formulae, stoichiometry and the 3 mole concept formulae, stoichiometry and the mole concept 3-3...

3 formulae, stoichiometry and the mole concept

3-1

3 Formulae, stoichiometry and the mole concept

Content

3.1 Symbols, Formulae and Chemical equations

3.2 Concept of Relative Mass

3.3 Mole Concept and Stoichiometry

Learning Outcomes

Candidates should be able to: (a) State the symbols of the elements and formulae of the compounds mentioned in the syllabus. (b) Deduce the formulae of simple compounds from the relative numbers of atoms present and

vice versa. (c) Deduce the formulae of ionic compounds from the charges on the ions present and vice versa. (d) Interpret chemical equations with state symbols. (e) Construct chemical equations, with state symbols, including ionic equations. (f) Define relative atomic mass, Ar. (g) Define relative molecular mass, Mr, and calculate relative molecular mass (and relative formula

mass) as the sum of relative atomic masses. (h) Calculate the percentage mass of an element in a compound when given appropriate

information. (i) Calculate empirical and molecular formulae from relevant data. (j) Calculate stoichiometric reacting masses and volumes of gases (one mole of gas occupies 24

dm3 at room temperature and pressure); calculations involving the idea of limiting reactants may be set. (The gas laws and the calculations of gaseous volumes at different temperatures and pressures are not required.)

(k) Apply the concept of solution concentration (in mol/dm3 or g/dm3) to process the results of

volumetric experiments and to solve simple problems. (Appropriate guidance will be provided where unfamiliar reactions are involved.)

(l) Calculate % yield and % purity.

Chemistry − Complete Guide themis

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3.1 Symbols, formulae and chemical equations

Symbols

Each element is represented by a symbol, based on its name.

• The symbol is generally represented by the first capital letter of the atom of that element.

• When more than one element start with the same letter, the first two letters are chosen or their first and third letters are chosen.

• Either the English name or Latin name is chosen to derive each symbol.

Examples

Phosphorus – P, Polonium – Po, Potassium – K (Latin – kalium).

1st Letter 1st and 2nd

Letters 1st and 3rd

Letters From Latin

Names

Boron (B) Aluminium (Al) Chlorine (Cl) Copper (Cu)

(Cuprum)

Carbon (C) Argon (Ar) Magnesium (Mg)Gold (Au)

(Aurum)

Hydrogen (H) Barium (Ba) Manganese (Mn)Sodium (Na)

(Natrium)

Uranium (U) Cobalt (Co) Zinc (Zn) Tin (Sn)

(Stannum)

Valency

Valency is the combining power of an atom. It is equal to the number of electrons that the atom uses to form bonds.

Valency of an atom may vary from 0 (Helium) to 8 (Osmium).

Noble gases have a stable electronic configuration with a valency of zero.

Every atom aims to achieve stable noble gas structure by loosing or gaining

electrons.

Various elements, such as transition metals, have variable valency. Their combining capacity changes with the conditions of the reaction.


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Examples

Iron can have valency of 2 or 3.

Copper can have valency of 1 or 2.

Phosphorus can have valency of 3 or 5.

Depending on the valency, the atoms form ions with different charges.

Examples

Cations of various charges.

Monovalent (1) Bivalent (2) Trivalent (3)

Hydrogen

(H+)

Calcium

(Ca2+)

Aluminium

(Al3+)

Potassium

(K+)

Copper

(Cu2+)

Chromium

(Cr3+)

Silver

(Ag+)

Iron

(Fe2+)

Iron

(Fe3+)

Ammonium

(NH4+)

Zinc

(Zn2+) -

Anions of various charges.

Monovalent (1) Bivalent (2) Trivalent (3)

Chloride

(Cl −)

Oxide

(O2−)

Nitride

(N3−)

Hydroxyl

(OH−)

Carbonate

(CO32−)

Phosphate

(PO43−)

Nitrate

(NO3−)

Sulphate

(SO42−)

-

Formulae

A chemical formula is a concise way of expressing information about the atoms that constitute a particular chemical compound.

Symbols are used in a chemical formula.

It denotes the number of atoms of an element to be found in each discrete

molecule of that compound. The number of atoms (if greater than one) is indicated as a subscript.


