f321 Mod 1 Atoms and Reactions1

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1.1.1 Atomic StructureDetails of the three Sub-atomic (fundamental) Particles

Particle Position Relative Mass Relative Charge

Proton Nucleus 1 +1

Neutron Nucleus 1 0

Electron Orbitals 1/1800 -1

An atom of Lithium (Li) can be represented as follows:

7

3Li

Atomic Number

Atomic SymbolMass Number

The atomic number, Z, is the number of protons in the nucleus.

The mass number ,A, is the total number of protons and neutrons in the atom.

Number of neutrons = A - Z

IsotopesIsotopes are atoms of same element with the same number of protons, but different numbers of neutrons.

Isotopes have similar chemical properties because they have the same electronic

structure. They may have slightly varying physical properties because they have

different masses.

There are various

models for atomic

structure

1N Goalby chemrevise.org

!"INITI#N$ Relative atomic mass is the average mass of one atom

compared to one twelfth of the mass of one atom of carbon%1&

!"INITI#N$ Relative Isotopic mass is the mass of one isotope compared to

one twelfth of the mass of one atom of carbon%1&

The relative atomic mass 'uoted on the periodic table is a weighted average of all the isotopes

R.A.M = Σ (isotopic mass x % abundance)

100

"or above e(ample of )g

*.A.) + -/. ( &0 2 -13.14 ( &5 2 -11.1 ( &67 8133 + &0.4

R.A.M = Σ (isotopic mass x relative abundance)

total relative abundanceIf relative abundance is used instead of

percentage abundance use this e'uation

Calculating the Relative Atomic Mass of an Element

9ercentage Abundance /. 13.14 11.1

*elative Isotopic mass &0.33 &5.33 &6.33

Isotope )g&0 )g&5 )g&6



&

1.1.2 Moles and Equations

The mole is the :ey concept for chemical calculations

!"INITI#N$ The mole is the amount of substance in grams that has the same

number of particles as there are atoms in 1& grams of carbon%1&.Avogadro;s Number

There are 6.3& ( 13&4 atoms in

1& grams of carbon%1&.Therefore e(plained in simpler

terms ;#ne mole of any specified

entity contains 6.3& ( 13&4 of

that entity;$

"or most calculations we will do at A< we will use the following 4 e'uations

amount = massMr

1. For pure solids and gases 3. For solutions

Concentration = amountvolume

=earn these e'uations carefully and what units to use in them.

>nit of )ass$ grams

>nit of amount $ mol

>nit of concentration$ mol dm%4 or )

>nit of ?olume$ dm3

Converting volumes

cm4 dm4 ÷ 1333

cm4 m4 ÷ 1333 333

dm4 m4 ÷ 1333

Tpical mole calculations

!"INITI#N$ Relative atomic mass is the average mass of one atom


!"INITI#N$ Relative molecular mass is the average mass of a molecule


<ome <imple calculations using above e'uationsE!ample 2$ @hat is the concentration of solution made by

dissolving 5g of Na&#4 in &53 cm4 waterB

amount + mass8)r

+ 5 8 -&4 (& 2 1& 216 (4

+ 3.30& mol

conc+ amount8?olume

+ 3.30& 8 ".2#

+ 3.1/C mol dm%4

E!ample 1$ @hat is the amount, in mol, in 45g of

u<#0B

amount + mass8)r

+ 458 -64.5 2 4& 216 (0

+ 3.&1C mol

It is usually best to give

your answers to 4sf

Remember the Mr mustbe calculated and quotedto 1dp

N Goalby chemrevise.org

2. For $ases

Gas Volume (dm3)= amount x 24

These e'uations give the volume of a

gas at room pressure -1atm and

room temperature &5o.

E!ample 3 $ @hat is the volume in dm4 at room temperature

and pressure of 53g of arbon dio(ide gas B

amount + mass8)r

+ 538 -1& 2 16 (&

+ 1.146 mol

Gas Volume (dm3)= amount x 24

+ 1.146 ( &0

+ &.&6 dm4 - or &.4 dm4 to 4 sig fig

E!ample % $ Dow many atoms of Tin are there in a 6g sample

of Tin metalB

amount + mass8Ar

+ 68 11/.

