AS 90171 revision

Level 1 Chemistry – AS 90171 Describe chemical reactions

Pages 4 - 7 give a brief summary of the content of this topic.

Become familiar with the Activity Series, Solubility Rules and Table of Ions

Activity Series

Na Ca Mg Al Zn Fe Pb (H) Cu

Solubility Rules

nitrates, NO3– all soluble

chlorides, Cl– all soluble except AgCl and PbCl2

sulfates, SO42– all soluble except BaSO4,

PbSO4, CaSO4

hydroxides, OH– all insoluble except NaOH and KOH

carbonates, CO32– all insoluble except (NH4)2CO3,

Na2CO3 and K2CO3

Table of Ions

+1 +2 +3 –3 –2 –1

NH4+

Na+

K+

Ag+

H+

Ca2+

Mg2+

Cu2+

Pb2+

Fe2+

Ba2+

Zn2+

Al3+

Fe3+

O2–

S2–

CO32–

SO42–

OH–

Cl–

NO3–

HCO3–

CH3COO–

You will need to refer to the Periodic Table.

Identify reactions as: 1. Precipitation - two solutions are mixed and a solid (precipitate) forms2. Oxidation-reduction - loss / gain of oxygen (or hydrogen) or loss / gain (transfer) of electrons3. Thermal Decomposition - one substance on heating decomposes producing two or more substances4. Neutralisation (acid-base)

either acid + base → salt + water or acid + metal carbonate (or metal hydrogen carbonate) → salt + water + carbon dioxide

Identify the type of reaction for each of the following. Justify your answer.(a) CaCl2(aq) + K2CO3(aq) → 2KCl(aq) + CaCO3(s) (h) Fe2+(aq) + Zn(s) → Fe(s) + Zn2+(aq)(b) Pb(OH)2(s) → PbO(s) + H2O(l) (i) MgCO3(s) → MgO(s) + CO2(g)(c) AgNO3(aq) + NaCl(aq) → NaNO3(aq) + AgCl(s) (j) PbO(s) + C(s) → Pb(s) + CO(g)(d) Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s) (k) CuCl2(aq) + Zn(s) → Cu(s) + ZnCl2(aq)(e) CO(g) + PbO(s) → Pb(s) + CO2(g) (l) C(s) + O2(g) → CO2(g)(f) 2AgNO3(aq) + BaCl2(aq) → 2AgCl(s) + Ba(NO3)2(aq) (m) Ca(OH)2(s) → CaO(s) + H2O(g)(g) CuCl2(aq) + K2CO3(aq) → CuCO3(s) + 2KCl(aq) (n) Cl2(aq) + 2I–(aq) → 2Cl–(aq) + I2(aq)

Precipitation:Aqueous solutions of ionic substances contain ions. Precipitates are produced when two solutions are mixed if one pair of ions forms an insoluble substance. States must be included in equations for precipitation reactions. Spectator ions should not be included the equation. NZQA does not accept the abbreviation ppt Use the Solubility Table to determine whether or not a precipitate forms, name the precipitate (if any)

and write a balanced equation to show the formation of any precipitate.(a) silver nitrate + calcium chloride (i) potassium sulfate + iron(II) nitrate(b) calcium nitrate + sodium sulfate (j) sodium chloride and copper nitrate(c) lead nitrate and potassium chloride (k) potassium hydroxide and magnesium sulfate(d) calcium nitrate and sodium sulfate (l) copper chloride and potassium hydroxide(e) magnesium sulfate and sodium chloride (m) magnesium sulfate and calcium nitrate(f) potassium carbonate and zinc sulfate (n) potassium carbonate + sodium hydroxide(g) lead nitrate + magnesium sulfate (o) sodium hydroxide + copper chloride(h) barium chloride + zinc sulfate (p) lead nitrate + sodium carbonate Write Word Equations

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AS 90171 revision

The elements contained in the reactants (on the left) must be contained in the products (on the right) Copy and complete the following word equations and state any observations which would be made.(a) heat

magnesium + oxygen → (g) heat

calcium carbonate → (b) lead nitrate + potassium chloride → (h) barium chloride + magnesium sulfate → (c) heat

sodium hydrogen carbonate → (i) heat

lead hydroxide →(d) aluminium + silver nitrate → (j) magnesium chloride + sodium hydroxide → (e) heat

copper hydroxide →(k) sodium hydroxide + …

→ sodium sulfate + magnesium hydroxide(f) magnesium + oxygen → (l) iron + copper sulfate →

Write and Balance EquationsThe elements contained in the reactants (on the left) must be contained in the products (on the right) Copy and complete the following equations.

(a) I–(aq) + Cl2(aq) → (f) Zn(s) + AgNO3(aq) →(b) CaCl2(aq) + … → CaSO4(s) + NaCl(aq) (g) Cu(NO3)2(aq) + NaOH(aq) → (c) Cl2(g) + KI(aq) → (h) Zn(s) + H+(aq) →(d) heat

NaHCO3(s) →(i) heat

Cu(OH)2(s) →(e) Ag+(aq) + Cu(s) → (j) CuO(s) + H2(g) →

Recall Observations and Link Observations to the Species InvolveBook learning. For the following reactions:

1. Write down all observations and link these observations with the species involved.2. Write a word equation for the reaction.3. Write a balanced equation for the reaction using formulae.4. State the type of reaction. Justify your answer.

(a) A small amount of zinc hydroxide is heated in a test tube over a Bunsen burner.(b) A solution of barium nitrate is added to a solution of iron(II) sulfate in a beaker. (c) Bromine water is added to aqueous sodium iodide.(d) A strip of magnesium ribbon is heated over a Bunsen burner. (e) Heat:

(i) sodium carbonate (ii) calcium carbonate (iii) copper carbonate (iv) zinc hydroxide(f) Some powdered magnesium is added to a solution of iron(II) sulfate in a beaker, and mixed well. (g) Aqueous chlorine is added to iron(II) chloride solution.(h) Iron(II) nitrate solution is added to sodium hydroxide solution in a test tube. (i) A piece of zinc foil is added to copper(II) nitrate solution and left. It is checked after 10 minutes and

then again after 24 hours. (j) A sample of copper carbonate is heated strongly in a boiling tube fitted with a delivery tube. The gas

formed is bubbled into limewater.(k) A solution of sodium hydroxide is added to a solution of iron(III) chloride.(l) A piece of copper is placed in a solution of silver nitrate.(m) Chlorine water is added to potassium iodide solution.(n) Sodium carbonate is added to a solution of copper sulfate and the resulting mixture is heated gently.(o) A strip of magnesium ribbon is added to hydrochloric acid in a test tube.(p) Sulfuric acid is added to barium nitrate solution.

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AS 90171 revision

Calculate Relative Molar MassesAdd up the individual molar masses to find the total.Note: relative molar mass has no units while molar mass has the unit: gram per mole (g mol-1)

(a) ZnO (d) CaCO3 (g) KNO3 (j) MgO (m) Fe(CH3COO)2

(b) CuSO4 (e) Al2O3 (h) Na2CO3 (k) Cu(NO3)2 (n) Ag2CO3

(c) Pb(NO3)2 (f) (NH4)2SO4 (i) Ba(NO3)2 (l) Na2SO4 (o) Fe2(CO3)3

Chemical Calculations Mmerm.C Show all working clearly.1. Determine the molar mass of substances involved in the question.2. Calculate amount (in mole) of the ‘known’ substance.3. Write the equation for the reaction. 4. Write down the amount ratio, i.e. n(known) : n(unknown)*5. Determine amount (in mole) of the ‘unknown’ substance.6. Complete the question, e.g. determine mass (in gram) of ‘unknown’ substance.

