1. Calculate the number of moles of the named chemical in each of the following.

a. 1.32g of ammonium sulfate, (NH4)2SO4

moles = mass/gfm gfm (NH4)2SO4 = 132 g

moles = 1.32/132 = 0.01 moles

b. 200cm3 of 3 mol l-1 sodium hydroxide, NaOH

moles = CxV = 3 x 200/1000 = 0.6 moles

2. 250 cm3 of a solution of magnesium chloride, MgCl2, is known to contain 0.5 moles of

chloride ions. What mass of magnesium does the solution contain?

Ratio of Mg: Cl = 1: 2.

If compound contains 0.5 moles Cl it must contain 0.5/2 = 0.25 moles Mg

Mass Mg = n x gfm = 0.25 x 24.3 = 6.075 g

3. In an experiment to prepare aspirin, 10 g of 2 – hydroxybenzoic acid, HOC6H4COOH,

was treated with excess ethanoyl chloride. The equation for the reaction is

CH3COCl + HOC6H4COOH CH3COOC6H4COOH + HCl

Calculate the mass of aspirin formed in this reaction.

gfm HOC6H4COOH = 138 g moles of HOC6H4COOH = mass/gfm = 10/138 = 0.556 mol

As g there is a one to one ratio moles of aspirin made = 0.556 mol

gfm of aspirin = 180 g mass of aspirin = n x gfm = 0.556 x 180 = 100.1 g

4. Calculate the percentage by mass of nitrogen in urea CO(NH2)2

%N = (mass of N / gfm of compound) x 100

gfm CO(NH2)2 = 60 g %N = (28/60) x 100 = 46.7%

5. Titanium is manufactured by heating titanium(IV) chloride with sodium.

TiCl4 + 4Na Ti + 4NaCl

If this reaction is only 75% efficient, calculate the mass of sodium is required to

produce 100 kg of titanium?

Moles of titanium produced = mass/gfm = 100000/47.9 = 2087.7 mol {for 75%}

Moles of titanium produced for100% efficiency = 100/75 x 2087.7 = 2783.6 mol

Mole ratio of Ti: 4Na = 1:4

Moles of Na required = 2783.6 x 4 = 11134.4 mol

Mass of Na required = n x gfm = 11134.4 x 23 = 256091g = 256.1 kg

6. 1.20g of magnesium was used to make hydrated magnesium sulfate crystals,

MgSO4.7H2O

The mass of crystals produced was 9.84 g. Calculate the percentage yield.

Moles of Mg = mass/gfm = 1.20/24.3 = 0.0494

Mole ratio of Mg:MgSO4.7H2O = 1:1

gfm of MgSO4.7H2O = 246.3 (remember to include the 7H2O)

Theoretical mass of MgSO4.7H2O produced = n x gfm = 0.0494 x 246.3 = 12.17 g

% yield = (actual yield/theoretical yield) x 100 = (9.84/12.17) x 100 = 80.9%

7. Zinc reacts with dilute hydrochloric acid according to the equation.

Zn + 2HCl ZnCl2 + H2

1.625 g of zinc is mixed with 250 cm3 of 2 mol l-1 hydrochloric acid.

Calculate which reactant is in excess.

Zn + 2HCl

1 mol 2 mol

Moles of Zn present = mass/gfm = 1.625/65.4 = 0.0248 mol

Moles of HCl required to react with zinc = 0.0248 x 2 = 0.0496 mol

Moles of HCl present = C x V = 2 x 250/1000 = 0.5 mol HCl is in excess

8. 25.0 cm3 of hydrogen peroxide, H2O2, was titrated with 0.040 mol l-1 acidified

potassium permanganate. The end - point was reached when 16.7 cm3 of the

permanganate solution had been added.

5H2O2 + 2MnO4- + 6H+ 2Mn2+ + 5O2 + 8H2O

Calculate the concentration of the hydrogen peroxide in moles per litre.

