Chemistry (ES)

Elements of the Sea Revision Cards

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  • Created on: 02-04-13 16:53

Structure of an ionic lattice

  • Sodium ions (Na+), each surrounded by 6 Cl- chloride ions and vice versa
  • Held together by attraction between oppositely charged ions.
  • Form giant ionic lattice (this case = simple cubic)
  • Similar electrostatic bonds in all lattices
  • Type of lattice depends on number and size of anions and cations

When ionic crystals containg water molecules, they are hydrated.

- water molecules sit within ionic lattic = water of crystallisation. e.g. blue copper sulphate = CuSO4 . 5H20 and anhydrous white copper sulphate = CuSO4.

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Ionic Substances in Solution

In water, oxygen is more electronegative than hydrogen = polar.

When ions are dissolved in water:

  • surrounded by water molecules
  • spread throughout solution
  • hydrated ions = randomly arranged and behave independantly.

This is because:

  • - water is a polar molecule with a bent shape.
  • - positive H atoms attracted to negative ions and vice versa.
  • - Each ion has own sphere of water molecules = hydration.

NB: always label charge on ion, show polarity of water and surround each ion w/ at least 5 water molecules.

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Ionic Equations

  • Ions in solutions behave independantly.
  • 2 solutions mixed form a solid = precipitation reaction

e.g. Ag+(aq) + NO3-(aq) + Na+(aq) + Br-(aq) --> AgBr(s) + NO3-(aq) + Na+(aq)

Remove spectator ions (same on left and right) to form:

Ag+(aq) + Br-(aq) --> AgBr(s)

  • Must always include state symbols for ionic equations.

Predicting whether two solutions form a precipitate:

  • All nitrates = soluble in water (s/w)
  • All chloride = s/w ex. AgCl and PbCl2
  • All sulphate = s/w ex. BaSO4, PbSO4 and SrSO4
  • All sodium, potassium and ammonium salts = s/w
  • All carbonates insoluble ex. (NH4)2CO3 and group 1 carbonates.

Neutralisation also summarised by ionic equation: H+(aq) + OH-(aq) --> H20(l)

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Calculations of Concentrations

  • Can be measured in g dm-3
  • e.g. 40g of sodium hydroxide in 2dm3 = 40/2 = 20gdm-3
  • Chemists prefer to use moles 
  • Molar mass = number of grams a mole of a substance weighs
  • Concentration in moles = moldm-3

1 mole of a substance in 1 dm3 solution = 1 moldm-3 concentration

Moles (mol) = concentration (moldm-3) x volume (dm3)

Example: 

  • 20cm3 of 0.1moldm NaOH. How many moles?
  • 0.1 x 0.020 = 0.002 moles

Remember to convert cm3 to dm3.

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Titration Method

Titration = quantitatively finding the concentration of a solution by reacting a known volume of that solution with another solution of a known concentration.

1) Fixed volume of solution w/unknown concentration into conical flask. - Use pipette.

2) Suitable indicator added and mixture placed on white tile (to see end point).

3) Solution w/known concentration added slowly from burette - Constant swirling.

4) As end point nears, add solution from burette dropwise.

5) Repeat for accuracy until concordant (consistant) results are obtained.

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Using Concentrations in Calculations

  • 25.0 cm3 of potassium hydroxide solution.
  • 0.020 moldm-3 solution of sulphuric acid.
  • Changed colour when 27.9 cm3 added.
  • Concentration of potassium hydroxide?

1) Write equation;

  • H2SO4 + 2KOH ---> K2SO4 + 2H2O

2) Ratio of reactants: 1:2

3) Find number of moles of H2SO4:

  • 0.020 x 0.0279 = 5.58 x 10-4

4) Multiply by ratio: x 2

5) Calculate concentration:

  • Volume of KOH = 25.0cm3 so 1.116 x 10-3 / 0.025 = 0.045moldm-3
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First Ionisation Enthalpy

First ionisatino enthalpy = energy needed to remove one electron from one mole of isolated gaseous atoms of an element.

= One mole of gaseous ions with one positive charge formed.

= kJmol-1

= X(g) --> X+(g) + e-

NB: Remember to write (g).

1st ion enthp always positive - energy put in to remove electron because it is attracted to nucleus (endothermic).

Elements show following patterns:

  • Elements at peaks on graph in Group 0 = High ion enthp, full outer shells, difficult to remove electrons, v. unreactive.
  • Elements at troughs in Group 1 = one outer shell electron, easy to ionise, v. reactive.
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Variations in First Ionisation Enthalpy Across a P

Period

  • General trend = increasing 1st ion. enthp. as atomic number increases.
  • Because nuclear charge increases left to right - electrons added into same shell.
  • Greater attraction between nucleus and electron = more energy needed to remove one.