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Every formula is unique to the compound. Thus, it can be used to identify the compound.

Examples

H2O represents water. The formula shows that in every water

molecule, there are two hydrogen atoms and one oxygen atom.

H2O2 is hydrogen peroxide. The formula shows that in every molecule, there are two hydrogen atoms and two oxygen atoms.

When writing a chemical formula, the positive ion is always written first.

Example

The chemical formula of barium sulphate is BaSO4. It is not considered

correct to write SO4Ba, even though it shows the same information.

Empirical formulae

The empirical formula of a compound is the simplest whole-number ratio of atoms of the elements in a molecule of the compound. It may be different from the actual number of atoms in the molecule.

For most ionic compounds, the empirical formula is the same as the ionic

formula for that compound.

Example

Sodium chloride, an ionic compound, has the same empirical and ionic

formula – NaCl. It is not considered correct to write the formula as Na2Cl2, even though the ratio of atoms is correct.

☺ Some compounds are never written in their empirical form. For example, hydrogen peroxide is always written as H2O2.

☺ An ionic formula represents the exact numbers of atoms of each element per formula unit in an

ionic compound. The formula unit of an ionic compound is the smallest possible integer number of different ions in the compound. (Note the formula unit is analogous to the molecule in a molecular compound.)

Molecular formulae

The molecular formula of a compound is the actual ratio of atoms in a molecule of that compound.

It is a positive integer multiple of the empirical formula.

• Molecular formula = (Empirical formula)n


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For some compounds, the molecular formula is the same as the empirical formula for that compound.

Example

Carbon dioxide has the same molecular and empirical formula – CO2.

Different compounds can have the same empirical formula, but different molecular formulae. They can also have the same empirical formula and the same molecular formula.

Examples

Benzene (C6H6) and acetylene (C2H2) have the same empirical formula

(CH) but have different molecular formulae.

Cis-dibromoethene and trans-dibromoethene have the same empirical formula (CHBr) and the same molecular formula (C2H2Br2). The only way that they can be distinguished is through their structural formulae.

Structural formulae

The structural formula of a compound is the arrangement of atoms in a molecule of that compound. It shows the way the atoms are bonded together.

Structural formulae are often used to represent organic compounds.

Example

Structural formula of ethane (C2H6) molecule:

Chemical equation

A chemical equation is a representation of a chemical reaction using the symbols for the participating particles (atoms, molecules, ions, etc).

A typical chemical equation has the following form:

• Reactant(s) catalyst⎯⎯⎯⎯→ Product(s)

H ― C ― C ― H

H ⎥

H ⎥

⎥ H

⎥ H


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There are three components in a chemical equation.

• Reactants are the substances which exist at the start of a chemical reaction.

• Products are the substances which are created during a chemical reaction.

• A catalyst is a substance that accelerates the rate of a chemical reaction, without itself being transformed or consumed by the reaction.

Reactants are always written on the left-hand side, while products are

written on the right-hand side.

Arrow ( → ): Indicates the direction in which the reaction proceeds.

• Double arrow ( ): This represents a reversible reaction, in which the

products formed can recombine to form the reactants.

The physical state of each reactant and product can also be shown in the equation using state symbols.

• (s) represents solid.

• (l) represents liquid.

• (g) represents gas.

• (aq) represents a solution in water.

Balanced equation

A chemical equation needs to be balanced in accordance to the law of conservation. There is a conservation of mass, energy, and charge.

• A balanced equation has same number of atoms of an element on the reactant side as well as on the product side.

The number in front of a chemical formula is known as a coefficient. It

describes the amount of the substance involved in a reaction, but not its identity.

The sum of the coefficients of reactants minus the sum of the coefficients of

the products is called the stoichiometric sum. The equation is balanced when the stoichiometric sum is equal to zero.

Example

When methane, CH4, burns in air, it produces carbon dioxide and water.

CH4 + O2 → CO2 + H2O The above equation is unbalanced as there are 4H atoms and 2O atoms on the reactant side and 2H atoms and 3O atoms on the product side. By putting suitable coefficients, the equation can be balanced with all coefficients as the smallest whole number.

CH4 + 2O2 → CO2 + 2H2O


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By adding in the states of the reactants and products, the balanced equation becomes complete.