Moles of permanganate = C x V = 0.040 x 16.7/100 = 0.000668 mol

Mole ratio of 2MnO4- : 5H2O2 = 2:5

Moles of H2O2 = 0.000668 x 5/2 = 0.00167 mol

Concentration of H2O2 = n/V = 0.00167/0.025 = 0.067 mol l-1

9. Seaweeds are a rich source of iodine in the form of iodide ions. The mass of iodine

in seaweed can be found using the procedure outlined below.

50.00 g of seaweed is dried in an oven and ground into a fine powder. Hydrogen

peroxide solution is then added to oxidise the iodide ions to iodine molecules.

Using starch solution as an indicator, the iodine solution is then titrated with sodium

thiosulfate solution to find the mass of iodine in the sample. The balanced equation

for the reaction is shown.

2Na2S2O3(aq) + I2(aq) 2NaI(aq) + Na2S4O6(aq)

In this analysis of seaweed, 14.9 cm3 of 0.00500 mol l–1 sodium thiosulfate solution was

required to reach the end-point.

Calculate the mass of iodine present in one gram of the seaweed sample.

Moles of thiosulfate = C x V = 0.00500 x 14.9/1000 = 0.0000745 mol

Mole ratio of 2Na2S2O3: I2 = 2:1

Moles of I2 = 0.0000745/2 = 0.00003725 mol

gfm I2 = 254 g

Mass of I2 = n x gfm = 0.00003725 x 254 = 0.0095 (in 50.00 g of seaweed)

Mass of I2 in 1 g of seaweed = 0.0095/50 = 0.00019 g

1. Titration with solutions of potassium bromate (KBrO3) can be used to determine the

concentration of arsenic (III) ions.

The balanced equation is:

3H3AsO3 + BrO3- Br- + 3H3AsO4

What is the concentration of As(III) in a solution if 22.35 cm3 of 0.100 mol l -1 KBrO3

is needed to titrate 50.00 cm3 of the As(III) solution?

Moles of bromate = C x V = 0.100 x 22.35/1000 = 0.002235 mol

Mole ratio = 3H3AsO3 : BrO3- = 3 : 1

Moles of As(III) = = 0.002235 x 3 = 0.006705 mol

Concentration of As(III) = n/V = 0.006705 /0.050 = 0.134 mol l-1

2. Alcohol(ethanol) levels in blood can be determined by a redox titration with potassium

dichromate according to the balanced equation:

C2H5OH(aq) + 2Cr2O72-(aq) + 16H+(aq) 2CO2(g) + 4Cr3+(aq) + 11H2O(l)

a. What is the blood alcohol level in mol l-1 if 8.76 ml of 0.050 mol l-1 K2Cr2O7 is required

for titration of a 10.026 cm3 sample of blood?

Moles of dichromate = C x V = 0.050 x 8.76/1000 = 0.000438 mol

Mole ratio C2H5OH : 2Cr2O72- = 1:2

Moles of C2H5OH = = 0.000438/2 = 0.000219 mol

Concentration of C2H5OH = n/V = 0.000219/0.01026 = 0.021 mol l-1

b. Suggest why an indicator is not required for this titration.

As the dichromate ion is orange and the chromium(III) ion is green, the reaction is

self – indicating.

3. To determine the concentration of chloride ions in seawater it is titrated with

silver(I) nitrate solution.

25 cm3 of raw seawater was diluted to 250 cm3 in a volumetric flask.

A 25 cm3 sample of the diluted seawater was pipetted into a conical flask and a few

drops of potassium chromate(VI) indicator solution was added.

On titration with 0.100 mol l-1 silver nitrate solution, 13.8 cm3 was required to

react with the chloride ions in the diluted sample.

The equation for this reaction is

Ag+(aq) + Cl-(aq) Ag+Cl- (s)

a. Silver(I) nitrate is a primary standard. What is meant by “primary standard”?

A chemical with high purity and gram formula mass which is stable in air. It

must be soluble and stable when dissolved.

b. What type of reaction is this?

Precipitation reaction.

c. Suggest the name of a chemical which could be used in a control experiment.

