Lots of definitions to learnLots of definitions to learn

• IsotopesIsotopes• The MoleThe Mole• Avogadro numberAvogadro number• Relative Atomic/Molecular MassRelative Atomic/Molecular Mass• Molecular FormulaMolecular Formula• Empirical FormulaEmpirical Formula

ALL MET AT GCSEALL MET AT GCSE

IsotopesIsotopes

Atoms (of the same element) that have Atoms (of the same element) that have the same number of protons, but a the same number of protons, but a different number of neutrons.different number of neutrons.SAME ATOMIC NUMBER, DIFFERENT MASS SAME ATOMIC NUMBER, DIFFERENT MASS NUMBERNUMBER

E.g. Carbon-12 and Carbon-14E.g. Carbon-12 and Carbon-14

6,6,6 and 6,6,86,6,6 and 6,6,8

The MoleThe Mole

• The The AMOUNT OF A SUBSTANCEAMOUNT OF A SUBSTANCE in in grams that has the grams that has the SAME NUMBER SAME NUMBER OF PARTICLESOF PARTICLES as there are as there are ATOMS ATOMS in in EXACTLY 12EXACTLY 12 grams of grams of CARBON-CARBON-1212..Avogadro's numberAvogadro's number

The number of atoms in exactly 12g of Carbon-12 = 6.023 X 1023

Relative atomic/molecular Relative atomic/molecular MassMass•The The AVERAGEAVERAGE mass of mass of ONE MOLEONE MOLE of of atoms atoms RELATIVE RELATIVE to the mass of to the mass of ONE MOLEONE MOLE of Carbon-12 of Carbon-12 (Which has a mass of 12g)(Which has a mass of 12g)

•The The AVERAGEAVERAGE mass of mass of ONE MOLEONE MOLE of of molecules molecules RELATIVE RELATIVE to the mass of to the mass of ONE ONE MOLEMOLE of Carbon-12 of Carbon-12 (Which has a mass of 12g)(Which has a mass of 12g)

IONIC COMPOUNDS - Can they have RMM?IONIC COMPOUNDS - Can they have RMM?

Relative Formula massRelative Formula mass

• The The AVERAGEAVERAGE mass of mass of ONE MOLEONE MOLE of the of the formula formula RELATIVE RELATIVE to the mass of to the mass of ONE ONE MOLEMOLE of Carbon-12 of Carbon-12 (Which has a mass of 12g)(Which has a mass of 12g)

• Average mass of one mole of formula/ (1/12(mass of 1 Average mass of one mole of formula/ (1/12(mass of 1 mole C-12))mole C-12))

Picture of MachinePicture of Machine

B

A

C

D

E

F

To computer

Determining RAM using Mass Determining RAM using Mass spectrometryspectrometryIn a vacuum

At A the sample is injected

At B the sample is heated to convert it to a gas

At C the atoms are ionized by electron bombardment

The heated cathode releases electrons which collide with the atoms

When the energy of the electrons equals the 1st IE of the atom X + e- X+ + e- + e- takes place

At D the ions are accelerated in an electric field

At E the ions path is deflected in a magnetic field

At F the ions are identified

Data ObtainedData Obtained

• The percentage composition of each The percentage composition of each isotope in the sample of the elementisotope in the sample of the element

RAM = RAM = (%composition x mass)(%composition x mass)) ) / / 100100

90

10

56 57 58

IRON DATA AS GRAPH

90% Fe = 56

8% Fe = 57

2% Fe = 58

RAM Fe =( (90*56)+(8*57)+(2*58))/100

RAM Fe = 5612/100 = 56.12 (NO UNITS)

Data obtainedData obtained• The isotopic abundancesThe isotopic abundances

• 50% of Bromine = 50% of Bromine = 7979Br Br • 50% of Bromine = 50% of Bromine = 8181BrBr

• 75% of Chlorine = 75% of Chlorine = 3535ClCl• 25% of Chlorine = 25% of Chlorine = 3737ClCl

