January 2012 – State the general trend in first ionisation energies for the Period 2 elements lithium to nitrogen. [1 mark]
Increases across Period 2.
January 2012 – Identify the element (from group 2) that deviates from the previously stated trend and explain your answer. [3 marks]
The electron configuration of boron is 1s2 2s2 2p1. The outermost electron is 2p1 is on its own. It is shielded from, the nucleus by the inner electrons and so isn’t held as tightly.
January 2012 – Explain, in terms of structure and bonding, why the melting point of carbon is high. [3 marks]
carbon exists as a giant macromolecular molecule. The carbon atoms are held together by strong covalent bonds. These strong covalent bonds require a lot of energy (thermal/heat) to overcome them.
January 2012 – Name the raw material from which ………. (alkane,alkene etc) is obtained. [1 mark]
January 2012 – Name the process use to obtain …………. (alkane,alkene etc) form this raw material. [1 mark]
January 2012 – State the type of cracking that produces a high percentage of alkenes. state the conditions needed for this type of cracking. [2 marks]
High Temperature / 400 degreesC – 900 degreesC
High Pressure / >= 10atm
January 2012 – Explain the main economic reasons why alkanes are cracked. [1 mark]
To produce shorter chain alkanes and short chain alkenes which are in greater demand.
January 2012 – Suggest the most suitable hazard label for hydrochloric acid. [1 mark]
January 2012 – Suggest one reason why the alkane/alkene combustion was incomplete.
There wasn’t enough O2 in the air to combust the octane.
January 2012 – Write the equation that occurs in the catalytic concerter. [1 mark]
2NO + 2CO –> N2 + 2CO2
January 2012 – Identify a metal used as a catalyst in a catalytic converter. Suggest one reason, other than cost, why the catalyst is coated on a ceramic honeycomb. [2 marks]
Platinum / Palladium / Rhodium / Iridium
– To increase the surface area
– Ceramic withstands high temperatures
– Increases reaction rate
– Ensures complete reaction
– Removes more of the gases
January 2012 – If a sample of fuel for a power station is contaminated with an organic sulphur compound, a toxic gas is formed by complete combustion of this sulphur compound.
State one environmental problem that can be caused by the release of this gas. [1 mark]
SO2 can react with water in the air to form acid rain.
2SO2 +O2 –> 2SO3
SO3+H20 –> H2SO4
January 2012 – Identify one substance that could be used to remove this gas. Suggest one reason, other than cost, why this substance is used. [2 marks]
CaO / lime
Neutralises the acid
January 2012 – The element nitrogen forms compounds with metals and non-metals.
Nitrogen forms a nitride ion with the electron configuration 1s2 2s2 2p6. Write the formula of the nitrode ion. [1 mark]
January 2012 – An element forms an ion Q with a single negative charge that has the same electron configuration as the nitride ion. Identify the ion Q. [1 mark]
January 2012 – Use the Periodic Table and your knowledge of electron arrangement to write the formula of lithium nitride. [1 mark]
January 2012 – Calcium nitride contains 81.1% by mass of the metal. Calculate the empirical formula of calcium nitride. Show your working. [3 marks]
whole number: 3
whole number: 2
Empirical formula = Ca3N2
January 2012 – Write an equation for the reaction between silicon and nitrogen to form silicon nitride, Si3N4. [1 mark]
3Si + 2N2 –> Si3N4
The metal lead reacts with warm dilute nitric acid to produce lead(II) nitrate, nitrogen monoxide and water according to the following equation.
3Pb(s) + 8HNO3(aq) –> 3Pb(NO3)2(aq) + 2NO(g) + 4H2O(l)
In an experiment, an 8.14g sample of lead reacted completely with a 2.00moldm-3 solution of nitric acid.
Calculate the volume, in dm3, of nitric acid required for complete reaction. Give your answer to 3 significant figures.
– Moles of lead = mass of lead diveded by Mr of lead
– Moles of lead = 8.14g divided by 207.2Mr
– Moles of lead = 0.039285714
– Moles of nitric acid = moles of lead divided by 3 x8
– Moles of nitric acid = 0.039285714 divided by 3 x8
– Moles of nitric acid = 0.1047619
– Volume of nitric acid = moles of nitric acid divided by concentration of nitric acid
– Volume of nitric acid = 0.1047619moles divided by 2.00moldm-3
– Volume of nitric acid = 0.0523809 dm3
– Volume of nitric acid = 0.0524dm3 to 3sf
In a second experiment, the nitrogen monoxide gas produced in the reaction occupied 638cm3 at 101kPa and 298K.
Calculate the amount, in moles, of NO gas produced.
(The gas constant R=8.31JK-1mol-1)
PV = nRT therfore n=PV divided by RT
n = (101,000 Pa x 0.000638cm3) divided by (8.31JK-1mol-1 x 298K)
n = 0.02602104685
moles of NO = 0.02602 to 3sf
January 2012 – When lead(II) nitrate is heated it decomposes to form lead(II) oxide, nitrogen dioxide, and oxygen.
Suggest one reason why the yield of nitrogen dioxide formed during this reaction is often less than expected.
Decomposition is not complete/side reactions/by products/some (NO2) escapes/not all reacts / impure Pb(NO3)2
January 2012 – Suggest one reason why it is difficult to obtain a pure sample of nitrogen dioxide from this reaction. [1 mark]
Hard to separate the O2 from the NO2 / hard to separate the two gases
State how krypton is ionised in the mass spectrometer.
Write an equation, including state symbols, to show the reaction that occurs when the first ionisation energy of Kr is measured.
Sometimes the mass spectrum of Kr has a very small peak with an m/z value of 42. Explain the occurence of this peak.
First of all, krypton is vapourised. An electron gun is fired at the Kr sample and an electron is knocked off the sample. This gives is a positive +1 charge.
Kr(g) –> Kr+(g) + e-
This means that two electrons have been knocked off. The m/z is the mass divided by the charge, The mass is 84 and the charge is +2 then 84 divided by 2 gives a peak of 42.
January 2012 – The mass spectrum is taken from a sample of …………. from a meteorite.
Explain why the value you have calculated is slightly different from the relative atomic mass given in the Periodic table. [1 mark]
This sample is from space. The relative atomic mass of ……….. in the Periodic table is calculated using samples from Earth. This is why they are different.
