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A LEVEL - Chemistry (Inorganic) > Topic 5 - Formulae, Equations & Amounts Of Substances > Flashcards

Topic 5 - Formulae, Equations & Amounts Of Substances Flashcards

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1
Q

What is the unit for the amount of a substance?

A

Mole

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2
Q

What is the symbol for the number of moles?

A

n

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3
Q

What is the Avogadro constant?

A
  • 6.02 x 10²³
  • It is the number of particles in a mole

(NOTE: The value is given in the exam!)

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4
Q

What is molar mass?

A

The mass per mole of something.

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5
Q

What is the definition of molar mass in terms of Mr?

A

It is the number of particles which weigh the same as the relative molecular mass, Mr.

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6
Q

What are the units for Mr?

A

g/mol

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7
Q

What is the formula for number of moles?

A

Moles = Mass / Mr

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8
Q

What are the two units for the concentration of a solution?

A
  • mol/dm³

* g/dm³

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9
Q

What is the equation for the concentration of a solution relative to the number of moles?

A

Concentration = Moles / Volume

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10
Q

How can you convert from cm³ to dm³?

A

Divide by 1000.

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11
Q

What is the empirical formula?

A

A formula with the smallest whole number ratio of atoms of each element.

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12
Q

What is the molecular formula?

A

A formula with the actual number of atoms of each element.

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13
Q

When a hydrocarbon is burnt in excess oxygen, 4.4g of CO₂ and 1.8g of water are made. What is the empirical formula of the hydrocarbon?

A
  • Moles of CO₂ = 4.4 / (12 + 2 x 16) = 0.10 moles
  • Therefore, there are 0.10 moles of C
  • Moles of H₂O = 1.8 / (2 x 10 + 16) = 0.10 moles
  • Therefore, there are 0.20 moles of H
  • Ratio of C:H = 1:2
  • So the empirical formula = CH₂
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14
Q

A compound is found to have percentage composition 56.5% potassium, 8.70% carbon and 34.8% oxygen by mass. Calculate its empirical formula.

A 
In 100g of compound:
• Moles of K = 56.5 / 39.1 = 1.45 moles
• Moles of C = 8.70 / 12.0 = 0.725 moles
• Moles of O = 34.8 / 16.0 = 2.18 moles
• Ratio of K:C:O = 2:1:3
• So the empirical formula = K₂CO₃
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15
Q

What is the percentage composition of H in CH₄?

A

(4 x 1.0) / (12.0 + 4 x 1.0) x 100% = 25%

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16
Q

When 4.6g of an alcohol, with molar mass 46g/mol, is burnt in excess oxygen, it produces 8.8g of carbon dioxide and 5.4g of water. Calculate the empirical formula for the alcohol and then its molecular formula.

A
  • Moles of CO₂ = 8.8 / 44 = 0.20 moles
  • Therefore, moles of C = 0.20 moles
  • Moles of H₂O = 5.4 / 18 = 0.30 moles
  • Mass of C = 0.20 x 12.0 = 2.4g
  • Mass of H = 0.60 x 1.0 = 0.60g
  • Mass of O = 4.6 - 2.4 - 0.60 = 1.6g
  • Number of moles of O = 1.6 / 16.0 = 0.10 moles
  • Ratio of C:H:O = 2:6:1
  • Therefore, empirical formula = C₂H₆O
  • Mass of empirical formula = 46.0g
  • In this case, the mass of the empirical formula equals the molecular mass, the the two formulae are the same.
  • Molecular formula = C₂H₆O
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17
Q

Remember to practise balancing equations.

A

Pg 58 of revision guide

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18
Q

What do ionic equations show?

A

Only the resting particles (without spectator ions).

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19
Q

Should charge balance in ionic equations?

A

Yes

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20
Q

Remember to practise writing ionic equations.

A

Pg 58 of revision guide

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21
Q

What is reaction stoichiometry?

