Calculations in Calculations in chemistry: stoichiometrychemistry: stoichiometry

Chapter 15Chapter 15

Return of the moleReturn of the mole

The mole is a very useful unit The mole is a very useful unit because it represents a ‘measurable’ because it represents a ‘measurable’ number of particles – atoms, ions or number of particles – atoms, ions or molecules.molecules.

The mole concept effectively acts as The mole concept effectively acts as a scale up from individual particles to a scale up from individual particles to measurable amounts of particles.measurable amounts of particles.

StoichiometryStoichiometry

In a chemical reaction, atoms are In a chemical reaction, atoms are neither created nor destroyed.neither created nor destroyed.

Consequently, given the amount of a Consequently, given the amount of a reactant consumed or a product reactant consumed or a product formed we can calculated the formed we can calculated the amounts of other reactants and amounts of other reactants and products involved.products involved.

We are now applying the mole to We are now applying the mole to chemical reactions and not just chemical reactions and not just chemicals.chemicals.

Balanced Chemical EquationsBalanced Chemical Equations

All particles that exist at the start of a chemical All particles that exist at the start of a chemical reaction must be accounted for after the reaction reaction must be accounted for after the reaction is completeis complete

ConsiderConsider

2H2H22(g) + O(g) + O22(g) (g) → → 2H2H22O(g) O(g)

The equation gives the following information The equation gives the following information about the reaction it representsabout the reaction it represents• It identifies the reactants and productsIt identifies the reactants and products• It identifies the states of the reactants and productsIt identifies the states of the reactants and products• Because it is balanced it gives the ratio in which the Because it is balanced it gives the ratio in which the

substances reactsubstances react

Reacting QuantitiesReacting Quantities

ConsiderConsider

2H2H22(g) + O(g) + O22(g) (g) → → 2H2H22O(g) O(g)

This reaction involves 2 moles of HThis reaction involves 2 moles of H22(g) reacting (g) reacting with 1 mole of Owith 1 mole of O22(g) to form 2 moles of H(g) to form 2 moles of H22O(g).O(g).

In more general terms the amount of oxygen used In more general terms the amount of oxygen used will always be equal to half the amount of will always be equal to half the amount of hydrogen used and half the amount of water hydrogen used and half the amount of water formedformed

What a chemical equation does not tell usWhat a chemical equation does not tell us

An equation does not tell us about the rate of a An equation does not tell us about the rate of a reactionreaction

It does not tell us whether heat is required or given It does not tell us whether heat is required or given offoff

It does not tell us what temperature or pressure is It does not tell us what temperature or pressure is requiredrequired

It gives no details as to how the individual atoms or It gives no details as to how the individual atoms or molecules are transformed from reacts to productsmolecules are transformed from reacts to products

Mass-Mass StoichiometryMass-Mass Stoichiometry

We generally measure substances by mass but chemical We generally measure substances by mass but chemical reactions depend on relative numbers of atoms, ions of reactions depend on relative numbers of atoms, ions of molecules as reflected in the chemical amount or number molecules as reflected in the chemical amount or number of moles. So the relationships between mass and number of of moles. So the relationships between mass and number of mole is critical.mole is critical.

Conversion of metric units of mass:Conversion of metric units of mass:

microgramµg

milligrammg

gramg

kilogramkg

tonnet

÷103 ÷103 ÷103 ÷103

x103 x103 x103 x103

m = n x M n =

Chemical amountmol

mass g molar massg mol-1

OR mM

mass g

molar massg mol-1

Chemical amountmol

Given the amount of one substance involved in a chemical Given the amount of one substance involved in a chemical reaction, the amounts of all other substances involved can be reaction, the amounts of all other substances involved can be calculated, provided a balance equation for the reaction is calculated, provided a balance equation for the reaction is known.known.

Mass-mass stoichiometry problems can be solved in four steps:Mass-mass stoichiometry problems can be solved in four steps:1.1. Write a balanced equation for the reaction.Write a balanced equation for the reaction.

2.2. Calculate the amount (in mol) of the substance with the Calculate the amount (in mol) of the substance with the know mass.know mass.

3.3. Use the mole ratio from the equation to calculate the Use the mole ratio from the equation to calculate the amount (in mol) of the required substance.amount (in mol) of the required substance.