* It is easier to find the amount of the ‘unknown’ if n(known) : n(unknown) is 1 : xFor example, if n(known) : n(unknown) is 2 : 1 rewrite as 1 : ½ or if it is 2 : 3 rewrite as 1 : 3/2 (a) Calculate the mass of zinc hydroxide that must be heated to produce 1.00 gram of water. Use the

equation: Zn(OH)2(s) → ZnO(s) + H2O(g)(b) Calculate the mass of sodium hydrogen carbonate, NaHCO3, required to form 5.40 grams of carbon

dioxide, CO2, when heated. Use the equation: 2NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g)(c) Ammonia, NH3(g), can be prepared by heating ammonium chloride, NH4Cl, with calcium hydroxide,

Ca(OH)2 according to the equation: 2NH4Cl + Ca(OH)2 → CaCl2 + 2NH3 + 2H2OCalculate the mass of calcium hydroxide required to react with 2.14 grams of ammonium chloride.

(d) In a reaction, 0.500 g of sodium hydrogen carbonate was decomposed in the following reaction, which went to completion: 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2OCalculate the mass of sodium carbonate expected to form in this reaction.

(e) Copper(II) oxide is reduced by excess hydrogen: CuO(s) + H2(g) → Cu(s) + H2O(l)Calculate the mass of copper(II) oxide that will be needed to give 18.0 g of copper.

(f) Baking powder contains sodium hydrogen carbonate, NaHCO3. When it is heated it decomposes:2NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g)

Calculate the mass of CO2 formed when 12.6 g of NaHCO3 is completely decomposed.

Determine Molecular FormulaShow all working clearly Units are essential.1. Draw up a table.2. Assume 100 g of compound and hence give mass (= percentage) of each element in compound.3. Determine amount (in mole) of each element.4. Divide amount of each element by the smallest amount to give amount ratio.5. Multiple (if necessary) by the appropriate factor so that the amount ratios are whole numbers.6. State the empirical formula.7. Determine the molar mass of the empirical formula and calculate:

molar mass of molecular formula molar mass of empirical formula8. Multiple empirical formula by the number found in 7. and this gives molecular formula.

(a) A compound was analysed and found to contain: 20.2% phosphorus, 10.4% oxygen, 69.4% chlorine. It has a relative molar mass of 153.5 g mol-1.Determine the molecular formula of this substance.

(b) Determine the formula of the compound made when 3.55 g of chlorine combines with 5.60 g of oxygen. The molar mass of the compound is 183.0 g mol-1.

(c) Determine the formula of the compound made when 8.65 g of iron combines with 3.72 g of oxygen. The molar mass of the iron oxide is 159.8 g mol-1.

(d) Calculate the empirical formula of a compound that is 43.7% phosphorus and 56.3% oxygen and determine the molecular formula for the compound given that its relative molar mass is 284.

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M(g mol-1)

mass(g)

n(mol)

AS 90171 revision

Revision Notes Oxidation – Reduction (Redox)Oxidation is: Gain of oxygen Loss of hydrogen Loss of electrons (LEO)

Reduction is: Loss of oxygen Gain of hydrogen Gain of electrons (GER)

Oxidation and Reducing Agents An oxidant (oxidising agent) brings about

oxidation – and is itself reduced. A reductant (reducing agent) brings about

reduction – and is itself oxidised.

Oxidising AgentsStrongest Cl2 + 2e– → 2Cl–

Br2 + 2e– → 2Br– halogensI2 + 2e– → 2I–

Ag+ + e– → Ag ions of Cu2+ + 2e– → Cu unreactive metalsFe3+ + e– → Fe2+ others

Weaker 2H+ + 2e– → H2

Reducing AgentsStrongest K → K+ + e– group 1 metals

Na → Na+ + e–

Ca → Ca2+ + 2e– group 2 metalsMg → Mg2+ + 2e–

Al → Al3+ + 3e– metals in Zn → Zn2+ + 2e– decreasing order Fe → Fe2+ + 2e– of reactivityPb → Pb2+ + 2e–

Weaker H2 → 2H+ + 2e–

- Chlorine is in aqueous solution is called aqueous chlorine, a very pale yellow colour. - Bromine water is liquid bromine dissolved in water, an orange-brown colour.- Iodine in solution is yellow-brown and shown as I2(aq) but if iodine is precipitated it forms a grey solid. - Metallic copper is a red-brown colour. Other metals are silver-grey.- Spectator ions (if present) are not involved in redox reactions.

Writing Redox Equations1. Write the reduction (GER) half equation I2 → I–

Balance the elements I2 → 2I–

Balance the charge by adding electrons I2 + 2e– → 2I– (to the most positive side - left hand side)2. Write the LEO half equation Fe2+ → Fe3+

Balance the elements not requiredBalance the charge by adding electrons Fe2+ + e– → Fe3+ (to the most positive side - right-hand side)

3. Multiply one or both equations until the electrons cancel. 2Fe2+ + 2e– → 2Fe3+

4. Add the equations, cancelling electrons. 2Fe2+ + I2 → 2Fe3+ + 2I–

The stronger oxidising agent always oxidises a weaker one, e.g.Chlorine will oxidise bromide (forming bromine) and iodide (forming iodine), but neither bromine nor iodine will not oxidise chloride.Bromine will oxidise iodide (forming iodine) but iodine will not oxidise bromide.

The stronger reducing agent always reduces a weaker one, e.g.Magnesium will reduce hydrogen ions (forming hydrogen gas), but hydrogen will not reduce magnesium ions.Magnesium will reduce lead ions (forming metallic lead), but lead will not reduce magnesium ions.

Redox – loss and gain of oxygenGain of oxygen = oxidationOxygen is a good oxidising agent and is itself reduced.1. Magnesium (silver-grey in colour) burns with a

bright white flame, a white powder (MgO) is produced and heat is evolved.

Mg + O2 → 2MgOMagnesium is oxidised, oxygen is reduced.

2. Copper (a red-brown colour) heated strongly in air becomes covered with a black coating (CuO).

2Cu + O2 → 2CuOCopper is oxidised, oxygen is reduced.

Loss of oxygen = reductionHydrogen and carbon are good reducing agents and are themselves oxidised.1. When black copper oxide is heated in a stream of

colourless hydrogen gas, red-brown metallic copper is produced and droplets of colourless water form on the inside of the vessel.

CuO + H2 → Cu + H2OCu2+ is reduced to Cu, hydrogen is oxidised.

2. When yellow lead oxide is heated strongly with black carbon, grey globules of metallic lead and colourless carbon dioxide are produced.

2PbO + C → 2Pb + CO2

Pb2+ is reduced to Pb, carbon is oxidised.

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AS 90171 revision

Redox – loss and gain of electrons (LEO and GER)1. Acid plus reactive metal → salt + hydrogenHydrochloric acid reacts with silver-grey magnesium metal.The metal disappears, bubbles of a colourless gas (H2) are evolved and the reaction mixture gets warm (an exothermic reaction).