Group

  • 1st ion enthp decreases down group.
  • Attraction betweeen nucleus and outer electrons decreases.
  • Filled electron shells shield nucleus from outermost electrons.
  • Easier (less energy) to remove electrons.
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Successive Ionisation Enthalpies

  • 1st = X(g) --> X+(g) + e-
  • 2nd = X+(g) --> X2+(g) + e-
  • 3rd = X2+(g) --> X3+(g) + e-
  • 4th = X3+(g) --> X4+(g) + e-

Removal of electron from positive ion.

Two patterns:

  • Ionisation enthps = increase as successive electrons removed.
  • Remaining electrons attracted to nucleus more strongly.
  • Sharp jump in ion. enthp. when electron removed from full outer shell.
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OILRIG and Assigning Oxidation States

Oilrig

Redox reaction = oxidation reaction and reduction reaction occuring simultaneously.

Oxidation = losing electrons (oxidation state increasing - more positive)

Reduction = gaining electrons (oxidation state decreasing - more positive)

Assigning oxidation states

  • Atoms in elements = oxidation state of zero.
  • Compounds or ions = oxidation states assigned to each atom or ion.
  • Compounds have no overall charge, oxidation states of constituents must add up to zero.
  • Ions, sum of oxidation states = charge on ion.
  • Some atoms rarely change oxidation states in reactions, helps assign oxidation states to other species e.g. F = -1, O = -2 (except O2- and OF2) and Cl = -1 (except when with F or O).
  • Example: CO2: O = -2, there are two O's, so total contribution of O to oxidation state = -4, so to add up to zero C must equal +4.
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Using Oxidation States and Oxidation States in Nam

Using Oxidation States

Cl2(aq) + 2I-(aq) --> 2Cl-(aq) + I2(aq)

0       -1 -1      0

  • Chlorine is reduced (oxidation state decreases in reaction).
  • Iodine is oxidised (oxidation state increases in reaction).

Oxidation states in names

Some compounds have elements that can exist in more than one oxidation state - systematic name includes oxidation state in these compounds.

e.g. FeO = Iron (II) oxide.

Oxyanions should also include an oxidation state.

e.g. ClO3- = Chlorate (V), oxidation state = +5

Oxyanions = negative ions, contain oxygen and another element, name ends in '-ate'.

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Electron Transfer, Half Equations and Other Redox

Electron Transfer and Half Equations

Sodium reacts with chlorine: 2Na + Cl2 --> 2NaCl

Written as two half equations:

  • 2Na --> 2Na+ + 2e- = oxidation (electron loss) - reducing agent.
  • Cl2 + 2e- --> 2Cl- = reduction (electron gain) - oxidising agent.

Other redox changes with halide ions

Displacement reaction occurs when more reactive halogen e.g. Cl2 passes over solution of less reactive halogen e.g. KI.

  • Cl2(g) + 2I-(aq) --> 2Cl-(aq) + I2(aq)
  • Cl2(aq) + 2e- --> 2Cl-(aq) = reduction.
  • 2I-(aq) --> I2(aq) + 2e- = oxidation.
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Physical Properties of Group 7 (P Block)

NB: Halogens = elements in group 7.

  • Appearance at RM temp: Flourine = yellow gas/ Chlorine = green gas/ Bromine = red liquid/ Iodine = shiny black solid.
  • Volatility: F = gas/ Cl = gas/ Br = liquid, forms brown gas on warming/ I = sublimes to give purple vapour on warming.
  • Solubility: F = reacts with water/ Cl = slightly soluble, pale green solution/ Br = slightly soluble, red-brown solution/ I = rarely solube, brown solution.
  • Solubility in organic solvents: F = soluble/ Cl = soluble, pale green solution/ Br = soluble, red/ I = soluble, violet.

Halogens exist as elements as diatomic non-polar molecules e.g. F2

Intramolecular bonds = covalent.

Intermolecular bonds = instantaneous dipole-induced dipole bonds.

Flourine = most volatile, smallest molecule, least electrons.

Strength of intermolecular bonds increases with size and no. of electrons - explains states.

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Chemical Properties of Halogens

Reactivity increases down the group.