A soluble chloride . Sodium chloride would be ideal.

d. Calculate the concentration of chloride ions in the undiluted seawater.

Moles of silver(I) nitrate = C x V = 0.10 x 13.8/1000 = 0.00138 mol

Mole ratio Ag+ : Cl- = 1:1

Moles of Cl- = 0.00138 mol (25 cm3 of diluted seawater)

Moles of Cl- in 250 cm3 flask = 0.00138 x 10 = 0.0138 mol

Moles of Cl- in 25 cm3 of undiluted seawater = 0.0138 mol

Concentration of Cl- = n/V = 0.0138/0.025 = 0.552 mol l-1

4. Brass is an alloy consisting mainly of copper and zinc. To determine the

percentage of copper in a sample of brass, 2.63 g of the brass was dissolved in

concentrated nitric acid and the solution diluted to 250 cm3 in a standard flask.

Excess potassium iodide solution was added to 25.0 cm3 of this solution, iodine

being produced according to the equation:

2Cu2+(aq) + 4I–(aq) 2CuI(s) + I2(aq)

The iodine formed was titrated with 0.10 mol l–1 sodium thiosulfate solution,

Na2S2O3(aq), the volume required for complete reaction being 24.8 cm3.

I2(aq) + 2S2O32–(aq) 2I–(aq) + S4O6

2–(aq)

colourless colourless

a. How could the end-point for the titration be made more obvious?

Add starch as an indicator.

b. Explain why the potassium iodide solution could be measured out in a measuring

cylinder instead of a pipette.

The potassium iodide is in excess which means the volume added need not be highly

accurate and so the accuracy provided by a measuring cylinder is sufficient.

b. How many moles of sodium thiosulfate were required in the titration?

Moles thiosulfate = C x V = 0.10 x 24.8/1000 = 0.00248 mol

c. Calculate the percentage by mass of copper in the sample of brass.

Mole ratio 2S2O32– : I2 = 2 : 1

Moles of I2 produced = 0.00248/2 = 0.00124 mol

Mole ratio I2 : 2Cu2+ = 1 :2

Moles of Cu2+ produced = 0.00124 x 2 = 0.00248 mol (25 cm3 sample)

Moles of Cu2+ in 250 cm3 flask = 0.00248 x 10 = 0.0248 mol

Mass of copper = n x gfm = 0.00248 x 63.5 = 1.5748 g

% copper in the brass = (1.5748/2.63) x 100 = 59.9%

5. In an experiment to determine the percentage by mass of ammonium sulfate,

(NH4)2SO4, in a fertiliser, 3.80 g of the fertiliser was dissolved in water and made up

to 250 cm3 in a standard flask..

To 25.0 cm3 portions of this solution, an excess of methanal was added.

4NH4+(aq) + 6HCHO(aq) C6H12N4(aq) + 4H+(aq) + 6H2O(l)

The H+(aq) ions produced were titrated with 0.100 mol l-1 sodium hydroxide solution.

The average volume required to neutralise these H+(aq) ions was 28.0 cm3.

a. Why was the methanol added in excess to the fertiliser solution?

To make sure all the ammonium ions reacted.

b. Calculate the number of moles of hydrogen ions produced by the methanol in the

25 cm3 of fertiliser solution.

Moles of hydroxide ions = C x V = 0.100 x 28/1000 = 0.0028 mol

Mole ratio OH- : H+ = 1 : 1

Moles of H+ = 0.0028 mol

c. Calculate then number of moles of ammonium ions in the 250 cm3 standard flask.

Mole ratio 4H+ : 4NH4+= 1 : 1

Moles of ammonium ions in 25 cm3 sample = 0.0028 mol

Moles of ammonium ions in 250 cm3 standard flask = 0.0028 x 10 = 0.028 mol

d. Calculate the number of moles of ammonium sulfate in the 250 cm3 standard flask.

Mole ratio NH4+ ammonium : (NH4)2SO4 ammonium sulfate = 2 : 1

Moles of ammonium sulfate = 0.028/2 = 0.014 mol

e. Calculate the percentage of ammonium sulfate in the fertiliser.