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)

Ionic equation

An ionic equation is an equation in which only those particles (atoms, ions and molecules) which participate in the reaction are listed.

• Spectator ions are ions that appear on both sides of the reaction that do not participate in the reaction. They are cancelled out.

Example

A neutralization reaction occurs when an acid reacts with an alkali.

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l) The Cl − and Na+ ions are spectator ions and do not change during the reaction. They are not listed in the ionic equation.

H+ (aq) + OH− (aq) → H2O (l)

Ionic equations can be written for any ion exchange reaction in solution. To

write them, follow the following steps:

• Write a balanced equation showing the reactant(s) and product(s) before dissociation.

• Re-write the equation with reactant(s) and product(s) dissociated where appropriate.

• Cancel all spectator ions and write the remaining ionic equation.

Example

When aqueous silver nitrate (AgNO3) is added to aqueous sodium

chloride (NaCl), a white solid forms and settles out of the solution.

AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq) Write the equation to show the ions separately.

Ag+(aq) + NO3−(aq) + Na+(aq) + Cl −(aq) → AgCl (s) + Na+(aq) + NO3

− (aq) Only the Ag+ ion and the Cl − ion are involved in the reaction. The sodium and nitrate ions are spectators and are unchanged in the reaction. Hence the ionic equation is:

Ag+(aq) + Cl −( (aq) → AgCl (s)


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Worked Examples

Example 1

Balance the following equation: Zn + HCl → ZnCl2 + H2.

Solution:

Zn is already balanced on both sides of the equation.

To balance H and Cl, add coefficient 2 in front of HCl.

State symbols can also be included.

Zn (s) + 2HCl (aq) → ZnCl2 (aq) + H2 (g)

Example 2

Which of the following equations show the balanced equation for the reaction of hydrogen with oxygen to form water?

(1) H2 (g) + O2 (g) → H2O2 (l)

(2) H2 (g) + O (g) → H2O (l)

Solution:

Both equations are not correctly balanced equations for the reaction.

(1) H2O2 is not the correct formula for water. (2) O is not the correct formula for oxygen. In balancing an equation the

chemical formula of each reactant and product is not changed, only coefficients are added.

The balanced equation for the reaction is:

2H2 (g) + O2 (g) → 2H2O (l)


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3.2 Concept of relative mass

Mass of particles

The masses of atoms and molecules are very small. In chemical calculations, the actual mass is not used because it is not practical.

Relative masses are used instead. By comparing the masses against one

another, the relationship between different substances can be easily calculated.

Relative atomic mass

The relative atomic mass (Ar) is the ratio of the average mass per atom of the

naturally occurring form of an element to 112 of the mass of a carbon-12

atom.

Mass of one atom of the element12

Mass of one atom of carbon-12rA = ×

The relative atomic mass of an atom is not the actual mass, but is the ratio of

two masses. Hence, it has no unit.

Since the hydrogen atom is the lightest known atom (one proton and one electron), the mass of other atoms is compared to it.

Example

A carbon atom has 12 times the mass of a hydrogen atom. Thus, it is

said to have a relative atomic mass of 12.

As it is not always convenient to compare masses with the mass of a

hydrogen atom, atomic masses have been based on the carbon atom instead.

• Carbon has more than one isotope. To be accurate, the isotope carbon-12 was chosen as a standard.

If an element is a mixture of isotopes, the relative atomic mass is the

average of all the atoms found in the element.

Example

Chlorine gas is a mixture of molecules containing 35Cl and 37Cl atoms.

Thus, the relative atomic mass of chlorine is not a whole number, but is calculated to be 35.5.


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Relative molecular mass

The relative molecular mass (Mr) is the ratio of the average mass per

molecule of the naturally occurring form of an element or compound to 112 of

the mass of a carbon-12 atom.

Mass of one molecule of the compound= 12

Mass of one atom of carbon-12rM ×

The relative molecular mass of a molecule is equal to the sum of the relative

atomic masses of the constituent atoms.

Since it is a ratio, the relative molecular mass has no unit.

Example

Sucrose has a formula C12H22O11. It is made up of 12 carbon atoms, 22

hydrogen atoms and 11 oxygen atoms (relative atomic masses of C, H and O are 12, 1 and 16 respectively).

