Section 1- Principles of Chemistry (part 3)

Section 1- Principles of Chemistry (part 3)

1. Empirical Formula

   • Empiricle formula is the simplest formula of a compound or molecule that you can get.
   • E.g. 'find the empiricle formula of the iron oxide used'
     • Mass of empty container= 32g Mass of container+iron oxide= 96g Mass of container+iron= 76.8g
       • Fe= 76.8 - 32 = 44.8g
       • O= 96 - 76.8 = 19.2g
       • 1. find amounts of each in Q
     • 2. Find their atomic mass from periodic table
       • Fe= 56g
       • O= 16g
     • 3. Amount in Q / Atomic mass
       • Fe= 44.8 / 56 = 0.8g
       • O= 19.2 / 16 = 1.2g
     • 4. Work out ratios
       • 0.8 / 1.2 = 2/3
         • 2:3
     • 5. Apply it to the empiricle formula
       • Fe2O3

2. Calculating masses in reactions

   • 1. Write out the balanced equation
     • 2Mg + O2 --> 2MgO
     • Q= Calculate the maximum mass of magnesium oxide that could be produced in the reaction when 10g of magnesium is burnt in air.
   • 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.
     • 2Mg= 48g O2= 32g 2MgO= 80g
       • 80 / 4.8= 16.67g
         • (put ? under this as 2MgO is the unknown in the Q)
       • 48 / 10= 4.8
         • (10 from Q)
   • Percentage Yield= actual yield (g) / theoretical yield (g) x100

3. Moles

   • The mole is simply the name given to a certain number
   • a mole= 602, 300, 000, 000, 000, 000, 000, 000 or 6.023 x 10^23
   • Mr= compound relative atomic mass Ar= element relative atomic mass
   • Carbon has an Ar of 12, so one mole of carbon weighs exactly 12g
   • Nitrogen gas has an Mr of 28 (2x14) so one mole of N2 weighs exactly 28g
   • 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.
   • Number of Moles=
     • Mass in g (of element or compound) ------------------------------------------------------------- Mr (of element or compound)

4. Water of Crystalisation

   • All solid salts consist of a lattice of positive and negative ions
     • In some salts, water molecules are incoorperated in the lattice too
       • Salts can be anhydrous (doesn't contain water) or hydrated (contains water)
   • A student removed the water of crystalisation from MgSO4xH2O to produce anhydrous salt MgSO4.
     • Mass of empty crucible= 42.000g Mass of crucible+MgSO4xH2O= 45.210g Mass of crucible+MgSO4= 43.567g
       • Calculate mass of hydrated salt MgSO4xH2O: 45.210 - 42= 3.21g
       • Calculate the mass of the anhydrous salt MgSO4: 43.567 - 42= 1.567 3.21 - 1.567= 1.643g
       • Work out the value of x in the formula MgSO4xH2O
         • (Q weights) MgSO4= 1.567g
           • (Periodic Table) 24+32+16+16+16= 120g
             • (ratio) 1.567 / 120 = 0.013
               • 1:7 ---> x=7
         • (Q weights) H2O= 1.643g
           • (periodic Table) 1+1+16= 18g
             • (ratio) 1.643 / 18 = 0.091
               • 1:7 ---> x=7

5. Moles and concentration

   • Concentration is the amount of stuff per unit of volume
   • Concentration= no. of moles / volume
   • E.g. What's the concentration of a solution with 2 moles of potassium iodide in 500cm3?
     • 2 moles of potassium iodide and 500cm3= 0.5 sm3
       • Concentration= 2 / 0.5 = 4 mol/dm3

6. Electrolysis

   • Splitting using electricity

Imagens e arquivos do mapa