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Chem - physical > 2. Amount of substance > Flashcards

2. Amount of substance Flashcards

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1
Q

Define relative atomic mass

A
  • The mean mass of an atom of an element divided by 1/12 of the mean mass of an atom of the carbon-12 isotope
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2
Q

Define relative molecular mass

A
  • The mean mass of a molecule of a compound, divided by 1/12 of the mean mass of an atom of the carbon-12 isotope
  • a.ka relative formula mass
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3
Q

Define a mole

A
  • A unit of measurement for substances
  • It always contains the same number of particles
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4
Q

State the Avogadro Constant

A
  • L = 6.022x10^23 particles
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5
Q

State the equation to find the number of particles present in a sample of a substance with known mass

A
  • Number of particles = nL
  • moles x L
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6
Q

State the equation that links moles, mass, and Mr

A
  • moles = mass / Mr
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7
Q

State the equation that links moles, conc, and volume

A
  • moles = conc x volume /1000
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8
Q
A

DEFINITION: The mole is the amount of substance in grams that has the same number of particles as there are atoms in 12 grams of carbon-12.

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9
Q
A

DEFINITION: Relative atomic mass is the average mass of one atom compared to one twelfth of the mass of one atom of carbon-12

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10
Q
A

vogadro’s Number
There are 6.022 x 1023 atoms in 12 grams of carbon-12. Therefore explained in simpler terms ‘One mole of any specified entity contains 6.022 x 1023 of that entity’:

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11
Q

how to caulker moles for pure solids liquids and gases

A

moles = mass / mr

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12
Q

how to calc moles for gses

A

PV = nRT

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13
Q

How to calculate moles for soulations

A

moles = conc x volume

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14
Q

Unit of mass

A

grams

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15
Q

unit of moles

A

mol

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16
Q

unit of pressure

A

pa

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17
Q

unit of volume

A

m3

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18
Q

unit of temp K

A
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19
Q

gas constant

A

8.31

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20
Q

how to convert Celsius to k

A

add 273

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21
Q

unit of conc

A

mol dm cube

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22
Q

unit of volume in conc

A

dm3

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23
Q

how to convert voulmes

A

cm3dm3 ÷1000 cm3  m3 ÷ 1000 000 dm3m3 ÷1000

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24
Q

how can molar mass for a comped be calculated

A

Molar mass (Mr) for a compound can be calculated by adding up the mass numbers (from the periodic table) of each element in the compound
eg CaCO3 = 40.1 + 12.0 +16.0 x3 = 100.1

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25
Q

convert grams

A

Many questions will involve changes of units 1000 mg =1g
1000 g =1 kg
1000 kg = 1 tonne

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26
Q

example

A

Example 1: Calculate the number of moles of CuSO4 in 35.0g of CuSO4
moles= mass Mr
= 35.0/ (63.5 + 32.0 +16.0 x4) = 0.219 mol

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27
Q
A

Example 2: What is the number of moles in 75.0mg of CaSO4.2H2O?
moles= mass Mr
= 0.075/ (40 + 32.0 +16.0 x4 + 18.0x2) = 4.36x10-4 mol

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28
Q
A

Avogadro’s Constant (L)
There are 6.022 x 1023 atoms in 12 grams of carbon-12. Therefore explained in simpler terms ‘One mole of any specified entity contains 6.022 x 1023 of that entity’:

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29
Q
A

Avogadro’s constant can be used for atoms, molecules and ions

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30
Q
A

No of particles = moles of substance (in mol) X Avogadro’s constant

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31
Q
A

Example 3 : Calculate the number of atoms of tin in a 6.00 g sample of tin metal.
moles = mass/Ar
= 6.00/ 118.7
= 0.05055 mol
number atoms = moles x 6.022 x 1023
0.05055 x 6.022 x 1023
= 3.04 x10

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32
Q
A

example 4

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33
Q
A

density = mass volume

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34
Q
A

Density is usually given in g cm-3
Care needs to be taken if different units are used.