A

The ratios of reactants to products in a reaction (i.e. how many moles of product are formed from a certain number of moles of reactants).

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22
Q

What are the 4 state symbols?

A
  • Solid (s)
  • Liquid (l)
  • Gas (g)
  • Aqueous solution (aq)
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23
Q

What happens in a displacement reaction?

A

A more reactive element replaces another in a compound.

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24
Q

What is produced in a reaction of an acid with a base?

A
  • Salt

* Water

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25
Q

What is formed in a precipitation reaction?

A

A solid

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26
Q

Which produced in a reaction of an acid with a carbonate?

A
  • Salt
  • Water
  • Carbon dioxide
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27
Q

What is molar gas volume?

A

The volume that one mole of a gas occupies at a certain temperature and pressure.

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28
Q

What are the units of molar gas volume?

A

dm³/mol

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29
Q

What can be said about the volume of any gas at the same temperature and pressure?

A

It is the same for any gas.

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30
Q

What is the molar gas volume at RTP (20°C and 101.3kPa)?

A

24 dm³/mol

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31
Q

What is the molar gas volume at STP (0°C and 101.3kPa)?

A

22.4 dm³/mol

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32
Q

What is the formula for the number of moles of a gas relative to the volume?

A

Moles = Volume / Molar gas volume

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33
Q

What are the two methods of finding the volume of gas evolved in a reaction?

A
  • Gas syringe

* Displacing water from an upturned measuring cycling we

34
Q

Explain how you could measure the molar volume of CO₂ in dm³ at room temperature using this reaction:
Na₂CO₃ + 2HCl -> NaCl + H₂O + CO₂

A

1) Measure out a set volume of HCl into a conical flask connected to a gas syringe
2) Add a known mass of sodium carbonate to the comical flask, replace the bung and allow the reaction to go to completion.
3) Record the volume of CO₂ collected in the gas syringe.
4) Repeat this with various masses of Na₂CO₃ each time.
5) Draw a graph of volume against mass of Na₂CO₃. Draw the line of best fit.
6) Read off the volume for a sensible mass of Na₂CO₃ (e.g. 0.20g of Na₂CO₃ produces 45cm³ of CO₂)
7) Do the calculations:
• 1 mole of Na₂CO₃ produces 1 mole of CO₂
• Moles of Na₂CO₃ = 0.20 / 106 = 0.00188… moles
• Therefore, molar volume of CO₂ = 0.045 / 0.00188… = 23.85 dm³
• This is almost 24dm³

35
Q

How much gas is produced when 15g of sodium is reacted with excess water at RTP?

2Na + 2H₂O -> 2NaOH + H₂

A
  • Moles of Na = 15 / 23.0 = 0.652… moles
  • Moles of H₂ = 0.326… moles
  • Volume of H₂ = 0.326 x 24 = 7.8 dm³ (to 2s.f.)
36
Q

Calculate the total volume of gas produced when 8.25 dm³ of dinitrogen pentoxide decomposes:

2N₂O₅(g) -> O₂(g) + 4NO₂(g)

A
  • 2 moles of N₂O₅ produce 1 mole of O₂ and 4 moles of NO₂, which is 5 moles in total
  • So 8.25 dm³ of N₂O₅ decomposes to produce 8.25 x (5/2) = 20.6 dm³
37
Q

What is the ideal gas equation?

A

pV = nRT

Where:
• p = Pressure (Pa)
• V = Volume (m³)
• n = No. of moles
• R = Gas constant (= 8.31 J/K/mol)
• T = Temperature (K)
38
Q

In the ideal gas equation, what are the units for volume?

A

m³

39
Q

In the ideal gas equation, what is the gas constant?

A

8.31 J/K/mol

40
Q

In the ideal gas equation, what are the units for temperature?

A

K

41
Q

What is the equation relating K and °C?

A

K = °C + 273

42
Q

At a temperature of 60°C and a pressure of 250 kPa, a gaseous hydrocarbon occupied a volume of 1100cm³ and had a mass of 1.60g. Find the molecular formula of the hydrocarbon.