• 92.58% of Lithium = 92.58% of Lithium = 77LiLi• 7.40% of Lithium = 7.40% of Lithium = 66LiLi• 0.02% of Lithium = 0.02% of Lithium = 88LiLi

RAM

79.5

35.5

6.93

CalculationCalculation• From RAMFrom RAM

•HH22OO

•NHNH33

•CC22HH44

•CC66HH66

•CC33HH77OHOH

1+1+ 16 = 18

14+ 1+1+1 =17

12+12+1+1+1+1=28

12+12+12+12+12+12+1+1+1+1+1+1=78

12+12+12+1+1+1+1+1+1+1+1+16=60

Formula determinationFormula determination

•Empirical formulaEmpirical formula

The simplest whole number ratio of elements The simplest whole number ratio of elements present in the compoundpresent in the compound

•Molecular formulaMolecular formula

The actual whole number ratio of elements The actual whole number ratio of elements present in the compoundpresent in the compound

BOTH ARE DETERMINED BY EXPERIMENTAL BOTH ARE DETERMINED BY EXPERIMENTAL DATADATA

FROM ELEMENTAL ANALYSISFROM ELEMENTAL ANALYSIS

• A compound containsA compound contains

a.a. 75% C and 25% H75% C and 25% H

b.b. 50% O, 37.5% C and 12.5% H50% O, 37.5% C and 12.5% H

c.c. 85.7% C, 14.3% H85.7% C, 14.3% H

d.d. 71.6% C, 23.9% O, 4.5%H71.6% C, 23.9% O, 4.5%H

e.e. 39.3% Na, 60.7% Cl39.3% Na, 60.7% Cl

To calculate formula 1To calculate formula 1• Make a column representing each element in compoundMake a column representing each element in compound

• C C HH• Write down the % comp under each elementWrite down the % comp under each element

• C C HH• 7575 2525• Write down the RAM under each elementWrite down the RAM under each element

• C C HH• 7575 2525• 1212 11• Divide the %comp by the RAM (to compensate for the Divide the %comp by the RAM (to compensate for the

different RAM)different RAM)

• C C HH• 7575 2525• 1212 11• 6.256.25 2525

To calculate formula 1To calculate formula 1• Divide the %comp by the RAM (to compensate for the different Divide the %comp by the RAM (to compensate for the different

RAM)RAM)

• C C HH

• 7575 2525

• 1212 11

• 6.256.25 2525• Ratio by dividing by the smallest of the figures (eg by 6.25 Ratio by dividing by the smallest of the figures (eg by 6.25

above)above)

• C C HH

• 7575 2525

• 1212 11

• 6.256.25 2525

• 1(6.25/6.25)1(6.25/6.25) 4(25/6.25)4(25/6.25)

• Empirical formulae = This ratio eg CHEmpirical formulae = This ratio eg CH44

To calculate formula if more than 2 elements in To calculate formula if more than 2 elements in compoundcompound

• Make a column representing each element in compoundMake a column representing each element in compound

• C C HH OO• Write down the % comp under each elementWrite down the % comp under each element

• C C HH OO• 37.537.5 12.512.5 5050• Write down the RAM under each elementWrite down the RAM under each element

• C C HH OO• 37.537.5 12.512.5 5050• 1212 11 1616• Divide the %comp by the RAM (to compensate for the Divide the %comp by the RAM (to compensate for the

different RAM)different RAM)

• C C HH OO• 37.537.5 12.512.5 5050• 1212 11 1616• 3.1253.125 12.512.5 3.1253.125

To calculate formula 1To calculate formula 1• Divide the %comp by the RAM (to compensate for the different Divide the %comp by the RAM (to compensate for the different

RAM)RAM)

• C C HH OO

• 37.537.5 12.512.5 5050

• 1212 11 1616

• 3.1253.125 12.512.5 3.1253.125• Ratio by dividing by the smallest of the figures (eg by 3.125 Ratio by dividing by the smallest of the figures (eg by 3.125

above)above)