2HCl(aq) + Mg(s) → MgCl2(aq) + H2(g)

Oxidation (LEO):Magnesium acts as a reducing agent and is itself oxidised.

Mg → Mg2+ + 2e–

Reduction (GER):Hydrogen ion acts as an oxidising agent and is itself reduced.

2H+ + 2e– → H2

Adding the two half equations: 2H+ + Mg → Mg2+ + H2

2. Water plus highly reactive metal → metal hydroxide compound + hydrogenWater reacts with sodium.The reaction is violent, the silver-grey metal disappears, bubbles of a colourless gas are evolved, the reaction mixture gets warm (exothermic reaction) and the hydrogen gas evolved may explode. Hydroxide ions and hydrogen gas are the products to the reduction half equation.

2H2O(l) + 2Na(s) → 2NaOH(aq) + H2(g)

Oxidation (LEO):Sodium acts as a reducing agent and is itself oxidised

Na → Na+ + e– x22Na → 2Na+ + 2e–

Reduction (GER):2H2O + 2e– → 2OH– + H2

Adding the two half equations:2H2O + 2Na → 2Na+ + 2OH– + H2

3. Metal plus ion of less reactive metal (refer to the reactivity series)Blue copper sulfate solution reacts with grey zinc metal. The blue solution goes colourless and a red-brown solid is formed.

CuSO4(aq) + Zn(s) → ZnSO4(aq) + Cu(s)

Oxidation of metallic zinc in which zinc acts as a reducing agent. Zn → Zn2+ + 2e–

Reduction of copper(II) ion in which copper ion acts as an oxidant Cu2+ + 2e– → CuBalanced redox equation: Cu2+ + Zn →Zn2+ + Cu

Colourless silver nitrate solution reacts with grey lead metal. The solution remains colourless and a shiny silver substance is formed.2AgNO3(aq) + Pb(s) → Pb(NO3)2(aq) + 2Ag(s)

Oxidation:Lead acts as a reducing agent. Pb → Pb2+ + 2eReduction:Silver ion acts as an oxidising agent. Ag+ + e → AgBalanced redox equation: 2Ag+ + Pb → Pb2+ + 2Ag

4. Halogen plus ion of less reactive halogen Pale yellow chlorine water reacts with colourless potassium iodide solution. A yellow-brown solution is formed.Cl2(aq) + 2KI(aq) → 2KCl(aq)+ I2(aq)

Oxidation:Chlorine is a strong oxidising agent and oxidises iodide to iodine.

2I– → I2 + 2e Reduction: Cl2 + 2e → 2Cl–

Balanced redox equation: Cl2 + 2I– → 2Cl– + I2

5. Halogen plus metal ion Plae yellow chlorine water reacts with iron(II) sulfate solution. The pale green solution turns yellow. 3Cl2(aq) + 6FeSO4(aq) → 2Fe2(SO4)3(aq) + 2FeCl3(aq)

Oxidation:Chlorine is a strong oxidising agent oxidising iron(II) to iron(III).

Fe2+ → Fe3+ + e–

Reduction: Cl2 + 2e– → 2Cl–

Balanced redox equation: Cl2 + 2Fe2+ → 2Cl– + 2Fe3+

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AS 90171 revision

Revision Notes Precipitation A solution comprises a solute (e.g. salt)

dissolved in a solvent (e.g. water). A solution using water as a solvent

gives a clear solution (e.g. sodium chloride solution is clear and colourless while copper sulphate solution is clear and blue).

A precipitation reaction occurs when two solutions are mixed and a solid, called a precipitate, forms.

A saturated solution is one in which no more solute will dissolve.

Precipitates may be separated from mixtures by filtration.

SolubilityAll nitrates are soluble.All group 1 metal and ammonium compounds are soluble.Most common chlorides are soluble – except silver chloride and lead chloride.Most common sulfates are soluble – except calcium sulfate, barium sulfate and lead sulphate.All carbonates are insoluble – except carbonates of group 1 metals and ammonium carbonate.All hydroxides are insoluble – except hydroxides of group 1 metals and aqueous ammonia (ammonium hydroxide). Barium hydroxide is slightly soluble and calcium hydroxide is very slightly soluble in water giving ‘limewater’ solution.

States are ImportantSolid (s):zinc hydroxideZn(OH)2 (s)

Liquid (l):waterH2O (l)

Gas (g):chlorineCl2 (g)

Aqueous solution (aq) = dissolved in water:sodium nitrate bromine water iodine solutionNaNO3 (aq) Br2 (aq) I2 (aq)

Colours are ImportantMost compounds groups 1 and 2 metals are white.Common coloured solids:

copper sulphate bluecopper carbonate greencopper oxide blackiron(II) sulphate pale green

Most solutions of group 1 and group 2 metals are colourless.Common coloured aqueous solutions:

copper, Cu2+ (aq) blueiron(II), Fe2+ (aq) pale greeniron(III), Fe3+ (aq) yellow

Most precipitates are white.Common coloured precipitates:

copper hydroxide blue*copper carbonate blueiron(II) hydroxide greeniron(III) hydroxide orange/brown*silver oxide muddy brown

Note: The ‘copper carbonate’ precipitated on addition of carbonate ion is more likely to be the basic copper carbonate (2CuCO3.Cu(OH)2) which is blue although pure dry copper carbonate (CuCO3) is green.Silver oxide, Ag2O, is likely to be precipitated on addition of hydroxide ion giving a muddy brown solid.DissolvingWhen ionic compounds ‘dissolve’, ions are formed. The solution contains ions.Dissolving:sodium chloride, NaCl, gives: Na+(aq) and Cl–(aq)copper chloride, CuCl2, gives: Cu2+ (aq) and 2Cl–(aq)iron(III) chloride, FeCl3, gives: Fe3+ (aq) and 3Cl–(aq)

Spectator IonsSpectator ions are not involved in precipitation reactions. Note: If a precipitate is filtered from the mixture and the remaining solution is evaporated the salt of these spectator ions may be recovered (providing correct amounts have been used).

Examples1. Mix solutions of sodium chloride and silver nitrate

Full equation Ionic equation (easier to balance)NaCl(aq) + AgNO3(aq) → NaNO3(aq) + AgCl(s)precipitate: silver chloride

Ag+(aq) + Cl–(aq) → AgCl (s)spectator ions: sodium, Na+ and nitrate, NO3

–

2. Mix solutions of zinc sulphate and barium chlorideZnSO4(aq) + BaCl2(aq) → ZnCl2(aq) + BaSO4(s)precipitate: barium sulfate

Ba2+(aq) + SO42–(aq) → BaSO4(s)

spectator ions: zinc, Zn2+ and chloride, Cl–

3. Mix solutions of sodium carbonate and iron(II) nitrateNa2CO3(aq) + Fe(NO3)2(aq) → 2NaNO3(aq) + FeCO3(s)precipitate: iron(III) carbonate

Fe2+(aq) + CO32–(aq) → FeCO3(s)

spectator ions: sodium, Na+ and nitrate, NO3–

4. Mix solutions of sodium hydroxide and iron(III) sulfate6NaOH(aq) + Fe2(SO4)3(aq) → 3Na2SO4(aq) + 2Fe(OH)3(s)

precipitate: iron(III) hydroxideFe3+(aq) + 3OH–(aq) → Fe(OH)3(s)spectator ions: sodium, Na+ and sulfate, SO4

2–

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AS 90171 revision

5. Mix solutions of sodium sulphate and zinc nitrate – no precipitate

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AS 90171 revision

Revision Notes Thermal Decomposition Compounds of metals high in the reactivity series are not easily split up by heating (thermal

decomposition). Compounds of metals low in the reactivity series are more easily decomposed. Some metal carbonate compounds decompose on give the metal oxide + carbon dioxide.