January 2013 – State the meaning of the term Mass Number of an isotope. [1 mark]
The mass number is the total number of protons and the total number of neutrons present in the nucleus of an atom.
January 2012 – In a mass spectrometer, the isotopes of an element are separated. Two measurements for each isotope are recorded on the mass spectrum.
State the two measurements that are recorded for each isotope. [2 marks]
The m/z (mass/charge ratio)
The abundance of each isotope
January 2013 – State why isotopes of an element have the same chemical properties. [1 mark]
They still have the same number of protons and thus the same number of electrons to counter the protons charge. The electron’s are in the same configuration and so react in the same manner.
January 2013 – Give one reason why the second ionisation energy of silicon is lower than the second ionisation energy of aluminum. [1 mark]
On the 2nd ionisation energy of Si, the electron is removed from the 3p sub-shell and so it is is less attracted to the nucleus as it is shielded by the inner electrons.
January 2013 – Predict the element in Period 3 that has the highest second ionisation energy. Give a reason for your answer. [2 marks]
January 2013 – Explain why the ionisation energy of every element is exothermic. [1 mark]
Because energy (in the form of heat) is needed in order to overcome the forces between the positively charged nucleus and the negatively charged electron.
January 2013 – State the meaning of the term electronegativity. [2 marks]
The power of an element to pull a pair of electrons (electron density) towards itself in a covalent bond.
January 2013 – Suggest why the electronegativity of the elements increases from lithium to fluorine. [2 marks]
As you go across the period. Electrons are added to the same shell, but the number of protons in the nucleus increases and so produces a greater attractive nuclear charge which pulls the electrons towards itself.
State the type of bonding in lithium fluoride.
Explain why a lot of energy is needed to melt a sample of lithium fluoride.
There are very strong electrostatic forces of attraction between the positively charged lithium ions (cations) and the negatively charged fluorine ions (anions). These forces need a lot of energy to be broken/overcome.
January 2013 – Deduce why the bonding in nitrogen oxide is covalent rather than ionic. [1 mark]
The difference in electronegativity is 0.5.
Oxygen forms several different compounds with fluorine.
Suggest the type of crystal shown by OF2.
January 2013 – Write an equation to show how OF2 reacts with steam to form oxygen and hydrogen fluoride. [1 mark]
OF2(s) + H2O(g) –> O2(g) + 2HF(aq)
January 2013 – State a process used to separate an alkane from a mixture of alkanes. [1 mark]
January 2013 – Explain why the complete combustion of an alkane may contribute to environment problems. [1 mark]
Complete combustion forms CO2(g) and H2O(g) which are both greenhouse gases that contribute to global warming.
January 2013 – Ethanethiol (C2H5SH), a compound with an unpleasant smell, is added to gas to enable leaks from gas pipes to be more easily detected.
Write an equation for the combustion of ethanethiol to form carbon dioxide, water and sulphur dioxide. [1 mark]
C2H5SH + 4.5O2 –> 2CO2 + 3H2O + SO2
January 2013 – Identify a compound that is used to react with the sulphur dioxide in the products of combustion before they enter the atmosphere. Give one reason why this compound reacts with sulphur dioxide. [1 mark]
It is a neutralisation reaction.
January 2013 – Ethanethiol and ethanol molecule have similar shapes. Explain why ethanol has the higher boiling point.
Ethanol has a hydrogen bonded to an oxygen. The difference in electronegativity means that ethanol molecules can hydrogen bond with one another (hydrogen bonding is the strongest type of intermolecular force). More energy is needed (in the form of heat) to break these hydrogen bonds.
January 2013 – X and Y are isomers with the same molecular formula. X is a straight chain molecule, Y is a branched molecule. Explain why the boiling point of Y is lower than that of X. [2 marks]
The straight chain molecule has a larger surface area to volume ratio and so has stronger Van der Waals forces between molecules.
January 2013 – State the type of cracking that produces a high proportion of ethene and propene. Give two conditions for this cracking process.
High Temperature / 400 degreesC – 900 degreesC
High Pressure / >= 10atm
Ammonia is used to make nitric acid (HNO3) by the Ostwald Process. Three reactions occur in the process.
Reaction 1: 4NH3(g) + 5O2(g) –> 4NO(g) + 6H2O(g)
Reaction 2: 2NO(g) + O2(g) –> 2NO2(g)
Reaction 3: 3NO2(g) + H2O(l) –> 2HNO3(aq) + NO(g)
In one production run, the gases formed in reaction 1occupied a total volume of 4.31m3 at 25degreesC and 100kPa.
Calculate the amouont, in moles, of NO produced.
Give your answer to 3 significant figures.
(The gas constant R = 8.31JK-1mol-1)
– PV=nRT therefore n=PV divided by RT
– n = (100,000 Pa x 4.31m3) divided by (8.31JK-1mol-1 x 298K)
– n = 174.0443712 moles of gas
– Ration of gaseous moles = (NO) 4:6 (H20)
– Therefore 174.0443712 divided by 10 x4 = moles of NO
– Moles of NO = 69.61774849
– Moles of NO = 69.6 to 3sf
In another production run, 3.00kg of ammonia gas were used in Reaction 1 and all the NO gas produced was used to make NO2 gas in Reaction 2.
Calculate the amount, in moles, of ammonia in 3.00kg.
– mass = Mr x moles therefore moles = mass divided by Mr
– moles = 3,000g divided by Mr of NH3
– moles = 3,000g divided by 17
– moles = 176.4705882352941176
– moles = 176.5 to 1dp
Calculate the mass of NO2 formed from 3.00kg of ammonia in Reaction 2 assuming an 80% yield.
Give your answer in kilograms
– moles of NO2 = moles of NH3
– moles of NO2 = 176.4705882352941178
– mass of NO2 = moles of NO2 x Mr of NO2
– mass of NO2 = 176.4705882352941178g x 46
– mass of NO2 = 8117.6470588235294117g
– mass of NO2 = 8117.6470588235294117g divided by 1000
– mass of NO2 = 8.1176470588235294117kg divided by 100 x80
– mass of NO2 = 6.49411764705882352kg
– mass of NO2 = 6.49kg to 3sf
Consider Reaction 3 in this process
Reaction3: 3NO2(g) + H20(l) –> 2HNO3(aq) + NO(g)
Calculate the concentration of nitric acid produced when 0.543mol of NO2 is reacted with water and solution is made up to 250cm3.
