Edexcel A Level Chemistry Unit 5 Example Questions and Answers

Unit 5: Questions and Answers

• Edexcel
• A Level

Empirical Formula

• Question 1:
• A compound has the percentage composition by mass Ca = 24.4%, N = 17.1% and O = 58.5% What is its empirical formula?
• Question 2:
• Combustion analysis of 2.16g of an organic compound produced 4.33g of carbon dioxide and 1.77g of water. What is its empirical formula?
• It is worth noting that these are worth considerably less marks than at GCSE so make sure you can get through them fairly quickly - certainly no more than 3 minutes or so.

Diapositiva 3

• Answer 1:
• First you need to divide the percentage mass of each element by its atomic mass.
• Ca = 24.4/40.1 = 0.60847880299
• N = 17.1/14 = 1.22142857143
• 0 = 3.65625
• Next you need to divide each value by the smallest, in this case calcium.
• Ca = 1
• N = 2.00734777519 = 2
• O = 2.99342105263 = 3
• So the empirical formula is...
• CaN2O2
• Note that I have not rounded any of my answers. On a scientific calculator these will appear as fractions - leave them that way! Rounding at this level can lead to significant errors and loss of easy marks. Make a note of which decimal is smallest by the side of each element though, that way you can tell easily which is smallest.
• Please remember to write any numbers in a formula as subtext!

Diapositiva 4

• Answer 2:
• First you need to work out the amount of carbon and hydrogen in their respective compounds.
• H = 1.77 x 2/18 = 0.19666666666g
• C = 4.33 x 12/44 = 1.18090909091
• From this you can work out the amount of oxygen by subtracting these masses from the overall mass of the combusted compound.
• O = compound - H - C = 0.78242424243
• Now proceed as usual with the empirical formula. First divide the masses by the elements' atomic numbers.
• H = 0.19666666
• C = 0.0984090909
• O = 0.04890151515
• Then divide by the smallest (oxygen).
• H = 4.02168861347
• C = 2.01239349329
• O = 1
• Thus the empirical formula is...
• H4C2O
• This one is a little harder because you have to work out the values of individual elements from compounds. As long as you remember that it is atomic number over atomic mass then you are good to go.
• Note that I have written hydrogen's atomic mass as 2. This is because there are two hydrogens in a water molecule so their atomic numbers should be added together for this particular calculation.

Molecular Formula

• Question 1:
• A 2.82g sample of a gas has a volume of 1.26dm3. measured at a pressure of 103 kPa and a temperature of 55 degrees celsius. Calculate the molar mass of the gas.
• Question 2:
• A compound has the percentage composition by mass C = 40.0%, H = 6.7%, O = 53.3%. A sample containing 0.146g of the compound had a volume of 69.5cm3 when measured at 98 kPa and 63 degrees celsius. What is the molecular formula of this compound?
• These questions both involve the ideal gas equation which is a new formula at A Level. Again these questions will not be worth many marks - many questions involving this topic are disappointingly 1 mark multiple choice questions. Although they are less marks than empirical formula, they will probably take longer. My advice is to try and spend only about 5 minutes on this type of question, particularly if you feel confident with them, as there are harder questions worth more marks that you can focus on until you have your timings sorted.

Diapositiva 6

• Answer 1:
• First you need to rearrange the equation.
• pV = nRt becomes n = pV/Rt
• Now you can insert the values you have been given to this new equation, thus working out the moles of the gas.
• n = 103,000 x 0.00126/8.31 x 328 = 0.04761380646
• Now you have the moles you can use the molar equation to find the molar mass of the gas.
• Mr = mass/moles = 0.00126/0.04761380646 = 26.462912623
• Try and learn both of these versions of the equation because it will make exams a lot easier!
• Note that I have converted the kPa into Pa and the dm3 to m3 - you will be expected to do so in your exams unless stated otherwise. You are also expected to use 8.31 as the gas constant. Sometimes questions will ask you to use a different value, in which case use that.
• Remember that the m at the top of the molar formula means mass, not grams, so you can substitute volume here in the absence of any other value of mass.