Carbonates of Group 1 metals (all white solids) are stable and do not decompose. There is no change in appearance of the white solidCarbonates of Group 2 metals (all white solids) require strong heating before they will decompose. There is no change in the appearance of the white solid although the mass will decrease. A colourless carbon dioxide gas evolved will not be noticeable. Carbonates of lead and copper are most easily decomposed.

Hydrogen carbonate compounds of group 1 metals decompose to give the metal carbonate + carbon dioxide + water.

Sodium and potassium hydrogen carbonate exist as solids. Calcium hydrogen carbonate exists only in solution and any attempt to isolate calcium hydrogen carbonate causes it to decompose. Other metals do not form hydrogen carbonate compounds.

Some metal hydroxide compounds decompose on heating to give metal oxide + water.Hydroxides of Group 1 metals are stable and do not decompose.

Test for carbon dioxide: Bubble carbon dioxide through clear limewater. The limewater goes milky or cloudy due to the presence of a precipitate of calcium carbonate.

CO2(g) + Ca(OH)2(aq) → CaCO3(s) + H2O(l)Note: if excess carbon dioxide is bubbled through the solution, the soluble calcium hydrogen carbonate is formed and the cloudy solution goes clear again.

CaCO3(s) + CO2(g) + H2O(l) → Ca(HCO3)2(aq) Summary – Thermal Decomposition1. (a) White sodium (or potassium) hydroxide and white sodium (or

potassium) carbonate do NOT decompose.(b) Thermal decomposition of white sodium hydrogen carbonate

(baking soda) allows cakes to rise. heat2NaHCO3 → Na2CO3 + H2O + 2CO2

2. (a) Strong heating of white calcium carbonate produces white calcium oxide and a colourless (undetectable) carbon dioxide.

heatCaCO3 → CaO + CO2

(b) Heating a colourless solution of calcium hydrogen carbonate produces a white precipitate (of calcium carbonate).

heatCa(HCO3)2 → CaCO3 + H2O + CO2

(c) Thermal decomposition of white calcium hydroxide produces white calcium oxide and colourless steam, H2O.

heatCa(OH)2 → CaO + H2O

3. (a) Thermal decomposition of green copper carbonate produces black copper(II) oxide and colourless CO2.

heatCuCO3 → CuO + CO2

(b) Thermal decomposition of the blue precipitate of copper hydroxide forms a black solid of copper(II) oxide

heatCu(OH)2 → CuO + H2O

4.. Thermal decomposition of the orange precipitate of iron(III) hydroxide produces reddish coloured ion(III) oxide

heat2Fe(OH)3 → Fe2O3 +3H2O

5. Strong heating of the white carbonates of magnesium and zinc produce the white oxide and colourless carbon dioxide.

heat carbonate compoundsMgCO3 → MgO + CO2

ZnCO3 → ZnO + CO2

6. Thermal decomposition of white hydroxides of magnesium, zinc and aluminium produce white oxides and colourless steam, H2O. Note: Zinc oxide is yellow when hot and white when cool.

heat hydroxide compoundsMg(OH)2 → MgO + H2OZn(OH)2 → ZnO + H2O2Al(OH)3 → Al2O3 + 3H2O

7. Thermal decomposition of white lead hydroxide and white lead carbonate produces lead oxide which is yellow when cold.

heat compoundsPb(OH)2 → PbO + H2OPbCO3 → PbO + CO2

* Steam, H2O(g) is likely to condense forming a colourless liquid, on the inside wall of the reaction vessel.

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* The mass of the reactant will decrease although this loss in mass may not be obvious.More Practice

1.(a)(b)(c)

Copper carbonate is heated strongly.What type of reaction is this?Write the chemical equation for the reaction.0.260 g of copper oxide was produced. Calculate the mass of copper carbonate that decomposed.

2.(a)(b)(c)

When copper oxide was heated with powdered carbon, copper and carbon dioxide are produced.What type of reaction is this?Write the chemical equation for the reaction.4.30 grams of copper were recovered. Calculate the mass of carbon used.

3.(a)(b)

The hydrated copper chloride used had a formula: CuCl2.2H2OWrite a chemical equation for the dehydration of copper chloride.Calculate the percentage of water in hydrated copper chloride.

4. 3.10 grams of a copper compound were formed when 2.50 grams of copper were heated in excess sulfur. Determine the formula of the sulfide compound.

5. 4.40 grams of an oxide of iron are formed when 3.08 g of iron are reacted with air (oxygen). Determine the formula for the oxide.

6.

(a)(b)

A compound has the following composition.percentage by mass: 54.5% carbon 9.1% hydrogen 36.4% oxygenDetermine the empirical formula for the compoundThe molar mass of the compound is 88.0 gmol-1. State the molecular formula.

7. Manganese forms five oxides.(a) 11.20 gram of manganese reacts with oxygen to form manganese(II) oxide, MnO, according to the

equation: 2Mn(s) + O2(g) → 2MnO(s)i) Calculate the mass of manganese(II) oxide produced.ii) Calculate the mass of oxygen gas required for this reaction.

(b) One oxide of manganese has the following percentage composition by mass: manganese, Mn = 49.5% and oxygen, O = 50.5% Determine the formula for this oxide.

(c) 9.40 gram of manganese reacts with oxygen to give 13.50 gram of an oxide of manganese.i) Calculate the mass of the oxygen required and state the conservation law used to find this mass.ii) Determine the formula for this oxide.

8. 1.35 mole of a gaseous compound containing sulfur and oxygen has a mass of 108 grams.Calculate the molar mass of the compound and determine whether the oxide is sulfur dioxide or sulfur trioxide.

9. Maleic acid has a molar mass of 116.0 g mol-1 and the following percentage composition by mass:C = 41.4% H = 3.4% O = 55.2%

Calculate the empirical formula of maleic acid and hence determine its molecular formula.

10. About 6 grams of lead in a crucible was weighed accurately and about 1 gram of sulphur (excess) was added. The crucible was strongly heated and reweighed until there was no change in the mass of the compound and crucible. Data: mass lead = 6.12 grams

mass compound = 7.04 gramsDetermine the empirical formula of the compound formed by the reaction between lead and sulphur.

11. Calcium phosphate occurs in animal bones as apatitie, Ca3(PO4)2.CaF2.Show that the molar mass of apatite is 388.0 gmol-1 and hence determine the percentage mass of calcium and of phosphorus in apatitie.

12. 2-nitroanaline has a mass composition:52.2% carbon, 4.3% hydrogen, 23.2% oxygen and 20.3% nitrogen. It has a molar mass of 138.0 gmol-1.Showing your working clearly, find the empirical formula and the molecular formula for 2-

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AS 90171 revision

nitroanaline.