• C C HH OO

• 37.537.5 12.512.5 5050

• 1212 11 1616

• 3.1253.125 12.512.5 3.125 3.125

• 1(3.125/3.125)1(3.125/3.125) 4(12.5/3.125)4(12.5/3.125) 1(3.125/3.125)1(3.125/3.125)

• Empirical formulae = CHEmpirical formulae = CH44OO

• Answers to other problemsAnswers to other problems

85.7% C, 14.3% H85.7% C, 14.3% H

Empirical formula = CHEmpirical formula = CH22

71.6% C, 23.9% O, 4.5%H71.6% C, 23.9% O, 4.5%H

Empirical formula = CEmpirical formula = C44HH33OO

39.3% Na, 60.7% Cl39.3% Na, 60.7% Cl

Empirical formula = NaClEmpirical formula = NaCl

NB NB

Ratio must be whole number so double if ratio is Ratio must be whole number so double if ratio is halfhalf

Eg 1.5 : 1Eg 1.5 : 1 becomes 3 : 2 becomes 3 : 2

Molecular formula from Empirical Molecular formula from Empirical formulaformula• Work out the Mass of the empirical formulaWork out the Mass of the empirical formula

• Look up the RMMLook up the RMM• Divide RMM / Mass of empirical formulaDivide RMM / Mass of empirical formula• Multiply each element ratio by the number Multiply each element ratio by the number

formed aboveformed above

• Eg Empirical formula = CHEg Empirical formula = CH22

• Empirical mass = 14 (12+2)Empirical mass = 14 (12+2)• If RMM = 28If RMM = 28• RMM/Empirical mass = 28/14 =2RMM/Empirical mass = 28/14 =2

• Molecular formula = CMolecular formula = C22HH44

Working out formulae from Working out formulae from experimental dataexperimental dataA Formula to rememberA Formula to remember

No of Moles =Mass present/Mass of 1 mole No of Moles =Mass present/Mass of 1 mole (RAM)(RAM)

Results in AirResults in Air•Mass of Iron turnings before heating = 1.9g

•Mass of iron turnings after heating = 2.38g

•Mass of oxygen combined with iron =

•Moles of Iron =

•Moles of Oxygen =

•FORMULA =

?

?

?

?

0.48g

0.034 (=1.9 /56)

0.03 ( = 0.48/16)

FeO

Results in oxygenResults in oxygen•Mass of Iron turnings before heating = 1.9g

•Mass of iron turnings after heating = 2.73g

•Mass of oxygen combined with iron =

•Moles of Iron =

•Moles of Oxygen =

•FORMULA =

?

?

?

?

0.83 g

0.034 (= 1.9 / 56)

0.051 (=0.83/16)

Fe2O3

Results in oxygenResults in oxygen•Mass of Mg turnings before heating = 2.1g

•Mass of Mg turnings after heating = 3.5g

•Mass of oxygen combined with Mg =

•Moles of Mg =

•Moles of Oxygen =

•FORMULA =

?

?

?

?

1.4g

0.0875 (=2.1/24)

0.0875 (=1.4/16)

MgO

Results in chlorineResults in chlorine•Mass of Al turnings before heating = 3.7g

•Mass of Al turnings after heating = 18.3g

•Mass of oxygen combined with Al =

•Moles of Al =

•Moles of Chlorine =

•FORMULA =

?

?

?

?

14.6 g

0.137 (= 3.7/27)

0.411 (= 0.411/35.5)

AlCl3

MOLE CALCULATIONSMOLE CALCULATIONS

Things you must get right

1)Correct formulae2)Balanced Equations3)The correct mole equation4)The numbers on your calculator

Ionic compoundsIonic compounds

• Write down the charges on the ionsWrite down the charges on the ions

• The charges become the subscripted The charges become the subscripted number for the other ion. number for the other ion.

