Formulas, Yield and Atom Economy

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  • Created by: chunks-42
  • Created on: 13-05-15 11:18

Empirical and Molecular Formulas

You have to know what's what with empirical and molecular formulas, so here goes...

1. The empirical formula gives just the smallest whole number ratio of atoms in a compound.

2. The molecular formula gives the actual numbers of atoms in a molecule.

3. The molecular formula is made up of a whole number of empirical units.

E.g. A molecule has an empirical formula of C4H3O2, and a molecular mass of 166g. Work out its molecular formula.

First find the empirical mass - (4x12) + (3x1) + (2x16) = 48 + 3 + 32 = 83g

But the molecular mass is 166g, so there are 166/83 = 2 empirical units in the molecule.

The molecular formula must be the empirical formula x 2, so the molecular formula = C8H6O4.

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Calculating Empirical Formula

You need to know how to work out empirical formula from the percentages of the different elements.

E.g. A compund is found to have percentage composition 56.5% potassium, 8.7% carbon and 34.8% oxygen by mass. Calculate the empirical formula.

In 100g of compound there are:

56.5/39 = 1.449 moles of K, 8.7/12 = 0.725 moles of C, 34.8/16 = 2.175 moles of O

Divide each number of moles by the smallest number - in this case it's 0.725

K: 1.449/0.725 = 2.0, C: 0.725/0.725 = 1.0, O: 2.175/0.725 = 3.0

The ratio of K:C:O = 2:1:3. So you know the empirical formula's got to be K2CO3

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Percentage Yield

1. The theortical yield is the mass of product that should be formed in a chemical reaction. It assumes no chemicals are 'lost' in the process. You can use the masses and a balanced equation to calculate the theoretical yield for a reaction.

E.g. 1.40g of iron filings is reacted with ammonia and sulphuric acid to make hydrated ammonium iron (II) sulphate. 

Fe + 2NH3 + 2H2SO4 + 6H2O --> (NH4)2Fe(SO4)2.6H2O +H2

Calculate the theoretical yield.

Number of moles of iron (Ar=56) reacted = mass / molar mass = 1.40/56 = 0.025 moles.

From the equation, 'moles of iron:moles of ammonium iron (II) sulphate' is 1:1, so 0.025 moles of product should form.

Molar mass of (NH4)2Fe(SO4)2.6H2O = 392, so theorectical yield - 0.025 x 392 = 9.8g

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Percentage Yield continued

2. For the reaction, the actual mass of product (the actual yield) will always be less than the theoretical yield. There are many reasons for this. For example, sometimes not all the 'starting' chemicals react fully. And some chemicals are always 'lost', e.g. some solution gets left on the filter paper, or is lost during transfers betwen containers.

3. Once you've found the theoretical yield and the actual yield, you can work out the percentage yield. Percentage Yield = Actual Yield/ Theoretical Yield x 100

4. So in the ammonium iron (II) sulphate example previuosly, the theoretical yield was 9.8g. Say you weighed the hydrated iron (II) sulphate crystals and found the actual yield was 5.2g. Then: Percentage yield = (5.2/9.80 x 100 = 53%

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Atom Economy

1. The efficiency of a reaction is often by the percentage yield. This tells you how wasteful the process is - it's based on how much of the product is lost because of things like reactions not completing or losses during collection and purification.

2. But percentage yield doesn't measure how wasteful the reaction itself is. A reaction that has a 100% yield could still be very wasteful if a lot of the atoms from the reactants wind up in by-products rather than the desired product.

3. Atom economy is a measure of the proportion of reactant atoms that become part of the desired product (rather than by-products) in the balanced chemical equation.

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Atom Economy continued

% atom economy = mass of desired product/total mass of reactants x 100

E.g. Bromoethane is reacted with sodium hydroxide to make methanol.

CH3Br + NaOH --> CH3OH + NaBr

Calculate the atom economy for this reaction.

% atom economy = mass of desired product/total mass of reactants x 100 

= (12+(3x1)+16+1)/ (12+(3x1)+80) + (23 +16+1) x 100 = 32/135 x 100 = 23.7%

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