist,
1 mole of hydrogen atoms corresponds to 6.022 × 10 23 hydrogen atoms. 2 moles of water molecules corresponds to 2 × 6.022 × 10^23 water molecules. 2 moles of eggs corresponds to 2 × 6.022 × 10 23 eggs.
The principle is the same: a dozen always means 12 of a thing and a mole always means 6.022 × 10^23 of a thing. The only difference is the scale.
The molar mass is the mass, in grams, of one mole of a substance. The unit for the mole is “ mol ”, just like the unit for the dozen is “doz”.
Example The formula of carbon dioxide is CO 2.
One molecule of CO 2 contains one atom of carbon and two atoms of oxygen One mole of CO 2 contains one mole of carbon atoms and two moles of oxygen atoms.
The mass of a mole of atoms of an element is called the atomic mass and is given in the periodic table, such as the one on the back cover of this manual and the one you will get on the data sheet in your exam.
The atomic masses of carbon and oxygen are 12.01 and 16.00 g respectively †.
W3-
- (^) “mole” is not an abbreviation for “molecule”! † (^) Check that you can find these numbers on your periodic table.
The molecular mass of a compound is obtained by the summing the atomic weights of the
constituent atoms multiplied by the number of those atoms present. For CO 2
molar mass of CO 2 = 12.01 (C) + 2 × 16.00 (O) = 44.01 g mol–
One mole of CO 2 has a mass of 44.01 g and contains one mole of carbon atoms and two moles of oxygen atoms.
Ionic solids consist of an infinite array of ions rather than individual molecules. For these compounds, the formula mass is used. It is calculated from the formula in the same way as the molecular mass.
The formula mass is the mass, in grams, of one mole of an ionic compound.
Example Magnesium chloride is an ionic solid with the formula MgCl 2. The atomic masses of magnesium and chlorine are 24.31 and 35.45 g respectively ‡:
formula mass of MgCl 2 = 24.31 (Mg) + 2 × 35.45 (Cl) = 95.21 g mol–
Notice that is does not matter if you do not know whether something is ionic or not – the mass of a mole is the same.
Molar mass is referred to as: atomic mass if the substance is an element , molecular mass if the substance is a molecule and formula mass if the substance is an ionic compound.
Some elements, like helium and argon, exist as separate gaseous atoms. Others like iron and carbon exist as very large collections of atoms, joined together. In both of these cases, they are written with just the atomic symbol in chemical equations, such as He, Ar, Fe and C and the molar mass is referred to as atomic mass. Other elements, like hydrogen, oxygen and nitrogen, exist as small molecules like H 2 , O 2 and N 2. Most commonly these molecules are made up of two atoms, in which case they are called diatomics. For these elements, the molar mass refers to the mass of the diatomic and the atomic mass refers to the mass of the atom.
Example Oxygen in the atmosphere is present as either O 2 or O 3. The diatomic O 2 is the substance that we need to breath and is what we almost always mean when we talk about oxygen. O 3 is ozone, which is nasty if we breathe it but is needed in the upper atmosphere.
atomic mass of oxygen = 16.00 g molar mass of oxygen = molecular mass of O 2 = 2 × 16.00 = 32.00 g mol – molar mass of ozone = molecular mass of O 3 = 3 × 16.00 = 48.00 g mol –
‡ (^) Check that you can find these numbers on your periodic table.
Amounts and Reactions
Chemistry is all about how and why substances react to form new substances. The symbols in a chemical equation show the substances that react (the reactants ) on the left and the substances that are formed (the products ) on the right. The physical state of the various species is indicated by (s) for a solid, (l) for a liquid or (g) for a gas. An arrow is used to show the transformation:
reactants → products
When carbon burns in air, it combines with oxygen to make carbon dioxide:
C(s) + O 2 (g) → CO 2 (g)
One atom of carbon reacts with one molecule of oxygen to make one molecule of carbon dioxide. One mole of carbon reacts with one mole of oxygen to make one mole of carbon dioxide.
In this example, one C reacts with one O 2 : a 1:1 ratio. Not all chemical reactions are this simple and most involve more complicated ratios. For example, nitrogen and hydrogen react in a 1:3 ratio to produce ammonia, NH 3. This is shown in the chemical equation:
N 2 (g) + 3H 2 (g) → 2NH 3 (g)
One molecule of nitrogen reacts with three molecules of hydrogen to make two molecules of ammonia One mole of nitrogen reacts with three moles of hydrogen to make two moles of ammonia
The numbers in front of the chemical formulae in the chemical equation are called the reaction coefficients : they tell us the ratios in which the substances react. Notice that if there is no number in front of the formula, the coefficient is just one.
Example The chemical equation for the combustion of glucose is:
C 6 H 12 O 6 (s) + 6O 2 (g) → 6CO 2 (g) + 6H 2 O(l)
One mole of glucose reacts with six moles of oxygen to produce six moles of carbon dioxide and six moles of water.
Q6. What are the reaction coefficients for the following reactions? (The first one has been completed as an example.)
