# 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

So we then reach an equilibrium

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## Making Ammonia - The Haber Process

Used to make FERTILISERS and other CHEMICALS

The Haber process also uses an IRON CATALYST

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