Section 1- Principles of Chemistry (part 3)

Empirical Formula
1. Empiricle formula is the simplest formula of a compound or molecule that you can get.
2. E.g. 'find the empiricle formula of the iron oxide used'
  1. Mass of empty container= 32g Mass of container+iron oxide= 96g Mass of container+iron= 76.8g
    1. Fe= 76.8 - 32 = 44.8g
    2. O= 96 - 76.8 = 19.2g
    3. 1. find amounts of each in Q
  2. 2. Find their atomic mass from periodic table
    1. Fe= 56g
    2. O= 16g
  3. 3. Amount in Q / Atomic mass
    1. Fe= 44.8 / 56 = 0.8g
    2. O= 19.2 / 16 = 1.2g
  4. 4. Work out ratios
    1. 0.8 / 1.2 = 2/3
      1. 2:3
  5. 5. Apply it to the empiricle formula
    1. Fe2O3
Calculating masses in reactions
1. 1. Write out the balanced equation
  1. 2Mg + O2 --> 2MgO
  2. Q= Calculate the maximum mass of magnesium oxide that could be produced in the reaction when 10g of magnesium is burnt in air.
2. 2. For the bit you want (in Q), put a ? under it and work out all relative formula masses. Then divide the known side by amount in Q (10) and do the same with the answer to the unknown.
  1. 2Mg= 48g O2= 32g 2MgO= 80g
    1. 80 / 4.8= 16.67g
      1. (put ? under this as 2MgO is the unknown in the Q)
    2. 48 / 10= 4.8
      1. (10 from Q)
3. Percentage Yield= actual yield (g) / theoretical yield (g) x100
Moles
1. The mole is simply the name given to a certain number
2. a mole= 602, 300, 000, 000, 000, 000, 000, 000 or 6.023 x 10^23
3. Mr= compound relative atomic mass Ar= element relative atomic mass
4. Carbon has an Ar of 12, so one mole of carbon weighs exactly 12g
5. Nitrogen gas has an Mr of 28 (2x14) so one mole of N2 weighs exactly 28g
6. One mole of an atom or molecules of any substance will have a mass in grams equal to the relative formula mass (Ar or Mr) for that substance.
7. Number of Moles=
  1. Mass in g (of element or compound) ------------------------------------------------------------- Mr (of element or compound)
Water of Crystalisation
1. All solid salts consist of a lattice of positive and negative ions
  1. In some salts, water molecules are incoorperated in the lattice too
    1. Salts can be anhydrous (doesn't contain water) or hydrated (contains water)
2. A student removed the water of crystalisation from MgSO4xH2O to produce anhydrous salt MgSO4.
  1. Mass of empty crucible= 42.000g Mass of crucible+MgSO4xH2O= 45.210g Mass of crucible+MgSO4= 43.567g
    1. Calculate mass of hydrated salt MgSO4xH2O: 45.210 - 42= 3.21g
    2. Calculate the mass of the anhydrous salt MgSO4: 43.567 - 42= 1.567 3.21 - 1.567= 1.643g
    3. Work out the value of x in the formula MgSO4xH2O
      1. (Q weights) MgSO4= 1.567g
        (Periodic Table) 24+32+16+16+16= 120g
        (ratio) 1.567 / 120 = 0.013
        1:7 ---> x=7
      2. (Q weights) H2O= 1.643g
        (periodic Table) 1+1+16= 18g
        (ratio) 1.643 / 18 = 0.091
        1:7 ---> x=7
Moles and concentration
1. Concentration is the amount of stuff per unit of volume
2. Concentration= no. of moles / volume
3. E.g. What's the concentration of a solution with 2 moles of potassium iodide in 500cm3?
  1. 2 moles of potassium iodide and 500cm3= 0.5 sm3
    1. Concentration= 2 / 0.5 = 4 mol/dm3
Electrolysis
1. Splitting using electricity

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Section 1- Principles of Chemistry (part 3)

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