– moles of HNO3 = moles of NO2 divided by 3 x2
– moles of HNO3 = 0.543mol divided by 3 x2
– moles of HNO3 = 0.362
-Concentration of HNO3 = moles of HNO3 divided by – volume of HNO3
– Concentration of HNO3 = 0.362 divided by 0.250
– Concentration of HNO3 = 1.448moldm-3
January 2013 – Suggest why a leak of NO2 gas form the Ostwald Process will cause atmospheric pollution. [1 mark]
NO2 will react with water in the atmosphere and form nitric acid which contributes to acid rain.
January 2013 – Give one reason why excess air is used in the Ostwald Process. [1 mark]
Ensure the ammonia is used up
January 2013 – Ammonia reacts with nitric acid as shown in this equation.
NH3 + HNO3 –> NH4NO3
Deduce the type of reaction occuring
Acid-Base reaction / neutralisation
January 2013 – Chlorine can form molecules and ions that contain only chlorine, or that contain chlorine combined with another element.
Use your understanding of the electron pair repulsion theory to draw the shape of the AsCl3 molecule and the shape of the Cl3+ ion. Name the shape made by the atoms inthe AsCl3 molecule and the Cl3+ ion. [4 marks]
Shape of the AsCl3 molecule = pyramidal
Shape of the Cl3+ ion = Bent/V-shaped
January 2013 – Explain why the AsCl4+ ion has a bond angle of 109.5degrees. [2 marks]
Because AsCl4+ ion is tetrahedral. AsCl4+ has four bond pairs (3 covalent, 1 dative covalent) and zero lone pairs, giving it a tetrahedral shape. Tetrahedral shapes have bond angles of 109.5degrees. These 4 electron pairs repel eachother equally.
May 2013 – Define the term relative atomic mass. [2 marks]
The average mass of an element compared to the mass of 1/12th of carbon 12.
May 2013 – Identify which one of the isotopes of ………. is deflected the most in the magnetic field of a mass spectrometer. Give a reason for your answer. [2 marks]
Isotope: lowest isotope
Reason: It has the lowest value for m/z
May 2013 – X and Y are different elements. Explain why the chemical properties of 70X and 70Y are different. [1 mark]
They have different electron configurations
May 2013 – State the meaning of term structural isomers. [2 marks]
Molecules with the same molecular formula but different arrangement of skeleton.
May 2013 – Write an equation for the combustion of but-1-ene. [1 mark]
C4H8 +2CO2 –> 4C + 4H2O
May 2013 – State one hazard associated with the solid product formed in the previous question. [1 mark]
It produces smog/global dimming/makes people cough/choke (especially those with asthma).
May 2013 – Identify a catalyst that can be used in a catalytic converter. [1 mark]
May 2013 – In 2009 a new material called graphane was discovered. The diagram shows part of a model of the structure of graphane. Each carbon atom is bonded to three other carbon atoms and one hydrogen atom.
Deduce the type of crystal structure shown by graphane [1 mark]
State how to carbon atoms form a carbon-carbon bond in graphane. [1 mark]
Suggest why graphane does not conduct electricity [1 mark]
Deduce the empirical formula of graphane. [1 mark]
They covalently bond with eachother. Each carbon atom shares an electron with the neighbouring carbon atom.
No delocalised/free/mobile electrons
May 2013 – State the strongest type of intermolecular force between one molecule of ammonia and one molecule of water. [1 mark]
Hydrogen bonding / H-bonding
May 2013 – Phosphine (PH3) has a structure similar to ammonia.
In terms of intermolecular forces, suggest the main reason why phosphine is almost insoluble in water. [1 mark]
PH3 cannot hydrogen bond with water molecules as there isn’t a H attached to an O,N,F.
May 2013 – An aluminium chloride molecule reacts with a chloride ion to form the AlCl4- ion.
Name the type of bond formed in this reaction. Explain how this type of bond is fomed in the AlCl4- ion.
Dative Covalent / Co-ordinate Bond
The Cl- ion donates a pair of electrons to the Al atom in the AlCl3 molecule.
May 2013 – What is the shape of the TlBr5 2- ion
May 2013 – The elements in Period 2 show periodic trends.
Identify the Period 2 element, from carbon to fluorine, that has the largest atomic radius. Explain your answer.
Explanation: All the outermost electrons in these elements are in the same energy level/sub-shell and so all experience the same amount of shielding but carbon has the fewest number of protons and so its nuclear charge is the lowest and so the electrons aren’t drawn as closely to the nucleus.
May 2013 – State the general trend in first ionisation energies from carbon to neon. Deduce the element that deviates from this trend and explain why this element deviates from the trend. [4 marks]
General Trend: Increases from left to right.
Element that deviates: Oxygen
Explanation: There is repulsion between the pair of electrons in the 2p sub-shell.
May 2013 – Explain why the second ionisation energy of carbon is higher than the first ionisation energy of carbon. [1 mark]
The second electron is drawn closer to the nucleus when the previous electron is removed. There is a greater effective nucleus charge. OR More energy needed to remove an electron from a more positive ion/cation OR next electron is closer to the nucleus as it is more positive (relatively)
May 2013 – Deduce the element in Period 2, from lithium to neon, that has the highest second ionisation energy. [1 mark]
Zinc forms many different slats including zinc sulphate, zinc chloride and zinc fluoride.
People who have a zinc deficiency can take hydrated zinc sulphate (ZnSO4.xH2O) as a dietry supplement.
A student heated 4.38g of hydrated zonc sulphate and obtained 2.46g of anhydrous zinc sulphate.
Use these data to calculate the value of the integer x in ZnSO4.XH2O. Show your working.
Empirical formula = ZnSO4.xH2O therefore x=7
Zinc chloride can be prepared in the laboratory by the reaction between zinc oxide and hydrochloric acid.
The equation for the reaction is
ZnO + 2HCl –> ZnCl2 + H2O
A 0.0830 mol sample of pure zinc oxide was added to 100cm3 of 1.20moldm-3 hydrochloric acid.
Calculate the maximum mass of anhydrous zinc chloride that could be obtained from the products of this reaction.
