Chemistry C2 GCSE AQA - UNIT 3

• 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

• 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

• 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%

• 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)

• 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

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

So we then reach an equilibrium

Used to make FERTILISERS and other CHEMICALS

The Haber process also uses an IRON CATALYST