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35
Q
A

ple 5 : Calculate the number of molecules of ethanol in a 0.500 dm3 of ethanol (CH3CH2OH) liquid.
The density of ethanol is 0.789 g cm-3
mass = density x volume ethanol = 0.789 x 500
= 394.5 g moles = mass/Mr
= 394.5/ 46.0
= 8.576 mol
number of molecules= moles x 6.022 x 1023
= 8.576 x 6.022 x 1023
= 5.16 x1024(to 3 sig fig

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36
Q
A

mple 6 : There are 980 mol of pure gold in a bar measuring 10 cm by 20 cm by 50 cm. Calculate the density of gold in kg dm−3
mass = moles x Mr = 980 x 197 = 193060 g
= 193.06 kg
volume = 10x20x50
= 10 000cm3
= 10 dm3
density = mass/volume = 193/10
= 19.3 kg dm-3

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37
Q
A

Definition: An empirical formula is the simplest ratio of atoms of each element in the compound.

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38
Q
A

Step 1 : Divide each mass (or % mass) by the atomic mass of the element
Step 2 : For each of the answers from step 1 divide by the smallest one of those numbers.
Step 3: sometimes the numbers calculated in step 2 will need to be multiplied up to give whole numbers.
These whole numbers will be the empirical formula.

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39
Q
A

The same method can be used for the following types of data:
1. masses of each element in the compound
2. percentage mass of each element in the compound

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40
Q
A

xample 7 : Calculate the empirical formula for a compound that contains 1.82g of K, 5.93g of I and 2.24g of O
Step1: Divide each mass by the atomic mass of the element to give moles K = 1.82 / 39.1 I = 5.93/126.9 O = 2.24/16
= 0.0465 mol = 0.0467mol = 0.14mol
Step 2 For each of the answers from step 1 divide by the smallest one of those numbers. K = 0.0465/0.0465 I = 0.0467/0.0465 O = 0.14 / 0.0465
=1 =1 =3
Empirical formula =KIO3

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41
Q

Definition: A molecular formula is the actual number of atoms of each element in the compound.

A
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42
Q
A

Example 8 : Deduce the molecular formula for the compound with an empirical formula of C3H6O and a Mr of 116
C3H6O has a mass of 58
The empirical formula fits twice into Mr of 116
So the molecular formula is C6H12O2

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43
Q
A

The Mr does not need to be exact to turn an empirical formula into the molecular formula because the molecular formula will be a whole number multiple of the empirical formula
Remember the Mr of a substance can be found out from using a mass spectrometer. The molecular ion ( the peak with highest m/z) will be equal to the Mr
from the mr work our how ma

44
Q

hydrated salt contains water of crystallisation

A
45
Q

heating in a crucible

A

The water of crystallisation in calcium sulfate crystals can be removed as water vapour by heating as shown in the following equation.
CaSO4.xH2O(s) → CaSO4(s) + xH2O(g)
Method.
*Weigh an empty clean dry crucible and lid .
*Add 2g of hydrated calcium sulfate to the crucible and weigh again
*Heat strongly with a Bunsen for a couple of minutes
*Allow to cool
*Weigh the crucible and contents again
*Heat crucible again and reweigh until you reach a constant mass ( do this to ensure reaction is complete).

46
Q
A

This method could be used for measuring mass loss in various thermal decomposition reactions and also for mass gain when reacting magnesium in oxygen.

47
Q
A

The lid improves the accuracy of the experiment as it prevents loss of solid from the crucible but should be loose fitting to allow gas to escape.

48
Q
A

Small amounts of the solid, such as 0.100 g, should not be used in this experiment as the percentage uncertainties in weighing will be too high.

49
Q
A

Large amounts of hydrated calcium sulfate, such as 50g, should not be used in this experiment as the decomposition is likely to be incomplete.

50
Q
A

The crucible needs to be dry otherwise a wet crucible would give an inaccurate result. It would cause mass loss to be too large as the water would be lost when heating.

51
Q
A

3.51 g of hydrated zinc sulfate were heated and 1.97 g of anhydrous zinc sulfate were obtained.
Calculate the value of the integer x in ZnSO4.xH2O
Calculate the mass of H2O = 3.51 – 1.97 = 1.54g
Calculate moles of = 1.97 ZnSO4 161.5
= 0.0122
Calculate ratio of mole of = 0.0122 ZnSO4 to H2O 0.0122
=1
Calculate moles of H2O
=7 X=7

52
Q
A

Na2SO4 . xH2O has a molar mass of 322.1. Calculate the value of x
Molar mass xH2O = 322.1 – (23x2 + 32.1 + 16x4)
= 180 X = 180/18
=10

53
Q
A

solution is a mixture formed when a solute dissolves in a solvent. In chemistry we most commonly use water as the solvent to form aqueous solutions. The solute can be a solid, liquid or a gas.