Identification of type of reaction: (a) CaCl2(aq) + K2CO3(aq) → 2KCl(aq) + CaCO3(s) Precipitation

All carbonates are insoluble except (NH4)2CO3, Na2CO3 and K2CO3

(b) Pb(OH)2(s) → PbO(s) + H2O(l) Thermal decompositionOne substance on heating decomposes producing two substances

(c) AgNO3(aq) + NaCl(aq) → NaNO3(aq) + AgCl(s) PrecipitationAll chlorides soluble except AgCl and PbCl2

(d) Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s) Oxidation-reductionCu2+ has gained 2 electrons and been reduced while Fe has lost 2 electrons and been oxidised. Reduction: Cu2+(aq) + 2e– → Cu(s) Oxidation: Fe(s) → Fe2+(aq) + 2e–

(e) CO(g) + PbO(s) → Pb(s) + CO2(g) Oxidation-reductionPbO has lost oxygen and been reduced while CO has gained oxygen and been oxidised.

(f) 2AgNO3(aq) + BaCl2(aq) → 2AgCl(s) + Ba(NO3)2(aq) PrecipitationAll chlorides soluble except AgCl and PbCl2

(g) CuCl2(aq) + K2CO3(aq) → CuCO3(s) + 2KCl(aq) PrecipitationAll carbonates are insoluble except (NH4)2CO3, Na2CO3 and K2CO3

(h) Fe2+(aq) + Zn(s) → Fe(s) + Zn2+(aq) Oxidation-reductionFe2+ has gained 2 electrons and been reduced while Zn has lost 2 electrons and been oxidised. Reduction: Fe2+(aq) + 2e– → Fe(s) Oxidation: Zn(s) → Zn2+(aq) + 2e–

(i) MgCO3(s) → MgO(s) + CO2(g) Thermal decompositionOne substance on heating decomposes producing two substances

(j) PbO(s) + C(s) → Pb(s) + CO(g) Oxidation-reductionPbO has lost oxygen and been reduced while C has gained oxygen and been oxidised.

(k) CuCl2(aq) + Zn(s) → Cu(s) + ZnCl2(aq) Oxidation-reductionCu2+ has gained 2 electrons and been reduced while Zn has lost 2 electrons and been oxidised. Reduction: Cu2+(aq) + 2e– → Cu(s) Oxidation: Zn(s) → Zn2+(aq) + 2e–

(l) C(s) + O2(g) → CO2(g) Oxidation-reductionCarbon has gained oxygen and been oxidised while oxygen has gained electrons (in forming a covalent bond with carbon) and been reduced.

(m) Ca(OH)2(s) → CaO(s) + H2O(g) Thermal decompositionOne substance on heating decomposes producing two substances

(n) Cl2(aq) + 2I–(aq) → 2Cl–(aq) + I2(aq) Oxidation-reductionChlorine has gained 1 electron and been reduced while iodide has lost 1 electron and been oxidised. Reduction: Cl2(aq) + 2e– → 2Cl–(aq) Oxidation: 2I–(aq) → I2(aq) + 2e–

Precipitation: (a)

silver chlorideAg+(aq) + Cl–(aq) → AgCl(s)

(i) potassium sulfate + iron(II) nitrateNo precipitate

(b) calcium sulfateCa2+(aq) + SO4

2–(aq) → CaSO4(s)(j) sodium chloride and copper nitrate

No precipitate(c) lead chloride

Pb2+(aq) + 2Cl–(aq) → PbCl2(s)(k) magnesium hydroxide

Mg2+(aq) + 2OH–(aq) → Mg(OH)2(s)(d) calcium sulfate

Ca2+(aq) + SO42–(aq) → CaSO4(s)

(l) copper hydroxideCu2+(aq) + 2OH–(aq) → Cu(OH)2(s)

(e) magnesium sulfate and sodium chlorideNo precipitate

(m) calcium sulfate Ca2+(aq) + SO4

2–(aq) → CaSO4(s)(f) zinc carbonate

Zn2+(aq) + CO32–(aq) → ZnCO3(s)

(n) potassium carbonate + sodium hydroxideNo precipitate

(g) lead sulfatePb2+(aq) + SO4

2–(aq) → PbSO4(s)(o) copper hydroxide

Cu2+(aq) + 2OH–(aq) → Cu(OH)2(s)

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(h) barium sulfate Ba2+(aq) + SO4

2–(aq) → BaSO4(s)(p) lead carbonate

Pb2+(aq) + CO32–(aq) → PbCO3(s)

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Write Word Equations(a) heat

magnesium + oxygen → magnesium oxide (b) lead nitrate + potassium chloride → lead chloride(s) + potassium nitrate(aq) (c) heat

sodium hydrogen carbonate → sodium carbonate + water + carbon dioxide (d) aluminium + silver nitrate → silver(s) + aluminium nitrate(aq) (e) heat

copper hydroxide → copper oxide + water(f) magnesium + oxygen → magnesium oxide(g) heat

calcium carbonate → calcium oxide + carbon dioxide (h) barium chloride + magnesium sulfate → barium sulfate(s) + magnesium chloride(aq) (i) heat

lead hydroxide → lead oxide + water(j) magnesium chloride + sodium hydroxide → magnesium hydroxide(s) + sodium chloride(aq)(k) sodium hydroxide + magnesium sulfate

→ sodium sulfate + magnesium hydroxide(l) iron + copper sulfate → iron(II) sulfate(aq) + copper(s)

Write and Balance Equations(a) 2I–(aq) + Cl2(aq) → 2Cl–(aq) + I2(aq) (f) Zn(s) + 2AgNO3(aq) → Zn(NO3)2(aq) + 2Ag(s)(b) CaCl2(aq) + Na2SO4(aq)

→ CaSO4(s) + 2NaCl(aq)(g) Cu(NO3)2(aq) + 2NaOH(aq)

→ Cu(OH)2(s) + 2NaNO3(aq)(c) Cl2(g) + 2KI(aq) → 2KCl(aq) + I2(aq) (h) Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)(d) heat

2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(i) heat

Cu(OH)2(s) → CuO(s) + H2O(g)(e) 2Ag+(aq) + Cu(s) → Cu2+(aq) + 2Ag(s) (j) CuO(s) + H2(g) → Cu(s) + H2O(g)

Recall Observations and Link Observations to the Species Involve

(a) A small amount of zinc hydroxide is heated in a test tube over a Bunsen burner.When white zinc hydroxide is heated, it turns yellow and as this product, zinc oxide, cools it turns white. Steam, also produced, may condense forming colourless water on the inside of the test tube. A loss in mass may be noticed. heat heatzinc hydroxide → zinc oxide + water Zn(OH)2(s) → ZnO(s) + H2O(g)This is thermal decomposition since one substance on heating decomposes producing two substances.

