chemical formulae and equations part 2

8/3/2019 chemical formulae and equations part 2

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Meaning of Chemical Formulay Chemical formula is used to represent a chemical

compound

y The formula of a molecule tells us

a) The name of the atoms present

b) The number of atoms of each element present

y Example : water ; H2O

-H represent hydrogen

-O represent oxygen

-the number 2 represent the number of hydrogen atomspresent

y Example : sulphuric acid ; H2SO4- one molecule of sulphuric acid contains 2 atomshydrogen, 1 atom sulphur and 4 atom oxygen


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Empirical Formula and Molecular

Formulay Empirical formula shows only the simplest ratio of

elements in it

y Molecular formula shows the actual number of each typeof element in it

y Example : Butenethe molecular formula C4H8the empirical formula CH2

y Empirical formula may or may not be the same as themolecular formula.

y Example : CO2in a molecule of carbon dioxide, there are 2 atomsoxygen and 1 atom of carbon. Its molecular formula isCO2


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Similarity Empirical formulaMolecular formula

Both types of formula containthe same type of elements

Differences

The number ofmole atom inthe formula

Formulamass

Shows only thesimplest ratio of the

constituent elements

Smaller mass unlessit is the same as the

molecular formula

Shows the exactmole atom of eachconstituent element

The mass isalways fixed


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Example 1:

1.69 g of iron combine with 0.72 g of oxygen. Calculate the empiricalformula of this oxide. [ relative atomic mass : Fe,56 ; O,16 ]

Steps Fe O

Mass of content 1.69 0.72

Relative atomic mass 56 16

Number of moles=

= 0.03 = 0.045

Ratio of the number of

moles = 1 = 1.5

X 2 to change ratio

into whole numbers

2 3

Empirical formula is Fe2O3


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y Example 2

0.91 g of aluminium burns in air to form 1.7 g ofaluminium oxide. What is the formula of aluminium

oxide ? [ relative atomic mass : Al,27 ; O,16 ]

= empirical formula is Al2O3

y Example 3

the following is the percentage composition of

hydrated magnesium sulphate : Mg,9.8% ; S,13% ;

O,26% ; H2O,51.2%. Calculate the empirical formulaof hydrated magnesium sulphate.

[ relative atomic mass : Mg,24 ; S,32 ; H,1 ; O,16 ]

= empirical formula is MgSO4.7H2O


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y Example 4

the following is the percentage composition of

calcium carbonate : Ca = 40%, C = 12%, O = 48%.Calculate the empirical formula of calciumcarbonate.

[ relative atomic mass : Ca,40 ; C,12 ; O,16 ]

= empirical formula is CaCO3y Example 5

the decomposition of 7.36 g of a compoundproduces 6.93 g of oxygen. The rest of the mass is

hydrogen. If the relative molecular mass of thiscompound is 34.0 g calculate its molecular formula.

[ relative atomic mass : H,1 ; O,16 ]

= molecular formula is H2O2


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y Example 6

the empirical formula of ethene is (CH3)n. Itsmolecular formula mass is 30. calculate the formula

of this compound.

= molecular formula for ethene is C2H6

y Example 7

the empirical formula of benzene is CH. If its relative

molecular mass is 78, what is its molecular formula ?

= the molecular formula for benzene is C6H6


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Ionic Formulaey In an ionic compound, positive and negative ions are

joined together by ionic bonds.Ions with 1+ Ions with 2+ Ions with 3+

Li + Mg2+ Al3+

Na+ Ca2+ Fe3+

K+ Ba2+ Cr3+

Ions with 1- Ions with 2- Ions with 3-

F- O2- N3-

Cl - CO PO43-

Br- SO42-


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y The charge on an ion is related to its position in the

Periodic Table

y During the formation of an ionic compound, thenumber of positive charges ( + ) must balance thenumber of negative charges ( - )

yWay to write the ionic compound formula :