Moles of HCl = concentration of HCl x volume of HCl
Moles of HCl = 1.20moldm-3 x 100/1000 dm3
Moles of HCl = 0.12
Moles of ZnCl2 = o.12 divided by 2
Moles of ZnCl2 = 0.06
Mass of ZnCl2 = moles x Mr
Mass of ZnCl2 = 0.06 moles x 136.4Mr
Mass of ZnCl2 = 8.184g
Zinc chloride can also be prepared in the laboratory by the reaction between zinc and hydrogen chloride gas.
Zn + 2HCl –> ZnCl2 + H2
An impure sample of zinc powder with a mass of 5.68g was reacted with hydrogen chloride gas until the reaction was complete. The zinc chloride produced had a mass of 10.7g.
Calculate the percentage purity of the zinc metal.
Give your answer to 3 significant figures.
– Moles of ZnCl2 = Mass of ZnCl2 divided by Mr of ZnCl2
– Moles of ZnCl2 = 10.7g divided by 136.4
– Moles of ZnCl2 = 0.07844564780058651026392961870832
– Moles of Zn = Moles of ZnCl2
– Moles of Zn = 0.07844564780058651026392961870832
– Mass of Zn = Moles of Zn x Mr of Zn
– Mass of Zn = 0.07844564780058651026392961870832 –moles x 65.4 Mr
– Mass of Zn = 5.13035190615835777126099700744868
– Percentage Purity =(5.1303519061583577712609970074 4868 divided by 5.68) x100
– Percentage Purity = 90.32309694%
– Percentage Purity = 90.3% to 3sf
May 2013 – Predict the type of crystal structure in solid zinc fluoride and explain why its melting point is high.
Crystal Structure: Giant ionic structure lattice
There are strong electrostatic forces of attraction between the positively charged zinc ions (cations) and the negatively charged fluoride ions (anions). A lot of (thermal) energy is needed to overcome these strong forces, hence why it has a high melting point.
May 2014 – A naturally occuring sample of the element boron has a relative atomic mass of 10.8. In this sample, boron exists as two isotopes, 10B and 11B.
Calculate the percentage abundance of 10B in this naturally occurring sample of boron. [2 marks]
10.8 = [(10×2)+(11×8)] divided by 10 therefore
Ratio = 8:2 , = 4:1
4+1=5, 100 divided by 5 = 20%
10B = 20%
May 2014 – State, in terms of fundamental particles, why the isotopes 10B and 11B have similar chemical reactions. [1 mark]
They have the same number of electrons and the same electron configuration.
May 2014 – Suggest a value for the third ionisation energy of boron. [1 mark]
May 2014 – Write an equation to show the process that occurs when the second ionisation energy of boron in measured. Include state symbol in your equation. [1 mark]
B+(g) –> B2+(g) + e-
May 2014 – Explain why the second ionisation energy of boron is higher than the first ionisation energy of boron. [1 mark]
The second electron comes from the 2s orbital which is closer the the nucleus and less shielded.
When heated, iron(III) nitrate (Mr = 241.8) is converted into iron(III) oxide, nitrogen dioxide and oxygen.
4Fe(NO3)3(s) –> 2Fe2O3(s) + 12NO2(g) + 3O2(g)
A 2.16g sample of iron(III) nitrate was completely onverted into the products shown.
Calcuate the amount, in moles, of iron(III) nitrate in the 2.16g sample. Give your answer to 3 significant figures. [1 mark]
Calculate the amount, in moles, of oxgen gas produced in this reaction. [1 mark]
Calculate the volume, in m3, of nitrogen dioxide gas at 293degreesC and 100kPa produced from 2.16g of iron(III) nitrate. The gas constant is R = 8.31JK-1mol-1. [4 marks]
Suggest a name for this type of reaction that iron(III) nitrate undergoes. [1 mark]
Suggest why hte iron(III) oxide obtained is pure. Assume a complete reaction. [1 mark]
– moles = mass divided by Mr
– moles = 2.16g divided by Mr of Fe(NO3)3
– moles = 2.16g divided by 241.8
– moles = 0.00893 to 3sf
– moles of O2 = moles of Fe(NO3)3 divided by 4 x3
– moles of O2 = 0.0066975
– moles of NO2 = moles of Fe(NO3)3 divided by 4 x12
– moles of NO2 = 0.0268 to 3sf
– PV = nRT therefore V = nRT divided by P
– V = (0.0268moles x 8.31JK-1mol-1 x 566K) divided by 100,000Pa
– V = 0.0012605272m3
– V = 0.00126m3 to 3sf
Other products are gases and can escape easily.
May 2014 – Nickel is a metal with a high melting point. State the block in the Periodic Table that contains nickel. [1 mark]
May 2014 – Explain, in terms of structure and bonding, why nickel has a high melting point. [2 marks]
There are strong electrostatic forces of attraction between the positively charged nickel ion (cations) and the negatively charged sea of delocalised electrons. A lot of (thermal/heat) energy is needed to overcome these strong forces/metallic bonds.
May 2014 – Explain why nickel is ductile (can be stretched into wires). [1 mark]
All the nickel ions are the same size and so the layers can slide over eachother.
May 2014 – Give the full electron configuration of the Ni2+ ion. [1 mark]
1s2 2s2 2p6 3s2 3p6 3d10
May 2014 – Balance the following equation to show hoe anhydrous nickel(II) chloride can be obtained from the hydrated alt using SOCl2.
Identify one substance that could react woth both gaseous products.
..NiCl2.6H2O(s) +..SOCl2(g) –> ..NiCl2(s) +..SO2(g) +..HCl(g)
1NiCl2.6H2O(s)+6SOCl2(g) –> 1NiCl2(s) +6SO2(g) +12HCl(g)
CaO / NaOH / NH3 / CaCO3
May 2014 – Ammonia gas readiky condenses to form a liquid when cooled.
Name the strongest attractive force between two ammonia molecules. [1 mark]
Hydrogen Bonding / H-Bonding
May 2014 – State how the bond between ammonia and boron trichloride is formed. [1 mark]
It is a dative covalent bond. The nitrogen donates a pair of electrons to the boron.
May 2014 – Give the meaning of the term electronegativity [2 marks]
The power of an atom to attract a pair of electron (electron density) towards itself, in a covalent bond.