Hence relative molecular mass of C12H22O11 = (12 × C) + (22 × H) + (11 × O) = (12 × 12) + (22 × 1) + (11 × 16) = 342

Relative formula mass

The relative formula mass is the sum of the relative atomic masses of the atoms present in the formula of the compound.

Relative formula mass is applied to compounds which are not composed of

molecules, such as ionic compounds.

Since it is a ratio, the relative formula mass has no unit.

Example

Sodium sulphate has a formula Na2SO4. It is made up of 2 sodium

atoms, 1 sulphur atom and 4 oxygen atoms (relative atomic masses of Na, S and O are 23, 32 and 16 respectively).

Hence relative formula mass of Na2SO4 = (2 × Na) + (1 × S) + (4 × O) = (2 × 23) + (1 × 32) + (4 × 16) = 142


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Percentage mass

Percentage mass of an element in a compound is the ratio of total relative atomic mass of all the atoms of that element to the relative formula mass of the compound, expressed as a percentage.

Total relative atomic mass of elementPercentage mass 100

Relative formula mass of compound= ×

Example

To calculate the percent composition by mass of H in H2O:

Relative mass of H in H2O = 2 × 1 = 2 Relative molecular mass of H2O = (2 × 1) + (1 × 16) = 18

Percentage mass of H in H2O = 2

10018

× = 11.1%

Worked Examples

Example 1

What is the percentage by mass of water in hydrated gypsum, CaSO4.2H2O?

Solution:

One CaSO4 contains 2 molecules of H2O.

Relative mass of 2H2O = (4 × 1) + (2 × 16) = 36

Relative formula mass of CaSO4.2H2O = 1 ×40 + 1 × 32 + (4 × 16) + 36 = 172 Percentage by mass of water in CaSO4.2H2O

= 36

100172

× = 20.9%

Example 2

How many grams of oxygen can be produced from the decomposition of 50g of H2O?

Solution:

Relative molecular mass of H2O = (2 × 1) + 16 = 18

Percentage composition of O in H2O = 16

10018

× = 88.89%

Amount of oxygen that can be produced from 50g of H2O = 88.89% × 50g = 44.4g


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Example 3

What is the relative molecular mass of (CH2Cl)2CHCH(CHCl2)2?

Solution:

Relative molecular mass of:

(CH2Cl)2 = 2 × [12 + (2 × 1) + 35.5] = 99 CHCH = 2 × (12 + 1) = 26 (CHCl2)2 = 2 × [12 + 1 + (2 × 35.5)] = 168 Total relative molecular mass = 99 + 26 + 168 = 293

Example 4

What is the empirical formula of a hydrocarbon which contains 85.7% of carbon and 14.3% of hydrogen by mass?

Solution:

Let the formula be CxHy. Ratio of mass of carbon to hydrogen

= 121

xy××

= 85.714.3

Thus, xy

= 12

Empirical formula is CH2.


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3.3 Mole concept and stoichiometry

The mole

Avogadro Number (or Avogadro constant) is defined as the number of atoms in 12g of the carbon-12 isotope.

Value of Avogadro Number is 6.023 × 1023.

The mole (mol) is defined as the amount of substance which contains the Avogadro Number (6.023 × 1023) of particles.

Number of moles =23

Number of particles

6.023 10×

Examples

One mole of copper contains 6.023 × 1023 atoms.

One mole of water contains 6.023 × 1023 molecules.

One mole of sodium chloride contains 6.023 × 1023 units.

One mole of any gas contains 6.023 × 1023 molecules.

Molar mass

The mass of one mole of any substance is called the molar mass.

• The molar mass of an element is the relative atomic mass in grams.

• The molar mass of a covalent compound is the relative molecular mass in grams.

• The molar mass of an ionic compound is the relative formula mass in grams.

Mass in grams = Number of moles × Ar or Mr of substance

Examples

Mr of H2O = (2 × 1) + (1 × 16) = 18

Thus, mass of 1 mole of H2O = 18g

Thus, mass of 4 moles of H2O = 4 × 18 = 72g

Mr of CO2 = (1 × 12) + (2 × 16) = 44

Thus, 2 288

88g CO 2 moles CO44

= =


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Molar volume

Avogadro’s Law states that equal volumes of gases contain the same number of molecules, under the same conditions of temperature and pressure.

