Edexcel International A Level Chemistry

Revision Notes

1.2.1 Concentration Calculations

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Concentration Calculations

Amount of substance calculations

  • As previously discussed, the amount of substance can be defined as the number of particles in a substance, n, measured in moles (often abbreviated to mol)
  • In reality, amount of substance is used as a blanket term to cover most chemical calculations, especially those that involve moles
  • The two main calculations for amount of substance are:

Moles, n fraction numerator m a s s comma space m over denominator m o l a r space m a s s comma space M end fraction

Moles, n = concentration x volume

  • Other common calculations for amount of substance include:

Moles, n fraction numerator n u m b e r space o f space p a r t i c l e s over denominator A v o g a d r o apostrophe s space c o n s t a n t end fraction

Moles, n =fraction numerator P V over denominator R T end fractionfrom the ideal gas equation (PV = nRT)

Concentration calculations

  • Concentration can be defined as the amount of a substance dissolved in a quantity of liquid
  • When a mass concentration is calculated, the units are usually g dm-3 
    • Other units are possible such as g cm-3, kg m-3 and sometimes in medicines you might see them as mg / ml
    • The equation to calculate mass concentration is:

mass concentration in g dm3fraction numerator mass space of space solute space in space straight g over denominator volume space of space solution space in space dm cubed end fraction

  • When a molar concentration is calculated, the units are mol dm-3 
    • Calculating molar concentrations requires the use of two equations:

number of moles (or amount) = fraction numerator mass over denominator molar space mass end fraction

    • Molar mass is the mass per mole of a substance in g mol-1

molar concentration in mol dm-3fraction numerator number space of space moles space left parenthesis or space amount right parenthesis over denominator volume space of space solution space in space dm cubed end fraction

Worked example

Mass concentration calculations

  1. What is the mass concentration when 6.34 g of sodium chloride is dissolved into a 0.250 dm3 solution?
  2. A sodium carbonate solution has a mass concentration of 5.2 g dm-3. What is the volume of the solution made when 250 g of sodium carbonate is used?
  3. The mass concentration of a solution is 26.7 g dm-3. What is the mass of sodium bromide in 500 cm3 of solution?

   Answer 1

    • Mass concentration equals fraction numerator mass space left parenthesis straight g right parenthesis over denominator volume space left parenthesis dm cubed right parenthesis end fraction equals fraction numerator 6.34 over denominator 0.250 end fraction equals25.4 g dm-3 (to 3 s.f.)

   Answer 2

    • Volume of solution equals fraction numerator mass space left parenthesis straight g right parenthesis over denominator mass space concentration space left parenthesis straight g space dm cubed right parenthesis end fraction equals fraction numerator 250 over denominator 5.2 end fraction equals48 dm3
    • This answer should be given to 2 s.f. as there is a value in the question with only 2 significant figures

   Answer 3

    • Mass = mass concentration (g dm-3) x volume of solution (dm3) = 26.7 x 0.5 = 13.4 g (to 3 s.f.)
       
      • The 500 cm3 in the question has to be converted into dm3 
      • 500 over 1000 = 0.5 dm3 

Worked example

Molar concentration calculations

  1. What is the molar concentration when 6.34 g of sodium chloride is dissolved into a 250 cm3 solution?
  2. A sodium carbonate solution has a molar concentration of 1.25 mol dm-3. What is the volume of the solution made when 250 g of sodium carbonate is used?
  3. The molar concentration of a sodium bromide solution is 0.250 mol dm-3. What is the mass of sodium bromide in 500 cm3 of this solution?