A
  • n = pV / RT = (250 x 10³ x 1.1 x 10⁻³) / (8.31 x 333) = 0.0993… moles
  • If 0.0993… moles is 1.60g, then 1 mole = 1.60 / 0.0993… = 16.1 g
  • So the molar mass is 16 g/mol
  • So this must be methane, CH₄.
43
Q

Describe how you can use the ideal gas equation to work out the molar mass of an unknown, volatile liquid.

A
  • Put a known mass of the liquid in a flask, then attach a sealed gas syringe.
  • Gently warm the apparatus in a water bath until the liquid completely evaporates.
  • Record the volume of the gas in the syringe and the temperature of the water bath.
  • Use the ideal gas equation to work out how many moles of the liquid were in your sample.
  • Molar mass = Mass / Moles to work out the molar mass.
44
Q

What are hazards and risk?

A
  • Hazard -> Something that has the potential to cause harm or damage
  • Risk -> The probability of someone being harmed if they are exposed to the given hazard
45
Q

What is a standard solution?

A

Any solution that you know the concentration of.

46
Q

Describe how to make 250 cm³ of a solution of benzoin acid (C₆H₅COOH) with a concentration of about 0.200 mol/dm³.

A

1) Work out the moles of dilute needed:
• Moles = Conc x Vol = 0.200 x 250/1000 = 0.0500 mol
2) Work out how many grams of solute is needed:
• Mass = Moles x Molar mass = 0.0500 x 122.0 = 6.10g
3) Carefully weigh out this mass of solute using a balance with a precision of at least 2d.p. First weigh the weighing vessel, note the weight, then add the correct mass.
4) Add the solid acid to a beaker containing about 100 cm³ of distilled water and stir until all the solute has dissolved.
5) Reweigh the weighing vessel, and use this value along with the combined mass of the vessel and the acid to calculate the exact mass of acid that has been added to the beaker.
6) Use this exact mass to calculate what the concentration of your standard solution will be (Conc = Moles / Vol).
7) Tip the solution into a volumetric flask, using a funnel.
8) Rinse the beaker and the stirring rod with distilled water and add that to the flask too.
9) Now, top the flask up to the correct volume (250 cm³) with more distilled water. Make sure the bottom of the meniscus reaches the line.
10) Stopper the bottle and turn it upside down a few times to make sure it’s all mixed.

47
Q

When weighing a solid, how can you ensure the highest accuracy?

A
  • Weigh the empty vessel
  • Weight the vessel and the solid together
  • Empty the vessel and reweigh it
  • Calculate the actual mass used by subtracting the final reading from the second one
48
Q

When filling water up to a line, what point on the water needs to reach the line?

A

The bottom of the meniscus has to reach the line.

49
Q

Describe how to do an acid-base titration.

A

1) Measure out some alkali of unknown concentration using a pipette and put it in a flask along with some indicator
2) Rinse the burette with some of your standard solution of acid. Then fill it with standard solution.
3) First, do a rough titration to get an idea where the end point is. Do this by reading how much alkali is in the burette and then adding the acid to the alkali, giving the flask a regular swirl. Stop when there is a permanent colour change. Record the final reading from the burette.
4) Now, do an accurate titration by running the acid to within 2cm³ of the endpoint. Then add the acid dropwise.
5) Work out the amount of acid used to neutralise the alkali (final reading - initial reading). This is the titre.
6) Repeat the titration a few times until you get concordant results, within 0.1cm³ of each other. Calculate a mean titre using just these results.

50
Q

What is the threshold for concordant results in a titration?

A

Within 0.1 cm³ of each other.

51
Q

What are the two main indicators used in acid-base titrations?

A
  • Methyl orange

* Phenolphthalein

52
Q

Describe the colour change in a titration with methyl orange when adding acid to alkali.

A

From yellow to red.