(b) A solution of barium nitrate is added to a solution of iron(II) sulfate in a beaker. When colourless barium nitrate solution is added to very pale green iron(II) sulfate solution a white precipitate forms. The iron(II) solution remains very pale green since iron(II) ions are spectator ions.barium nitrate(aq) + iron(II) sulfate(aq) → barium sulfate(s) + iron(II) nitrate(aq) Ba2+(aq) + SO4

2 –(aq) → BaSO4(s)This is precipitation since all sulfates are soluble except BaSO4, PbSO4, CaSO4

(c) Bromine water is added to aqueous sodium iodide.When orange-brown bromine water is added to colourless sodium iodide solution, a yellow-brown solution of aqueous iodine forms.This is oxidation-reduction since bromine has gained electrons to become bromide and iodide has lost electrons to become iodine. reduction: Br2 + 2e– → 2Br– oxidation: 2I–(aq) → I2(aq) + 2e– redox equation: 2I–(aq) + Br2(aq) → 2Br–(aq) + I2(aq)

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(d) A strip of magnesium ribbon is heated over a Bunsen burner. Magnesium burns with a bright white flame in oxygen and a white powder of magnesium oxide is produced. Heat is given out (since the reaction is exothermic). The magnesium loses 2 electrons forming Mg2+ and oxygen gains 2 electrons forming O2–. These oppositely charged ions attract each other producing magnesium oxide, MgO.oxidation: Mg → Mg2+ + 2e– x 2reduction: O2 + 4e– → 2O2–

redox equation: 2Mg + O2 → 2Mg2+ + 2O2– leading to 2Mg + O2 → 2MgO

(e) (i) Heat sodium carbonate: The white sodium carbonate does not change in appearance since it carbonate compounds of group 1 metals do not decompose on heating.

(ii) Heat calcium carbonate: The white calcium carbonate does not change in appearance since the product of the decomposition, calcium oxide, is also white. A loss in mass of the calcium carbonate may be obvious. A colourless gas which may not be obvious is given off.

heat heatcalcium carbonate → calcium oxide + water CaCO3(s) → CaO(s) + CO2(g)(iii) Heat copper carbonate: The green copper carbonate decomposes to produce black copper oxide.

A loss in mass of the copper carbonate may be obvious. A colourless gas which may not be obvious is given off.

heat heatcopper carbonate → copper oxide + water CuCO3(s) → CuO(s) + CO2(g)(iv) Heat zinc hydroxide: The white zinc oxide decomposes to produce yellow zinc oxide. The

yellow oxide turns white when cool. A loss in mass of the zinc hydroxide may be obvious. The colourless steam given off and may condense of the side of the reaction vessel forming water.

heat heatzinc hydroxide → zinc oxide + water Zn(OH)2(s) → ZnO(s) + H2O(g)

(f) Some powdered magnesium is added to a solution of iron(II) sulfate in a beaker, and mixed well. The magnesium loses 2 electrons to the Fe2+ in solution forming Mg2+ which is soluble, so some of the grey powder disappears. The Fe2+ gains 2 electrons from the Mg forming Fe metal, so the green colour of the solution caused by Fe2+ fades and the new dark grey solid forms at the bottom of the beaker.oxidation: Mg(s) → Mg2+(aq) + 2e–

reduction: Fe2+(aq) + 2e– → Fe(s)redox equation: Mg(s) + Fe2+(aq) → Mg2+(aq) + Fe(s)

(g) Aqueous chlorine is added to iron(II) chloride solution.When pale yellow chlorine water is added to iron(II) chloride solution a pale yellow solution forms. This is an oxidation-reduction reaction since electrons have been transferred from the iron(II) ions to chlorine. Iron(II) ions lose 1 electron to become iron(III) ions and chlorine gains electrons to become chloride ions.oxidation: Fe2+(aq) → Fe3+(aq) + e– x2reduction: Cl2(aq) + 2e– → 2Cl–(aq)redox equation: 2Fe2+(aq) + Cl2(aq) → 2Fe3+(aq) + 2Cl–(aq)

(h) Iron(II) nitrate solution is added to sodium hydroxide solution in a test tube. When the colourless solutions are mixed a green precipitate forms. Iron(II) ions from the iron(II) nitrate solution combine with the hydroxide ions from the sodium hydroxide solution to form an insoluble substance, hence this is a precipitation reaction.

Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s)

(i) A piece of zinc foil is added to copper(II) nitrate solution and left. After 10 minutes a red-brown coating of copper would have formed over the zinc foil. After 24 hours the blue solution, due to the presence of Cu2+(aq), would have become colourless (or the zinc foil would have disappeared) This is an oxidation-reduction reaction since electrons have been transferred from the zinc to the copper ions. Zinc loses 2 electrons to become zinc ions and copper ions gain 2 electrons to become metallic copper.reduction: Cu2+(aq) + 2e– → Cu(s)oxidation: Zn(s) → Zn2+(aq) + 2e– redox equation: Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s)

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(j) Copper carbonate heated strongly and the gas formed is bubbled into limewater:When green copper carbonate is heated it decomposes to form black copper oxide and a colourless gas which is carbon dioxide. This is a thermal decomposition reaction.

heatCuCO3(s) → CuO(s) + CO2(g)

When carbon dioxide gas is bubbled into limewater, the limewater goes milky due to the formation of insoluble calcium carbonate. This is a precipitation reaction.

CO2(aq) + Ca2+(aq) + 2OH–(aq) → CaCO3(s) + H2O(l)

(k) A solution of sodium hydroxide is added to a solution of iron(III) chloride.When the colourless solutions are mixed an orange-brown precipitate forms. Iron(III) ions from the iron(III) nitrate solution combine with the hydroxide ions from the sodium hydroxide solution to form an insoluble substance, hence this is a precipitation reaction.

Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s)

(l) A piece of copper is placed in a solution of silver nitrate.A silver coating of silver metal forms over the piece of copper. The colourless solution becomes blue due to the presence of Cu2+(aq). This is an oxidation-reduction reaction since electrons have been transferred from the copper to the silver ions. Copper loses 2 electrons to become copper ions and silver ions gain 1 electron to become metallic silver.reduction: Ag+(aq) + e– → Ag(s) x2oxidation: Cu(s) → Cu2+(aq) + 2e– redox equation: 2Ag+(aq) + Cu(s) → Cu2+(aq) + 2Ag(s)

(m) Chlorine water is added to potassium iodide solution.When pale green chlorine water is added to colourless potassium iodide solution, a yellow-brown solution of aqueous iodine forms.This is oxidation-reduction since chlorine has gained electrons to become chloride and iodide has lost electrons to become iodine. reduction: Cl2 + 2e– → 2Cl– oxidation: 2I–(aq) → I2(aq) + 2e– redox equation: 2I–(aq) + Cl2(aq) → 2Cl–(aq) + I2(aq)

(n) Sodium carbonate is added to a solution of copper sulfate and the resulting mixture is heated gently.When colourless sodium carbonate solution is added to blue copper sulfate solution, a pale blue precipitate of copper carbonate forms. The copper ions and hydroxide ions combine to form an insoluble substance, hence this is a precipitation reaction. Note: pure, dry copper carbonate is green.

Cu2+(aq) + CO32–(aq) → CuCO3(s)

When the blue copper carbonate formed is heated it decomposes to form black copper oxide, hence this is a thermal decomposition reaction. A colourless gas, carbon dioxide, is given off but this is not obvious. There is a loss in mass of the copper carbonate since some of the solid decomposes.

CuCO3(s) → CuO(s) + CO2(g)

(o) A strip of magnesium ribbon is added to hydrochloric acid in a test tube.When a strip of grey magnesium ribbon is added to hydrochloric acid, the mixture fizzes and a colourless gas, which is hydrogen, is given off. The reaction mixture gets hot and some of the magnesium ribbon disappears. The resulting solution is colourless.This is an oxidation-reduction reaction since electrons have been transferred from the magnesium to the hydrogen ions in the acidic solution. Magnesium loses 2 electrons to become magnesium ion and hydrogen ion gains electrons to become gaseous hydrogen.reduction: 2H+(aq) + 2e– → H2(g) oxidation: Mg(s) → Mg2+(aq) + 2e– redox equation: 2H+(aq) + Mg(s) → Mg2+(aq) + H2(g)

(p) Sulfuric acid is added to barium nitrate solution.When colourless sulfuric acid solution is added to colourless barium nitrate solution, a white precipitate forms. Barium ions from the barium nitrate solution combine with the sulfate ions from the sulfuric acid solution to form an insoluble substance, hence this is a precipitation reaction.