+ 3.35355 mol

Number atoms + amount ( 6.3& ( 13&4

+ 3.35355 ( 6.3& ( 13&4

+ 4.30 (13&&



Molecular formula from empirical formula

Definition: A molecular formula is the actual number of atoms of each element in the compound.

From the relative molecular mass (Mr) work out how many timesthe mass of the empirical formula fits into the Mr.

Example 6 : work out the molecular formula for thecompound with an empirical formula of C3H6O anda Mr of 116

C3H6O has a mass of 58

The empirical formula fits twice into Mr of 116

So molecular formula is C6H12O2

The Mr does not need to be exact to turn anempirical formula into the molecular formulabecause the molecular formula will be awhole number multiple of the empiricalformula

&e'inition( An empirical formula is the simplest ratio of atoms of each element in the compound.Empirical formulae

Step 1 : Divide each mass (or % mass) by the atomic mass of the element

Step 2 : For each of the answers from step 1 divide by the smallest one ofthose numbers.

Step 3: sometimes the numbers calculated in step 2 will need to be multipliedup to give whole numbers.

These whole numbers will be the empirical formula.

General method

The same method can beused for the following typesof data:

1. masses of each elementin the compound

2. percentage mass of eachelement in the compound

Example 5 : Calculate the empirical formula for a compound that contains 1.82g ofK, 5.93g of I and 2.24g of O

Step1: Calculate amount, in mol, by dividing each mass by the atomic mass of the element

K = 1.82 / 39.1 I = 5.93/126.9 O = 2.24/16

= 0.0465 mol = 0.0467mol = 0.14 mol

Step 2 For each of the answers from step 1 divide by the smallest one of those numbers.

K = 0.0465/0.0465 I = 0.0467/0.0465 O = 0.14 / 0.0465

=1 = 1 = 3

Empirical formula =KIO3




N& 2 4D& &ND4

onverting 'uantities between different substances using a balanced e'uation

The balancing -stoichiometric numbers are mole ratios

e.g. 1 mol of N& reacts with 4 mol of D& to produce &mol of ND4

Typically we are given a 'uantity of

one substance and are as:ed to

wor: out a 'uantity for another

substance in the reaction. Any of

the above three e'uations can be

used.

<tep 1$>se one of the above 4 e'uations to

convert any given 'uantity into amount

in mol

)ass amount

?olume of gas amount

onc and vol of solution amount

<tep &$

>se balanced e'uation to convert

amount in mol of initial substance

into amount in mol of second

substance

<tep 4

onvert amount, in mol, of secondsubstance into 'uantity 'uestion

as:ed for using relevant e'uation

e.g. amount ,)r mass

Amount gas vol gas

amount, vol solution conc

E!ample 1"$ @hat mass of opper would react completely with

153 cm4 of 1.6) nitric acidB

4u 2 /DN#4 4u-N#4 & 2 &N# 2 0D&#

<tep 1$ wor: out amount, in mol, of nitric acid

amount + conc ( vol

+ 1.6 ( 3.15

+ 3.&0 mol

<tep &$ use balanced e'uation to give moles of u

/ moles DN#4 $ 4 moles u

<o 3.&0 DN#4 $ 3.3C -3.&0 ( 48/ mol u

<tep 4$ wor: out mass of u

)ass + amount ( )r

+ 3.3C ( 64.5

+5.1g

E!ample )( &4.6cm4 of D&<#0 neutralised &5.3cm4 of 3.15)

