2.1.3 (Amount of substance)

• Amount of substance:

  The number of particles within a given space/concentration.

  (See "Moles")
• Moles (mol):

  The unit for the amount of substance.
• Avogadro's constant:

  6.02 x 10^23 mol^-1
• Molar mass:

  The mass in grams of one mole of a substance.

  (Units: g mol^-1)
• Molar gas volume:

   The volume occupied by one mole of any gas.

  (At room temp. and pressure, 24dm^3 is the standard.)
• Calculate empirical formula:

  1. Calculate the number of moles of each element presented in the question. (n=m/mr)
  2. Divide all results by the smallest value obtained.
  3. Use the lowest whole-number ratio of the moles as subscripts. (If decimal, multiply by a number to make it into a whole number.)
• Calculate molecular formula:

  Divide the molar mass of the compound by the empirical formula molar mass.

  Multiply all subscripts in the empirical formula by the value found in previous step.
• Anhydrous:

  An anhydrous substance is one that doesn't contain water.
• Hydrated:

  Hydrated means that the solid crystals contain water of crystallisation.
• Water of crystallisation:

  Water of crystallisation is when some compounds can form crystals which have water as part of their structure.
• Calculating the formula of a hydrated salt: (PAG 1)

  1. Weigh the hydrated salt.
  2. Evaporate the water from the hydrated salt.
  3. Weigh the anhydrous salt.
  4. Compare the masses of the original hydrate to the resulting anhydrous solid.
  5. The mass of water evaporated is obtained by subtracting the mass of the anhydrous solid from the mass of the original hydrate.
  6. Use n=m/mr to find the moles of the water evaporated and the moles of the anhydrous solid.
  7. To get the formula of the hydrate, divide the moles of water by the moles of anhydrous solid.
• Calculation to find mass using moles:

  n x mr = m
• Calculation to find gas volume using moles:

  n x 24 = v
• Calculation to find concentration using moles:


  n x v = c
• Ideal gas equation:

  pV=nRT

  (p = pressure, Pa)

  (V = volume, m^3)

  (n = moles, mol)

  (R = gas constant [8.314 jk^-1 mol^-1])

  (T = temperature, K)
• Conversions:

  1kPa = 1000Pa

  1 m^3 = 1000 dm^3

  K = C + 273(.15)
• Standard conditions:

  298 K (25 °C) a pressure of 1 bar (100 kPa).
• Use of stoichiometric relationships in calculations:

  Using stoichiometric relationships in calculations allow us to determine the ratio of the reactant and products.

  It also helps us balance chemical equations.
• Calculation of percentage yield:

  percent yield = (actual yield / theoretical yield) x 100%.
• Calculation of atom economy:

  [mr(desired product)/mr(actual product)] x100.
• Percentage yield:

  A way to measure how much of a product is made compared to the amount of reactants used.
• Percentage yields and atom economy

  the techniques and procedures required during

  experiments requiring the measurement of mass,

  volumes of solutions and gas volumes

  PAG1

  HSW4 Many opportunities to carry out experimental

  and investigative work.

  (FIX THIS CARD WHEN YOU GET HOME!!!)
• Benefits of high atom economy in a chemical reaction:

  1. Reducing the amount of waste produced.
  2. Conserving resources.
  3. Minimizing environmental impact.
  4. Less energy is needed to carry out the reaction-->more cost-effective