Ba2+(aq) + SO42–(aq) → BaSO4(s)

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Calculate Relative Molar Masses(a) M(ZnO)

= 81.4 g mol-1(d) M(CaCO3)

= 100.1 g mol-1

(g) M(KNO3)= 101.1 g mol-1

(j) M(MgO)= 40.3 g mol-1

(m) M(Fe(CH3COO)2)= 173.9 g mol-1

(b) M(CuSO4)= 159.7 g mol-1

(e) M(Al2O3)= 102.0 g mol-1

(h) M(Na2CO3)= 106.0 g mol-1

(k) M(Cu(NO3)2)187.6 g mol-1

(n) M(Ag2CO3)= 276.0 g mol-1

(c) M(Pb(NO3)2)= 331.0 g mol-1

(f) M((NH4)2SO4)=68.1 g mol-1

(i) M(Ba(NO3)2)= 261.0 g mol-1

(l) M(Na2SO4)= 142.1 g mol-1

(o) M(Fe2(CO3)3)= 291.8 g mol-1

Chemical Calculations Mmerm.C

Rewrite your final answer using 3 significant figures(a) Calculate the mass of zinc hydroxide that must be heated to produce 1.00 gram of water.

The equation is: Zn(OH)2(s) → ZnO(s) + H2O(g)M(Zn(OH)2) = 99.4 g mol-1 M(H2O) = 18.0 g mol-1

n(H2O) = mass / M = 1.00 / 18.0 = 0.05555 moln(H2O) : n(Zn(OH)2) is 1:1 hence n(Zn(OH)2) = 0.05555 molmass(Zn(OH)2) = n x M = 0.05555 x 99.4 = 5.5222 g mass Zn(OH)2 = 5.52 g

(b) Calculate the mass of sodium hydrogen carbonate required to form 5.40 grams of carbon dioxide.The equation is: 2NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g)M(NaHCO3) = 84 g mol-1 M(CO2) = 44 g mol-1

n(CO2) = mass / M = 5.40 / 44 = 0.1227 moln(CO2) : n(NaHCO3) is 1 : 2 hence n(NaHCO3) = 2 x 0.1227 = 0.2454 molmass(NaHCO3) = n x M = 0.2454 x 84 = 20.618 g mass NaHCO3 = 20.6 g

(c) Calculate the mass of calcium hydroxide required to react with 2.14 grams of ammonium chloride.The equation is: 2NH4Cl + Ca(OH)2 → CaCl2 + 2NH3 + 2H2OM(Ca(OH)2) = 74.1 g mol-1 M(NH4Cl) = 53.5 g mol-1

n(NH4Cl) = mass / M = 2.14 / 53.5 = 0.04 moln(NH4Cl) : n(Ca(OH)2) is 2 : 1 or 1 : ½ n(Ca(OH)2) = ½ x 0.04 = 0.02 molmass(Ca(OH)2) = n x M = 0.02 x 74.1 = 1.482 g mass Ca(OH)2 = 1.48 g

(d) Calculate the mass of sodium carbonate expected to form when, 0.500 g of sodium hydrogen carbonate was decomposed. 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2OM(Na2CO3) = 106.0 g mol-1 M(NaHCO3) = 84.0 g mol-1

n(NaHCO3) = mass / M = 0.500 / 84.0 = 5.952 x 10-3 moln(NaHCO3) : n(Na2CO3) is 2 : 1 or 1 : ½ n(Na2CO3) = ½ x 5.952 x 10-3 = 2.976 x 10-3 molmass(Na2CO3) = n x M = 2.976 x 10-3 x 106.0 = 0.3154 g mass Na2CO3 = 0.315 g

(e) Calculate the mass of copper(II) oxide that will be needed to give 18.0 g of copper.The equation is: CuO(s) + H2(g) → Cu(s) + H2O(l)M(CuO) = 79.6 g mol-1 M(Cu) = 63.6 g mol-1

n(Cu) = mass / M = 18.0 / 63.6 = 0.2830 moln(Cu) : n(CuO) is 1 : 1 n(CuO) = 0.2830 molmass(CuO) = n x M = 0.2830 x 79.6 = 22.52 g mass CuO = 22.5 g

(f) Calculate the mass of CO2 formed when 12.6 g of NaHCO3 is completely decomposed.The equation is: 2NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) M(NaHCO3) = 84.0 g mol-1 M(CO2) = 44.0 g mol-1

n(NaHCO3) = mass / M = 12.6 / 84.0 = 0.15 moln(NaHCO3) : n(CO2) is 2 : 1 or 1 : ½ n(CO2) = ½ x 0.15 = 0.075 molmass(CO2) = n x M = 0.075 x 44.0 = 3.3 g mass CO2 = 3.30 g

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Determine Molecular Formula(a) A compound was analysed and found to contain: 20.2% phosphorus, 10.4% oxygen, 69.4% chlorine.

It has a relative molar mass of 153.5 g mol-1.element P O Clmass 20.2 10.4 69.4 amount 20.2 / 31.0 = 0.6516 10.4 / 16 = 0.65 69.4 / 35.5 = 1.9549amount ratio 1 : 1 : 3 giving POCl3

M(POCl3) = 153.5 g mol-1 molecular formula is POCl

(b) Determine the formula of the compound made when 3.55 g of chlorine combines with 5.60 g of oxygen. The molar mass of the compound is 183.0 g mol-1.M(Cl) = 35.5 g mol-1 M(O) = 16 g mol-1

n(Cl) = mass / M = 3.55 / 35.5 = 0.1 mol n(O) = mass / M = 5.60 / 16.0 = 0.35 moln(Cl) : n(O) is 0.1 : 3.5 or 1 : 3.5 or using whole numbers ratio is 2 : 7M(Cl2O7) = 183 g mol-1 (which is the molar mass given) hence formula is Cl2O7

(c) Determine the formula of the compound made when 8.65 g of iron combines with 3.72 g of oxygen. The molar mass of the iron oxide is 159.8 g mol-1. M(Fe) = 55.9 g mol-1 M(O) = 16 g mol-1

n(Fe) = mass / M = 8.65 / 55.9 = 0.1547 mol n(O) = mass / M = 3.72 / 16.0 = 0.2325 moln(Fe) : n(O) is 0.1547 : 0.2325 or 1 : 1.5 or using whole numbers ratio is 2 : 3M(Fe2O3) = 159.8 g mol-1 (which is the molar mass given) hence formula is Fe2O3

(d) Calculate the empirical formula of a compound that is 43.7% phosphorus and 56.3% oxygen and determine the molecular formula for the compound given that its relative molar mass is 284.element P Omass 43.7 56.3 amount 43.7 / 31.0 = 1.409 56.3 / 16 = 3.51875amount ratio 1 : 2.49or 2 : 5 empirical formual is P2O5