Mass of ammonium sulfate = n x gfm

gfm ammonium sulfate = 132 g

Mass = 0.014 x 132 = 1.848 g

% ammonium sulfate = (1.848/3.8) x 100 = 48.6%

6. Some bleaches use hypochlorite ions (OCl-) as the bleaching agent. The concentration

of the hypochlorite can be found by adding a diluted sample of the bleach to an excess

of ethanoic acid and potassium iodide. This releases iodine, the concentration of which

can be determined using a standard thiosulfate solution with starch indicator.

OCl-(aq) + 2H+(aq) + 2I-(aq) I2(aq) + Cl-(aq) + H2O(l)

2S2O32-(aq) + I2(aq) 2I-(aq) + S4O6

2-(aq)

25.0 cm3 of bleach was diluted to 250 cm3 in a standard flask. 10.0 cm3 samples of

diluted bleach were pipetted into conical flasks containing an excess of ethanoic acid

and potassium iodide. Each sample was titrated against sodium thiosulfate solution

(concentration 0.075 mol l-1). The results are shown in the table below.

a. Why was the titration carried out three times?

To increase the reliability of the result. It is not correct to say “increase accuracy”.

b. Why was ethanoic acid added to the bleach/iodide mixture?

To provide the hydrogen ions needed for the reaction.

c. Calculate the concentration of hypochlorite ions in the undiluted bleach.

Average titre (from titration 2 and 3) = (13.55 + 13.45)/2 = 13.50 cm3

Moles of thiosulfate = C x V = 0.075 x 13.50/1000 = 0.0010125 mol

Burette reading 1 2 3

Initial/cm3 0.00 14.00 28.00

Final/cm3 13.70 27.55 41.45

Titration

Mole ratio 2S2O32- : I2 = 2 : 1

Moles of iodine produced = 0.0010125/2 = 0.00050625 mol

Mole ratio I2 : OCl- = 1 : 1

Moles of OCl- = 0.00050625 mol (in 10 cm3 of diluted bleach)

Moles of OCl- in 250 cm3 standard flask = = 0.00050625 x 250/10 = 0.01266 mol

1. Before 1947, ‘silver’ coins were made from an alloy of silver ,copper and nickel. To

determine the metal composition, a coin weighing 10.00 g was dissolved in nitric

acid and the resulting solution diluted to 1000 cm3 in a standard flask.

A 100 cm3 portion was treated in the following way.

Hydrochloric acid (0.20 mol l–1) was added to this solution until precipitation of

silver(I) chloride was complete. The precipitate was recovered by filtration.

It was washed and dried and found to weigh 0.60 g.

.

a. (i) Calculate the percentage by mass of silver in the coin.

GFM AgCl = 143.4g

Mass of Ag in precipitate = 107.9/143.4 × 0.6 = 0.451 g {in 100 cm3 of sample}

Mass of Ag in coin = 1000/100 × 0.451 = 4.51 g

Percentage of Ag in coin = 4.51/10 × 100 = 45.1%

(ii) How could you tell when precipitation was complete?

Add some AgNO3 to filtrate. There should be no more precipitate.

b. The filtrate was treated to reduce the copper(II) ions to copper(I) ions. Ammonium

thiocyanate solution was added to precipitate the copper as copper(I) thiocyanate:

Cu+(aq) + CNS–(aq) CuCNS(s)

After filtration, drying and weighing, the precipitate was found to weigh 0.31 g.

Calculate the percentage by mass of copper in the coin.

GFM CuCNS = 121.6 g

Mass of Cu = 63.5/121.6 × 0.31 = 0.1618 {in 100 cm3 of sample}

Mass of Cu in coin = 1000/100 × 0.1618 = 1.62 g

Concentration of OCl- in undiluted bleach (25 cm3) = n/V = 0.01266/0.025 = 0.51 mol l-1

Percentage of Cu in coin = 16.2%

2. Crystals of hydrated sodium carbonate left exposed to the atmosphere gradually

lose some of their water of crystallisation. The formula of the crystals may be given

by Na2CO3. xH2O, where x has a numerical value between 0 and 10.