• This results in the charges balancing out.This results in the charges balancing out.

E.g. Sodium NaE.g. Sodium Na++ Oxide O Oxide O-2-2

You need 2 NaYou need 2 Na++ to one O to one O-2-2

Formula = NaFormula = Na22OO

Covalent MoleculesCovalent Molecules

• LEARN THE FOLLOWINGLEARN THE FOLLOWING

MethaneMethane Hydrochloric Hydrochloric acidacid

Carbon dioxideCarbon dioxide Sulphuric acidSulphuric acid

WaterWater AmmoniaAmmonia

Sulphur dioxideSulphur dioxide Nitric acidNitric acid

Sulphur trioxideSulphur trioxide All elementsAll elements

Complex ionsComplex ionsAmmonium =Ammonium =

Sulphate =Sulphate =

Nitrate = Nitrate =

Carbonate = Carbonate =

Hydroxide = Phosphate Hydroxide = Phosphate ==

Ethanoate =Ethanoate =

Iron (II) =Iron (II) =

Iron (III) = Iron (III) =

NH4+

SO42-

NO3-

CO32-

OH-

PO43-

CH3COO-

Fe2+

Fe3+

• Sodium Carbonate =Sodium Carbonate =

• Sodium Oxide = Sodium Oxide =

• Lithium Sulphate =Lithium Sulphate =

• Calcium Nitrate =Calcium Nitrate =

• Aluminium Sulphate =Aluminium Sulphate =

• Potassium Phosphate Potassium Phosphate ==

• Ammonium Sulphate Ammonium Sulphate ==

• Sodium Ethanoate =Sodium Ethanoate =

• Iron (II) Hydroxide =Iron (II) Hydroxide =

• Iron (III) Chloride =Iron (III) Chloride =

Na2CO3

Na2OLi2SO4

Ca(NO3)2

Al2(SO4)3

K3PO4

(NH4)2SO4

CH3COONaFe(OH)2

FeCl3

Putting formulae together to get an Putting formulae together to get an equationequationA balanced equation will haveA balanced equation will have• Same number of atoms of each element on both Same number of atoms of each element on both

sidessides

Eg Mg + HEg Mg + H22O O Mg(OH) Mg(OH)22 + H + H22 isn’t balanced as there are 2 O on one side and 1 O on the otherisn’t balanced as there are 2 O on one side and 1 O on the other

Mg + 2HMg + 2H22O O Mg(OH) Mg(OH)22 + H + H22 • Same charge on both sidesSame charge on both sidesZnZn+2+2 + Na + Na Na Na+1+1 + Zn + Znisn’t balanced as there are +2 charge on one side and +1 charge on isn’t balanced as there are +2 charge on one side and +1 charge on

the otherthe other

ZnZn+2+2 + 2Na + 2Na 2Na2Na+1+1 + Zn + Zn

BALANCED EQUATIONSBALANCED EQUATIONS• CuCu+2+2 + K + K Cu + K Cu + K+ +

• CHCH44 + O + O22 CO CO22 + H + H22OO

• Ammonia + sulphuric acid Ammonia + sulphuric acid ammonium ammonium sulphatesulphate

• Calcium hydrogen carbonate + Nitric acid Calcium hydrogen carbonate + Nitric acid Calcium nitrate + carbon dioxide + waterCalcium nitrate + carbon dioxide + water

• Calcium ions + Aluminium Calcium ions + Aluminium Aluminium ions Aluminium ions and calciumand calcium

• Iron(II) chloride + Chlorine Iron(II) chloride + Chlorine Iron (III) chloride Iron (III) chloride

• Copper (II) sulphate + Sodium hydroxide Copper (II) sulphate + Sodium hydroxide Copper (II) hydroxide + sodium chlorideCopper (II) hydroxide + sodium chloride