Halide ions and silver ions react to form a precipitate (except flourine):

  • White precipitate = silver chloride.
  • Cream precipitate = silver bromide.
  • Yellow precipitate = silver iodide.
  • General reaction = Ag+(aq) + X-(aq) --> AgX(s)

More reactive halogens displace less reactive e.g. Cl2(aq) + 2Br-(aq) --> 2Cl-(aq) + Br2

Halogens = oxidising agents, oxidation state decreases, remove electrons from other elements.

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Redox Reactions involving Halogens

E.g. solution containing chlorine added to solution containing iodided ions, brown colour appears as iodine is produced:

Cl2(aq) + 2I-(aq) --> I2(aq) + 2Cl-(aq)

Chlorine is a stronger oxidising agent than iodine.

Flourine = strongest oxidising agent (small so attraction of electrons to nucleus is stronger as they are closer, explains reactivity down group).

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Electrolysis of Solutions with Halide Ions and Sto

Electrolysis of solutions with halide ions

Electric current passed through sodium chloride solution:

  • Chlorine gas bubbles off at anode.
  • Chlorine ions lose electrons to anode, become oxidised: 2Cl- --> Cl2 + 2e-
  • Also works with other halogens.

This is the basis of the industrial manufacture of chlorine.

Storage and Transport of Halogens

  • Flourine = too reactive to store
  • Made in situ when needed by electrolysing liquid hydrogen flouride.
  • Chlorine = highly toxic gas, transported by rail or road tanker as a liquid.
  • Bromine transported in lead lined steel tanks supported by strong metal frames.
  • Transport routes for bromine planned to minimise accidents e.g. away from residential areas.
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Electronic Structure: Sub-shells and Orbitals

Electrons exist in shells e.g. 1, 2, 3... etc.

Shells divided into sub-shells (s,p,d,f) and sub-shells further divided into atomic orbitals.

NB: Each orbital holds a maxiumum of two electrons.

  • First shell = 1 sub-shell, s
  • Second shell = 2 sub-shells, s and p.
  • Third shell = 3 sub-shells, s,p and d.
  • Fourth shell = 4 sub-shells, s,p,d and f.
  • S sub-shell = 1 orbital, 2 electrons max.
  • P sub-shell = 3 orbitals, 6 electrons max.
  • D sub-shell = 5 orbitals, 10 electrons max.
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Electronic Configuration and Representing Electron

Electronic configuration = arrangement of electrons in shells and orbitals.

1) Orbitals filled in order of increasing energy.

2) More than one orbitals with same energy = orbitals first occupied singly by electrons. When each orbital is singly occupied then electrons pair up in orbitals.

3) Electrons in singly occupied orbitals have parralleled spins.

4) Electrons in doubly occupied orbitals = opposite spins.

NB: 4s orbital = lower energy level than 3d so fills up before 3d e.g. 3p6 4s2 3d10

Each orbital can be represented as a box and each electron as an arrow.

Electronic configuration for nitrogen e.g. 1s2 2s2 2p3

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Deducing Electron Configurations and s, p, d and f

Example:

1) Magnesium = atomic number 12, therefore has 12 protons.

2) Therefore magnesium has 12 electrons - 2 in shell 1, 8 in shell 2 and 2 in shell 3.

3) Therefore its electronic configuration = 1s2 2s2 2p6 3s2

NB: Some elements have abbreviations e.g. electronic configuration of potassium abbreviated to (Ar) 4s1, (Ar = argon).

  • Group 1 and 2 elements = 1-2 electrons in outer shell = s orbital is outermost sub-shell = s block elements.
  • Group 3, 4, 5, 6, 7 and 0 = 3-8 electrons in outer shell = p orbital outermost = p block elements.
  • First transition block elements = electrons filling d sub-shell = d block elements.
  • Further transition metals = electron filling f sub-shell = f block elements.
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Polar Molecules

Intermolecular bonds = bonds between molecules in any liquid or solid.

Dipole = occurs when molecule or part of a molecule has a positive end and a negative end.

Molecule has a dipole = polarised.

Molecules with a permanent dipole = polar molecules.

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Instantaneous Dipoles

If a molecule hasn't got a permenant dipole, electron density in the molecule may be unevenly distributed at any point (more electrons at one end of the molecule than the other).

If this happens it has an instantaneous dipole.

Electron density distribution changes so polarity will change.

Electrons on the second molecule get attracted to the positive end of the first molecule's instantaneous dipole (with less electrons). The second molecule's electrons then move to one end of the molecule, causing that molecule to have an induced dipole

This is an instantaneous dipole-induced dipole attraction.

