SWGS Chemistry
Name:………………………………
Starting AS – Summer Work
In the first unit of the course you will extend your current equation writing and calculation skills in addition
to your knowledge of atomic structure and bonding. This work will give you chance to practise what you
already know.
Answer each question in the space provided. Some examples are given. Use the Periodic Table provided to
help you and the table of formulae of ions.
You will be handing in a hard copy of this work at the start of the course, so make sure you have it with
you.
1. Writing Formulae of ionic compounds: complete the table below using the pdf from
“creative chemistry” to help
sodium
carbonate
potassium
chloride
sodium
hydroxide
aluminium oxide
copper
oxide
magnesium
chloride
ammonium
nitrate
calcium
sulfate
(8)
2. Writing formulae of common covalent compounds. Some of these you can work out from
their name, some you will know, others you need to look up and learn.
nitrogen dioxide
sulphuric acid
ammonia
ethanoic acid
propane
carbon tetrachloride
ethane
dihydrogen monoxide
(8)
3. Writing Chemical Equations: write balanced symbol equations for each reaction:
1. sodium
2Na
+
+
sulfuric acid
H2SO4
 sodium sulfate

Na2SO4
+
+
hydrogen
H2
2. hydrogen
+
 steam
oxygen
3. sodium hydroxide
+
copper II sulphate 
4. calcium carbonate
+
hydrochloric acid 
5. zinc
+
nitric acid

(8)
Formula Mass:
Fill in the gaps in the first two columns and calculate the formula mass of each compound.
name
carbon dioxide
CO2
formula
ammonia
NH3
formula mass
12 + (16*2) = 44
NaCl
calcium sulfate
butane
C4H10
C2H4
(8)
Reacting Masses: use the equation to work out the mass of reactant used or product formed .
Mg
+
2HCl

MgCl2
+
H2
Example: What mass of magnesium chloride is formed when 6g magnesium react?
1 mole of magnesium makes 1 mole of magnesium chloride, so we can use the ratios of
mass in grams:relative formula mass of each substance
so for Mg: 6/24 = x/95 for MgCl2
so x = (6*95)/24
so x = 23.75 g. This answer is sensible because the relative formula mass of MgCl2 is much greater than Mg
Example 2: What mass of oxygen is needed to react with 8.5 g of hydrogen sulphide (H 2S)?
2 H2S +
3 O2

2 SO2 +
2 H2O
Her the reacting mole ratio is 2H2S : 3O2 so we must multiply the relative formula masses (rfm) by these
coefficients:
So for H2S: 8.5/(2*34) = x/(3*32)
so x = (8.5*96)/68
So x = 12g. This answer is sensible because although the rfm of H2S is higher than for O2, 3 moles of O2 are
needed to react with 2 moles of H2S so mre grams of O2 will be required
1)
What mass of potassium oxide is formed when 7.8 g of potassium is burned in oxygen?
4 K + O2
(3)
2O
2)
Railway lines are welded together by the Thermitt reaction, which produces molten iron. What
mass of iron is formed from 1 kg of iron oxide?
Fe2O3
2O3
(3)
3)
What mass of aluminium oxide is produced when 135 g of aluminium is burned in oxygen?
2 Al + 3 O2
(3)
2O3
5)
5.00 g of hydrated sodium sulphate crystals (Na2SO4.nH2O) gave 2.20 g of anhydrous sodium
sulphate on heating to constant mass. Work out the relative molecular mass (M r) of the
hydrated sodium sulphate and the value of n.
Na2SO4.nH2
(3)
2SO4 + n H2O
Percentage Composition: calculate the percentage (by mass) of the named element in the compound
given.
Percent composition =
mass of the element
X 100
formula mass of compound
1. Show that the percentage Nitrogen in ammonium chloride (NH4Cl) is 26.2%
2. Carbon in ethanol (C2H5OH)
3. Oxygen in zinc sulphate (ZnSO4)
(6)
Percentage Yield: work out the percentage yield for these reactions.
% Yield =
Mass of product actually made
x 100
Maximum mass of product that could be made
1. Barium sulfate can be made by reacting barium chloride with ammonium sulfate.
BaCl2 + 2NH4SO4  BaSO4 + 2NH4Cl
If 2.00g of barium chloride are used the mass of barium sulfate produced should be 2.24g if the
yield is 100%.
Only 1.68g of barium sulfate was formed. Show that the percentage yield is 75%?
(2)
2. Copper II carbonate decomposes when heated.
Copper oxide and carbon dioxide are made.
The mass of copper carbonate heated was 5.72g.
The mass of copper II oxide produced was 3.32g.
Write the symbol equation for the reaction.
(1)
Calculate the formula mass for
a. Copper II carbonate
b. copper II oxide
(2)
Calculate the mass of copper II oxide that would have been made if the yield was 100%.
(2)
Calculate the percentage yield.
(1)
Titration Calculations. As you saw on the taster day, a titration can be used to find the
concentration of an unknown solution. There are many steps to these calculations:
1. Convert volumes in decimetres cubed to centimetres cubed
2. Use n = c x v (no. moles = concentration in moles per decimetre cubed x volume) for one solution
3. Calculate the number of moles required of the second solution using the mole ratios from the
equation
4. Use c = v/n to find the unknown concentration
Example: see attached file and try the exercise.
(7)
Follow this step and
the ones on the
next page to
complete the
calculation for the
titration.
Structure and Bonding:
1. Draw a labelled diagram of an atom, stating the name, mass and charge of each of the sub-atomic
particles
(3)
2. Complete the table below about bonding in the 3 different substances
(12)
TOTAL
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