• FeFe22OO33 + Al + Al Al Al22OO33 + Fe + Fe

ANSWERS

Cu2+ + 2K Cu + 2K+

CH4 + 2O2 CO2 + 2H2O

2NH3 + H2SO4 (NH4)2SO4

Ca(HCO3)2 + 2HNO3 Ca(NO3)2 + CO2 + H2O

3Ca2+ + 2Al 2Al+3 + 3Ca

FeCl2 + ½ Cl2 FeCl3

CuSO4 + 2NaOH Na2SO4 + Cu(OH)2

Fe2O3 + 2Al Al2O3 + 2Fe

WHAT DOES A BALANCED EQUATION TELL US?WHAT DOES A BALANCED EQUATION TELL US?

•The reaction taking placeThe reaction taking place

•The state the reactants / products are inThe state the reactants / products are in

•The Mole ratio of reactants / ProductsThe Mole ratio of reactants / Products

•The equation should not include any The equation should not include any spectator ions unless the full equation is spectator ions unless the full equation is specifically requestedspecifically requested

• ..

Let’s write some ionic Let’s write some ionic equationsequations• HCl + NaOH HCl + NaOH NaCl + H NaCl + H22OO

• 2 KI + Cl2 KI + Cl2 2 2 KCl + I 2 KCl + I22

• Mg + CuSOMg + CuSO44 MgSO MgSO44 + Cu + Cu

• Mg + 2 HCl Mg + 2 HCl MgCl MgCl22 + H + H22

• NaNa22COCO3 3 + H+ H22SOSO44 Na Na22SOSO44 + H + H22O + O + COCO22

• HH22SOSO44 + Ca(OH) + Ca(OH)22 CaSO CaSO44 + 2H + 2H22OO

• Ca + 2 HNOCa + 2 HNO33 Ca(NO Ca(NO33))22 + H + H22

H+ + OH- H2O

2I- + Cl2 2Cl- + I2

Mg + Cu2+ Mg2+ + Cu

Mg + 2 H+ Mg2+ + H2

CO3-2 + 2H+ CO2 + H2O

H+ + OH- H2O

Ca + 2 H+ Ca2+ + H2

Mole CalculationsMole CalculationsTHREE EQUATIONS TO LEARN AND THREE EQUATIONS TO LEARN AND USEUSE

• Moles = mass / rfmMoles = mass / rfm

• Moles = volume/24dmMoles = volume/24dm33

• Moles = Moles = (volume*concentration)/1000(volume*concentration)/1000

In any question use a marker pen to highlight the key information given in the questions so you

can work out which equation to use

Solving mole equationsSolving mole equations1.1. Use the appropriate mole equation to Use the appropriate mole equation to

calculate the number of moles of one of the calculate the number of moles of one of the substances in the equationsubstances in the equation

2.2. Use the balanced equation to work out the Use the balanced equation to work out the mole ratiomole ratio

3.3. Using the mole ratio calculate the number of Using the mole ratio calculate the number of moles of the other substance.moles of the other substance.

4.4. Rearrange the appropriate mole equation to Rearrange the appropriate mole equation to answer the question.answer the question.

SEEMS COMPLICATED BUT IT ISN’T WITH PRACTICE!!!!!SEEMS COMPLICATED BUT IT ISN’T WITH PRACTICE!!!!!

Moles = mass / rfmMoles = mass / rfmWhat mass of Calcium oxide is formed when 25g of What mass of Calcium oxide is formed when 25g of

Calcium Carbonate is decomposed?Calcium Carbonate is decomposed?

CaCOCaCO33 CaO + CO CaO + CO22

Using the equation moles = mass/rfm calculate the moles of CaCO3

Moles of CaCO3 = 25/100 (from 40+12+16+16+16) = 0.25

Now use the balanced equation to work out the mole ratio

1 CaCO3 1 CaO + CO2 ( Hence mole ratio is 1:1)

Moles of CaO = Moles of CaCO3 = 0.25

Using the Using the equation moles = mass/rfm calculate the mass of CaO

0.25 = Mass/Rfm = mass / 56 (from 40+16)

Mass = 0.25 x 56 = 14 g

Moles = mass / rfmMoles = mass / rfmUSE WHEN ASKED ABOUT THE MASS OF USE WHEN ASKED ABOUT THE MASS OF REACTANT REQUIRED.REACTANT REQUIRED.What mass of Potassium is required to form 0.94g of Potassium What mass of Potassium is required to form 0.94g of Potassium oxide?oxide?

