Amount of substance

• Mole


  The amount of substance in grams that has the same number of particles as there are atoms in 12 grams of carbon-12
• Relative atomic mass


  The average mass of one atom compared to one twelfth of the mass of one atom of carbon-12
• Molar mass


  The mass in grams of 1 mole of a substance and is given the unit of g mol-1
• Molar gas volume


  The volume of 1 mole of a gas at a given temperature and pressure. All gases have this same volume. At room pressure (1atm) and room temperature 25oC the molar gas volume is 24 dm3 mol–1
• For pure solids, liquids and gases, the unit of mass is grams and the unit of amount is mol
• Example 1: Calculate the amount, in mol, in 35.0g of CuSO4


  1. amount = mass/Mr

  2. = 35/ (63.5 + 32 +16 x4)

  3. = 0.219 mol
• Avogadro's constant


  There are 6.02 x 1023 atoms in 12 grams of carbon-12. Therefore explained in simpler terms 'One mole of any specified entity contains 6.02 x 1023 of that entity'
• Avogadro's constant can be used for atoms, molecules and ions
• Calculating molar mass for a compound


  1. Add up the mass numbers (from the periodic table) of each element in the compound

  2. e.g. CaCO3 = 40.1 + 12.0 +16.0 x3 = 100.1
• Significant Figures


  Give your answers to the same number of significant figures as the number of significant figures for the data you given in a question. If you are given a mixture of different significant figures, use the smallest
• Example 2: Calculate the amount, in mol, in 75.0mg of CaSO4.2H2O

  1. amount = mass/Mr

  2. = 0.075/ (40 + 32.0 +16.0 x4 + 18.0x2)

  3. = 4.36x10-4 mol
• Molecular formula


  The actual number of atoms of each element in the compound
• Empirical formula


  The simplest ratio of atoms of each element in the compound
• General method for determining empirical formula


  1. Step 1 : Divide each mass (or % mass) by the atomic mass of the element

  2. Step 2 : For each of the answers from step 1 divide by the smallest one of those numbers.

  3. Step 3: sometimes the numbers calculated in step 2 will need to be multiplied up to give whole numbers.
• Example 3: Calculate the empirical formula for a compound that contains 1.82g of K, 5.93g of I and 2.24g of O

  1. Step1: Calculate amount, in mol, by dividing each mass by the atomic mass of the element

  2. K = 1.82 / 39.1 = 0.0465 mol

  3. I = 5.93/126.9 = 0.0467mol

  4. O = 2.24/16 = 0.14 mol

  5. Step 2 For each of the answers from step 1 divide by the smallest one of those numbers.

  6. K = 0.0465/0.0465 = 1

  7. I = 0.0467/0.0465 = 1

  8. O = 0.14 / 0.0465 = 3

  9. Empirical formula = KIO3
• Heating hydrated calcium sulfate in a crucible

  1. Weigh an empty clean dry crucible and lid

  2. Add 2g of hydrated calcium sulfate to the crucible and weigh

  3. Heat strongly with a Bunsen for a couple of minutes

  4. Allow to cool

  5. Weigh the crucible and contents again

  6. Heat crucible again and reweigh until you reach a constant mass ( do this to ensure reaction is complete)
• Small amounts the solid , such as 0.100 g, should not be used in this experiment as errors in weighing are too high
• Large amounts of hydrated calcium sulphate, such as 50g, should not be used in this experiment as the decomposition is like to be incomplete
• The lid improves the accuracy of the experiment as it prevents loss of solid from the crucible but should be loose fitting to allow gas to escape
• The crucible needs to be dry otherwise a wet crucible would give an inaccurate result. It would cause mass loss to be too large as water would be lost when heating
• Example 6: 3.51 g of hydrated zinc sulfate were heated and 1.97 g of anhydrous zinc sulfate were obtained. Calculate the value of the integer x in ZnSO4.xH2O

  1. Calculate the mass of H2O = 3.51 – 1.97 = 1.54g

  2. Calculate moles of ZnSO4 = 1.97/161.5 = 0.0122 mol

  3. Calculate moles of H2O = 1.54/18 = 0.085 mol

  4. Calculate ratio of mole of ZnSO4 to H2O = 0.0122/0.0122 = 1/7

  5. X = 7
• Hydrated salt


  Contains water of crystallisation
• Example 5: Na2SO4 . xH2O has a molar mass of 322.1, Calculate the value of x

  1. Molar mass xH2O = 322.1 – (23x2 + 32.1 + 16x4)

  2. = 180

  3. X = 180/18

  4. =10
• Concentration of solutions


  • Concentration = amount/volume
  • Unit of concentration: mol dm-3 or M
  • Unit of Volume: dm3
• Example 7: Calculate the concentration of solution made by dissolving 5.00 g of Na2CO3 in 250 cm3 water

  1. amount = mass/Mr

  2. = 5 / (23.0 x2 + 12 +16 x3)

  3. = 0.0472 mol

  4. conc = amount/volume

  5. = 0.0472 / 0.25

  6. = 0.189 mol dm-3
• Converting volumes


  • cm3 to dm3: divide by 1000
  • cm3 to m3: divide by 1000 000
  • dm3 to m3: divide by 1000
• Example 8: Calculate the concentration of solution made by dissolving 10 kg of Na2CO3 in 0.50 m3 water

  1. amount = mass/Mr

  2. = 10 000 / (23.0 x2 + 12 +16 x3)

  3. = 94.2 mol

  4. conc = amount/volume

  5. = 94.2 / 500

  6. = 0.19 mol dm-3
• Mass Concentration


  • Concentration measured in terms of mass of solute per volume of solution
  • Unit of mass concentration: g dm-3
  • Unit of Mass: g
  • Unit of Volume: dm3
• Converting concentration from mol dm-3 to g dm-3


  conc in g dm-3 = conc in mol dm-3 x Mr
• Ions dissociating


  When soluble ionic solids dissolve in water they will dissociate into separate ions. This can lead to the concentration of ions differing from the concentration of the solute.
• Concentration of ions


  Differing from the concentration of the solute
• Dissolving sodium chloride (NaCl) in water


  1. NaCl(s) +aq Na+(aq) + Cl- (aq)

  2. 0.1mol

  3. 0.1mol

  4. 0.1mol
• Dissolving magnesium chloride (MgCl2) in water


  1. MgCl2(s) +aq Mg2+(aq) + 2Cl- (aq)

  2. 0.1mol

  3. 0.1mol

  4. 0.2mol
• Making a solution


  1. Weigh sample bottle

  2. Transfer to beaker and reweigh

  3. Record difference in mass

  4. Add 100cm3 distilled water

  5. Stir to dissolve

  6. Heat if needed

  7. Pour into 250cm3 volumetric flask

  8. Rinse beaker and funnel, add washings

  9. Make up to mark with distilled water

  10. Invert flask to ensure uniform solution
• Diluting a solution


  1. Pipette 25cm3 of original solution

  2. Add to 250cm3 volumetric flask

  3. Make up to mark with distilled water

  4. Invert flask to ensure uniform solution
• Diluting a solution


  Will not change the amount of moles of solute present but increase the volume of solution and hence the concentration will lower
• Hazardous substances in low concentrations or amounts will not pose the same risks as the pure substance
• Irritant


  Dilute acid and alkalis - wear goggles
• Corrosive


  Stronger acids and alkalis - wear goggles
• Flammable


  Keep away from naked flames