WJEC Chemistry for AS Level Student Book: 2nd Edition (Draft)

WJEC Chemistry for AS Level

Atom economy and percentage yield When a reaction occurs, the compounds formed, other than the product needed, are a waste. An indication of the efficiency of a reaction can be given as its atom economy or its percentage yield . Atom economy is obtained from the chemical equation for the reaction. The higher the atom economy, the more efficient the process. Worked example Titanium used to be manufactured from titanium(IV) oxide by converting titanium(IV) oxide to titanium(IV) chloride then reducing it by magnesium. The process can be represented by: TiO 2 + 2Cl 2 + 2Mg + 2C →  Ti + 2MgCl 2 + 2CO

Key terms

Atom economy = mass of required product total mass of reactants × 100% or total M r of required product total M r of reactants × 100% Percentage yield = mass (or moles) of product obtained maximum theoretic mass (or moles)  × 100%

Calculate the atom economy for this reaction. Atom economy = total M r of required product total M r of reactants

× 100

47.9 79.9 + 142 + 48.6 + 24

 = 

× 100

 = 16.3%

Percentage yield is calculated from the mass of product actually obtained in an experiment.

Worked example 32.1g of ethanoic acid are obtained from the oxidation of 27.6g of ethanol. The reaction can be represented by the equation: C 2 H 5 OH + 2[O] →  CH 3 CO 2 H + H 2 O Calculate the percentage yield for this reaction. × 100 To calculate the maximum mass the three steps on page 37 must be followed: Step 1 Calculate the amount, in mol, of 27.6g C 2 H 5 OH n  =  m M  = 27.6 46.0  = 0.600 Step 2 Use the equation to calculate the amount in moles of CH 3 CO 2 H formed 1mol C 2 H 5 OH gives 1mol CH 3 CO 2 H 0.600mol C 2 H 5 OH give 0.600mol CH 3 CO 2 H Step 3 Calculate the mass of CH 3 CO 2 H m  =  nM  = 0.600 × 60.0 = 36.0g % yield = 32.1 36.0 × 100 = 89.2% DRAFT % yield = mass of product obtained maximum theoretical yield 16 Knowledge check Calculate the atom economy for the production of iron in the reduction of iron(III) oxide by carbon monoxide. Fe 2 O 3 (s) + 3CO(g)  →  2Fe(s) + 3CO 2 (g) 44

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