4.4. Calculate the mass required.Calculate the mass required.

Mass-Mass StoichiometryMass-Mass Stoichiometry

Worked ExampleWorked ExampleDetermine the mass of oxygen used when 2.00 kg of water is Determine the mass of oxygen used when 2.00 kg of water is

produced according to the equation 2Hproduced according to the equation 2H22(g) + O(g) + O22(g) (g) → → 2H2H22O(l).O(l).

Suggested SolutionSuggested Solution1.1. Tag the equation with the data supplied and the quantity Tag the equation with the data supplied and the quantity

you have to determineyou have to determine

2H2H22(g) + O(g) + O22(g) (g) → → 2H2H22O(l)O(l)

2.2. Identify the starting point (the data you’ve got) and the Identify the starting point (the data you’ve got) and the finishing point (the quantity you want) of the calculations.finishing point (the quantity you want) of the calculations.

Mass-Mass StoichiometryMass-Mass Stoichiometry

m(H2O) m(O2)

? m 2.00kg

3.3. Link the data you’ve got and the quantity you Link the data you’ve got and the quantity you want want to to their number of mole.their number of mole.

4.4. The link between n(OThe link between n(O22) -what you want to find- and n(H) -what you want to find- and n(H22O) O) –what you’ve got- is obtained from the equation.–what you’ve got- is obtained from the equation.

The equation tells us that n(OThe equation tells us that n(O22)/n(H)/n(H22O) = ½, so n(OO) = ½, so n(O22) = ½ ) = ½ x n(Hx n(H22O)O)

So the calculation flow chart can now be completedSo the calculation flow chart can now be completed

Mass-Mass StoichiometryMass-Mass Stoichiometry

m(H2O) m(O2)n(H2O) n(O2)÷ M(H2O) x M(O2)

m(H2O) m(O2)n(H2O) n(O2)÷ M(H2O) x M(O2)X 1/2

It is then relatively quick to follow through the It is then relatively quick to follow through the calculation sequence.calculation sequence.n(Hn(H22O) produced = m(HO) produced = m(H22O) / M(HO) / M(H22O)O)

= 2.00x10= 2.00x103 3 g / 18.0 g molg / 18.0 g mol-1-1

= 1.11x10= 1.11x102 2 molmol

n(On(O22) reacting ) reacting = ½ x n(H= ½ x n(H22O) producedO) produced= ½ x 111 mol= ½ x 111 mol= 55.6 mol= 55.6 mol

m(Om(O22) required ) required = n(O= n(O22) x M(O) x M(O22))= 55.6 mol x (2 x 16.0) g mol= 55.6 mol x (2 x 16.0) g mol-1-1

= 1.78x10= 1.78x103 3 gg= 1.78 kg= 1.78 kg

Mass-Mass StoichiometryMass-Mass Stoichiometry

QuestionQuestionA phosphorous manufacturer is to extract 1.00 tonne of A phosphorous manufacturer is to extract 1.00 tonne of

phosphorous per day by the process given by the phosphorous per day by the process given by the equation:equation:

2Ca2Ca33(PO(PO44))22(s) + 6SiO(s) + 6SiO22(s) + 10C(s) (s) + 10C(s) → P→ P44(s) + 10CO(g) + (s) + 10CO(g) + 6CaSio6CaSio33(s)(s)

Calculate the mass required daily of:Calculate the mass required daily of:

a)a) Calcium phosphateCalcium phosphate

b)b) Silicon dioxideSilicon dioxide

Mass-Mass StoichiometryMass-Mass Stoichiometry

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Page 263Page 263 Question 1 - 3Question 1 - 3

Concentrating on SolutionsConcentrating on Solutions

Many chemical reactions take place between Many chemical reactions take place between substances that have been dissolved in a liquid, substances that have been dissolved in a liquid, most commonly water, to form a solution. most commonly water, to form a solution.

The liquid in which the substances are dissolved The liquid in which the substances are dissolved is called the is called the solvent.solvent.

The dissolved substance is called the The dissolved substance is called the solutesolute..

When working with solutions, the most easily When working with solutions, the most easily measured quantity is their volume. However, for measured quantity is their volume. However, for the volume of a solution to provide useful the volume of a solution to provide useful information, the solution’s concentration must be information, the solution’s concentration must be known.known.