54
Q
A

Molar concentration can be measured for solutions. This is calculated by dividing the amount in moles of the solute by the volume of the solution. The volume is measure is dm3. The unit of molar concentration is mol dm-3 ; it can also be called molar using symbol M

55
Q

converting volumes

A

cm3dm3 ÷1000 cm3  m3 ÷ 1000 000 dm3m3 ÷1000

56
Q
A

Example 11 Calculate the concentration of solution made by dissolving 5.00 g of Na2CO3 in 250 cm3 water.
moles = mass/Mr
= 5 / (23.0 x2 + 12 +16 x3)
= 0.0472 mol conc= moles/volume
= 0.0472 / 0.25
= 0.189 mol dm-3

57
Q
A

Example 12 Calculate the concentration of solution made by dissolving 10 kg of Na2CO3 in 0.50 m3 water.
moles = mass/Mr
= 10 000 / (23.0 x2 + 12 +16 x3)
= 94.2 mol
conc= moles/volume
= 94.2 / 500
= 0.19 mol dm-3

58
Q
A

The concentration of a solution can also be measured in terms of mass of solute per volume of solution

59
Q
A

To convert concentration measured in mol dm-3 into concentrationmeasuredingdm-3 multiplybyMrofthe substance
concingdm-3 =concinmoldm-3 xMr
The concentration in g dm-3 is the same as the mass of solute dissolved in 1 dm3

60
Q
A

mass concentration= mass / volume

61
Q

Ions dissociating
When soluble ionic solids dissolve in water they will dissociate into separate ions. This can lead to the concentration of ions differing from the concentration of the solute.

A
62
Q
A

Example 13
If 5.86g (0.1 mol) of sodium chloride (NaCl) is dissolved in 1 dm3 of water then the concentration of sodium chloride solution would be 0.1 mol dm-3 .
However the 0.1mol sodium chloride would split up to form 0.1 mol of sodium ions and 0.1 mol of chloride ions. The concentration of sodium ions is therefore 0.1 mol dm-3 and the concentration of chloride ions is also 0.1 mol dm-3

63
Q
A

Example 14
If 9.53g (0.1 mol) of magnesium chloride (MgCl2) is dissolved in 1dm3 of water then the concentration of magnesium chloride solution (MgCl2 aq) would be 0.1mol dm-3 .
However the 0.1mol magnesium chloride would split up to form 0.1 mol of magnesium ions and 0.2 mol of chloride ions. The concentration of magnesium ions is therefore 0.1 mol dm-3 and the concentration of chloride ions is now 0.2 mol dm-3

64
Q

Diluting a solution

A

*Pipette 25cm3 of original solution into a 250cm3 volumetric flask
*make up to the mark with distilled water using a dropping pipette for last few drops.
* Invert flask several times to ensure uniform solution.

65
Q
A

Using a volumetric pipette is more accurate than a measuring cylinder because it has a smaller uncertainty
Use a teat pipette to make up to the mark in volumetric flask to ensure volume of solution accurately measured and one doesn’t go over the line.

66
Q
A

Calculating Dilutions
Diluting a solution will not change the amount of moles of solute present but increase the volume of solution and hence the concentration will lower.
Moles = volume x concentration
If amount of moles does not change then:
original volume x original concentration = new diluted volume x new diluted concentration

67
Q
A

new diluted conc = conc x original / new diluted volume

68
Q

The new diluted volume will be equal to the original volume of solution added + the volume of water added

A
69
Q
A

Example 15
50 cm3 of water are added to 150 cm3 of a 0.20 mol dm-3 NaOH solution. Calculate the concentration of the diluted solution.
new diluted concentration = original concentration x original volume
new diluted concentration = 0.20 x 0.150 0.200
= 0.15 mol dm-3

70
Q
A

Example 16
Calculate the volume of water in cm3 that must be added to dilute 5.00 cm3 of 1.00 mol dm−3 hydrochloric acid so that it has a concentration of 0.050 mol dm−3
Moles original solution = conc x vol
= 1.00 x 0.005 = 0.005
New volume = moles /conc = 0.005/0.05
= 0.1 dm3 = 100 cm3
Volume of water added = 100-5 = 95 cm3

71
Q
A

Safety and hazards
Irritant - dilute acid and alkalis- wear googles
Corrosive- stronger acids and alkalis wear goggles
Flammable – keep away from naked flames
Toxic – wear gloves- avoid skin contact- wash hands after use Oxidising- Keep away from flammable / easily oxidised materials

72
Q
A

Hazardous substances in low concentrations or amounts will not pose the same risks as the pure substance.

