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Summary IAL: UNIT 1_ Structure, bonding and introduction to organic chemistry (NEW specification)

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It might be challenging to ace Chemistry:( This is my condensed revision notes that cover the different areas as listed in the specification. Try to get as many marks in your Chemistry Unit 1 paper by understanding the basics and building up your knowledge. I achieved an A* in my Chemistry IAL and an A in my Chemistry As Level. I am sure you can do this too!

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UNIT 1: Structure, bonding and
introduction to organic chemistry
(NEW specification)

, Formulae, equations and amount of substances

1.1 atom, element, ion, molecule, compound, empirical formula and molecular formula
Atom - Smallest part of an elements (v properties of element)

element - An element is substance that contains atom of only one type
- Element are chemically the simplest substances ∴cannot be
broken down by chemical mean into other substance

Molecule - Particle made of ≥ 2 atoms bonded together
Smallest part of covalent compound or elements that can exist on its own
Monoatomic: atom (noble gas)

Compound Two or more elements are chemically bonded together

ion - One or more atoms joined together and have a +ve or -ve charge
- Atom gains or loses one or more electrons → charged particles

Molecular Actual no of atoms of each type of atom
formula (def.)

empirical simplest whole number ratio of atoms of each element in the compound
formula (def).


1.2 mole (mol) is unit for amount of a substance and able to perform calculations using the
23 −1
Avogadro constant L (6.02 x10 𝑚𝑜𝑙 )

Mole (def) Amount of substances in grams that has the same no of particles as
atoms in 12 g of carbon-12

Avogadro constant Number of atoms contained in one mole of any monatomic element
23 −1
6.02 x 10 𝑚𝑜𝑙

Avogadro’s Law Equal volume of any gas at same temperature and pressure contains
same number of moles of gas// some number of gas molecules

Number of particles = Avogadro’s number x no. of moles


1.3 balanced full and ionic equations (incl state symbols) for chemical reactions
12
1.4 (i) “relative atomic mass” based on the 𝐶 scale

Relative atomic weighted average mass of the element’s isotope (atom) relative to
mass one-twelfth of the mass of a carbon-12 atom

, Relative isotopic Mass of an atom of isotope, compared to 1/12th of the mass of an
mass atom of carbon-12
(ii) “relative molecular mass” and “relative formula mass” incl calculating the values from
relative atomic masses

Relative molecular mass Covalent molecules
Sum of relative atomic masses of all its atoms

Relative formula mass Sum of all relative atomic masses in formula

−1
(iii) “molar mass” is mass per mole of a substance [ g𝑚𝑜𝑙 ]

(iv) parts per million (ppm) incl gases in atmosphere

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 𝑚𝑖𝑥𝑡𝑢𝑟𝑒
1 ppm = x 1 000 000
𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑚𝑖𝑥𝑡𝑢𝑟𝑒
1 ppm = 1mg / 1 L(dm-3)


−3 −3
1.5 calculate conc of a solution in mol𝑑𝑚 and g𝑑𝑚

1.6 experimental data to calculate empirical and molecular formulae

Oxide of Copper

1. Place a known mass of oxygen od copper in the tube
2. Heat oxide in a steam of hydrogen gas
3. Gas reacts with oxygen in copper oxide to form steam
4. Colour of solid → orange orange brown
5. Excess hydrogen gas is burned off at the end of the tube
6. After cooling, remove and weigh the solid copper
7. Measure weight of mass before experiment and after experiment and calculate the
difference
CuO + H2 → Cu + H2O


1.7 calculate reacting mass, using concepts of amount of substance and molar mass

1.8 calculate volumes of gases:
(i) concepts of amount of substance

(ii) molar volume of gases
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
Molar volume =
𝑑𝑒𝑛𝑠𝑖𝑡𝑦
Volume occupied by 1 mole of substance

, In gas: particles are far apart → large empty space
Molar volume of gas (at room temp: 25°// 1 atm): 24 dm3 mol-1
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑔𝑎𝑠
Number of mole =
𝑚𝑜𝑙𝑎𝑟 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑔𝑎𝑠




(iii) PV= nRT for gases and volatile liquids
Pressure Nm-2 or Pa [1atm = 101kPa 101x 103]

Volume m3 [1m3 = 106 cm3 = 103 dm3 ]

Temperature K [ °C +273 = K]

no of moles mol

Universal gas constant (R) 8.31 J mol-1 K-1


1.9 calculate percentage yields and percentage atom economies (by mass)

𝑎𝑐𝑡𝑢𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡
percentage yield = x 100%
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡
In experiment (lab) → requires skills

reasons 1. Reaction may be reversible (ie. incomplete reaction) [50% -70%]
2. Some of the reactants // products may be left behind in apparatus
(e.g. filtering or pouring) [90%]
3. Some of the reactants react in ways different to expected reactants
(side reaction → by-products) [<50%]


𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑝𝑟𝑜𝑑𝑢𝑐𝑡
Atom economy = x 100%
𝑠𝑢𝑚 𝑜𝑓 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠𝑒𝑠 𝑜𝑓 𝑎𝑙𝑙 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠
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