3.1.2 Amount of Substance - Balanced equations and associated calculations

Specification

 Students should: be able to write balanced equations (full and ionic) for reactions studied be able to balance equations for unfamiliar reactions when reactants and products are specified. (This is an important skill that applies in all units.) be able to calculate reacting volumes of gases be able to calculate concentrations and volumes for reactions in solutions, limited to titrations of monoprotic acids and bases and examples for which the equations are given know that % atom economy = mass of desired product x 100 /total mass of reactants be able to calculate reacting masses, % yields and % atom economies from balanced equations

Word equations

Word equations simply show the names of the reacting chemical and products. They are of limited use except for giving an overall description of the chemical reaction. They give no indication of the relative amounts of the reactants or products involved.

 Example: Sodium hydroxide + sulfuric acid sodium sulfate + water

To make equations useful, they must show the individual formulae of the reactants and products and indicate the relative quantities in which they react. These 'formula equations' are dealt with below.

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Formula equations and coefficients

Formula equations show the formulae of the reactants and the products on either side. Balancing numbers are used - called the coefficients of the reaction - to ensure that the numbers of particles on both sides of the equation are equal.

In this equation one nitrogen molecule is needed to react with every three hydrogen molecules to produce 2 molecules of ammonia. The coefficient of nitrogen is 1, that of hydrogen is 3, and that of ammonia is 2.

To balance a chemical equation it is important to remember that the formula of the reactants and products cannot be changed and that coefficients may only be placed before the formulae, multiplying them by whole numbers.

 Example: One stage in the manufacture of nitric acid is the oxidation of ammonia, as shown below: 4NH3 + _O2 _NO + _H2O What is the coefficient for O2 when the equation is balanced? Solution: Balance the nitrogen by counting up the nitrogen atoms on both sides: 4NH3 + _O2 4NO + _H2O Balance the hydrogen atoms: 4NH3 + _O2 4NO + 6H2O Now balance the oxygen atoms: 4NH3 + 5O2 4NO + 6H2O Correct response : Coefficient = 5

These show the actual numbers of reacting particles in a chemical reaction. The reaction must be BALANCED to give the correct number of particles on each side of the reaction arrow.

 Example: 2NaOH + H2SO4 Na2SO4 + 2H2O

These equations are constructed by writing the formula of each of the compounds in the reaction, and then by counting up the number of atoms on each side to make sure they are equal. If they are not equal, balancing numbers (coefficients) are added in front of each chemical formula (where needed), so that the numbers of each type of particle on each side of the equation are the same

Step 1 - write the chemical equation

ammonia + oxygen nitrogen monoxide + water

Step 2 - write the formula of each of the reaction components

ammonia + oxygen nitrogen monoxide + water

NH3 + O2 NO + H2O

Step 3 - add coefficients IN FRONT OF the formulae to balance the equation

4NH3 + 5O2 4NO + 6H2O

Note Whenever an exam question asks for an equation, it is the balanced formula equation that is required, unless specified otherwise.

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Ionic equations

When ionic solutions react, the reaction only usually involves some of the ions and not others. An ionic equation shows just the ions implicated in the reaction. The other ions are often called "spectator ions".

 Example: 2NaOH + H2SO4 Na2SO4 + 2H2O The ionic equation is written as: H+ + OH- H2O

Although the formula equation represents the overall process, the sodium ions start off in solution as Na+(aq) and at the end of the reaction they are still Na+(aq), nothing has changed, they are merely spectator ions. The same applies to the sulfate ions, SO42-. The only particles that actually react are the OH- ions from the sodium hydroxide and the H+ ions from the sulfuric acid.

State symbols

These are used to show the states of the various compounds that constitute the equation for the reaction.

(s) means that the compound is in the solid state.

(l) means that the compound is in the liquid state.

(g) means that the compound is in the gaseous state.

(aq) means that the compound is dissolved in water, i.e. it is in solution.

 Example: 2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l)

The sodium hydroxide and the sulfuric acid are in solution. These make sodium sulfate in solution and water liquid.

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Reacting volumes of gases - Gay Lussac's law

In 1808 the Freanch chemist Gay Lussac investigating the reactions of gases came to the conclusion that when gases combine chemically they do so in volumes that have a simple ratio to one another, and to any gaseous product, provided that all gases are measured at the same temperature and pressure. This became known as Gay Lussac's law.