Diapositiva 7

• Answer 2:
• First we need our old friend the empirical formula.
• Compound = CH2O
• Next step is to work out the amount in moles.
• n = pV/Rt = 98000 x 0.00000695/8.31 x 336 = 0.000243933012
• Then the molar mass...
• Mr = 0.146/0.000243933012 = 598.524975
• Finally the molecular formula...
• Formula mass of EF = C+H2+O = 12+2+16 = 30

Calculating using the Avogadro constant

• Question 1:
• What is the amount of substance in each of the following?
• a) 8.0g of sulphur, S
• b) 8.0g of sulphur dioxide, SO2
• c) 8.0g of sulfate ions, SO42-
• Question 2:
• How many particles are there of the specified substance?
• a) atoms in 2.0g of sulphur, S
• b) molecules in 4.0g of sulphur dioxide, S02
• c) ins in 8.0g of sulfate ions, S042-
• Avogadro is another new face this year. Luckily this only really applies to a few, multiple choice calculations thus far. Still needs learning though!

Diapositiva 9

• Question 1:
• For these all we need is the simple molar equation.
• a) 8/32.1 = 0.249221184 = 0.249
• b) 8/64.1 = 0.124804992 = 0.125
• c) 8/64.1 = 0.125
• Question 2:
• For these we need to use Avogadro's constant.
• a) 2/32.1 = 0.0623052959501558
• 6.02x10^23 x 0.0623052959501558 = 3.750778816x10^22
• b) 4/64.1 = 0.062402496099844
• 6.02x10^23 x 0.062402496099844 = 3.756630265x10^22
• c) 8/64.1 = 0.125
• 6.02x10^23 x 0.125 = 7.525x10^22
• These are fairly easy - just make sure you know the constant! Also remember to use a scientific calculator (which you should have anyway) but DO NOT use at online calculator - this will make your calculations extremely difficult as it will round automatically and could cost you marks.

Writing Chemical Equations

• Question 1:
• Sodium thiosulfate (Na2S203) solution reacts with dilute hydrochloric acid to form a precipitate of sulfur, gaseous sulfur dioxide and a solution of sodium chloride. Write an equation, including state symbols, for this reaction.
• Question 2:
• Solutions of ammonium sulfate and sodium hydroxide are warmed together to form sodium sulfate solution, water and ammonia gas. Write the simplest ionic equation for this reaction.
• Initially familiar these questions get harder before they get easier. If you're struggling with these, the only way to get better is to practice.

Diapositiva 11

• Answer 1:
• Na2S2O3 + HCl --> SO2 + NaCl + S + H2O
• Na2S2O3 (aq) + HCl (aq) --> SO2 (g) + NaCl (aq) + S (s) + H2O (l)
• Na2S2O3 (aq) + 2HCl (aq) --> SO2 (g) + 2NaCl (aq) + S (s) + H2O (l)
• Notice how the equation has evolved to reach the full answer. Personally I can't just look at an equation and see the number of moles for each compound/element, so writing it out a few times really helps me. I don't put the state symbols in the first equation because at that point I am just recalling the basic reaction without worrying about what state each compound/element is in. There are many flaws with this method and I would not recommend it - find whatever works for you personally. Other than that this is basically just applying skills from GCSE for more visually complex reactions.
• Another thing to bear in mind is that water is a product of this reaction, although the question has not mentioned it. If you remove the water where has the hydrogen gone from the hydrochloric acid? Hydrogen MUST be included in the product; if there's oxygen involved it is a safe bet that they will bond to form good old H2O.

Diapositiva 12

• Answer 2:
• Na2S2O3 (aq) + 2HCl (aq) --> S (s) + SO2 (g) + 2NaCl (aq) + H2O (l)
• 2Na+ + 2S2- + 3O2- + 2H+ + 2Cl- --> S + S2- + 2O2- + 2Na+ + 2Cl- + H20
• 2S2- (aq) + 3O2- (aq) + 2H+ (aq) --> S (s) + S2- (g) + 2O2- (g) + H2O (l)
• WIth a question like this it is worth writing out the full equation before embarking on the ionic one. This way you can clearly see the different state symbols, which are the most important things in these reactions. With ionic equations you are basically looking for anything that changes state as it goes from reactant to product - these are the only ones included in your final equation.