53
Q

Describe the colour change in a titration with phenolphthalein when adding acid to alkali.

A

From red to colourless.

54
Q

25.0 cm³ of 0.500 mol/dm³ HCl was used to neutralise 35.0 cm³ of NaOH solution. Calculate the concentration of the sodium hydroxide solution.

A

First, write a balanced equation and decide what you need to know:
• HCl + NaOH -> NaCl + H₂O
Now work out how many moles of HCl you have:
• Moles of HCl = 0.500 x (25/1000) = 0.0125 moles
• Therefore, moles of NaOH = 0.0125 moles
• Conc of NaOH = 0.0125 / (35 x 10⁻³) = 0.360 mol/dm³

55
Q

A student carried out an experiment to find the concentration of a solution of HCl. He first dissolved 0.987 g of NaOH in 250cm³ of distilled water to make a standard solution. He then titrated this standard solution against 15.0 cm³ of the solution of unknown concentration. Given that the mean titre of NaOH required to neutralise this volume of HCl solution was 21.7 cm³, calculate the concentration of the solution of HCl.

A

First, calculate the concentration of the standard solution of NaOH:
• Moles of NaOH dissolved = 0.987 / 40.0 = 0.024675 moles
• Conc of standard solution = (0.024675 x 1000) / 250 = 0.0987 mol/dm³
Now write out a balanced equation:
• HCl + NaOH -> NaCl + H₂O
Now work out the number of moles of NaOH:
• Moles of NaOH = 0.0987 x (21.7/1000) = 0.00214… moles
• Therefore, moles of HCl = 0.00214… moles
• Conc of HCl = 0.00214… / (15.0 x 10⁻³) = 0.143 mol/dm³

56
Q

20.4 cm³ of a 0.500 mol/dm³ solution of sodium carbonate reacts with 1.50 mol/dm³ nitric acid. Calculate the volume of nitric acid required to neutralise the sodium carbonate.

A

Write a balanced equation:
• Na₂CO₃ + 2HNO₃ -> 2NaNO₃ + H₂O + CO₂
Work out how many moles of Na₂CO₃ there are:
• Moles of Na₂CO₃ = 0.500 x (20.4/1000) = 0.0102 moles
• Therefore, moles of HNO₃ = 0.0204 moles
Work out the volume:
• Volume of HNO₃ = 0.0204 / 1.50 = 13.6 x 10⁻³ dm³ = 13.6 cm³

57
Q

Remember to practise doing titration calculations.

A

Pg 65 of revision guide

58
Q

What is uncertainty?

A

The maximum amount of error your measurements might have using an instrument.

59
Q

What is the uncertainty for an instrument?

A

+/- half of the smallest increment the equipment can measure

60
Q

What is the uncertainty for a 50cm³ burette?

A
  • It has readings every 0.1 cm³

* So the uncertainty is +/- 0.05 cm³

61
Q

Aside from error due to increments on an instrument, what is the other type of error?

A
  • Error based on how accurately the equipment has been made.

* These are usually written on the equipment somewhere.

62
Q

A student is using a set of electronic scales that measures to the nearest 0.05 g. He zeros the scales and measures out 1.35 g of solid. Calculate the total uncertainty of the mass reading.

A
  • There are two readings - the initial reading is 0.00 g and the final reading is 1.35 g.
  • The certainty of each reading is 0.05 / 2 = 0.025 g.
  • So the total uncertainty = 0.025 + 0.025 = 0.05 g.
63
Q

What is the equation for the percentage uncertainty of a reading?

A

Percentage uncertainty = (Uncertainty / Reading) x 100%

64
Q

What is the equation for the percentage error of a combined uncertainty? (e.g. The weighing of a vessel with acid and then without it)

A

Percentage uncertainty = (Total uncertainty / Difference in readings) x 100%

65
Q

A student is using a set of electronic scales that measures to the nearest 0.05 g. He zeros the scales and measures out 1.35 g of solid. Calculate the percentage uncertainty of the mass reading.

