Chemistry C2 GCSE AQA - UNIT 3

Follow up to 'Chemistry C2 GCSE AQA - UNIT 1 and 2' (higher)

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Mass Numbers

  • Number of neutrons = mass number - atomic number
  • Protons/Neutals EQUAL MASSES
  • Electrons have VERY SMALL MASSES
  • Mass number = TOTAL of protons+neutrons
  • Atoms of the same element have the same ATOMIC NUMBER
  • Number of protons and electrons must ALWAYS BE THE SAME
  • Atoms of the same element with different numbers of neutrons are called ISOTOPES
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Masses of Atoms and Moles

  • Atoms are too small to weigh, so therefore 'relative atomic masses
  • Relative Atomic Mass (RAM - to remember it), is an average value that depends on the isotopes the element contains - when rounded to a whole number.
  • Relative Formula Mass (RFM) is found by adding up the relative atomic masses of the atomics in its formula -
    • Mr of CaCl(2), solution: Ar of Ca = 40, Ar of Cl= 35.5, so 40+35.5 = 111
  • Relative Formula Mass is called ONE MOLE.
  • Allows us to calculate and weigh out in grams masses of substances with the same number of particles.
    • Mass of NaOH?. solution: Ar of Na=30, Ar of O=16, Ar of H=1, so 23g+16g+g+1g = 40g
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Percentages

  • Percentage of any of the elements in a COMPOUND
  • RELATIVE ATOM MASS / RELATIVE FORMULA MASS x 100 = PERCENTAGE
    • Example- Mr of CO(2) = 12+(16x2)=44 
    • Therefore percentage of carbon = (12/44) x 100 = 27.3%
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Empirical Formula

  • EMPIRICAL FORUMULA is the simplest ratio of the atoms or ions in a compound.
  • We work this out by dividing the MASS of each element in 100g of the compound by its MASS NUMBER to give the ratio of atoms, then convert this to a WHOLE NUMBER RATIO
  • Example- 100g of hydrocarbon contains 80g of C and 20g of H
  • Number of moles of carbon- 80/12 = 6.67
  • Number of moles in hydrogen = 20/1 = 20
  • Ratio of atoms = 6.67C:20H
  • Simplest ratio is 1C:3H
  • So EMPIRICAL FORMULA is CH(3)
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Equations and Calculations

  • Chemical equations show the REACTANTS and PRODUCTS of a reaction.
    • E.g. 2Mg + O(2) > 2MgO
    • Shows TWO magnesium react with ONE molecule of oxygen to form TWO magnesium ions and TWO oxide ions
  • In RELATIVE MASS this becomes
    • (2xAr of Mg)+(2xAr of O) gives (2xMr of MgO) or (2x24+2x16=2x40)
  • In MOLES this tells us TWO moles of Mg react with ONE mole of O(2), to produce TWO moles of MgO
  • This means 48g of Mg react with 32g of O(2) to give 80g of MgO
  • If the known mass of Mg is 5g, we can work out the MASS
    • (5/25) x 40 = 8.33g of MgO
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Making As Much As We Want

Making As Much As We Want

  • PERCENTAGE YIELD = (amount of product collected / maximum amount of product possible) x 100
  • PERCENTAGE ATOM ECONOMY = (relative formula mass of useful product / relative formula mass of all products) x 100
  • To avoid waste both percentage yield and atom economy should be as HIGH as possible
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Reversible Reactions

(http://i49.tinypic.com/2i6pg1.png)

So we then reach an equilibrium

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Making Ammonia - The Haber Process

Used to make FERTILISERS and other CHEMICALS (http://i48.tinypic.com/5bo0ls.png)

The Haber process also uses an IRON CATALYST

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