Table of electronegativity values.
H 2.1 Li 1.0 B 2.0 C 2.5 O 3.5 F 4.0
Suggest the formula of an ionic compound that is formed by the chemical combination of two different elements from the table. [1 mark]
Suggest the formula of the compound that has the least polar bond and is formed by chemical combination of two of the elements from the table. [1 mark]
May 2014 – Some oil-fired heaters use paraffin as a fuel. One of the compounds in paraffin is the straight-chain alkane, dodecane (C12H26).
Give the name of the substance from which paraffin is obtained. State the name of the process used to obtain paraffin from this substance. [2 marks]
Substance: Crude Oil
Process: Fractional Distillation
May 2014 – The combustion of dodecane produces several products. Write an equation for the incomplete combustion of dodecane to produce gaseous products only. [1 mark]
C12H26 + 12.5O2 –> 12CO + 13H2O
May 2014 – Oxides of nitrogen are also produced during the combustion of paraffin in air.
Explain how thee oxides of nitrogen are formed. [2 marks]
The high temperatures and spark present in the engine is enough to split the NtriplebondN bond and separate the N2 into 2N. These then react with oxygen in the air to produce nitrous oxides.
May 2014 – Write an equation to show how nitrogen monoxide in the air is converted into nitrogen dioxide. [1 mark]
2NO + O2 –> 2NO2
May 2014 – Nitric acid (HNO3) contributes to acidity in rainwater.
Deduce an equation to show how nitrogen dioxide reacts with oxygen and water to form nitric acid. [1 mark]
2NO2 + H2O + 1.5O2 –> 2HNO3
May 2014 – Give the general formula for the homologous series that contains ………..(alkane). [1 mark]
May 2014 – Write an equation for the cracking of one molecule of dodecane into equal amounts of two different molecules each containing the same number of carbon atoms. State the empirical formula of the straight-chain alkane that formed. Name the catalyst used in this reaction. [3 marks]
Equation: C12H26 –> C6H12 + C6H14
Empirical Formula of the alkane: C3H7
Catalyst: Zeolite Catalyst / alumino sillicate
May 2014 – Explain why the melting point of dodecane is higher than the melting point of the straight-chain alkane produced by cracking dodecane.
Dodecane is a bigger molecule than hexane and so has more electrons meaning that the Van der Waals forces between the dodecane molecules are greater than those between the hexane molecules.
May 2014 – Deduce the formula of a substance that could be reacted with dodecane to produce 1-chlorododecane and hydrogen chloride only. [1 mark]
Calcium phosphate reacts with aqueous nitric acid to produce phosphoric acid and calcium nitrate as shown in the equation.
Ca3(PO4)2 + 6HNO3 –> 2H3PO4 + 3Ca(NO3)2
A 7.26g sample of calcium phosphate reacted completely when added to an excess of aqueous nitric acid to form 38.0cm3 of solution.
Calculate the concentration, in moldm-3, of phosphoric acid in this solution. Give your answer to 3 significant figures. [5 marks]
Calculate the percentage atom economy for the formation of calcium nitrate in this reaction. Give your answer to 1 decimal place. [2 marks]
– moles of Ca3(PO4)2 = mass of Ca(PO4)2 divided by Mr of Ca(PO4)2
– moles of Ca3(PO4)2 = 7.26g divided by 310.3 Mr
– moles of Ca3(PO4)2 = 0.02339671286
– moles of H3PO4 = moles of Ca3(PO4)2 x2
– moles of H3PO4 = 0.02339671286 x2
– moles of H3PO4 = 0.04679342572
– concentration of H3PO4 = moles of H3PO4 divided by the volume of H3PO4
– concentration of H3PO4 = 0.04679342572 divided by 0.038dm3
– concentration of H3PO4 = 1.23140594 moldm-3
– concentration of H3PO4 = 1.23moldm-3 to 3sf
– Percentage Atom Economy = (Mr of desired product divided by Mr of the reactants) x100
– Percentage Atom Economy = (492.3 divided by 688.3) x100
– Percentage Atom Economy = 71.52404475%
– Percentage Atom Economy = 71.5% to 1dp
May 2014 – Write an equation to show the reaction between calcium hydroxide and phosphoric acid to produce calcium phosphate and water. [1 mark]
3Ca(OH)2 + 2H3PO4 –> Ca3(PO4)2 + 6H2O
Caclium dihydrogenphosphate can be represented by the formula Ca(H2PO4)x, where x in an interger.
A 9.76g sample of calcium dihydrogenphosphate contains 0.17g hydrogen, 2.59g of phosphorous and 5.33g of oxygen.
Calculate the empirical formula and hence the value of x.
Show your working. [4 marks]
– Ca mass = 1.67g
– Ca Mr = 40.1
– Ca moles = mass divided by Mr, 1.67g divided by 40.1
– Ca moles = 0.0416458829
– H mass = 0.17g
– H Mr = 1.0
– H moles = mass divided by Mr, 0.17g divided by 1.0
– H moles = 0.17
– P mass = 2.59g
– P Mr = 31.0
– P moles = mass divided by Mr, 2.59g divided by 31.0
– P moles = 0.08354838709677419
– O mass = 5.33g
– O Mr = 16.0
– O moles = mass divided by Mr, 5.33g divided by 16.0
– O moles = 0.333125
– Ca = 0.0416458829 divided by 0.0416458829 = 1
– H = 0.17 divided by 0.0416458829 = 4.082035928
– P = 0.08354838709677419 divided by 0.0416458829 = 2.00616187
– O = 0.333125 divided by 0.0416458829 = 7.99898952
– Ca = 1
– H = 4
– P = 2
– O = 8
Molecular formula = Ca(H4P2O8) which, the above formula = Ca(H2PO4), therefore x=2
May 2014 – What is the shape and bond angle of TlBr3 2-? [4 marks]
May 2014 – Thallium(I) bromide (TlBr) is a crystalline solid with a melting point of 480degreesC.
Suggest the type of bonding present in thallium(I) bromide and state why the melting point is high. [3 marks]
Thallium(I) bromide posses ionic bonding. It has a high melting point because there are strong electrostatic forces of attraction between the positively charged thallium ions (cations) and the negatively charged bromine ions (anions). A lot of (thermal/heat) energy is needed to overcome these strong ionic/electrostatic forces.