One mole of any gas contains the Avogadro number of particles and

occupies a volume of 24 dm3 (24 000 cm3) at room temperature and pressure.

• The molar volume at r.t.p. (room temperature and pressure) is said to be 24dm3.

r

mass of gasVolume of gas molar gas volume

of gasM= ×

Examples

Molar volume at r.t.p. = 24dm3

Thus, volume of 2 mol oxygen gas = 2 × 24 = 48dm3

Thus, volume of 0.5 mol carbon dioxide gas = 0.5 × 24 = 12dm3

Mr of O2 = (2 × 16) = 32

Thus, volume of 32g oxygen gas = 24dm3

Thus, volume of 48g oxygen gas = 48

2432

× = 36dm3

Mr of N2 = (2 × 14) = 28

Thus, mass of 24dm3 nitrogen gas = 28g

Thus, mass of 12dm3 nitrogen gas = 12

28 14g24

× =

Empirical and molecular formulae calculation

The empirical formula can be determined once the percentage or mass of each element in a compound is known.

• Divide the percentage or mass of each element by its relative atomic mass to get the number of moles of the element.

• Divide by the smallest number to convert to the simplest ratio.

• The number of atoms of different elements is the empirical formula.

Examples

When iron is heated in a stream of dry chlorine, it produces a chloride

that contains 34.5% by mass of iron. Given that the relative molecular mass of this chloride (Mr) is 325, the molecular formula can be calculated.


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Ratio of Fe : Cl = 34.5 (100 34.5)

:56 35.5

−

= 0.616 : 1.845

= 1 : 3

Thus, empirical formula is FeCl3. Relative formula mass of FeCl3 = (1 × 56) + (3 × 35.5) = 162.5

Let the molecular formula be (FeCl3)n.

Thus, n = 325

162.5 = 2

Therefore, molecular formula is Fe2Cl6.

Given that a compound is composed of 7.20g of carbon, 1.20g of hydrogen, and 9.60g of oxygen, the empirical formula can be calculated.

Number of moles of each element:

Carbon: 7.20 ÷ 12 = 0.6 moles Hydrogen: 1.20 ÷ 1 = 1.2 moles Oxygen: 9.60 ÷ 16 = 0.6 moles

Ratio of C : H : O = 0.6 1.2 0.6

: :0.6 0.6 0.6

= 1 : 2 : 1

Thus, empirical formula is CH2O.

Stoichiometry

Stoichiometry is the measure of relative proportions in which elements form compounds or in which substances react.

Stoichiometric coefficients are the numbers which appear in front of a

chemical formula in a chemical equation.

• They are the numbers of moles of reactant or product which take part in the reaction.

Molar ratio is the ratio of the coefficients of a balanced equation.

☺ The word stoichiometry derives from two Greek words: stoicheion (meaning "element") and

metron (meaning "measure”).

Example

In the equation: 2H2 + O2 → 2H2O

The molar ratio between H2 and O2 = 2:1

The molar ratio between O2 and H2O = 1:2

The molar ratio between H2 and H2O = 2:2 = 1:1


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Limiting reactants

Limiting reactant is the reactant that gives the smallest amount of product.

The limiting reactant is the substance in a chemical reaction that runs out

first. When the limiting reactant is used up, the reaction stops.

There is a difference between the amounts of reactants used in chemical reactions and the number of moles actually required (by stoichiometry) for the reaction to proceed.

In the laboratory, it is often more convenient to have one or more reactants

present in excess, so that only one reactant, the limiting reactant, will be completely depleted during the reaction.

• The amount of limiting reactant can then be used to calculate the amount of product formed.

Stoichiometric calculations

To calculate the mass of product formed from a limited quantity of reactant.

• Balance the equation.

• Using the molar mass of the given substance, convert the mass in grams to moles.

• Based on the molar ratio of reactant to product, calculate the moles of the unknown.

• Using the molar mass of the unknown substance, convert the results in moles to mass in grams.

Examples

To find the mass of NH3 in grams that will be produced if 10g of H2 is

reacted with enough N2 using the Haber process:

N2 + 3H2 → 2NH3 Check that the chemical equation is balanced.

N2 + 3H2 → 2NH3

Convert the known substance from grams to moles.

