Chemistry - EDEXCEL IGCSE - Separation techniques and calculations involving moles

SEPARATION TECHNIQUES

Solids and Liquid:

1. If solid has not dissolved then use filtration. The liquid (filtrate) passes through leaving the solid (residue). This works because of different sizes of particles.
2. If the solid in a solution is required use evaporation.
3. If both are needed then simple distillation is required. The solution is heated and the liquid evaporates and passes into the condenser where it cools and condenses. The solid remains in the flask as residue, the liquid (distillate) can be collected.

Liquid and Liquid:

1. For immiscible liquids (liquids that don’t mix) use a separating funnel. This works because of differences in densities.
2. For miscible liquids then use fractional distillation. This is where a temperature gradient is created in a fractionating column meaning the liquid with the lower boiling point condenses at the top and vice versa. This occurs as the liquid with the lower boiling point boils first.

Solid and Solid:

1. Chromatography:

• This takes advantage of different solubility. 
• Draw a pencil line on paper.
• Place a concentrated dot of the mixture on the paper.
• Place the paper in a beaker of solvent which dissolves the mixture.
• Allow the solvent to move up the paper and dry the paper.
• The solids will have been separated.

Retention factor = distance travelled by the solid / distance travelled by the solvent.

RELATIVE FORMULA MASS AND MOLAR VOLUMES OF GASES

Relative formula mass (Mr) - relative atomic mass of a formula. e.g. the Mr of H2O is 18

A mole is one of chemistry’s counting unit. One mole is equal to 6.022 * 10

For molar volume of gas, one mole always equates to 24dm3 or 24,000 cm3 at room temperature and pressure.

Example question: 6g of a substance is 4.8dm3, what is its molar mass?

1 mole = 24dm3

0.2 mole = 4.8dm3

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0.2 moles = 6g

1 mole = 30g

Molar mass = 30g/mol

Chemical formulae and equations:

How to test if a formulae is correct:

1. Weigh substance (A)
2. Remove one element through a reaction
3. Weigh again (B)
4. Start weight (A-B) = weight of element removed (C) 
5. Repeat until all the elements have been separated
6. Divide weight C by the atomic mass of the element removed
7. The ratio tells you the formula

Calculating empirical formulae:

1. Find the mass of all the different elements
2. Divide the masses by the molar mass of the element
3. Divide through by the number of moles which is the lowest
4. Multiply to get the smallest whole numbers 

Calculating molecular formulae:

1. Work out the empirical formula
2. Do molecular mass / formula mass
3. Times the result by the empirical formula

Calculations with reacting masses:

What mass of oxygen is needed to burn 3kg of propane, C₃H₈?

C₃H₈ + 5O₂ —>  3CO₂ + 4H₂O

The relative formula mass of propane = 3*12+8*1 = 44

So the molar mass = 44g/mol

3000 / 44 = 68.2 moles

5O₂ = 68.2 * 5 = 341 moles of O₂

O₂ = 32g/mol

341*32 = 10.9kg

Therefore answer = 10.9kg

Calculating percentage yield:

Percentage yield = total amount obtained * 100 / maximum theoretical amount 

2.8g of Fe reacts with S to form 4.1g of FeS, what is its percentage yield?

Atomic mass of Fe = 56

2.8 / 56 = 0.05

0.05 moles of FeS = 0.05*88 = 4.4g

4.1*100/4.4 = 93.2(3s.f.)

Therefore answer = 93.2% 

Molar Concentration (molarity) = Number of moles / Volume