 Example: The reaction between hydrogen and oxygen making water hydrogen + oxygen water 2H2 + O2 2H2O The volume of hydrogen needed for complete reaction is always double the volume of oxygen, provided the temperature and pressure of the two gases are the same. 2 volumes of hydrogen 1 volume of oxygen

Extending Gay Lussac's work in 1811, Avogadro suggested that "equal volumes of all gases contain equal number of molecules (the gases being measured at the same temperature and pressure). This became known as Avogadro's hypothesis or Avogadro's law.

This means that in the reaction:

N2 + 3H2 2NH3

For a given volume of nitrogen, three times the volume of hydrogen is needed for complete reaction. The volume of nitrogen contains a certain number of molecules and there are three times as many molecules in the volume of hydrogen.

N2 + 3H2 2NH3

1 molecule + 3 molecules 2 molecules

1 volume + 3 volumes 2 volumes

Avogadro's hypothesis allows us to substitute the coefficients of any balanced gaseous equation for volumes of gas, for example:

 Example: Find the volume of hydrogen required to react completely with 200 cm3 of nitrogen according to the equation: N2 + 3H2 2NH3 The equation tells us that 1 volume of nitrogen reactrs completely with 3 volumes of hydrogen: Therefore volume of hydrogen = 3 x volume of nitrogen Therefore volume of hydrogen = 3 x 200 cm3 = 600 cm3
 Example: Using the equation: N2 + 3H2 2NH3 if 100cm3 of nitrogen was used (this is now equivalent to 1 'volume') it would need 3 times as much hydrogen for complete reaction i.e. 3 volumes = 300cm3. The reaction would produce 2 volumes of ammonia i.e. 2 x 100cm3 = 200cm3

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Concentrations of solutions

The concentration of a solution is the quantity of solute that it contains per unit volume.

This may be given in grams per 100cm3 or grams per litre, but it is usually given in terms of molarity as this gives a direct measure of the number of solute particles contained by the solution.

Molarity

The concept of molarity arises from the need to know the amount of solute present in a solution in moles. 1 mole of any substance contains an Avogadro number of particles of that substance = 6.02 x 1023. A 1 molar solution contains 1 mole of solute, dissolved in 1 litre of solution.

Note: the definition is not per 1 litre of solvent, but per 1 litre of solution. This allows us to measure a volume of solution and work out the number of moles, and hence the number of particles, that it contains.

Molarity = number of moles of solute per litre of solution (1 litre = 1000cm3)

molarity = moles/litres

The molarity is denoted by the capital letter M, or the units mol dm-3

 Example: Calculate the molarity of a solution containing 0.15 moles of potassium nitrate in 100cm3 of solution. Molarity = moles/litres 100cm3 = 0.1dm3 Molarity = 0.15/0.1 = 1.5 M

The molarity of a specific ion within an ionic solution may also be considered separately. In a 1M solution of copper sulfate (CuSO4) the copper ions are separate from the sulfate ions. The solution may be said to be both 1 molar in terms of copper 2+ ions and 1 molar in terms of sulfate 2- ions. A 1 molar solution of copper nitrate (Cu(NO3)2), however, is 1 molar with respect to copper 2+ ions but 2M with respect to nitrate ions.

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Reactions in solutions

The dissolution process

Substances dissolve because they form bonds with the solvent molecules. This bond formation (an exothermic process), coupled with the entropy increase on forming a solution, makes the process thermodynamically favourable.

For insoluble substances, the opposite is true; either the bonding within the solid is too strong and requires too much energy to break it, or the bonds formed with the solute molecules are too weak to make the process favourable.

Dissolution, then, is a balance between several factors, making prediction of solubility difficult. Unfortunately, solubility is something that has to be learned, but fortunately, there are a few facts that make the task a little easier. See precipitation reactions.

Reactions occur easily in solution, as the particles are in constant motion and can come into contact with one another. The majority of solutions dealt with in chemistry are ionic. The ions in the structure of an ionic solid are broken apart by interaction with the water molecules. Energy is required to overcome the strong electrostatic forces within the ionic lattice.

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Titration

The procedure involves adding one solution to another solution, in the presence of an indicator that shows when the two solutions have completely reacted. If the concentration and volume of one of the solutions is known and the volume of the other solution is known, then the unknown concentration can be calculated providing the stoichiometry of the reaction is known.

The experimental procedure is very accurate and liquids are carefully measured, using specially designed glass apparatus with low margins of error.

Titrations are intended for analytical and determination/estimation purposes. If the compound being used for the titration cannot be obtained in a very pure state then it first must be standardised against a solution whose concentration can be determined to a high degree of accuracy. Such substances are called primary standards.