1. Write the ions side by side

2. The number of the charges on the ion is exchangedwith each other

Group 1 2 13 14 15 16 17 18

Charge 1 + 2 + 3 + 2 - 1 - 0

example Na+ Mg2+ Al3+ O2- Cl - Ne


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y Example 1

the ions of zinc chloride ; Zn2+ and Cl- = ZnCl2

y Example 2

the ions of sodium oxide ; Na+ and O2- = Na2O

y Example 3

the ions of aluminium sulphate ; Al3+ and SO42-

= Al2(SO4)3

y Example 4

the ions of magnesium sulphate ; Mg2+ and SO42-

= MgSO4


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IUPAC Nomenclaturey For compounds with simple ions, the positive half

has the same name as the metal.

y The negative half always ends in ide

Oxygen changes to oxide

Chlorine changes to chloride

Sulphur changes to sulphide

y

Example 1 ; NaClthe positive half is a metal called sodium

the negative half is chloride from chlorine

therefore the name of NaCl is sodium chloride


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y Ions containing oxygen always end in ate or ite

y The ions with the higher proportion of oxygen isnamed ate

example ; SO42- sulphate SO3

2- sulphite

y Exception this rule is the ion OH- called hydroxide

y Roman numerals used for ionic compounds of metal

ions with more than one type of charge

FeCl2 is called iron(II) chloride

FeCl3 is called iron(III) chloride


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Chemical Equationsy Equation are chemical sentences. They tell us :

a) The types of chemicals that are reacting

b) The products of the reaction

c) How much of the different chemicals are reacting

with each other

y Example ; nA + mB pC + qD

a) A and B are called reactants.b) C and D are called products of the reaction

c) n and m shows the number of moles that are react

d) p and q shows the number of moles that are form


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y 2H2 + O2 2H2O

two molecules of hydrogen react with one moleculeof oxygen to form two molecules of water

- hydrogen and oxygen are reactants and water is

the product

y

Chemical equation also show the state of eachchemical in the reaction

( s ) solid state ( l ) liquid state

( aq ) aqueous state ( g ) gaseous state

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

1 mole of methane molecules (16g) react with 2

moles of oxygen molecules (64g) to give 1 mole of

CO2 (44g) and 2 moles of water molecules (36g)


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Balancing Chemical Equations

y Example 1

Mg + HCl MgCl2 + H2

Mg(s)+2HCl (aq) MgCl2 (aq)+H2 (g)

y Example 2

potassium hydroxide reacts with sulphuric acid to

form potassium sulphate and water.

KOH + H2SO4 K2SO4 + H2O

2KOH(aq)+H2SO4(aq) K2SO4(aq)+2H2O(l)


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Solving Numerical Problemsy Steps used for Quantitative Calculations

1) Write down a balanced equation and write down thenumber of moles of each reactant and product

2) Get the information from the question. Changes the

quantities of the reactants to moles if necessary.3) Find the ratio of the number of moles related to the

questions

4) Calculate according to the ratio of the number ofmoles

y Step 1 : equationStep 2 : change to moles

Step 3 : use the ratio of moles

Step 4 : calculate according to the ratio of moles


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y Example 1

how much water is produced if 4.0 g of methane CH4is burnt in excess supply of oxygen ?

mass of H2O = 9g

y Example 2

calculate the volume of oxygen produced when 1.7 g

of hydrogen peroxide (H2O2) are decomposed atSTP

volume of oxygen = 0.56 dm3


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y In an experiment 12 dm3 of nitrogen dioxide wereproduced from the decomposition of lead(II) nitrate.

During the decomposition, lead(II) oxide and oxygenwere also formed. If the experiment was carried atroom temperature and pressure of 1 atmosphere,how many grams of lead(II) oxide were produced inthis reaction?

mass of PbO = 55.75g

y In an experiment aluminium powder is used toreduced copper(II) oxide to form aluminium oxideand copper. How many atoms of copper will beformed if 2.7 g of aluminium are used?

[Av:6.02 x 1023 ; relative atomic mass:Al,27]

number of copper atoms = 9.03 x 1022

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chemical formulae and equations part 2