Chemistry - Hardenhuish School

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Chemistry - Hardenhuish SchoolChemis

    Hardenhuish School

    ‘A High Performing Specialist Academy’

    AS Level Chemistry

    Induction Task

    Introduction: Welcome to AS Chemistry!

    Your first module is made up of two topics:

    1. Elements for Life; where we will study the following areas

    ; atomic structure

    ; radioactivity: fission and fusion

    ; chemical equations and amount of substance

    ; the Periodic Table and Group 2 chemistry

    ; bonding and the shapes of molecules

    As you can see some of these topics are new but mostly we will be extending your

    knowledge from GCSE chemistry.

    2. Developing Fuels; we will look at:

    ; Thermochemistry

    ; organic chemistry: alkanes, structural isomers

    ; introduction to entropy

    ; dealing with polluting exhaust emissions

    The environmental issues you are probably all aware of from watching the news and reading papers, you will understand the chemistry behind the headlines.

To get you stated with the first topic, your summer task is concerning the history

    behind the atom and how its structure was developed. It’s essential that you complete this task. There will be testing early in the new term on GCSE content such

    as Mole calculations and atomic structure and bonding. To help your understanding of the calculations there are a set of help sheets attached with

    exemplars and questions.

    Year 12 Chemistry teachers will include: Dr Ovens (Curriculum Leader for Chemistry), Mr. Gingell and Mr. Wiggall. Enjoy your holidays and we look forward to seeing you in a few weeks!

    Task: Research on the following scientists and produce a poster or leaflet on…

    …Who were they, what did we think the atom looked like before, what was their

    experiment, what did they expect to see and what did they actually see? What did

    they think the atom looked like after they’ve made their discoveries?

    ; Ernest Rutherford

    ; Joseph John Thompson

    ; Niels Bohr

    ; John Dalton

    ; James Chadwick

Additional: Help sheets: Moles helpsheet

Useful Websites:

    Due: To be handed in to your teacher at the start of your first chemistry lesson in


    Set by: If you have any queries regarding the task set please contact Chemistry teacher

    Mr. Gingell email to:

Expected Time 2-3 hours



    The Mole

    ; The standard unit of amount of a substance

    ; The number of particles in a mole is known as Avagadro’s Constant (L)

    23-1; Avagadro’s Constant has a value of 6.023x10 mol

    -23Example: If one atom has a mass of 1.234x10 g

    -2323 Then on mole of atoms will have a mass of 1.234x10 g x 6.023x10 = 7.432g

    ; Q1. Calculate the mass of one mole of carbon-12 atoms.

    -24-24-28[mass of 1 proton = 1.672x10g, mass of 1 neutron = 1.674x10g, mass of 1 electron = 9.109x10]

    Mole Calculations

Substances mass g or kg

    -1-1 Molar mass g mol or kg mol

Example: Calculate the number of moles of oxygen molecules in 4g

    -1 Oxygen molecules have the formula O, therefore the relative mass will be 16 x 2= 32 g mol2

Moles = mass / molar mass

    -1 = 4g / 32 g mol

     = 0.125 mol

    ; Q2.

    a) Calculate the number of moles in b) Calculate the mass of

    o 10g of Ca atoms o 2 mol of CH 4

    o 10g of CaCO o 0.5 mol of NaNO 33

    o 36g of water molecules o 6 mol of nitrogen atoms

    o 4g of hydrogen atoms o 6 mol of nitrogen molecules

    Solutions n = number of mol

    c = concentration

    v = volume 3(convert dm)

    -3Molarity/Concentration mol dm 3 3 Volume dm or cm

    3-3 Example 1: Calcuate the number of mol of sodium hydroxide in 25cm of 2mol dmNaOH

    3Mol = concentration x volume (in cm)


     -3 = 2mol dmx 25


     = 0.05 mol

    3Example 2: 4.25g of NaCOare dissolved in water and the solution made upto 250cm. What is the 23

    -3concentration of the solution in mol dm?

    -1 Concentration= molar mass: 106g mol

     3-1no. of mol in 250 cm: 4.25g / 106 g mol = 0.04 mol

    3 -3 no. of mol in 1000cm0.04 x 4 = 0.16 mol dm

    ; Q3. Calculate the number of mol in

    3-3o 1 dm of 2mol dmNaOH

    3 -3o 5 dmof 0.1mol dm HCl

    3-3o 250cm of 2mol dmNaOH

    3-3o 25cm of 0.2mol dmHSO 24

Empirical formulae and molecular formulae

Description Expresses the elements in a simple ratio (e.g. CH) 2

     It can sometimes be the same as the molecular formula (e.g. HO) 2

Calculations You are essentially using m = n x M

     You will use; -Actual mass; in grams or percentage for each element in the molecule.

     - Relative atomic masses of each element

Example Calculate the empirical formula of a compound containing 48% carbon,4% hydrogen

    and 48% oxygen

    Set out as a table; atoms as headings. C H O Write mass (g/%) 48 4 48 Divide by A12 1 16 r

    This is the molar ratio 4 4 3 Ensure all numbers are whole (divide all by n/a n/a n/a smallest number, may need to multiply up too)

    Express as a formula CHO443

    ; Q4. Calculate the empirical formula of a compound containing 1.8g carbon, 0.48g oxygen and

    0.3g hydrogen.

Molecular formula

Description Exact number of atoms of each element in the formula

    Calculations compare the empirical formula with the relative molecular mass, M. The M will be an rr exact multiple of the empirical formula mass.

    Example Calculate the molecular formula of a compound of empirical formulaCHand M 84. 2r

     Mass of CH unit = 14 2

     Divide M (84) by 14 = 6 r

     Molecular formula = empirical formula x 6 = CH612

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