Calculating concentrations of solutionsCalculating concentrations of solutions Amounts of solutions are usually measured by volume.Amounts of solutions are usually measured by volume. Common units for volume are litres (L) and millilitres (mL), Common units for volume are litres (L) and millilitres (mL),

• where 1L = 10where 1L = 1033 mL mL Just as chemical amount (mol) and mass (g) are linked by Just as chemical amount (mol) and mass (g) are linked by

molar mass (g molmolar mass (g mol-1-1), then chemical amount (mol) and ), then chemical amount (mol) and volume (L) are linked by molar concentration (mol Lvolume (L) are linked by molar concentration (mol L-1-1))

Molar concentration, symbol c, which has units mol per litre Molar concentration, symbol c, which has units mol per litre (mol L(mol L-1-1), is often expressed as molarity (M).), is often expressed as molarity (M).

So 1.0 M HCl(aq), ie a 1 molar solution of hydrochloric acid is So 1.0 M HCl(aq), ie a 1 molar solution of hydrochloric acid is the same as saying that c(HCl) = 1.0 mol Lthe same as saying that c(HCl) = 1.0 mol L-1-1

n = cV

Chemical amountmol

Concentrationmol L-1

VolumeL

c = nV

VolumeL

Chemical amountmol

Concentrationmol L-1

Preparation of a solution of known Preparation of a solution of known concentrationconcentration

A solution of known concentration is known as a A solution of known concentration is known as a standard standard solutionsolution..

In order to prepare a particular volume of solution of known In order to prepare a particular volume of solution of known concentration, the following five steps should be followedconcentration, the following five steps should be followed

1.1. Calculate the number of moles of solute that are needed Calculate the number of moles of solute that are needed to obtain correct concentration of solution for the to obtain correct concentration of solution for the volume of solvent to be used, according to the formula volume of solvent to be used, according to the formula n = cVn = cV

2.2. Calculate the mass of the solute needed, using the Calculate the mass of the solute needed, using the formula formula m = nMm = nM

1.1. Partially fill a volumetric flask with water, and add the Partially fill a volumetric flask with water, and add the correct mass of solutecorrect mass of solute

2.2. Dissolve the soluteDissolve the solute3.3. Add water to the required volumeAdd water to the required volume

ExampleExample

Calculate the number of moles of sodium chloride needed to Calculate the number of moles of sodium chloride needed to prepare 500 mL of a 0.0800 mol Lprepare 500 mL of a 0.0800 mol L-1-1 salt solution. What salt solution. What mass of sodium chloride would be weighed out.mass of sodium chloride would be weighed out.

SolutionSolution

1.1. List the known informationList the known information

volume (V) = 500 mL = 0.5 Lvolume (V) = 500 mL = 0.5 L

concentration (c) = 0.0800 mol Lconcentration (c) = 0.0800 mol L-1-1

Preparation of a solution of known Preparation of a solution of known concentrationconcentration

c(NaCl) m(NaCl)n(NaCl)x V(NaCl) x M(NaCl)

2.2. Calculate number of moles (n) of NaCl neededCalculate number of moles (n) of NaCl needed

n = cVn = cV

= 0.0800 x 0.500= 0.0800 x 0.500

= 0.0400 mol= 0.0400 mol

3.3. Calculate the mass represented by the number Calculate the mass represented by the number of moles.of moles.

m = n x Mm = n x M

= 0.0400 x 58.5= 0.0400 x 58.5

= 2.34 g= 2.34 g

Preparation of a solution of known Preparation of a solution of known concentrationconcentration

Worked Example 15.1dWorked Example 15.1d

What volume of 0.100 M sulfuric acid What volume of 0.100 M sulfuric acid reacts completely with 17.8 ml of reacts completely with 17.8 ml of 0.150 M potassium hydroxide?0.150 M potassium hydroxide?

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Page 263Page 263 Questions 6 - 8Questions 6 - 8

Limiting Reactant CalculationsLimiting Reactant Calculations In many chemical reactions an excess of one reactant is added In many chemical reactions an excess of one reactant is added

to ensure complete reaction of another – generally more to ensure complete reaction of another – generally more valuable – reactant.valuable – reactant.