Na#D. @hat is the concentration of the D&<#0B

D&<#0 2 &Na#D Na&<#0 2&D&#

<tep 1$ wor: out amount, in mol, of sodium hydro(ide

amount + conc ( vol+ 3.15 ( 3.3&5

+ 3. 3345 mol

<tep &$ use balanced e'uation to give moles of D&<#0

& moles Na#D $ 1 moles D&<#0

<o 3.3345 Na#D $ 3.331/5 mol D&<#0

<tep 4 wor: out concentration of D&<#0

conc+ amount8?olume

+ 3.331/5 8 3.3&46

+ 3.3C0 mol dm%4

E!ample *$ @hat mass of arbon dio(ide would be produced

from heating 5.5 g of sodium hydrogencarbonateB

&NaD#4 Na&#4 2 #& 2 D&#

<tep 1$ wor: out amount, in mol, of sodium hydrogencarbonate

amount + mass 8 )r+ 5.5 8/0

+ 3.3655 mol

<tep &$ use balanced e'uation to give amount in mol of #&

& moles NaD#4 $ 1 moles #&

<o 3.3655 DN#4 $ 3.34&/mol #&

<tep 4$ wor: out mass of #&

)ass + amount ( )r

+ 3.34&/ ( 00.3

+1.00g


E!ample +$ @hat volume in cm4 of o(ygen gas would be

produced from the decomposition of 3.54& g of potassium

chlorate-?B

&El#4 & El 2 4#&

<tep 1$ wor: out amount, in mol, of potassium chlorate-?B

amount + mass 8 )r

+ 3.54& 81&&.6

+ 3.33040 mol

<tep 4$ wor: out volume of #&

Gas ?olume -dm4+ amount ( &0

+ 3.33651 ( &0

+ 3.156 dm4

+ 156 cm4

<tep &$ use balanced e'uation to give amount in mol of #&

& moles El#4 $ 4 moles #&

<o 3.33040 DN#4

$ 3.33651mol #&



5

Common Neutralisation Reaction Equations

D&<#0 2 &Na#D Na&<#0 2&D&#

Dl 2 Na#D Nal2D&#

&Dl 2 Na2CO3 &Nal 2 #& 2 D&#

&Dl 2 a#4 al& 2 #& 2 D&#


An Alkali is a soluble base that releases OH- ions in aqueoussolution;The most common alkalis are sodium hydroxide (NaOH),potassium hydroxide (KOH) and aqueous ammonia (NH3)

1.1.3 Acids

An acid releases H+ ions in aqueous solution There are several definitions of acidswe use in chemistry. One we use laterin the course is the Bronsted- Lowryacid which is a defined as a proton (H+)donor

The most common strong acids are :Hydrochloric ( HCl), sulphuric (H2SO4) and nitric (HNO3) acid;

A Salt is formed when the H+ ion of an acid is replaced bya metal ion or an ammonium ion

#bservations $ In carbonate reactions there will be

!ffervescence due to the #& gas evolved and the

solid carbonate will dissolve

Bases and Alkalis

Bases neutralise acids. Common bases are metaloxides, metal hydroxides and ammonia.

The Bronsted- Lowry base is definedas a proton (H+) acceptor

A base readily accepts H+ ions from an acid:eg OH- ions accepts an H+ ion forming H2ONH3 accepts an H+ ion forming NH4

+ ion

Neutralisation Reactions

Neutralisation reaction form salts

,CI& - ,/E /,0T - ,TER

&DN#4 2 )g-#D& )g-N#4& 2 &D&#

&Dl 2 a# al& 2D&#

,cid - Carbonate /alt - ater - Carbon &io!ide

D&<#0 2 E&#4 E&<#0 2 #& 2 D&#

Dl 2 ND4 ND0l



6

The water of crystallisation in calcium sulphate crystals can beremoved as water vapour by heating as shown in the followingequation.CaSO4.x H2O(s)→ CaSO4(s) + x H2O(g)

Method.•Weigh an empty clean dry crucible and lid .•Add 2g of hydrated calcium sulphate to the crucible andweigh again•Heat strongly with a Bunsen for a couple of minutes•Allow to cool•Weigh the crucible and contents again•Heat crucible again and reweigh until you reach a constantmass ( do this to ensure reaction is complete).

Small amounts the solid , such as0.100 g, should not be used inthis experiment as errors inweighing are too high.

Large amounts of hydrated calcium sulphate, such as50g, should not be used in this experiment as thedecomposition is like to be incomplete.

The lid improves the accuracy of theexperiment as it prevents loss of solidfrom the crucible but should be loosefitting to allow gas to escape.