Mr of H2 = 2 × 1 = 2

No. of moles H2 = 102

= 5 mol

Find no. of moles of unknown using the coefficients in the balanced equation.

Molar ratio H2 : NH3 = 3 : 2

No. of moles NH3 = 5

23× = 3.333 mol


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Convert the unknown substance from moles to grams.

Mr of NH3 = 1 × 14 + (3 × 1) = 17

Mass of NH3 = 3.333 × 17 = 56.6g

To find the mass of H2 required to obtain 68g of NH3 from the Haber process.

Check that the chemical equation is balanced.

N2 + 3H2 → 2NH3

Convert the known substance from grams to moles.

Mr of NH3 = 1 × 14 + (3 × 1) = 17

No. of moles NH3 = 6817

= 4 moles

Find no. of moles of unknown using the coefficients in the balanced equation.

Molar ratio H2 : NH3 = 3 : 2

No. of moles H2 = 4

32× = 6 mol

Convert the unknown substance from moles to grams.

Mr of H2 = 2 × 1 = 2

Mass of H2 = 6 × 2 = 12g

Concentration of solutions

A solution is a homogeneous mixture of a liquid (solvent) with a gas or solid (solute).

• A homogeneous mixture means that the composition of the mixture is the same throughout.

• A solution is made up of a solute and a solvent.

• Aqueous solutions are those in which the solvent is water.

Concentration = 3

Number of grams (or moles) of solute

Volume of solution (in dm )

Molarity (or molar concentration) is defined as the amount (in moles) of solute dissolved per unit volume of solution. It is used to describe the concentration of a solution and how much solute is dissolved.

• A concentrated solution contains a high amount of solute, while a dilute solution contains a small amount of solute.

A 1.00 molar solution contains 1.00 moles of solute in every litre of solution

(not solvent). It can be expressed as 1.00 M or 1.00 mol dm−1.

Examples

To calculate the molarity of a solution containing 10.0g of NaCl

dissolved in 200ml of water.


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Relative formula mass of NaCl = 1 × 23 + (1 × 35.5) = 58.5

No. of moles NaCl = 10

58.5 = 0.1709 mol

Molarity of NaCl = 0 17090 200..

= 0.855 M

To calculate the mass in grams of CuSO4 required to make up 250ml of a

1.00 M CuSO4 solution.

Number of moles CuSO4 = 1.00 M × 0.250 = 0.250 mol

Relative formula mass of CuSO4 = 64 + 32 + (4 × 16) = 160

Mass of CuSO4 = 0.250 × 160 = 40g

Percentage yield

Most chemical reactions do not proceed to 100% completion due to a variety of reasons, including side chemical reactions, experimental error and environmental factors.

The products which are obtained are less than theoretically expected from

the reaction stoichiometry.

• The maximum amount of product that can be obtained from the reaction is called the theoretical yield.

• The actual amount of product that is obtained is the actual yield.

• The percentage yield relates the actual yield to the theoretical yield as a percentage and is always less than 100%.

actual yieldPercentage yield 100%

theoretical yield= ×

Example

In the equation: 2H2 + O2 → 2H2O

When 20g of H2 and 220g of O2 were reacted, the theoretical yield of H2O was 180g, but the actual yield of H2O was only 78g.

2

78Percentage yield of H O 100%

180= × = 43.3%


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Percentage purity

Samples of chemicals may not be totally pure. If the purity of the chemical sample is known, the purity can be accounted for in stoichiometric calculations.

Percentage purity can be determined by measuring the amount of product

obtained from a reaction. This approach assumes 100% yield of the product.

Percentage purity = mass of pure compound in the impure sample

100total mass of impure sample

×

Example

For the reaction of magnesium hydroxide with phosphoric acid:

3Mg(OH)2 + 2H3PO4 → Mg3(PO4)2 + 6H2O

The titration of 2.568g of the magnesium hydroxide sample required 38.45ml of 0.6695 M H3PO4.