A
  • Total uncertainty = 0.025 + 0.025 = 0.05 g.

* Percentage uncertainty = (0.05 / (1.35 - 0.00)) x 100% = 4%

66
Q

How can you minimise percentage uncertainty?

A
  • Use more precise equipment

* Use larger volumes/masses of everything

67
Q

What are the two types of error?

A
  • Systematic

* Random

68
Q

What are systematic errors?

A
  • They are the same every time you repeat the experiment

* They may be caused by the setup or equipment used

69
Q

What are random errors?

A
  • Errors that are a bit different each time you repeat the experiment
  • This are things like having to estimate the volume of a liquid between two markings
70
Q

How can systematic and random errors be reduced?

A
  • Systematic -> Adjusting your setup

* Random -> Repeating the experiment many times so that they cancel out

71
Q

How can you find the total uncertainty in an experiment?

A
  • Find the percentage uncertainty for each bit of equipment
  • Add the percentage uncertainties together
  • This gives the percentage uncertainty in the final result
  • Use this to work out the actual total uncertainty in the final result
72
Q

10.00 cm³ of KOH is neutralised by 27.30 cm³ of HCl of known concentration.
The volume of KOH has an uncertainty of 0.060 cm³.
The volume of HCl has an uncertainty of 0.10 cm³.
The concentration of the KOH is calculated to be 1.365 mol/dm³
What is the uncertainty in this concentration?

A
  • % uncertainty in KOH = (0.060 / 10.00) x 100 = 0.60%
  • % uncertainty in HCl = (0.10 / 27.30) x 100 = 0.36…%
  • Total % uncertainty = 0.60 + 0.36… = 0.96…%
  • Uncertainty in final answer = 0.96…% of 1.365 mol/dm³ = 0.013 mol/dm³
73
Q

What is the theoretical yield of a product?

A

The maximum mass of a product that could be obtained from a reaction if no chemicals are lost in the process.

74
Q

What is percentage yield?

A

The actual amount of the product you collect, relative to the theoretical yield.

75
Q

What is the formula for percentage yield?

A

Percentage yield = (Actual yield / Theoretical yield) x 100%

76
Q

Ethanol can be oxidised to form ethanal:
C₂H₅OH + [O] -> CH₃CHO + H₂O
9.2g of ethanol was reacted with an oxidising agent in excess and 2.1g of ethanol was produced. Calculate the theoretical yield and the percentage yield.

A
  • Moles = Mass / Mr
  • Moles of C₂H₅OH = 9.2 / (2 x 12 + 5 x 1 + 16 + 1) = 9.2 / 46 = 0.20 moles
  • Therefore, moles of CH₃CHO = 0.20 moles
  • Theoretical yield = 0.20 x (2 x 12 + 4 x 1 + 16) = 0.20 x 44 = 8.8g
  • Percentage yield = 2.1 / 8.8 = 24%
77
Q

What is atom economy a measure of?

A

How efficient a reaction is at producing a given reactant relative to the by-products.

78
Q

What is the formula for atom economy?

A

% Atom Economy = (Molar mass of desired products / Sum of molar masses of all products) x 100%

79
Q

Does atom economy require an experiment to be calculated?

A

No, it is all theoretical from the equation.

80
Q

What is the atom economy in an addition reaction?

A

100%, since only one product is formed.

81
Q

Aluminium oxide is formed by heating aluminium hydroxide until it decomposes. Calculate the atom economy.

2Al(OH)₃ -> Al₂O₃ + 3H₂O

A
  • % Atom economy = (M(Al₂O₃) / (M(Al₂O₃) + 3 x M(H₂O))) x 100%
  • % Atom economy = (102 / (102 + 54)) x 100% = 65.4%
82
Q

Can a reaction have a high percentage yield and low atom economy?

A

Yes

A LEVEL - Chemistry (Inorganic) (21 decks)

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