The crucible needs to be dry otherwise a wet cruciblewould give an inaccurate result. It would cause mass lossto be too large as water would be lost when heating.

Heating in a crucibleThis method could be used for measuring mass loss in variousthermal decomposition reactions and also for mass gain whenreacting magnesium in oxygen.

Example 12. 3.51 g of hydrated zinc sulphate were heated and 1.97 gof anhydrous zinc sulphate were obtained.Use these data to calculate the value of the integer x in ZnSO4.x H2O

Calculate the mass of H2O = 3.51 – 1.97 = 1.54g

Calculate molesof ZnSO4

Calculate molesof H2O

1.97161.5

=1.5418

=

= 0.0122 mol =0.085 mol

Calculate ratio of moleof ZnSO4 to H2O

0.01220.0122

= 0.0850.0122

=1=7

=

X = 7


A Hydrated salt contains water of crystallisation

Hydrated salt

Cu(NO3)2 .6H2O

hydrated copper (II) nitrate(V).

Cu(NO3)2

Anhydrous copper (II) nitrate(V).

Example 11Na&<#0 . (D&# has a molar mass of 4&&.1, alculate

the value of (

)olar mass (D&# + 4&&.1 F -&4(& 2 4&.1 2 16(0

+ 1/3

+ 1/381/

+13



Potential errors in using a gas syringe•gas escapes before bung inserted•syringe sticks• some gases like carbon dioxide or sulphurdioxide are soluble in water so the true amount ofgas is not measured.

Using a gas syringe

If drawing a gas syringe makesure you draw it with somemeasurement markings on the

barrel to show measurementscan be made.

The volume of a gas depends on pressureand temperature so when recording volumeit is important to note down the temperatureand pressure of the room.

Make sure you don’t leavegaps in your diagram wheregas could escape

Gas syringes can be used for a variety of experiments where the volume of a gas is measured, possibly towork out moles of gas or to follow reaction rates.

Moles of gas can be calculated from gasvolume (and temperature and pressure)using ideal gas equation PV = nRT

•Weigh required mass of solute in a weighing bottle.

• Tip contents into a beaker and add100cm3 of distilledwater. Use a glass rod to stir to help dissolve the solid.

•Sometimes the substance may not dissolve well in coldwater so the beaker and its contents could be heatedgently until all the solid had dissolved.

•Pour solution into a 250cm3 graduated flask via a funnel.Rinse beaker and funnel and add washings from thebeaker and glass rod to the volumetric flask.•make up to the mark with distilled water using adropping pipette for last few drops.• Invert flask several times to ensure uniform solution.

Making a solution

Weighing can be made moreaccurate by weighing bottle againafter it has been emptied into thebeaker- or alternatively the weighingbottle could be washed andwashings added to the beaker.

Remember to fill so the bottom of the

meniscus sits on the line on the neck ofthe flask. With dark liquids like potassiummanganate it can be difficult to see themeniscus.




/

Titrations

The method for carrying out the titration

•rinse equipment (burette with acid, pipette with alkali, conicalflask with distilled water)•pipette 25 cm3 of alkali into conical flask•touch surface of alkali with pipette ( to ensure correct amountis added)•adds acid solution from burette•make sure the jet space in the burette is filled with acid•add a few drops of indicator and refer to colour change at endpoint•phenolphthalein [pink (alkali) to colourless (acid): end point pinkcolour just disappears] [use if NaOH is used]•methyl orange [yellow (alkali) to red (acid): end point orange]

[use if HCl is used]•use a white tile underneath the flask to help observe the colourchange•add acid to alkali whilst swirling the mixture and add aciddropwise at end point

•note burette reading before and after addition of acid•repeats titration until at least 2 concordant results areobtained- two readings within 0.1 of each other

Recording results•Results should be clearly recorded in a table

•Result should be recorded in full (i.e. both initial and finalreadings)•Record titre volumes to 2dp (0.05 cm3)

Safety precautions

Acids and alkalis are corrosive(at low concentrations acids areirritants)Wear eye protection and glovesIf spilled immediately wash affectedparts after spillage

If substance is unknown treat it aspotentially toxic and wear gloves.