May 2014 – Write an equation to shoe the formation of thallium(I) bromide from its elements. [1 mark]
2Tl(s) + Br2(g) –> 2TlBr(s)
May 2015 – State the element in Period 3 that has the highest melting point. Explain your answer. [3 marks]
Explanation: It is held together by many strong covalent bonds what need a lot of energy to be broken/overcome.
May 2015 – State the element in Period 3 that has the highest forst ionisation energy. Explain your answer. [3 marks]
Explanation: It has the largest number of protons in its nucleus/has the largest nuclear charge of any of the elements in Period 3 but has the same amount of shielding ans the electrons are in the same energy levels. This means that the electrons are more attracted to the nucleus and thus held more tightly.
May 2015 – Suggest the element in Period 3 that has the highest electronegativity value. [1 mark]
May 2015 – Name the shape of CCl2. [1 marl]
Bent / v-shape
May 2015 – Write an equation to show the formation of one mole of ClF3 from its elements. [1 mark]
0.5Cl2(g) + 1.5F2(g) –> ClF3(g)
May 2015 – Tellurium is the element with atomic number of 52. Using information from the Periodic table, complete the electron configuration of tellurium. [1 mark]
[Kr] 5s2 4d10 5p4
May 2015 – Suggest what might cause the relative atomic mass of this sample to be different from the relative atomic mass given in the Periodic Table. [1 mark]
Other isotopes present/some isotopes absent/different abundances of isotope
May 2015 – Write an equation for the reaction that occurs when a tellurium ion hits the detector. [1 mark]
Te+(g) + e- –> Te(g)
May 2015 – State the m/z value of the ions that produce the biggest current of the detector when the spectrum in …….. was recorded. Give a reason for your answer. [2 marks]
m/z value: whichever is the most abundant
Reason: …….. is the most abundant ans so generates a greater current.
May 2015 – The mass spectrum of tellurim (128) also has a small peak at m/z 64. Explain the existence of this peak. [2 marks]
Some of the 128Te atoms get 2 electrons knocked off to form Te2+ ions. This makes the m/z value 128/2 which is 64.
May 2015 – Predict whether the atomic radius of 124Te is larger than, smaller than or the same as the atomic radius of 130Te. Explain your answer. [2 marks]
Atomic radius of 124Te compared to 130Te: The same.
Explanation: Have the same number of protons in the nucleus and so have the same nuclear charge. Also have the same number of electrons and so the shielding is the same/the electron configuration is the same.
May 2015 – Silicon dioxide (SiO2) has a crystal structure similar to diamond. Give the name of the type of crystal structure shown by silicon dioxide. [1 mark]
Macromolecular / Giant Covalent / Giant Molecule
May 2015 – Suggest why silicon dioxide does not conduct electricity when molten.
No delocalised electrons to flow through the macromolecular crystal and conduct electricity.
No free ions/no free charged particles.
May 2015 – Silicon dioxide reacts woth hydrofluoric acid (HF) to produce hexafluorosilicic acid (H2SiF6) and one other substance. Write an equation for this reaction. [1 mark]
SiO2(s) + 6HF(aq) –> H2SiF6(aq) + 2H2O(l)
May 2015 – A sample of hydrated nickel sulfate (NiSO4.xH2O) with a mass of 2.287g was heated to remove all water of crystallisation. The solid remaining had a mass of 1.344g. Calculate the value of the interger x. Show your working.
– Mass: 1.344g
– Mr: 154.8
– Moles = mass/Mr, 1.344g/154.8
– Moles = 0.00868217054263565891472
– Mass: 2.287g-1.344g = 0.943g
– Mr: 18.0
– Moles = 0.943g/18.0
– Moles = 0.05238
– NiSO4 = 0.00868217054263565891472 divided by 0.05238 = 6.034077380952
– xH2O = 0.05238 divided by 0.05238 = 1
– NiSO4 = 6
– xH2O = 1
May 2015 = Suggest how a student doing this experiment could check that all the water had been removed. [2 marks]
Re-heat the sample and record the mass. Keep reheating the sample until the mass no longer changes, this ensures that all the water is removed. Alternatively you could put the contents into a vacuum filter.
May 2015 – State the strongest type of intermolecular force that occurs between molecules of hydrogen peroxide and water. [1 mark]
Hydrogen Bonding / H-Bonding
May 2015 – Explain, in terms of electronegativity, why the boiling point og H2S2 is lower than H2O2. [2 marks]
The elecgtronegativity difference between S and H is lower than that between O and H. This means that no hydrogen bonds can form between H2S2 molecules (H bonds can only from when the molecule contains a H bonded to a N, O or F). This means that the only intermolecular forces between the H2S2 molecule is Van der Walls which are a lot weaker than the hydrogen bonding in H2O2, and so less energy is needed to overcome them.
May 2015 – Give meaning of the terms saturated and hydrocarbon. [2 marks]
Saturated – Only contains single bonds (no double bonds)
Hydrocarbon – Only contains carbon and hydrogen
May 2015 – If the boiler for a central heating system is fauklty, a poisonous gas may be produced during the combustion of C16H34.
Write an equation for the reaction that forms this poisonous gas and one other product only. [1 mark]
C16H34 + 16.5O2 –> 16CO + 17H2O
May 2015 – Explain why the sulphur compounds found in crude oil should be removed from the fractions before they are used for central heating fuel. [2 marks]
When the sulphur compounds are combusted they form SO2 which contributes to acid rain and also causes asthmatic people to choke.
May 2015 – A hydrocarbon C16H34 can be cracked to form C8H18, ethene and propene. Write an equation to show this cracking reaction. [1 mark]
C16H34 –> C8H18 + C2H4 + 2C3H6
May 2015 – Suggest one important substance manufactured on a large scale from propene. [1 mark]
polypropene / propan-1-ol / propan-2-ol / propane-1-diol / propane-2-diol / isopropanol / propanone / propanal
May 2015 – Some airbags in cars contain sodium azide (NaN3). Sodium azide is made by reacting dinitrogen monoxide gas with sodium amide (NaNH2) as shown by the equation.