  • Weakest type of intermolecular bond.
  • Happens in all types of molecules, even those already with a permenant dipole.
  • Electrons continually moving so instantaneous dipoles constantly being formed and broken.
  • More electrons an atom or molecule has = greater attraction = higher boiling point.
  • E.g. only intermolecular bonds between polythene are instantaneous dipole-induced dipole bonds but polythene is solid at rm temp = chains are long, pack closely together, lots of bonds even though they are weak.
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Permanent Dipole-Permanent Dipole Bonds

Molecules with permanent dipoles = atoms with different electronegatively values.

Slightly positively charged end of one molecule attracts slightly negatively charged end of next molecule - intermolecular bond occurs.

Stronger than instantantous dipole-induced dipole bonds but weaker than hydrogen bonds.

Responsible for holding polyester molecules together.

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Electronegativity and Bond Polarity and Polar Mole

Electronegativity

Degree to which atom of element attracts electrons = electronegativity.

More electronegative an element is, the greater its attraction for electrons is.

Electronegativity generally increases towards the top and right of the periodic table.

Order of electronegativity for some elements:

F > O > Cl > Br and N > I > S > C > H

Difference between electronegativities of C and H so small = considered to be non-polar.

Bond polarity and polar molecules.

  • Polar molecule has permenant dipole e.g. ethanoic acid.
  • Differences in electronegatives v. small = dipole is negligible.
  • Even if bonds are polar, molecule might not have a dipole e.g. tetrachloromethane.
  • Due to arrangement of polar bonds in the molecule e.g. symmetrical arrangement of chlorine atoms in tetrachloromethane - no overall dipole.
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Naming Halogenoalkanes and Physical Properties

Homologous series of halogenoalkanes = alkane series w/ hydrogen atoms substituted by one or more halogen atoms - often shown as R-Hal.

Alkane chain name = prefixed with name of halogen - halogens listed in alphebetical order with number indicating position e.g. 3-bromo-1-chlorobutane.

Physical Properties

  • Boiling points increase w/ heavier halogen atom (down the group)
  • BP increases with number of halogen atoms.
  • Both happen because overall number of electrons increases.
  • Increases strength of instantaneous dipole-induced dipole bonds.
  • Stronger intermolecular bonds = more energy needed to pull atoms apart = higher BP.
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Bond Enthalpies and Reactivity of Halogens

C-Hal bond becomes weaker as the size of the halogen atom increases.

Makes bond easier to break so compounds become more reactive.

Although C-F bonds are most polar, fluoroalkanes are very unreactive.

Bond strength rather than bond polarity affects reactivity the most.

  • Fluoro- compounds are very unreactive.
  • Chloro- compounds = reasonably stable in trophosphere, can react to produce chlorine radicals that deplete ozone.
  • Bromo- and iodo- compounds are reactive = useful as intermediates in chemical synthesis.
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Reactions of Halogenoalkanes

1) Homolytic fission = Breaks down into two radicals, with one electron from bond going to each radical (use curly arrows).

Conditions for homolytic fission = gas phase w/ high temperatures or presence of UV radiation e.g. in stratosphere.

2) Heterolytic fission = Carbon-halogen bond breaks to give ions. If polar C-Hal bond is brokend completely, negative halide ion moves away leaving C group positively charged.

This is now a carbocation.

Conditions for heterolytic fission = dissolved in polar solvent e.g. ethanol or water.

3) Substitution reactions = R-Hal + X- --> R-X + Hal-

X- is an example of a nucleophile, C-Hal bond breaks and Hal is replaced by another functional group.

Hal being replaced by nucleophile therefore = nucleophilic substitution.

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Mechanism for the Nucleophilic Substitution Reacti

Worked example

1) Carbon from C-Hal attacked by nucleophile.

2) Lone pair of electrons on nucleophile (nucleophiles have one or more lone pairs so can donate to form new bonds) forms a new bond with carbpn.

3) At the same time carbon-halogen bond breaks, giving a halide ion.

Reaction conditions for Hal with HOH nucleophile = heat under reflux - forms alcohol.

Reaction conditions for Hal with OH- nucleophile = heat under reflux with NaOH(aq) with ethanol as solvent - produces alcohol.

Reaction conditions for Hal with NH3 nucleophile = heated with concentrated ammonia solution in sealed tube - forms amine.

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Preparation of Halogenoalkanes

A tertiary alcohol like 2-methylpropan-2-ol reacts at room temp w/ concentrated hydrochloric acid to form 2-chloro-2-methylpropane.

Reaction is carried out in separating funnel, chloroalkane is immischible in water, forms a layer above aqueous products.