Working out average titre resultsOnly make an average of the concordant titre results

lf 2 or 3 values are within 0.10cm3

and therefore concordant or closethen we can say results are accurate

and reproducible and the titrationtechnique is good/ consistent

Testing batchesIn quality control it will be necessary to do titrations/testing onseveral samples as the amount/concentration of the chemical beingtested may vary between samples.

Titrating mixturesIf titrating a mixture to work out the concentration of an activeingredient it is necessary to consider if the mixture contains othersubstances that have acid base properties.If they don’t have acid base properties we can titrate withconfidence.

Safely dealing with excess acidSodium hydrogen carbonate (NaHCO3) and calcium carbonate (CaCO3) are good for neutralising excessacid in the stomach or acid spills because they are not corrosive and will not cause a hazard if used inexcess. They also have no toxicity if used for indigestion remedies but the CO2 produced can cause wind.Magnesium hydroxide is also suitable for dealing with excess stomach acid as it has low solubility in waterand is only weakly alkaline so not corrosive or dangerous to drink (unlike the strong alkali sodiumhydroxide). It will also not produce any carbon dioxide gas.


A conical flask is used in preferenceto a beaker because it is easier toswirl the mixture in a conical flaskwithout spilling the contents.

Common Titration Equations

D4#&D 2 Na#D D4#&%Na2 2 D&#

D&<#0 2 &Na#D Na&<#0 2&D&#

Dl 2 Na#D Nal2D&#

NaD#4 2 Dl Nal 2 #& 2 D&#

Na&#4 2 &Dl &Nal 2 #& 2 D&#

Distilled water can be added to the

conical flask during a titration to washthe sides of the flask so that all theacid on the side is washed into thereaction mixture to react with the alkali.It does not affect the titration readingas water does not react with thereagents or change the number ofmoles of acid added.



N Goalby chemrevise.org C

Calculating apparatus errorsEach type of apparatus has a sensitivity error

•balance ± 0.001 g•volumetric flask ± 0.1 cm3

•25 cm3

pipette ±

0.1 cm3

•burette (start & end readings and end point ) ± 0.15 cm3

Calculate the percentage error for each piece of equipment used by

% error = ± sensitivity error x 100Measurement made on apparatus

e.g. for burette% error = 0.15/average titre result x100

To calculate the maximum percentage apparatus error in the finalresult add all the individual equipment errors together.

Reducing errors in a titration

Replacing measuring cylinders with pipettes or burettes which havelower apparatus sensitivity error will lower the error

To reduce the error in a burette reading it is necessary to make thetitre a larger volume. This could be done by: increasing the volumeand concentration of the substance in the conical flask or bydecreasing the concentration of the substance in the burette.

To decrease the apparatus errors youcan either decrease the sensitivity errorby using apparatus with a greater

resolution (finer scale divisions ) or youcan increase the size of themeasurement made.

If looking at a series of measurementsin an investigation the experiments withthe smallest readings will have thehighest experimental errors.

Reducing errors in measuring massUsing a more accurate balance or a larger mass willreduce the error in weighing a solidWeighing sample before and after addition and thencalculating difference will ensure a more accuratemeasurement of the mass added.

If the jet space is not filled properlyprior to commencing the titration itwill lead to errors if it then fills duringthe titration, leading to a larger thanexpected titre reading.

Calculating the percentage difference between theactual value and the calculated value

If we calculated an Mr of 203 and the real value is 214,then the calculation is as follows:Calculate difference 214-203 = 11% = 11/214 x100

=5.41%

If the %error due to the apparatus < percentage

difference between the actual value and the

calculated value then there is a discrepancy inthe result due to other errors.

If the %error due to the apparatus > percentage

difference between the actual value and thecalculated value then there is no discrepancyand all errors in the results can be explained bythe sensitivity of the equipment.