   Answer 1

    • Number of moles of NaCl equals fraction numerator mass over denominator molar space mass end fraction equals fraction numerator 6.34 over denominator left parenthesis 23.0 plus 35.5 right parenthesis end fraction equals0.1084 moles
    • Molar concentration in mol dm-3 equals fraction numerator number space of space moles space left parenthesis or space amount right parenthesis over denominator volume space of space solution space in space dm cubed end fraction equals fraction numerator 0.1084 over denominator 0.250 end fraction equals0.434 mol dm-3 

   Answer 2

    • Number of moles of Na2CO3 equals fraction numerator mass over denominator molar space mass end fraction equals fraction numerator 250 over denominator left parenthesis 23.0 cross times 2 right parenthesis plus 12.0 plus left parenthesis 16.0 cross times 3 right parenthesis end fraction equals2.358 moles
    • Volume of solution in dm3

equals fraction numerator number space of space moles space left parenthesis or space amount right parenthesis over denominator Molar space concentration space in space mol space dm to the power of negative 3 end exponent end fraction equals fraction numerator 2.358 over denominator 1.25 end fraction equals1.89 dm3 

   Answer 3

    • Number of moles of NaBr = molar concentration x volume of solution

= 0.250 x 0.500 = 0.125 moles 

    • Mass of NaBr = number of moles of NaBr x molar mass

= 0.125 x (23.0 + 79.9) = 12.9 g

Parts per million

  • When expressing extremely low concentrations a unit that can be used is parts per million or ppm
  • This is useful when giving the concentration of a pollutant in water or the air when the absolute amount is tiny compared the the volume of water or air
  • 1 ppm is defined as
    • A mass of 1 mg dissolved in 1 dm3 of water

  • Since 1 dm3 weighs 1 kg we can also say it is
    • A mass of 1 mg dissolved in 1 kg of water, or 10-3 g in 103 g which is the same as saying the concentration is 1 in 10or 1 in a million

Worked example

The concentration of chlorine in a swimming pool should between between 1 and 3 ppm. Calculate the maximum mass, in kg, of chlorine that should be present in an olympic swimming pool of size 2.5 million litres.

Answer:

Step 1: calculate the total mass in mg assuming 3ppm(1 litre is the same as 1 dm3)

    • 3 x 2.5 x 106 = 7.5 x 106 mg

Step 2: convert the mass into kilograms (1 mg = 10-6 kg)

    • 7.5 x 106  x 10-6  kg = 7.5 kg

Atmospheric gas concentration

  • The concentration of atmospheric gases, particularly pollutants, can be measured in parts per million, ppm
  • Instead of using mass, the comparison of gas is done by volume
    • You might see the values quoted in ppmv - the v shows that the value relates to concentration by volume
  • A concentration of 1 ppmv means that there is 1 cm3 of a particular gas in 1000000cm3 or 1000 dm3 
  • The equation to calculate the concentration of a gas in ppm is:

concentration space in space ppm equals fraction numerator volume space of space gas cross times 1000000 over denominator volume space of space air end fraction

    • The volumes can be given in any units but they must be the same units, otherwise one of them will need to be converted 

Worked example

Atmospheric gas concentration calculations

Calculate the concentration, in ppm, of the following:

  1. A volume of 2.5 dm3 of carbon dioxide in 10000 dm3 of air
  2. A volume of 2.5 dm3 of sulfur dioxide in 4000 dm3 of air
  3. A volume of 152 cm3 of ozone in 112 dm3 of air

   Answer 1

    • Concentration in ppm

equals fraction numerator volume space of space gas cross times 1000000 over denominator volume space of space air end fraction equals fraction numerator 2.5 cross times 1000000 over denominator 10000 end fraction equals 250 ppm 

   Answer 2

    • Concentration in ppm

equals fraction numerator volume space of space gas cross times 1000000 over denominator volume space of space air end fraction equals fraction numerator 2.5 cross times 1000000 over denominator 4000 end fraction equals 625 ppm 

   Answer 3

A volume of 152 cm3 of ozone in 112 dm3 of air

    • Concentration in ppm equals fraction numerator volume space of space gas cross times 1000000 over denominator volume space of space air end fraction equals fraction numerator 0.152 cross times 1000000 over denominator 112.0 end fraction equals1360 ppm 

Exam Tip

When completing atmospheric gas calculations, the gas involved does not affect the calculation as shown by worked examples 1 and 2

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