16.0 g of the crystals was dissolved in water and made up to 250 cm3 of solution in a

standard flask. To determine the value of x in the formula, 25 cm3 of the sodium

carbonate solution was titrated with 1.0 mol l–1 hydrochloric acid. 15.0 cm3 of the acid

was required for neutralisation.

{Hint: Carbonate ions react with hydrochloric acid in a 1:2 ratio}

a. Calculate the mass of sodium carbonate (Na2CO3) in 16.0 g of the crystals.

Moles of HCl = 15/1000 × 1.0 = 0.015

Moles of Na2CO3 {in 25 cm3 of solution} = ½ x 0.015 = 0.0075

Moles of Na2CO3 {in 250 cm3 of solution} = 0.0075 x 10 = 0075

Mass of in 16 g of crystals = n x gfm = 0.075 x 106 = 7.95 g

b. Find the value of x in the formula Na2CO3.xH2O.

Mass of water = 16 - 7.95 = 8.05 g

Moles of water = mass/gfm = 8.05/18 = 0.447

Moles of Na2CO3 = 7.95/106 = 0.075

So Na2CO3 : H2O

Moles 0.075 : 0.447

Divide by smallest 0.075/0.075 : 0.447/0.075

1 : 5.96 {this is rounded to 6}

Formula is therefore Na2CO3. 6H2O

3. A 5.02 g sample of a silver alloy was analysed as follows. The sample was completely

dissolved in excess dilute nitric acid and then treated with an excess of sodium

chloride solution, causing a white precipitate of silver(I) chloride.

The precipitate had a mass of 2.37 g.

Calculate the percentage mass of silver in the alloy

Formula of silver(I) chloride is AgCl GFM = 143.5 g

Moles of silver(I) chloride = mass/gfm = 2.37/143.5 = 0.0165 mol

Mass of silver(1:1 ratio of Ag:AgCl) = n x gfm = 0.0165 x 108 = 1.78 g

Percentage of silver in alloy = (1.78/5.02) x 100 = 35.5%

Alternatively mass of silver = (108/143.5) x 2.37 = 1.78 g then calculate

percentage.

1. An experiment was carried out to determine the % manganese in a sample of stainless

steel. A series of standard permanganate solutions were prepared from a stock

solution of 0.001 mol l-1 solution of permanganate and used to produce the

calibration graph shown below.

a. One of the standard solutions used to prepare the calibration graph had a

concentration of 1 x 10-4 mol l-1. Describe, giving appropriate volumes and apparatus,

how this solution could be prepared from the 0.001 mol l-1 solution of permanganate.

As 1x10-4 (0.0001) is ten times as dilute as 0.001, the original solution must be diluted

with deionised water by a factor of 10. Accuracy is very important here so pipettes,

burettes and standard flasks should be used. – not measuring cylinders.l

Pipette 10 cm3 of 0.001 mol l-1 into a 100 cm3 standard flask

{could also use 25 cm3 into 250 cm3 etc}.

Fill the flask to 1 cm below the graduation line with deionised water.

Use a dropper to add more deionised water up to the graduation mark.

Stopper the flask and invert it to ensure thorough mixing.

b. The results of the experiment are shown below.

Mass of steel used = 0.19 g

Absorbance of permanganate solution = 0.25

Total volume of permanganate solution = 100 cm3

Use the graph and the results to calculate the percentage, by mass, of manganese in

the sample of stainless steel.

From the graph, 0.25 absorbance gives a concentration of 1.4 x 10-4 mol l-1 MnO4-.

Moles of MnO4- = C x V = 1.4 x 10-4 x 0.1 = 1.4 x 10-5

Mass of Mn = n x gfm = 1.4 x 10-5 x 54.9 = 0.0007686 g

Percentage Mn = (0.0007686/0.19) x 100 = 0.405 = 0.41%

2. Margaret analysed a sample of 100 cm3 of contaminated water for its copper content.

Suitable chemicals were added to the water to produce a blue coloured copper

compound, and the intensity of the colour was measured using a colorimeter. A series

of 5 solutions with known copper concentrations were prepared in a similar way and the

colorimeter reading produced by each was recorded as shown in the table below.