73
Q
A

The ideal gas equation applies to all gases and mixtures of gases. If a mixture of gases is used the value n will be the total moles of all gases in the mixture.

74
Q
A

Example 17: Calculate the mass of Cl2 gas that has a pressure of
100 kPa, temperature 20 oC , volume 500 cm3.
moles = PV/RT
= 100 000 x 0.0005 / (8.31 x 293)
= 0.0205 mol
mass = moles x Mr
= 0.0205 (35.5 x2)
= 1.46 g

75
Q
A

Example 18: 0.150g of a volatile liquid was injected into a sealed gas syringe. The gas syringe was placed in an oven at 70oC at a pressure of 100kPa and a volume of 80.0cm3 was measured. Calculate the Mr of the volatile liquid (R =8.31)
moles = PV/RT 100 kPa = 100 000 Pa
= 100 000 x 0.00008 / (8.31 x 343) = 0.00281 mol
Mr = mass/moles
= 0.15 / 0.00281
= 53.4 g mol-1

76
Q
A

Gas syringes can be used for a variety of experiments where the volume of a gas is measured, possibly to work out moles of gas or to follow reaction rates.

77
Q
A

The volume of a gas depends on pressure and temperature, so when recording volume it is important to note down the temperature and pressure of the room.

78
Q
A

Moles of gas can be calculated from gas volume (and temperature and pressure) using ideal gas equation PV = nRT

79
Q
A

Potential errors in using a gas syringe:
*gas escapes before bung inserted
*syringe sticks
* some gases like carbon dioxide or sulfur dioxide are soluble in water so the true amount of gas is not measured.

80
Q
A

Make sure you don’t leave gaps in your diagram where gas could escape If drawing a gas syringe make sure you draw it with some measurement markings on the barrel to show measurements can be made

81
Q

hanging the conditions of a gas

A

Questions may involve the same amount of gas under different conditions.
Example 19
40 cm3 of oxygen and 60 cm3 of carbon dioxide, each at 298 K and 100 kPa, were placed into an evacuated flask of volume 0.50 dm3. Calculate the pressure of the gas mixture in the flask at 298 K.
As temperature is the same can make the above equation P1V1 = P2V2 P2 = P1V1 /V2
= 100000 x 1x 10-4 / 5x10-4 = 20 000 Pa
There are two approaches to solving this
1. Work out moles of gas using ideal gas equation then put back into ideal gas equation
with new conditions
2. Or combine the equation n= PV/RT as on right
We can do this as the moles of gas do not change

82
Q
A

Reacting volumes of gas
Equal volumes of any gases measured under the same conditions of temperature and pressure contain equal numbers of molecules (or atoms if the gas in monatomic).
Volumes of gases reacting in a balanced equation can be calculated by simple ratio.

83
Q
A

1 mole of any gas at room pressure (1atm) and room temperature 25oC will have the volume of 24dm3

84
Q
A

Example 20 500 cm3 of methane is combusted at 1atm and 300K. Calculate the volume of oxygen needed to react and calculate the volume of CO2 given off under the same conditions.
CH4(g) + 2O2(g)CO2(g) + 2H2O(l) 1 mole 2 mole 1 mole
500cm3 1dm3 500cm3
Simply multiply gas volume x2

85
Q
A

Example 21 An important reaction which occurs in the catalytic converter of a car is: 2CO(g) + 2NO(g)2CO2(g) + N2(g)
In this reaction, when 500 cm3 of CO reacts with 500 cm3 of NO at 650 °C and at 1 atm. Calculate the total volume of gases produced at the same temperature and pressure.
2CO(g) + 2NO(g)2CO2(g) + N2(g) total volume of gases produced = 750cm3 500cm3 500cm3 500cm3 250cm3