M(P2O5) = 142.0 g mol-1 284 / 142 = 2 molecular formula is P4O10

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More Practice1. Heating copper carbonate – a thermal decomposition reaction

heatCuCO3 → CuO + CO2

M(CuCO3) = 123.5 gmol-1 M(CuO) = 79.5 gmol-1

n(CuO) = mass / M = 0.26 / 79.5 = 0.003270 moleamount ratio CuCO3 : CuO is 1 : 1n(CuCO3) = 0.003270 molemass(CuCO3) = n x M = 0.003270 x 123.5 = 0.4038 g mass CuCO3 = 0.404 g

2. Copper oxide and carbon – an oxidation-reduction reaction (transfer of electrons)Copper ions have gained electrons to become metallic copper (reduction) or copper oxide has lost oxygen (reduction) while carbon has gained oxygen (oxidation)2CuO + C → 2Cu + CO2 M(C) = 12.0 gmol-1 M(Cu) = 63.5 gmol-1 n(Cu) = mass / M = 4.30 / 63.5 = 0.06771 moleamount ratio C : Cu is 1 : 2 or ½ : 1n(C) = ½ x 0.06771 = 0.03385 molemass(C) = n x M = 0.03385 x 12 = 0.4062 g mass carbon = 0.406 g

3. dehydration of copper chlorideCuCl2.2H2O → CuCl2 + 2H2OM(CuCl2.2H2O) = 170.5 gmol-1 M(H2O) = 18.0 gmol-1

percentage water = (2 x 18.0 / 170.5) x 100 = 21.11% percentage water = 21 %

4. Formula for the copper and sulfur compoundM(Cu) = 63.5 gmol-1 M(S) = 32.0 gmol-1

mass(Cu) = 2.50 g n(Cu) = mass / M = 2.50 / 63.5 = 0.03937 molemass(S) = 3.10 – 2.50 = 0.60 g n(S) = mass / M = 0.60 / 32 = 0.01875 moleamount ratio Cu : S is 0.03937 : 0.01875

2.09 : 1 (about 2 : 1) formula is Cu2S

5. Formula for the iron and oxygen compoundM(Fe) = 56.0 gmol-1 M(O) = 16.0 gmol-1

Mass(Fe) = 3.08 g n(Fe) = mass / M = 3.08 / 56 = 0.055 molemass(O) = 4.40 – 3.08 = 1.32 g n (O) = mass/A = 1.32 / 16 = 0.0825 moleamount ratio n(Fe) : n(O) is 0.055 : 0.0825

1 : 1.5 or 2 : 3 formula is Fe2O3

6. Consider a 100 g sample of the compoundElement C H Omass in 100 g 54.5 9.1 36.4amount (mole) 54.5 / 12 = 4.54 9.1 / 1 = 9.10 36.4 / 16 = 2.275amount ratio 1.996 : 4 : 1 empirical formula is C2H4OM (C2H4O) = 44.0 gmol-1 88.0 / 44.0 = 2 molecular formula is C4H8O2

7. Manganese forms five oxides.(a) 2Mn(s) + O2(g) → 2MnO(s)

M(Mn) = 54.9 gmol-1 M(MnO) = 70.9 gmol-1 M(O2) = 32.0 gmol-1

n(Mn) = mass / M = 11.2 / 54.9 = 0.2040 mol(i) amount ratio n(Mn) : n(MnO) is 2 : 2 or 1 : 1n(MnO) = 0.2040 molmass(MnO) = n x M = 0.2040 x 70.9 = 14.464 g mass(MnO) = 14.5 gram(ii) amount ratio n(Mn) : n(O2) is 2 : 1 or ½ : 1 n(O2) = ½ x 0.2040 = 0.1020 molmass(O2) = n x M = 0.1020 x 32.0 = 3.2641 g mass(O2) = 3.3 gram(iii) conservation of massmass(Mn) + mass(O2) = 11.2 + 3.3 = 14.5 g mass(MnO) = 14.5 gSince mass products = mass reactants, mass is conserved in the chemical reaction.

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(b) Determine the formula for an oxide of manganese, composition: Mn = 49.5% and oxygen, O = 50.5%Assume a 100 g sample.element Mn Omass (g) 49.5 50.5amount (mol) 49.5 / 54.9 = 0.9016 50.5 / 16.0 = 3.1562amount ratio 1 : 3.5 or 2 : 7 formula is Mn2O7

(c) Determine the formula for an oxide of manganese where: 9.40 gram of manganese reacts with oxygen to give 13.50 gram of the oxide of manganese.Mass is conserved in a chemical reaction hence mass oxygen required = 13.5 – 9.4 = 4.1 gFor the 13.5 gram sampleelement Mn Omass (g) 9.4 4.1amount (mol) 9.4 / 54.9 = 0.17122 4.1 / 16.0 = 0.25625 amount ratio 1 : 1.496 or 2 : 3 formula is Mn2O3

8. Find the molar mass of a compound containing sulfur and oxygen given: 1.35 mole has mass of 108 gmolar mass = mass / amount M = 108 / 1.35 = 80.0 gmol-1

M(SO2) = 32.0 + (16.0 x 2) = 64.0 gmol-1 M(SO3) = 32.0 + (16.0 x 3) = 80.0 gmol-1

The oxide is sulfur trioxide.

9. Maleic acid (molar mass of 116.0 g mol-1) C = 41.4% H = 3.4% O = 55.2% Assume a 100 g sample.element C H Omass (g) 41.4 3.4 55.2amount (mol) 41.4 / 12.0 = 3.45 3.4 / 1.0 = 3.45 55.2 / 16.0 = 3.45amount ratio 1 : 1 : 1 empirical formula: CHOM(CHO) = 29.0 g mol-1

M(maleic acid) / M(CHO) = 116.0 / 29.0 = 4 molecular formula: C4H4O4

10. Determine the empirical formula of compound: mass(Pb) = 6.12 g mass(compound) = 7.04 gxPb + yS → PbxSY n(Pb) = 6.12/207.2 = 0.02953 molemass sulphur reacted = 7.04 - 6.12 = 0.92 g n(S) = 0.92/32 = 0.02875 mole n(Pb) : n(S)

0.02953 : 0.02875 or 1 : 1empirical formula is PbS

11. Determine the percentage mass of calcium and of phosphorus in apatitie.apatitie, Ca3(PO4)2.CaF2. M(Ca3(PO4)2.CaF2) = 4xCa + 2xP +8xO + 2xF = 160.0 + 62.0 + 128.0 + 38.0 = 388.0 gmol-1

% composition Ca = 160.0 / 388.0 x 100 = 41.237 41.2% calcium% composition P = 62.0 / 388.0 x 100 = 15.979 16.0% phosphorus

12. 2-nitroanaline composition: 52.2% carbon, 4.3% hydrogen, 23.2% oxygen and 20.3% nitrogen. molar mass = 138.0 gmol-1.Consider a 100 gram sampleelement C H O Nmass in 100 g 52.5 4.3 23.2 20.3amount (mole) 52.2 / 12 = 4.35 4.3 / 1 = 4.3 23.2 / 16 = 1.45 20.3 / 14 = 1.45ratio 3 : 3 : 1 : 1 empirical formula: C3H3ONM(C3H3ON) = 69 gmol-1 M(2-nitroanaline) / M(C3H3ON) = 138.0 / 69 = 2 molecular formula: C6H6O2N2

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