Errors



13

1.1.4 Redox

oxidation is the process of electron loss:Zn Zn2+ + 2e-

It involves an increase in oxidation number

reduction is the process of electron gain:Cl2 + 2e- 2Cl-

It involves a decrease in oxidation number

Rules 'or assigning o!idation numbers

1. All uncombined elements have an o(idation number of Hero eg . Zn, l&, #&, Ar all have o(idation numbers of Hero

&. The o(idation numbers of the elements in a compound

add up to HeroIn Nal Na+ 21 l+ %1

<um + 21 %1 + 3

4. The o(idation number of a monoatomic ion is e'ual to

the ionic charge e.g. Zn&2 + 2& l% + %1

0. In a polyatomic ion -#4

&% the sum of the individual

o(idation numbers of the elements adds up to the charge

on the ion

e.g. in #4

&% + 20 and # + %&

sum + 20 2 -4 ( %& + %&

5. <everal elements have invariable o(idation numbers in theircommon compounds.

Group 1 metals + 21

Group & metals + 2&

Al + 24

D + 21 -e(cept in metal hydrides where it is F1 eg NaD

" + %1

l, r, I + F1 e(cept in compounds with o(ygen and fluorine

# + %& e(cept in pero(ides -D&#

& where it is F1 and in compounds with fluorine.

@e use these rules to

identify the o(idation

numbers of elements that

have variable o(idation

numbers.

Note the o(idation number of l

in al&

+ %1 and not %& because

there are two lJs

Always wor: out the o(idation

for one atom of the element

@hat is the o(idation number of "e in "el4

>sing rule 5, l has an #.N. of F1>sing rule &, the #.N. of the elements must add up to 3

"e must have an #.N. of 24

in order to cancel out 4 ( F1 + %4 of the lJs


Naming using Roman Numerals

In IUPAC convention the various forms of sulphur, nitrogen and chlorine compounds whereoxygen is combined are all called sulfates, nitrates and chlorates with relevant oxidation numbergiven in roman numerals. If asked to name these compounds remember to add the oxidationnumber.

NaClO: sodium chlorate(I)NaClO3: sodium chlorate(V)K2SO4 potassium sulfate(VI)K2SO3 potassium sulfate(IV)

Use a Roman numeral to indicate the magnitude of the oxidation state of anelement, when a name may be ambiguous.

NaNO2: Sodium nitrate(III)NaNO3: Sodium nitrate(V)

FeCl2 Iron(II) ChlorideFeCl3 Iron(III) Chloride



N Goalby chemrevise.org 11

Redox Reactions

metals generally form ions by losingelectrons with an increase in oxidationnumber to form positive ions:

Zn Zn2+ + 2e-

non-metals generally react by gainingelectrons with a decrease in oxidationnumber to form negative ions

Cl2 + 2e- 2Cl-

4Li + O2 2Li2O

0

0

+1

-2

Lithium is oxidising because itsoxidation number is increasing from 0to +1

Oxygen is reducing because

its oxidation number isdecreasing from 0 to -2

WO3 + 3H2 W + 3H2O

0

0

+1

+6

Hydrogen is oxidisingbecause its oxidation numberis increasing from 0 to +1

Tungsten is reducing because

its oxidation number isdecreasing from +6 to 0

2Sr(NO3)2 2SrO + 4NO2 + O2

-2

+4

0

+5

Oxygen is oxidising because its oxidationnumber is increasing from -2 to 0

Nitrogen is reducing becauseits oxidation number isdecreasing from +5 to+4

Redox Reactions of Metals and acid

&Dl 2 )g )gl& 2D&

ACID + METAL SALT + HYDROGEN

0

0

+2

+1

Magnesium is oxidisingbecause its oxidation number isincreasing from 0 to +2

Hydrogen is reducingbecause its oxidation numberis decreasing from +1 to 0

"e 2 D&<#0 "e<#0 2D&

Be able to write equations for reactions ofmetals with hydrochloric acid and sulphuricacid

Observations: These reaction willeffervesce because H2 gas is evolvedand the metal will dissolve

Note that not all the oxygen atoms are

changing oxidation number in this reaction

2 NH3 + NaClO N2H4 + NaCl + H2O

+1

-2

-1

-3

Nitrogen is oxidising because its oxidationnumber is increasing from -3 to -2

Chlorine is reducing becauseits oxidation number isdecreasing from +1 to -1

f321 Mod 1 Atoms and Reactions1

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