Amount of H3PO4 = 0.6695 × 0.03845 = 0.02574 mol

Molar ratio Mg(OH)2 : H3PO4 = 3 : 2

No. of moles Mg(OH)2 = 0.02574

32

× = 0.03861 mol

Relative formula mass of Mg(OH)2 = 24 + 2(16 + 1) = 58

Mass of Mg(OH)2 = 0.03861 × 58 = 2.240g

Percentage purity = 2.240

1002.568

× = 87.2%

Worked Example

Example 1

Given that: 8Fe + S8 → 8FeS

(a) What mass of iron is needed to react with 16.0 grams of sulphur?

(b) How many grams of FeS are produced?

Solution:

(a) Relative molecular mass of S8 = 8 × 32 = 256

No. of moles S8 = 16

256 = 0.0625 mol

Molar ratio Fe : S8 = 8 : 1

No. of moles Fe needed = 8 × 0.0625 = 0.500 mol

Relative atomic mass of Fe = 56 Mass of Fe needed = 0.500 × 56 = 28.0g


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(b) Molar ratio S8 : FeS = 1 : 8

No. of moles FeS produced = 8 × 0.0625 = 0.500 mol

Relative formula mass of FeS = 56 + 32 = 88 Mass of FeS produced = 0.500 × 88 = 44.0g


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Worked Problems

Example 1

Given that a sample of sulphuric acid (H2SO4) contains 5 moles, answer the following questions.

(a) What is the mass of sulphuric acid in the sample?

(b) How many molecules of sulphuric acid are there in the sample?

(c) How many hydrogen atoms are there in the sample?

(d) How many oxygen atoms are there in the sample?

Solution:

(a) Relative molecular mass of H2SO4 = (2 × 1) + (1 × 32) + (4 × 16) = 98

Mass of H2SO4 in sample = 5 × 98 = 490g

(b) 1 mol H2SO4 contains 6.023 × 1023 molecules of H2SO4

No. of H2SO4 molecules in sample = 5 × 6.023 × 1023 = 3.01 × 1024

(c) There are 2 hydrogen atoms in every molecule of H2SO4.

No. of hydrogen atoms in sample = 2 × 3.01 × 1024 = 6.02 × 1024

(d) There are 4 oxygen atoms in every molecule of H2SO4.

No. of oxygen atoms in sample = 4 × 3.01 × 1024 = 1.20 × 1025

Example 2

Phosphine (PH3) burns in air according to the equation:

PH3 (g) + O2 (g) → P4O10 (s) + H2O (g)

(a) Balance the equation.

(b) What is the mass of 48dm3 of phosphine at r.t.p.?

(c) What volume of oxygen at r.t.p. is needed for complete combustion of 48dm3 of phosphine?

Solution:

(a) 4PH3 (g) + 8O2 (g) → P4O10 (s) + 6H2O (g) (b) Molar volume at r.t.p. = 24dm3

No. of moles in 48dm3 PH3 = 4824

= 2 mol

Relative molecular mass of PH3 = 31 + 3 = 34

Mass of 48dm3 PH3 = 2 × 34 = 68g

(c) Molar volume at r.t.p. = 24dm3

Molar ratio PH3 : O2 = 4 : 8 = 1 : 2

Volume of O2 needed = 2 × 48 = 96dm3


3-22

Example 3

13.0g of zinc granules is reacted with 7.3g of dilute hydrochloric acid.

Zn (s) + 2HCl (aq) → ZnCl2 (aq) + H2 (g)

(a) Calculate the number of moles of zinc and dilute hydrochloric acid present.

(b) The reaction stops before all the substances have reacted. Explain why the reaction stops.

(c) What is the volume of H2 obtained at r.t.p.?

Solution:

(a) Relative atomic mass of Zn = 65

No. of moles Zn = 1365

= 0.2 mol

Relative molecular mass of HCl = 1 + 35.5 = 36.5

No. of moles HCl = 7.3

36.5 = 0.2 mol

(b) Molar ratio Zn : HCl = 1 : 2

There is not enough HCl to react with all the zinc. Thus, HCl is the limiting reactant in this case. When the limiting reactant has reacted completely, the reaction stops.

(c) Molar ratio HCl : H2 = 2 : 1

No. of moles H2 = 0.22

= 0.1 mol

Molar volume at r.t.p. = 24dm3

Volume of H2 obtained = 0.1 × 24 = 2.4dm3

formulae, stoichiometry and the 3 mole concept formulae, stoichiometry and the mole concept 3-3...

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