2NaNH2 + N2O –> NaH3 + NaOH + NH3
Calculate the mass of sodium amide needed to obtain 550g of sodium azide, assuming there is a 95.0% yield of sodium azide. Give your answer to 3 significant figures. [5 marks]
550g = 95% therefore 550g/95 x100 = 578.9g (100%)
Moles of NaN3 = mass/Mr
Moles of NaN3 = 578.9g/65
Moles of NaN3 = 8.90615384615
Moles of NaNH2 = Moles of NaN3 x2
Moles of NaNH2 = 8.90615384615 x2
Moles of NaNH2 = 17.8123076923
Mass of NaNH2 = moles of NaNH2 x Mr of NaNH2
Mass of NaNH2 = 17.8123076923 x 39
Mass of NaNH2 = 694.68g
Mass of NaNH2 = 695g to 3sf
If a car is involved in a serious collision, the sodium azide decomposes to from sodium and nitrogen as shown in the equation.
2NaN3(s) –> 2Na(s) + 3N2(g)
The nitrogen produced then inflates the airbag to a volume of 7.50×10-2m3 at a pressure of 150kPa and temperature of 35degreesC.
Calculate the minimum mass of sodium azide that must decompose. (The gas constant R = 8.31JK-1mol-1). [6 marks]
Moles of N2 = n=(PV/RT)
Moles of N2 = (150,000Pax7.50×10-2m3)/(8.31JK-1mol-1x308K)
Moles of N2 = 4.395424071
Moles of NaN3 = moles of N2 divided by 3 x2
Moles of NaN3 = 4.395424071/3 x2
Moles of NaN3 = 2.930282714
Mass of NaN3 = moles of NaN3 x Mr of NaN3
Mass of NaN3 = 2.930282714 x 65
Mass of NaN3 = 190.4683764g
Mass of NaN3 = 190.5g to 3sf
Sodium azide is toxic. It can be destroyed by reaction with an acidified solution of nitrous acid (HNO2) as shown in the equation.
2NaN3+2HNO2+2HCl –> 3N2+2NO+2NaCl+2H2O
A 500cm3 volume of the nitrous acid solution was used to destroy completely 150g of the sodium azide.
Calculate the concentration, in moldm-3, of the nitrous acid used. [3 marks]
Moles of NaN3 = Mass of NaN3 / Mr of NaN3
Moles of NaN3 = 150g / 65
Moles of NaN3 = 2.307692
Moles of HNO2 = Moles of NaN3
Moles of HNO2 = 2.307692
Concentration of HNO2 = moles / volume
Concentration of HNO2 = 2.307692 / 0.5dm3
Concentration of HNO2 = 4.615384 moldm-3
Concentration of HNO2 = 4.62 moldm-3 to 3sf
May 2015 – Nitrous acid decomposes on heating. Balance the following equation for this reaction.
…HNO2 –> …HNO3 + …NO + …H2O [1 mark]
3HNO2 –> 1HNO3 + 2NO + 1H2O
May 2015 – Sodium azide has a high melting point. Predict the type of bonding in a crystal of sodium azide. Suggest why its melting point is high. [3 marks]
Type of Bonding: Ionic Bonding
Reason for high melting point: There are stong electrostatic forces of attraction between the positevly charged sodium ions (cations) and the negatively charged nitride ions (anions). These strong electrostatic forces need a lot of (thermal/heat) energy to be overcome/broken.
May 2015 – The azide ion has the formula N3 1-. Give the formula of a molecule that has the same number of electrons as the azide ion. [1 mark]
May 2015 – Which is the correct formula for magnesium azide?
January 2010 – Complete the electron configuration of the Mg+ ion. [1 mark]
1s2 2s2 2p6 3s1
January 2010 – State the meaning of the term first ionisation energy. [2 marks]
The enthalpy change for the removal of one mole of electrons from one mole of atoms of the element in the gas phase. X(g) –> X+(g) + e-
January 2010 – Write an equation, including state symbols, to show the reaction that occurs when the second ionisation energy of magnesium is measured. [1 mark]
Mg+(g) –> Mg2+(g) + e-
January 2010 – Explain why the second ionisation energy of magnesium is greater than the first ionisation energy of magnesium. [1 mark]
Electron being removed from a positive ion (therefore need more energy) / electron being removed is closer the the nucleus / Mg+ smaller then Mg / Mg+ more positive than Mg.
January 2010 – Suggest a value for the 3rd ionisation energy of magnesium. [1 mark]
5000 – 9000 (approx 8000)
January 2010 – State and explain the general trend in the first ionisation energies of the Period 3 elements sodium to chlorine. [3 marks]
Explanation: There is a general increase in ionisation energy across Period 3 (sodium to argon). Across the period from Na (11 protons) to Ar (18 protons) the nuclear charge in each element increases so the electrons are attracted more strongly to the nucleus and it takes more energy to remove one from the atom.
January 2010 – State how the element sulphur deviates from the general trend in first ionisation energies across Period 3. Explain your answer. [3 marks]
How sulphur deviated from the trend: It decreases
Explanation: In sulphur the 4th 3p electron is paired. There is some repulsion between paired electrons in the same sub-level, which increases there energy. Therefore it is easier to removes one of these paired 3p electrons from sulphur that it is to remove an unpaired 3p electron from phosphorous.
January 2010 – A general trend exists in the first ionisation energies of the Period 2 elements lithium to fluorine. Identify one element which deviates from this general trend. [1 mark]
Ammonium sulphate reacts with sodium hydroxide to form ammonia, sodium sulphate and water as shown in the equation below.
A 3.14g sample of ammonium sulphate reacted completely with 39.0cm3 of a sodium hydroxide solution.
Calculate the amount, in moles, of (NH4)2SO4 in 3.14g of ammonium sulphate. [2 marks]
Hence calculate the amount, in moles, of sodium hydroxide which reacted. [1 mark]
Calculate the concentration, in moldm-3, of the sodium hydroxide which reacted. [1 mark]
Calculate the percentage atom economy for the production of ammonia in the reaction between ammonium sulphate and sodium hydroxide. [2 marks]
Ammonia is manufactured by the Haber Process.