. Margaret analysed a sample of 100 cm3 of contaminated water for its copper content.

Suitable chemicals were added to the water to produce a blue coloured copper

compound, and the intensity of the colour was measured using a colorimeter.

A series of 5 solutions with known copper concentrations were prepared in a similar

way and the colorimeter reading produced by each was recorded as shown in the table

below.

a. The solutions appeared blue because they were absorbing part of the visible region of

the electromagnetic spectrum. Which colours of the spectrum were being absorbed?

Blue and green.

b. Suggest the most appropriate coloured filter for use in the colorimeter and explain

your choice.

A yellow filter as this is the colour most absorbed by the blue coloured solution.

c. Draw a calibration graph of the colorimeter results.

Concentration of copper/

milligrams per litre (mg l-1)

Absorbance

0 0.00

50 0.12

100 0.27

150 0.40

200 0.53

Absorbance

0.2

0.3

0.4

0.5

0.6

d. The colorimeter reading for the contaminated water sample was 0.35. Calculate the

number of moles of copper in the water sample.

Red line on graph (which is too thick – but had to be so you can see it) gives a

concentration of 132 mg l-1( your value may not identical but it should be close to this)

Mass of copper in 100 cm3 = 0.1 x 132 = 13.3 mg

Moles of copper = mass/gfm = 0.0132/63.5 = 2.08 x 10-4 mol

3. The diagram shows white light passing through a cyan coloured solution.

a. The steps shown below outline how the concentration of the cyan solution could be

determined by colorimetry. Put the steps in the correct order.

Order is b > d > a > e > c

b. Explain why the absorbance graph shown below is NOT that of the cyan coloured

solution.

Concentration Cu /mg l-1

0

0.1

0 100 200

The graph shows that the solution has maximum

absorbance at 510 nm (green light) and low absorbance

at around 625nm (red light). This means the solution is

transmitting red and so will appear red in colour.

A list of “learning outcomes” for the topic is shown below. When the topic is

complete you should review each learning outcome.

Your teacher will collect your completed notes, mark them,

and then decide if any revision work is necessary.

State that chemical reaction can occur at different rates.

Give examples of both fast and slow reactions.

State that decreasing the particle size of a solid will increase its surface area.

State that increasing the surface area of a solid will increase reaction rate.

State that increasing the concentration of a reactant will increase reaction rate.

State that increasing the temperature of a reaction will increase reaction rate.

Need Help

Understand

Revise

State that a catalyst is a chemical which increases the rate of a reaction but

remains unchanged at the end of the reaction {E.g. If a catalyst weighs 1g at the

start of the reaction, 1g will be present at the end of the reaction}

State that the “collision theory” says that a reaction will take place if particles

collide with sufficient energy.

State that the ACTIVATION ENERGY is the minimum energy required for a

reaction to occur.

Be able to draw a labelled diagram of the apparatus used to generate, collect and

measure a gas.

Be able to relate the rate of a reaction to the slope of the line in a reaction

progress graph.

Be able to calculate the reaction rate from the gradient of a line in a reaction

progress graph and give the correct rate units.

Be able to determine the time a reaction is complete from a reaction progress

graph.

state that a catalyst saves money by lowering the energy needed for a reaction

State that the finishing volume/mass/concentration in a reaction progress graph

is related to the initial quantity of reactants used.

Be able to draw lines on a reaction progress graph which identify changes in the

initial reaction conditions.

State that in all chemical reactions the reaction rate slows down from the start

of the reaction to the end of the reaction.

State that, in general, reactions slow down due to the reactants getting used up.

I have discussed the learning outcomes with my teacher.

My work has been marked by my teacher.

Teacher Comments.

Date. __________________________________

Pupil Signature. __________________________

Teacher Signature. _______________________