The reaction stops when one reactant is used up (the limiting The reaction stops when one reactant is used up (the limiting reagent), even though some of the other substance is reagent), even though some of the other substance is unreacted. The other reactant is said to be in excess (the unreacted. The other reactant is said to be in excess (the excess reagent) – some of it remains when the reaction has excess reagent) – some of it remains when the reaction has finishedfinished

In associated problems it is often necessary to determine which In associated problems it is often necessary to determine which reactant is in excess before amounts of product can be reactant is in excess before amounts of product can be determined. Again there are logical calculation techniques, determined. Again there are logical calculation techniques, which will lead to efficient solutions of such problemswhich will lead to efficient solutions of such problems

Chemical reactions may be considered to involved three Chemical reactions may be considered to involved three stages: an stages: an initialinitial stage where the reactants are added, a stage where the reactants are added, a reactingreacting stage where the reactants combine in the mole ration stage where the reactants combine in the mole ration suggested by the equation, and a suggested by the equation, and a finalfinal stage where the stage where the reactions appear to be completereactions appear to be complete

Example 1Example 1

A shop advertises the sale of dining settings, A shop advertises the sale of dining settings, consisting of a table and four chairs. Using consisting of a table and four chairs. Using chemical terminology, a dining setting would be chemical terminology, a dining setting would be written as TaChwritten as TaCh44. The arrangement can be set out . The arrangement can be set out as an equation:as an equation:

Ta + 4Ch Ta + 4Ch → → TaChTaCh44

The shop does some stocktaking and discovers it The shop does some stocktaking and discovers it has 16 chairs and five tables, how many complete has 16 chairs and five tables, how many complete settings can it sell?settings can it sell?

Limiting Reactant CalculationsLimiting Reactant Calculations

Example 2Example 2

A container has a mixture of eight molecules of A container has a mixture of eight molecules of hydrogen gas and six molecules of oxygen gas. hydrogen gas and six molecules of oxygen gas. When ignited, hydrogen gas and oxygen gas When ignited, hydrogen gas and oxygen gas react to form water according to the equation:react to form water according to the equation:

2H2H22(g) + O(g) + O22(g) (g) → → 2H2H22O(g)O(g)

How many water molecules can be formed?How many water molecules can be formed?

A real exampleA real example

A gaseous mixture of 25.0g of hydrogen gas and 100.0g of A gaseous mixture of 25.0g of hydrogen gas and 100.0g of oxygen gas are mixed and ignited. The water produced is oxygen gas are mixed and ignited. The water produced is collected and weighed. What is the expected mass of collected and weighed. What is the expected mass of water produced?water produced?

Suggested answerSuggested answer

1.1. Tag the equation with the data supplied and the quantity Tag the equation with the data supplied and the quantity you have to determineyou have to determine

2H2H22(g) + O(g) + O22(g) (g) → → 2H2H22O(g)O(g)25.0 g 100.0 g ? g

Limiting Reactant CalculationsLimiting Reactant Calculations

2.2. Calculate the amount of each reactant.Calculate the amount of each reactant.n(Hn(H22) = m(H) = m(H22)/M(H)/M(H22)) andand n(On(O22) = m(O) = m(O22)/M(O)/M(O22)) = 25.0 g / 2 x 1.01 g mol= 25.0 g / 2 x 1.01 g mol-1-1 = 100.0 g / 2 x 16.0 g mol = 100.0 g / 2 x 16.0 g mol-1-1

= = 12.4 mol12.4 mol = = 3.13 mol3.13 mol

3.3. i) Work out which reactant is fully used (the other must be i) Work out which reactant is fully used (the other must be in excess). This requires comparison of initial amounts of in excess). This requires comparison of initial amounts of reactants with the mole ratios suggested in the equation.reactants with the mole ratios suggested in the equation.Assume all the HAssume all the H22 reacts, then reacts, then

2H2H22(g) + O(g) + O22(g) (g) → → 2H2H22O(g)O(g)InitiallyInitiallyReactingReacting

This is not possible because n(OThis is not possible because n(O22) = 2 x n(H) = 2 x n(H22) – according to the ) – according to the equation – and there is simply not enough Oequation – and there is simply not enough O22 present. So present. So HH22 is in excess and O is in excess and O22 will be fully used. will be fully used.