N2 + 3H2 –> 2NH3
Calculate the percentage atom economy for the production of ammonia in this process. [1 mark]
A sample of ammonia gas occupied a volume of 1.53×10-2m3 at 37degreesC and a pressure at 100kPa. (The gas constant R = 8.31JK-1mol-1). Calculate the amount, in moles, of ammonia in this sample. [3 marks]
Glauber’s salt is a form of hydrated sodium sulphate that contains 44.1% by mass of sodium sulphate. Hydrated sodium sulphate can be represented by the formula Na2SO4.xH2O where x is an integer. Calculate the value of x. [3 marks]
Moles of (NH4)2SO4 = Mass / Mr
Moles of (NH4)2SO4 = 3.14g / 132.1
Moles of (NH4)2SO4 = 0.02376987130961392884178
Moles of (NH4)2SO4 = 0.0238 to 3sf
Moles of NaOH = Moles of (NH4)2SO4 x2
Moles of NaOH = 0.023769871309613928841786525
Moles of NaOH = 0.047539742619227857683573050
Moles of NaOH = 0.0475 to 3sf
Concentration of NaOH = Moles / Volume
Concentration of NaOH = 0.0475397426192278576835
Concentration of NaOH = 1.209662662 moldm-3
Concentration of NaOH = 1.25 moldm-3 to 3sf
Percentage Atom Economy = (Mr of desired product / Mr of reactants) x100
Percentage Atom Economy = (34/203.1) x100
Percentage Atom Economy = 16.74052191
Percentage Atom Economy = 16.7% to 3sf
PV = nRT therefore n = PV/RT
n = (100,000Pa x 0.0153m3) / (8.31JK-1mol-1 x 310K)
n = 0.5939210434
n = 0.594 to 3sf
Na2SO4: Mass = 44.1, Na2SO4: Mr = 142.1
Na2SO4 Moles: = 44.1/142.1 Na2SO4 Moles = 0.3103
.xH2O: Mass = 55.9 .xH2O: Mr = 18.0
.xH2O: Moles = 55.9/18.0 .xH2O = 3.105
Ratio: Na2SO4 = 0.3103 / 0.3103 = 1
Ratio: .xH2O = 3.105 / 0.3103 = 10.00823578
Rounded Ratio: Na2SO4:.xH2O = 1:10 therefore x=10
January 2010 – State the strongest type of intermolecular force in water and hydrogen sulphide (H2S). [2 marks]
Water: Hydrogen Bonding
Hydrogen Sulphide: Van der Walls / dipole-dipole / London forces / temporarily induced dipole / dispersion forces
January 2010 – Explain why the boiling point of water is much higher than the boiling pouint of hydrogen sulphide. [1 mark]
Water can hydrogen bond whereas H2S only posses Van der Waals which are much weaker and need less (thermal/heat) energy to be overcome.
January 2010 – Explain why the boiling points increase from H2S to H2Te. [2 marks]
The size of the molecules increases, so the number of electrons increases, and so the Van der Waals intermolecular forces become stronger.
January 2010 – When H+ ions react with H2O molecules, H3O+ ions are formed. Name the type of bond formed when H+ ions react with H2O molecules. Explain how this type of bond is formed in the H3O+ ion. [2 marks]
Type of Bond: Dative Covalent Bond / Co-ordinate Bond
Explanation: The oxygen atom in the H2O molecule donates a pair of electrons to the H+ ion and they share them to form a covalent bond.
January 2010 – Sodium sulphide (Na2S) has a melting point of 1223K. Predict the type of bonding in sodium sulphide and explain why its melting point is high. [3 marks]
Type of Bonding: Ionic Bonding
Explanation: There are strong electrostatic forces of attraction between the positively charged Na+ ions (cations) and the negatively charged S2- ions (anions). A lot of (thermal/heat) energy is needed to overcome/break these strong ionic bonds / electrostatic forces.
January 2010 – State the meaning of the term saturated and the term hydrocarbon. [2 marks]
Saturated: Only contains single bonds (no double bonds / no C=C bonds).
Hydrocarbon: Only contains the elements carbon and hydrogen.
January 2010 – Give the general formula for the alkanes. [1 mark]
January 2010 – Write an equation for the complete combustion of pentane. [1 mark]
C5H12 + 8O2 –> 5CO2 + 6H2O
January 2010 – State how the products of this reaction may affect the environment. [1 mark]
CO2(g) and H2O(g) are greenhouse gases that contribute to global warming.
January 2010 – Give the name of a solid pollutant which may form when pentane burns incompletely in air. [1 mark]
Carbon / C(s) / Soot / Carbon Particulate
January 2010 – One molecule of C9H20 can be cracked to form one molecule of pentane and one other product. Write an equation for this cracking reaction. [1 mark]
C9H20 –> C5H12 + C4H4
C9H20 –> C5H12 + 2C2H4
January 2010 – Suggest a type of compound that can be manufactured from the other products of this cracking reaction. [1 mark]
Poly(alkenes)/plastics / alcohols / aldehydes / ketones
January 2010 – State why a high temperature is needed for cracking reactions to occur. [1 mark]
A lot of energy is needed to break the strong covalent bond between to carbon atoms.
January 2010 – Define the term relative atomic mass of an element. [2 marks]
The average mass of one atom of one element compared to 1/12th the mass of one atom of 12C.
January 2010 – Element X has a relative atomic mass of 47.9. Identify the block in the Periodic Table to which element X belongs and give the electron configuration of an atom of element X. Calculate the number of neutrons in the isotope of X which has a mass number 49. [3 marks]
[Ar] 3d10 4s2
27 (49-22 = 27)
January 2010 – State how vapourised atoms of Z are converted into Z+ ions in a mass spectrometer. State and explain which of the Z+ ions formed from the isotopes of Z will be deflected the most in a mass spectrometer. [4 marks]
– First of all the sample is vapourised.
– The sample is then bombarded with high energy electrons via an electron gun.
– This knocks an electron of the sample Z forming Z+ ion. Equation: Z(g) –> Z+(g) + e-
– Whichever isotope has the smallest m/z value will the deflected the most, as it has the smallest m/z value.
January 2010 – Explain breifly how the relative abundance of an ion is measured in a mass spectrometer. [2 marks]
– Ions hit the detector and cause current/ions accept electrons/cause electron flow
– Bigger current = more of that isotope / current proportional to abundance
January 2010 – Shape and Bond Angles of ClF2+ and AsF5. [4 marks]
Bond Angle: 104.5 degrees
Shape: Trigonal Bipyramidal
Bond Angle: 120 degrees and 90 degrees.