2K + ½ O2K + ½ O22 K K22OOUse the equation Moles = mass/rfm calculate the moles of K2O present

Rfm K2O = 39+39+16 = 94

Moles K2O = 0.94 /94 = 0.01

Use the equation to work out the mole ratio

2K + ½ O2 1K2O Mole ratio = 1:2 hence moles of K = 2xmoles of K2O

Use the equation Moles = mass/rfm calculate the mass of K present

0.02 = mass / 39

Mass K = 0.02 x 39 = 0.78

Moles = volume/24dmMoles = volume/24dm33

• USE WHEN ASKED ABOUT THE VOLUME OF A GAS BEING FORMED. USE WHEN ASKED ABOUT THE VOLUME OF A GAS BEING FORMED. • You will almost certainly have to use it along side another mole equation.You will almost certainly have to use it along side another mole equation.

What volume of carbon dioxide is formed when 25g of Calcium Carbonate is decomposed?What volume of carbon dioxide is formed when 25g of Calcium Carbonate is decomposed?

CaCOCaCO33 CaO + CO CaO + CO22Using the equation moles = mass/rfm calculate the moles of CaCO3

Moles of CaCO3 = 25/100 (from 40+12+16+16+16) = 0.25

Now use the balanced equation to work out the mole ratio1 CaCO3 CaO + 1 CO2 ( Hence mole ratio is 1:1)

Moles of CO2 = Moles of CaCO3 = 0.25Using the Using the equation moles = Volume of gas / 24dm3 calculate the volume of CO2

0.25 = Volume of gas / 24dm3 Volume = 0.25 x 24= 6dm3

Moles = volume/24dmMoles = volume/24dm33

• USE WHEN ASKED ABOUT THE VOLUME OF A GAS required. USE WHEN ASKED ABOUT THE VOLUME OF A GAS required. • You will almost certainly have to use it along side another mole equation.You will almost certainly have to use it along side another mole equation.

What volume of oxygen is required to form 18.8g of Potassium oxide?What volume of oxygen is required to form 18.8g of Potassium oxide?

2K + ½ O2K + ½ O22 K K22OO

Moles of KMoles of K22O = 18.8 / 94 = 0.2O = 18.8 / 94 = 0.2

Moles of OMoles of O22 = 0.1 (mole ratio = ½ O = 0.1 (mole ratio = ½ O2 2 : 1 K: 1 K22OO

Volume of OVolume of O22 = 2.4dm = 2.4dm33

Moles = volume/24dmMoles = volume/24dm33

• You need to make sure that the units of the You need to make sure that the units of the gas has been converted to dmgas has been converted to dm33

• 1dm1dm33 = 1000cm = 1000cm33 so 1cm so 1cm33 = 0.001dm = 0.001dm33

How many moles of ClHow many moles of Cl22 are present in 24cm are present in 24cm33 ? ?

Volume = 24cmVolume = 24cm33 = 0.024dm = 0.024dm33

Moles = volume /24 = 0.024/24 = 0.001 Moles = volume /24 = 0.024/24 = 0.001 molesmoles

Moles = Moles = (volume*concentration)/1000(volume*concentration)/1000•USE WHEN GIVEN INFORMATION REGARDING THE USE WHEN GIVEN INFORMATION REGARDING THE CONCENTRATION OF A SOLUTIONCONCENTRATION OF A SOLUTION

Key questionsKey questionsWhat is concentration?What is concentration?

Whatt units are concentration measured in?Whatt units are concentration measured in?