5 stages in purification of chloroalkane:

  • Upper layer containing chloroalkane run off into clean beaker.
  • Chloroalkane shaken w/ sodium hydrogencarbonate solution - remove any acidic impurities.
  • Chloroalkane layer run off for second time.
  • Anhydrous sodium sulphate (drying agent) added to remove any water.
  • Chloroalkane purified by distillation.
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Batch or Continuous Process?

Chemical plant = place where chemicals are manufactured.

Plant also describes site equipment e.g. reaction vessels, storage facilities and pipework

Typical chemical process = feedstock (reactants) prepared --> Reaction (temperature, pressure, catalyst, energy in or out) --> Separation (unused solvents and feedstock recycled) --> Products.

With chemicals manufactured using batch process, reactants are placed in reaction vessel, allowed to react. Products then removed, vessel is cleaned, ready for net batch.

In continuous process, starting materials = regularly fed in at one end of plant, product made at other.

Input or removal of energy may be required at any stage.

Dyes, pharmaceuticals and pesticides = batch process.

Ethene, ammonia and sulphuric acid = continuous process.

Main construction element for plants = mild steel, more expensive is used where corrosion resistance is required e.g. glass-lined steel.

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Advantages and Disadvantages of Batch and Continuo

Batch

Advantages:

  • Cost effective for small amounts
  • Low capital costs for plant
  • Same vessel can make different products.

Disadvantages:

  • Filling, emptying and cleaning vessel is time-consuming.
  • Larger workforce may be required.
  • Contamination of different products may be possible.

Continuous

Advantages:

  • Suited for high tonnage products.
  • Can operate for months at a time without shutdown
  • More easily automated (less workforce needed)

Disadvantages:

  • Less flexible - makes one product normally.
  • Higher capital costs for plant.
  • Not cost effective if run below full capacity.
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Raw Materials

Raw materials = materials that feedstocks are prepared from, usually obtained from the ground, e.g. rock salt or crude oil, or from the atmosphere.

Feedstocks = processed raw materials, reactants fed in at start of process.

E.g. Electrolysis of brine:

Raw material (rock salt) ---> feedstock (brine) ---> products (chlorine, sodium hydroxide and hydrogen).

Chlorine is main desired product - sodium hydroxide and hydrogen are co-products (produced at the same time via same reaction, sometimes useful). 

As amount of desired product increases, so does co-product.

By-products = result of unwanted side reaction.

Conditions designed to increase desired product and decrease by-products.

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Cost and Efficiency

Amount of profit chemical company makes depends on income from sales of product and costs.

Fixed costs = same. / Variable costs = depend on amount of product manufactured.

Efficiency of chemical process depends on factors including operating temperature and pressure.

High temps increase rate of reaction but may affect position of equilibrium and reduce yields.

High pressure can improve yield for some reactions but needs more expensive reaction vessels and can be dangerous. Recycling unreacted feedstock reduces costs.

Efficient energy use is essential - energy costs are high, any heat produced during exothermic reaction is converted by heat exchanger into hot water or steam for use elsewhere in plant.

Fixed costs = capital costs (buying land, building plant) and ongoing costs (labour, research and development, insurance).

Variable costs = raw materials, distribution of product.

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Plant Location, Health and Safety and Waste Dispos

Chemical plants traditionally sited near source of raw materials, however nowadays factors include:

  • Good transport network - sea terminal, good rail and road links to bring in and distribute raw materials and products.
  • Labour - new plants built near old ones to use available skilled labour.
  • Shared facilities - product from one may be feedstock for another.
  • Cheap energy
  • Water - often important for high tonnage materials.

Safety = major concern, companies in UK must abide by national and EU legislation, regular safety training provided, process checkes to reduce exposure of employees to hazardous chemicals or procedures.

Waste disposal = strict limits on amount of hazardous waste that can be released, fines for those disobeying limits, chemical waste must be treated before disposal.

Potential pollutants can be dealt with e.g. sulphuric acid from SO4, sulphuric acid sold to increase profits.

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Percentage Yield and Atom Economy

Percentage yield = (actual mass of product / theoretical mass of product) x 100.

E.g. moles of ethanol used = mass/Mr = 23.0/46.0 = 0.50 moles.

Look at ratio, should get 0.50 moles of ethene or 14g but we have 6.0. so % yield = 6.0/14.0 = 43%.

Atom economy = (relative formula mass of useful product / relative formula mass of reactants used) x 100.

Can be used to decide whether chemical process is economically viable.

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