C 6 H 12 O 6 (s) + 6O 2 (g) → 6CO 2 (g) + 6H 2 O(l) C 6 H 12 O 6 : 1 O 2 : 6 CO 2 : 6 H 2 O: 6
(a) 2H 2 (g) + O 2 (g) → 2H 2 O(l) H 2 : O 2 : H 2 O:
(b) 4Fe(s) + 3O 2 (g) → 2Fe 2 O 3 (s) Fe: O 2 : Fe 2 O 3 :
(c) AgNO 3 + NaCl → AgCl + NaNO 3 AgNO^3 :^ NaCl:^ AgCl:^ NaNO 3 :
If we react one mole of glucose with six moles of oxygen, we can expect to get six moles of carbon dioxide and six moles of water. By multiplying each coefficient in the reaction by 0.5, it can be seen that if we react 0.5 mol of glucose with 3 mol of oxygen, we can expect to get 3 mol of carbon dioxide and 3 mol of water.
In the laboratory, we usually begin by weighing our reactants and end by weighing our products. This means we will need to convert these masses to moles, and vice versa using the method outlined previously.
Example If 10.0 g of glucose is burnt in a vessel which is open to the air, how many grams of water will be produced?
The calculation follows the sequence:
mass of glucose we start with
Æ
moles of glucose we start with
Æ
moles of water we expect to make
Æ
mass of water we expect to make
From Q1(b), the molar mass of C 6 H 12 O 6 is 180.16 g mol –1. This corresponds to,
moles of glucose =
mass
molar mass
= 0.0555 mol
The chemical equation, C 6 H 12 O 6 (s) + 6O 2 (g) → 6CO 2 (g) + 6H 2 O(l), shows that:
1 mol of C 6 H 12 O 6 produces 6 mol of H 2 O. 0.0555 mol of C 6 H 12 O 6 produces 6 × 0.0555 mol of H 2 O.
Therefore 0.333 mol of water will be produced. From Q1(a), the molar mass of H 2 O is 18.02 g mol–1, so the mass of water produced in the reaction is,
mass of water = number of moles × molar mass = 0.333 mol × 18.02 g mol –1^ = 6.00 g
Q7. What mass of Fe 2 O 3 is produced when 25.0 g of iron filings is burnt in air?
chemical equation:
moles of Fe at the start =
moles of Fe 2 O 3 expected at the end =
formula mass of Fe 2 O 3 =
mass of Fe 2 O 3 expected at the end =
Q8. N 2 and H 2 react to produce ammonia, NH 3 , according to the following chemical equation.
N 2 (g) + 3H 2 (g) → 2NH 3 (g)
What amounts of N 2 and H 2 will remain at the end of the reaction if you start with 5.0 mol of H (^2) and 2.0 mol of N 2 and react these to form NH 3?
Limiting reactant is
Moles of NH 3 that could be produced =
Moles of N 2 and H 2 left at the end of the reaction =
Q9. When solutions of copper(II) sulfate and barium chloride are mixed they react to produce solid barium sulfate according to the chemical equation:
CuSO 4 + BaCl 2 → CuCl 2 + BaSO (^4)
If the solutions contain 5.3 g of CuSO 4 and 8.2 g of BaCl 2 , which is the limiting reactant and what mass of BaSO 4 will be produced?
Hint: for each reactant, your calculation should follow the steps:
mass of reactant Æ moles of reactant Æ moles of product Æ mass of product
Demonstrator's Initials
Reaction Yield
As shown in the examples above, working out the limiting reactant enables us to calculate the mass of product we should obtain. This amount is called the theoretical yield. It is unlikely that this will be the actual yield. There are many reasons why the amount that we obtain is less than the theoretical yield. One limiting factor will always be the skill of the chemist. Sometimes the nature of the reaction means that even the best chemist cannot achieve a perfect yield. Many reactions lead to by-products that inevitably lower the yield. By working out the theoretical yield and measuring the actual yield, we can work out the percentage yield:
percentage yield =
actual yield
theoretical yield
×
Example In Q9, solutions containing 5.8 g of CuSO 4 and 8.2 g of BaCl 2 are mixed to produce a theoretical yield of 7.7 g of BaSO 4. If, in a real experiment, only 5.2 g of BaCl 2 is made:
percentage yield =
5.2 g
7.7 g
×
In this example, the actual and theoretical masses were used. The yields in moles can also be used as the units cancel.
Example In the example on page W3-6, solutions containing 2.0 mol of AgNO 3 and 3.0 mol of NaCl are mixed to produce a theoretical yield of 2.0 mol of AgCl. If in an experiment, only 1.7 mol of AgCl is made,
percentage yield =
1.5 mol
2.0 mol
×
Q10. If the reaction discussed in Q8 produces 60 g of ammonia, what is the percentage yield?
The actual yield must be less than or equal to the theoretical yield. If the actual yield is higher, it can mean that measurements or calculations are wrong. However, it is actually quite common for yields to be higher than is theoretically possible without any measurement or calculation errors having been made. It is important to consider why this might occur.
Q11. Suggest a reason why the yield might be higher than the theoretical yield.
Demonstrator's Initials