Concentration is defined as the number of moles (or the mass) of a substance dissolved in 1dm3 (1000cm3) of water.

A concentration of NaCl of 0.5Molsdm-3 means

0.5 moles of NaCl dissolved in 1dm3 H2O

The units are Moles / dm3 = moldm-3 = Molar = M

OR g/dm3 or gdm-3

Calculating concentrationsCalculating concentrations

• A student added 5 g of NaOH into 250mls of A student added 5 g of NaOH into 250mls of water. What was the concentration of NaOH inwater. What was the concentration of NaOH in

• A) g/dmA) g/dm33

B) mols/dmB) mols/dm33

• Another student added 0.0074g of Ca(OH)Another student added 0.0074g of Ca(OH)22 into 50cminto 50cm33 of water. What was the of water. What was the concentration ofconcentration of

• A) Ca(OH)A) Ca(OH)22 in g/dm in g/dm33

B) Ca(OH)B) Ca(OH)22 in mols/dm in mols/dm33

C) OHC) OH-- ions in mols/dm ions in mols/dm33

It is easy to determine the conc in g / dm3 using simple ratios

5 grams in 250cm3 so in 1000 cm3 there must be 20g

Hence conc = 20g/dm3

Use the equation Moles = mass/rfm to calculate the number of NaOH dissolved in the water

Rfm NaOH = 23+16+1 = 40

Hence 5g of NaOH = 5 / 40 = 0.125 moles

Volume of solution = 250 Moles = 0.125

Use the equation Moles = Volume x conc /1000 to calculate the concentration of the NaOH

0.125 = 250xconc/1000 hence 0.125x1000/250 = conc

Conc = 0.5Moldm-3

It is easy to determine the conc in g / dm3 using simple ratios0.0074 grams in 50cm3 so in 1000 cm3 there must be 0.148g Hence conc = 0.148g/dm3

To convert from g/dm3 to molsdm-3 use the equation moles = mass/rfm

0.148gdm3 = 0.148/(40+16+16+1+1)molsdm3 = 0.148/74molsdm-3 = 0.002moldm-3

ASIDE

To convert from mols/dm3 to g/dm3 you use the equation mass = moles x rfm

Hence 0.1Molddm-3 Ca(OH)2 = 0.1x74gdm3 = 7.4gdm3

In 1 mole of Ca(OH)2 there are 2 moles of OH- hence the conc of the OH- ions is twice that of the Ca(OH)2 = 0.004M

• A student took 10cmA student took 10cm33 of 1.08M HCl. 13cm of 1.08M HCl. 13cm33 of NaOH of NaOH neutralised the acid. What is the concentration of the neutralised the acid. What is the concentration of the acid?acid?

• A student took 10cmA student took 10cm33 of 0.1M Ca(OH) of 0.1M Ca(OH)22 solution and solution and neutralised it with 0.25M HCl. What volume of HCl did neutralised it with 0.25M HCl. What volume of HCl did he add?he add?

• Another student took 5cmAnother student took 5cm33 of an unknown concentration of an unknown concentration of Hof H22SOSO4 4 andand diluted it to 1000cmdiluted it to 1000cm3. 3. 10cm10cm3 3 of the diluted of the diluted solution was exactly neutralised with 18cmsolution was exactly neutralised with 18cm33 of 0.05M of 0.05M NaOH.NaOH.

• What was the concentration of the diluted acid?What was the concentration of the diluted acid?• What was the concentration of the undiluted acid?What was the concentration of the undiluted acid?