Limiting Reactant CalculationsLimiting Reactant Calculations

12.4 mol 3.13 mol12.4 mol 6.2 mol

3.3. ii) Show that all the Oii) Show that all the O22 does react and find does react and find amounts present at end of reactionamounts present at end of reaction

2H2H22(g) + O(g) + O22(g) (g) → → 2H2H22O(g)O(g)Initially Initially 12.4 mol 3.13 mol12.4 mol 3.13 molReacting 6.26 mol 3.13 mol Reacting 6.26 mol 3.13 mol → 6.26mol→ 6.26molFinally Finally 6.14 mol - 6.26 mol6.14 mol - 6.26 mol

4.4. Complete the calculationsComplete the calculations

m(Hm(H22O) = n(HO) = n(H22O) x M(HO) x M(H22O)O) = 6.26 mol x 18.02 g mol= 6.26 mol x 18.02 g mol-1-1

= 113 g= 113 gTherefore, 113 g of water is produced from this Therefore, 113 g of water is produced from this

hydrogen – oxygen mixturehydrogen – oxygen mixture

Limiting Reactant CalculationsLimiting Reactant Calculations

Worked Example 15.2bWorked Example 15.2b

2.50 g of aluminium is mixed with 2.50 g of aluminium is mixed with 5.00 g of iodine and allowed to 5.00 g of iodine and allowed to react according to the equation:react according to the equation:

2Al(s) + 3I2Al(s) + 3I22(s) ―› 2AlI(s) ―› 2AlI33(s)(s)

(a)(a) What mass of aluminium iodide What mass of aluminium iodide would be produced.would be produced.

(b)(b) What is the mass of the reactant in What is the mass of the reactant in excess?excess?

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Page 267Page 267 Question 9, 10 and 11Question 9, 10 and 11

Volumetric analysisVolumetric analysis

This is what your first outcome is all This is what your first outcome is all about.about.

There are many situations when it is There are many situations when it is essential to know the exact amount essential to know the exact amount of acid or base in a substance.of acid or base in a substance.

The concentration of solutions of The concentration of solutions of acids and bases can be determined acids and bases can be determined accurately by a technique called accurately by a technique called volumetric analysis.volumetric analysis.

Volumetric Analysis.Volumetric Analysis.

This involves reacting the solution of This involves reacting the solution of an unknown concentration with a an unknown concentration with a solution of accurately known solution of accurately known concentration.concentration.

A solution of accurately known A solution of accurately known concentration is a concentration is a standard standard solutionsolution..

Volumetric AnalysisVolumetric Analysis

If we want to find the concentration If we want to find the concentration of a solution of hydrochloric acid, the of a solution of hydrochloric acid, the hydrochloric acid can be reacted with hydrochloric acid can be reacted with a basic standard solution. a basic standard solution.

This is more of an exact procedure This is more of an exact procedure than anything you have done to date than anything you have done to date in the lab so we need to use in the lab so we need to use precisely calibrated equipment.precisely calibrated equipment.

Precise EquipmentPrecise Equipment

A volumetric flask – A volumetric flask – is used to prepare the is used to prepare the standard solution. An standard solution. An accurately weighed accurately weighed sample is placed in the sample is placed in the flask and dissolved in flask and dissolved in de-ionised water to de-ionised water to form a specific volume form a specific volume of solution.of solution.

A pipetteA pipette

Is used to measure Is used to measure accurately a specific accurately a specific volume of the solution. volume of the solution. This known volume, or This known volume, or aliquot, is then poured aliquot, is then poured into a conical flask ready into a conical flask ready for analysis.for analysis.

A BuretteA Burette

A burette delivers accurately known, A burette delivers accurately known, but variable volumes of solution but variable volumes of solution (titres).(titres).

A volume of acid or base is slowly A volume of acid or base is slowly titrated to the solution of base or acid titrated to the solution of base or acid in the conical flask until the reactants in the conical flask until the reactants are present in stoichiometrically are present in stoichiometrically equivalent amounts. This is called the equivalent amounts. This is called the equivalent point.equivalent point.

The equivalent pointThe equivalent point

For the reaction:For the reaction:NaNa22COCO33(aq) + 2HCl(aq) ―› 2NaCl (aq) + H(aq) + 2HCl(aq) ―› 2NaCl (aq) + H22O(l) + COO(l) + CO22(g)(g)

The equivalence point is reached when The equivalence point is reached when exactly 2 mol HCl has been added for each 1 exactly 2 mol HCl has been added for each 1 mol Namol Na22COCO33

This whole process is called a titration.This whole process is called a titration. When the equivalence point is reached the When the equivalence point is reached the

reaction is complete.reaction is complete.

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Page 268Page 268 Question 12Question 12