86
Q

pg10 chem revise

A
87
Q

limting and excess reactants

A
88
Q

percentage yield

A

= actual yield / theoretical yield x 100

89
Q
A

percentage atom economy = mass of useful prdoctus / mass of all reactants x 100

90
Q

Example 27: Calculate the % atom economy for the following reaction where Fe is the desired product assuming the reaction goes to completion.
Fe2O3 + 3CO2Fe+3CO2
% atom economy = (2 x 55.8)
x 100 (2 x 55.8 + 3x16) + 3 x (12+16)
=45.8%

A
91
Q
A

Example 28: 25.0g of Fe2O3 was reacted and it produced 10.0g of Fe. Calculate the percentage yield.
Fe2O3 + 3CO2Fe + 3 CO2
First calculate maximum mass of Fe that could be produced.
Step 1: work out moles of iron oxide Moles = mass / Mr
=25.0/ 159.6 = 0.1566 mol
Step 2: use balanced equation to give moles of Fe 1 moles Fe2O3 : 2 moles Fe
So 0.1566 Fe2O3 : 0.313moles Fe
Step 3: work out mass of Fe Mass = moles x Mr
= 0.313 x 55.8 =17.5 g
% yield = (actual yield/theoretical yield) x 100
= (10/ 17.5) x 100 =57.1%

92
Q
A

Chemists want a high percentage yield as means there has been an efficient conversion of reactants to products.

93
Q
A

Chemists want a high percentage atom economy so that the maximum mass of reactants ends up in the desired product (so minimising the amount of by-product).

94
Q
A

Readings
the values found from a single judgement when using a piece of equipment

95
Q
A

Measurements
the values taken as the difference between the judgements of two values (e.g. using a burette in a titration)

96
Q
A

he uncertainty of a reading (one judgement) is at least ±0.5 of the smallest scale reading.
The uncertainty of a measurement (two judgements) is at least ±1 of the smallest scale reading.

97
Q
A

Calculating apparatus uncertainties
Each type of apparatus has a sensitivity uncertainty
*balance  0.001 g (if using a 3 d.p. balance) *volumetric flask  0.1 cm3
*25 cm3 pipette  0.1 cm3
*burette (start & end readings and end point )  0.15 cm3
Calculate the percentage error for each piece of equipment used by
% uncertainty =  uncertainty x 100 measurement made on apparatus
e.g. for burette
% uncertainty = 0.15/average titre result x 100
To calculate the maximum total percentage apparatus uncertainty in the final result add all the individual equipment uncertainties together.

98
Q
A

To decrease the apparatus uncertainties you can either decrease the sensitivity
uncertainty by using apparatus with a greater resolution (finer scale divisions ) or you can increase the size of the measurement made.

99
Q
A

Uncertainty of a measurement using a burette. If the burette used in the titration had an uncertainty for each reading of  0.05 cm3 then during a titration two readings would be taken so the uncertainty on the titre volume would be  0.10 cm3 . Then often another 0.05 is added on because of uncertainty identifying the end point colour change

100
Q
A

Reducing uncertainties in a titration
Replacing measuring cylinders with pipettes or burettes which have lower apparatus uncertainty will lower the % uncertainty.
To reduce the % uncertainty in a burette reading it is necessary to make the titre a larger volume. This could be done by: increasing the volume and concentration of the substance in the conical flask or by decreasing the concentration of the substance in the burette.

101
Q
A

If looking at a series of measurements in an investigation, the experiments with the smallest readings will have the highest experimental uncertainties.

102
Q
A

Reducing uncertainties in measuring mass
Using a balance that measures to more decimal places or using a larger mass will reduce the % uncertainty in weighing a solid.
Weighing sample before and after addition and then calculating difference will ensure a more accurate measurement of the mass added.

103
Q
A

Calculating the percentage difference between the actual value and the calculated value
If we calculated an Mr of 203 and the real value is 214, then the calculation is as follows:
Calculate difference 214-203 = 11
% = 11/214 x100
=5.41%

104
Q
A

If the % uncertainty due to the apparatus < percentage difference between the actual value and the calculated value then there is a discrepancy in the result due to other errors.

105
Q
A

If the % uncertainty due to the apparatus > percentage difference between the actual value and the calculated value then there is no discrepancy and all the difference between values can be explained by the sensitivity of the equipment.

106
Q
A

Chem - physical (12 decks)

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