Calculations in organic chemistry

Empirical n molecular formular

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Empirical and molecular formulae

Empirical formula- simplest whoe number ratio of atoms in each element in a compound.

Calculate- 

firstly convert the mass into the amount of moles e.g 1.5g of carbon and 0.5g of H.

amount in moles- Mass (G) / molar mass

H= 0.5 / 1= 0.5    C= 1.5 / 12= 0.125 mol

Therefore find the ratio... 0.125 : 0.5 ( there is 4x more hydrogen than carbon) you can do this by dividing by the smallest mole. empirical formula of the compound is CH4.

Example-

A 2.8g sample of a hydrocarbon is found to contain 2.4g of carbon and 0.4g of hydrogen.

Carbon- 2.4 / 12 =0.2 mol   Hydrogen- 0.4/ 1= 0.4 mol                               

Ratio  0.2 / 0.4              .for every 1 carbon there is 2 hydrogen.

Empirical formula is CH2. (you know this can not be a molecules as c has 4 bonds) 

Molecular Formulae- actual number of atoms of each element in a molecule.

.. Calculate the empirical formula from the percentage composition by mass.

.. Use empirical formula and the relative molecular mass to deduce the molecular formula.

Example

Compound A has a relative molecular mass 62. Iit composition mass is C- 38.7%; H- 9.7%; O 51.5%

percentage divided by R.A.M- C-38.7 / 12=3.23, H-9.7/1= 9.7, O-51.6/16= 3.23

the divided each one by the smallest value, 3.23 / 3.23= 1  9.7 / 3.23= 3  3.23 / 3.23= 1

therefore C:H:O   1:3:1  Empirical formula is CH3O.  empirical mass =31

As it said in the question the relative molecular mass is 62.  62 / 31= 2

you can deduce the molecular formula C2H6O2.

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Percentage yield

% yield= (amount (moles) of product obtained / theoretical amount (moles) of product) x 100

This is used as not all reactions result in the complete conversion of reactants to products.

How.... Calculate the number of moles used. then calculate the number of moles produced. The calculate % yield which is the actual yield / maximum yield.

Example..

! make sure the equation is balanced !

Ethanol, C2H5OH, can be oxidised to form ethanoic acid, CH3COOH. 2.3g of ethanol is oxidised to produce 2.4g of ethanoic acid.

C2H5OH + 2 (O) ---> CH3COOH +H2O

number of moles of ethanol used - 2.3(G) / 46 = 0.050 mol

there is ratio of 1:1 on the moles of ethanol and athanoic acid therefore the amount of moles of ethanoic acid is also 0.050 mol.

mole of ethanoic acid produced= 2.4 / 60 = 0.040mol

percentag yield= (actual / max. ) x100        (0.040 / 0.050) x100 = 80%

This calculation can also be reversed likewise...

In a reaction between methane and chlorine, the % yield of tetrachloromethane is 72%. Calculate the mass of tetrachloromethane that could be obtained by reacting 10g of methane with excess chlorine.

calculate the number of moles of methane present, then calculate the amount in moles of tetrachloromethane produced, finally calculate the mass of tetrachloromethane produced

mole of methane= 10 / 16 =0.625 mol

mole of tetrachloromethane produced  (72 / 100) x 0.625 = 0.45 mol

mass of ttrachloromethane produced mass of 1 mol = 154g mass of 0.45 mol = 0.45x 154 = 69.3g

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Atom economy

Atom economy= (molecular mas of desire product / sum of the molecular masses of all products) x100.

When there is less waste in industrial procedures they have high atom economy. THerefore it is more likey to be used to consere resources and avoid any unwanted by-products.

This measurement can be applied to reactions to take into account of wastage.

How ...

calculate the Mr of the desired product

then Calculate the Mr of all the products.

then divid the desired product by all products and times that value by 100.

C4H9Br  +  NaOH ---> C4H9OH + NaBr

Mr of butanol (C4H9OH) = (4x12) +(9x1) + 16 +1 = 74

Mr of NaBr = 23+ 80 = 103

atom economy = (74 / 74+103) x 100 = 42%

Therefore from this we know that the reaction is not ideal and is inefficient.

Addition reactions of alkenes have 100% atom economy because they are only forming one products,

Example...

H2C=CH2 + H-H ----> H3C-CH3.

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Summary question and past exam questions

June 09

 

 

(c)

Alkenes can be prepared by the dehydration of alcohols with an acid catalyst.

 

Cyclohexene can be prepared by the dehydration of cyclohexanol,

A student reacted 7.65 g of cyclohexanol, C6H12

O, and obtained 0.0268 mol of cyclohexene.

(i)

What is the molecular formula of cyclohexene?

...................................................................................................................................

[1]

(ii)

Calculate the percentage yield of cyclohexene.

answer = .................................................... %

[3]

(d)

Percentage yield has been used for many years to measure the ‘success’ of a reaction.

Recently, chemists have turned their thoughts also to the atom economy of a reaction.

(i)

Explain the term atom economy

.

..........................................................................................................................................

...................................................................................................................................

[1]

(ii)

Cyclohexene can also be prepared by the reaction below (see paper)

Explain why the atom economy of this cyclohexene preparation is higher than that from cyclohexanol in (c)

[2]

----- Jan 10

 

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