Use the equation Moles = Vol x conc to determine the number of moles of HCl present

Moles = 10 x1.08/1000 = 0.0108 moles

Use the balanced equation to work out the mole ratio

1HCl + 1NaOH NaCl + H2O (Mole ratio = 1:1)

Moles NaOH = 0.0108

Rearrange the equation Moles = Vol x conc to determine conc of NaOH

Moles = Vol x conc / 1000 so Moles x 1000/ vol = conc

Conc = 0.0108x1000/13 = 0.831M

Use the equation Moles = Vol x conc to determine the number of moles of Ca(OH)2 present

Moles = 10 x0.1/1000 = 0.001 moles

Use the balanced equation to work out the mole ratio2HCl + 1Ca(OH)2 CaCl2 + 2H2O (Mole ratio = 2:1)Moles HCl = 0.002 (twice that of the Ca(OH)2)

Rearrange the equation Moles = Vol x conc to determine vol of HClMoles = Vol x conc / 1000 so Moles x 1000/ conc = vol

Vol = 0.002 x 1000 / 0.25 = 8cm3

Use the equation Moles = Vol x conc to determine the number of moles NaOH presentMoles = 18 x 0.05 / 1000 = 0.0009

Use the balanced equation to work out the mole ratio

1H2SO4 + 2NaOH Na2SO4 + 2H2O (Mole ratio = 1:2)

Moles diluted H2SO4 = 0.000045 (half of 0.0009)

Rearrange the equation Moles = Vol x conc to determine conc of the diluted H2SO4

Conc = 0.000045*1000/10 = 0.045Mol/dm3

The student originally diluted the sulphuric acid from 5cm3 to 1000cm3 making it 200 times more diluteHence the original undiluted acid must be 200 times more concentrated than the original Hence conc of undiluted acid = 200x0.045 = 9Mol/dm-3

All the equations in same questionAll the equations in same question

• A student took 2.3g limestone and reacted it with A student took 2.3g limestone and reacted it with excess 0.5M HCl. He collected 0.5dmexcess 0.5M HCl. He collected 0.5dm33 of CO of CO22

• Assuming that the limestone was pure CaCOAssuming that the limestone was pure CaCO33. . What is the minimum volume of HCl needed to What is the minimum volume of HCl needed to react with the limestone? react with the limestone?

• What is the percentage of CaCOWhat is the percentage of CaCO33 in the limestone in the limestone given the results obtained by the student.given the results obtained by the student.

A lot of information here highlight relevant and think which mole equations you can useA lot of information here highlight relevant and think which mole equations you can use

This might seem a complicated question, but is quite easy if the data This might seem a complicated question, but is quite easy if the data given is used sensiblygiven is used sensibly

You are given a mass of limestone (2.3g) (and hence CaCO3 if pure) so the moles of CaCO3 can be determined using moles = mass/rfm

RFm CaCO3 = 100 (40 + 12 + 16 + 16 + 16)

Moles of CaCO3 = 2.3/100 = 0.023

Balanced equation 1 CaCO3 + 2 HCl CaCl2 + CO2 + H2O

Mole ratio = 1:2 Moles of HCl = 2 x moles CaCO2 = 0.046

You are given a conc of HCl so you can work out the volume of HCl required by rearranging Moles = vol x conc / 1000

Vol = 0.046*1000 / 0.5 = 92 cm3

As Limestone IS NOT pure CaCO3 this amount of HCL was be an excess (assuming other materials in Limestone do not themselves react with HCl.

The student collected 0.5dm3 of CO2 in the experiment so the number of moles of CO2 can be determined

Moles = volume of gas/24

Moles = 0.5 / 24 = 0.020833

1 CaCO3 + 2 HCl CaCl2 + 1 CO2 + H2OMole ratio = 1:1 Moles of CaCO3 present = Moles CO2

Moles CaCO3 = 0.020833

Mass CaCO3 present in the limestone = 100 x 0.020833Mass of CaCO3 present in the limestone = 2.0833Mass of limestone = 2.3g% CaCO3 in limestone = 2.0833/2.3 x 100 = 90.6%

• 1.1. Limestone (CaCOLimestone (CaCO33) is thermally ) is thermally decomposed to CaOdecomposed to CaO

a.a. Write a balanced equation (Write a balanced equation (with with state symbols)state symbols) for this reaction for this reaction