Experimental procedure:

• Measure 25cm3 of one solution into a conical flask
• Add a few drops of indicator solution
• Add a second solution very slowly from a burette until the indicator just changes colour
• Repeat the process until concordant results are obtained (results within 0.1 cm3 of one another)

Calculations

The number of moles of solute in the known solution is calculated using the relationship:

moles = molarity x volume

Then the stoichiometric relationship between this substance and the other reacting solution is used to determine the number of moles of solute in the unknown solution.

 Example: Equation stoichoimetry: Sodium hydroxide + nitric acid sodium nitrate + water NaOH + HNO3 NaNO3 + H2O 1 mole of NaOH is equivalent to 1 mole of HNO3

In the last example the stiochiometric relationship is:

moles of base = moles of acid

The molarity of the unknown solution can now be calculated from the number of moles present and the volume used:

molarity = moles/volume(litres)

 Example - Determination of the concentration of an unknown sodium hydroxide solution. Procedure 25cm3 of the unknown base is measured into a conical flask using a pipette. 5 drops of phenolphthalein indicator is added A standard solution of potassium hydrogen phthalate (an acid) is added from a burette until the red colour of the inicator has completely disappeared. The burette reading is noted. The process is repeated several times until concordant results are obtained (results within 0.1 cm3 of one another) The potassium hydrogen phthalate reacts in a 1:1 ratio with sodium hydroxide Therefore at the end-point (equivalence point) Moles of potassium hydrogen phthalate = moles of sodium hydroxide molarity(acid) x volume(acid) = molarity(base) x volume(base) The molarity of the unknown sodium hydroxide solution may now be calculated.

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Atom economy

This term refers to the efficiency of a chemical process in terms of the atoms that are lost as by-products to the intended product. For example in the reaction between nitrogen and hydrogen to produce ammonia there are no other products so the atom economy is 100%

However, in the reaction between ammonia and oxygen to produce nitrogen(II) oxide and water the equation is:

4NH3 + 5O2 4NO + 6H2O

Of the original reactants with 26 atoms we produce only 8 atoms in the nitrogen(II) oxide and we lose 18 atoms in the water which is formed as a by -product. This is poor atom economy, unless we go on to use the water produced in a subsequent part of the overall process.

Industrial processes with poor atom economy are inefficient in terms of resources and often produce undesirable waste.

The atom ecomomy is calculated as a percentage of desired product divided by the total mass of products and scaled up by multiplying by 100 to make a percentage.

For example, in the above reaction the total mass of nitrogen(II) oxide is 4 x 30 = 120, while the total mass of products is 120 + 108 = 228

Hence the percentage atom economy is 100 x 120/228 = 52.6%

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The reagent in excess

When there is more of one of the reactants present than the required amount, the extra will not have anything to react with. This should be apparent bearing in mind the particulate nature of matter. If two molecules of hydrogen react with one molecule of oxygen to make two molecules of water, then any 'extra' molecules of hydrogen (or oxygen for that matter) will be unable to react.

Three hydrogen molecules reacting with one oxygen molecule

3H2 + O2 2H2O + ?

From the equation you can see that one hydrogen molecule remains unreacted. We call this the excess reagent.

3H2 + O2 2H2O + H2

An extra amount of chemical over and above that which is needed for complete reaction is called the excess. A reagent in excess (i.e. one of which there is more than the amount needed) cannot completely react. Some of it can react, but the rest simply remains unreacted after the reaction has finished.

Calculating the excess

To find the excess reagent, the first stage is to calculate the number of moles of each reagent in the reaction. Then the stoichiometry of the equation shows the relative number of moles reacting in an ideal situation. The excess is found by substituting the number of moles of the first reagent (reacting chemical) in the given situation and seeing how many moles of the second reagent is required for complete reaction.

If there is more than enough of the second reagent then it is in excess. If there is not enough of the second reagent then the first reagent is in excess.

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The Limiting reagent

In the above example it can be seen that the component that determines the amount of sulfur that can react is the iron. All of the iron manages to react and it is this that determines the quantity of products formed. The iron is said to LIMIT the reaction. It is called the limiting reagent.

To determine the limiting reagent (and to find out which of the reactants is in excess) the stoichiometry of the reaction must be considered.

Procedure

• A - Firstly find the relative number of moles of each component in the balanced equation.
• B - Then convert the data given in the question under study into moles.
• C - Now by inspection see which one of the components will completely react and which one will be in excess.

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