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A-Level Chemistry > Chapter 3 Amount Of Substance > Flashcards

Chapter 3 Amount Of Substance Flashcards

1
Q

Avogadro’s Constant (NA)

A

6.02*10^23 mol^-1 ; number of particles in each mole of carbon 12

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2
Q

Mass of 1 mol of atoms of any element

A

Equals to the relative atomic mass in grams

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3
Q

1 mol of Carbon

A

12g

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4
Q

1 mol of Hydrogen

A

1g

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5
Q

Define the mole

A

The amount of any substance containing as many particles as there are carbon atoms in exactly 12g of carbon-12

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6
Q

Molar mass (M)

A

Gives the mass in grams in each mole of the substance - units are g mol^-1 (mass/mol)

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7
Q

Mol equation

A

Amount (mol - n) = mass (m - g) / molar mass (M - g mol^-1)

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8
Q

What is a molecule?

A

Two or more atoms held together by covalent bonds

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9
Q

Molecular formula

A

Number of atoms of each element in a molecule

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10
Q

Elements that exist as molecules

A

H2, N2, O2, F2, Cl2, Br2, I2, P4 and S8

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11
Q

Empirical formula

A

Simplest whole number ratio of atoms of each element in a compound

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12
Q

Why is empirical formula important?

A

It helps in simplifying elements that do not exist as molecules (ionic/non-metals) - where giant lattices are formed with billions of atoms so this shows the simplest ratio that never changes

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13
Q

Relative molecular mass

A

The mass of a molecule relative to the mass of an atom of carbon 12 - ONLY FOR SIMPLE MOLECULES

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14
Q

Relative formula mass

A

Compares the mass of a formula unit with the mass of an atom of carbon 12 ; INLY EMPIRICAL FORMULA (ionic lattices etc)

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15
Q

What is analysis?

A

Investigating the chemical composition of a substance is called analysis - results of an experiment

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16
Q

How to work out empirical formula from mass?

A

CONVERT TO MOLES AND THEN SIMPLEST RATIO (DIVIDE BY SMALLEST NUMBER - DO NOT ALWAYS ROUND AND SEE WHAT RATIO IS POSSIBLE)

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17
Q

How to determine molecular mass?

A

Percentage composition/molar mass (replace mass (g) with percentage in equation) and then empirical formula divided by total molar mass
Multiply whole empirical by factor to give MOLECULAR FORMULA

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18
Q

Hydrated salts

A

Water part of crystalline structure - water of crystallisation is what makes CuSO4.H2O blue and when heated bonds are broken and white anhydrous CuSO4 left behind

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19
Q

Formula of hydrated copper sulfate to anhydrous

A

CuSO4.5H2O -> CuSO4 + 5H2O

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20
Q

Experiment to find out formula of hydrated salt

A

Weigh an empty crucible
Weigh crucible + hydrated salt
Heat the crucible gently using a Bunsen Bruner clay pipe triangle and tripod - heat to a constant mass till consecutive values are the same (make sure not to decompose)
Weigh mass of crucible + anhydrous after cooling

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21
Q

ASSUMPTIONS when heating hydrated salt

A

1) Not always going to be a distinct colour change therefore must heat to a constant mass - crystals reheated until the mass of residue does not change ; ALL water removed ; if not then value of x is smaller than true value
2) Many salts may further decompose when heated so blue CuSO4 may turn to black CuSO4 - then value of x higher than real value

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22
Q

1cm^3 into ml

A

1 ml

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23
Q

1dm^3 into ml

A

1000ml = 1000cm^3 = 1 litre

24
Q

Cm3 to dm3

A

Divide by 1000

25
Q

Equation linking moles and volume and concentration

A

n= c*v

26
Q

What are standard solutions and how are they prepared?

A

Known concentration - dissolve an exact mass of solute in a solvent and make up the solution to an exact volume => INVERT FLASK SLOWLY AND TAKE IN ALL THE WASHINGS

27
Q

How to convert from mol/dm^3 to g/dm^3 or vice versa?

A

Use mass = mr*moles so moles = mass/mr

28
Q

What about gases - how to measure the volume of gases?

A

Measure gas volumes - at same temperature and pressure equal volumes of different gases contain the same number of molecules

29
Q

What is molar gas volume?

A

Volume per mole of gas molecules at a stated temperature and pressure

30
Q

RTP

A

20 degrees Celsius and 101000 Pa (1 atm)

31
Q

Molar gas volume at RTP

A

24 dm^3/mol

32
Q

Volume of a gas at RTP Equation

A

V = n*24dm^3/mol

33
Q

Why do we use ideal gas equation?

A

RTP will be approximate so when gases are at different temperatures or you need to be accurate use the ideal gas equation

34
Q

Assumptions for molecules in an ideal gas

A
  • No intermolecular forces between particles
  • All collisions are perfectly elastic (no KE lost)
  • Temperature is proportional to KE
  • Moving in random motion in straight lines
35
Q

Ideal gas equation

A

pV = nRT

36
Q

Pressure

A

Measured in Pa

37
Q

Volume (V)

A

Measured in m^3

38
Q

Cm^3 to m^3

A

*10^-6

39
Q

n

A

Moles (mass/mr)

40
Q

R

A

Constant of 8.314 J/mol/K

41
Q

T

A

Temperature in Kelvin +273 TO GET KELVIN

42
Q

Experiment to find RFM of a liquid

A

Must be liquid at RTP but boiling point below 100

1) Weigh mass of gas syringe
2) Inject sample info a gas syringe through the self-sealing rubber cap and reading the to find mass of volatile liquid added
3) Place in boiling water bath at 100 and the liquid vaporises producing a gas with atmospheric pressure and volume of gas in syringe recorded

43
Q

Stoichiometry

A

Ratio of the amount in moles of each substance (balancing numbers)

44
Q

What do we use balanced equations for?

A

Quantities of reactants required to prepare a required quantity
Quantities of products that should be formed from certain quantities of reactants

45
Q

Experiment to identify an unknown metal

A

1) Conical flask containing acid + metal attached to gas syringe
2) Weigh sample of metal and add to acid
3) Using measuring cylinder add standard solution (in excess) of acid and put on bung
4) Measure maximum volume of gas created in syringe

46
Q

Standard solution of acid

A

25 cm^3 = 0.025 dm^3

1 moldm^-3

47
Q

Theoretical yield

A

The maximum possible amount of product created (assumption) - all of the reactants turned into products

48
Q

Why is maximum theoretical yield not always achieved?

A

Reaction may not have gone to completion
Side reactions may have taken place alongside main reaction
Purification/filtration may lead to loss of product

49
Q

Actual yield

A

The actual amount of product created - usually less than theoretical yield

50
Q

Percentage yield

A

Conversion of starting materials into a desired product

Actual yield/theoretical yield * 100

51
Q

Limiting reagent

A

Reactant not in excess that is completely used up first and stops the reaction

52
Q

How to find out limiting reagent?

A

Compare available moles with the ratio (stoichiometry - balanced equation)
2H2 + O2 -> 2H2O
If equal amounts of hydrogen and oxygen, hydrogen will be used up first because 2:1 ratio and half of the oxygen would be unreacted (excess)

53
Q

Atom economy

A

(Sum of molar masses of desired / sum of molar masses of all products) * 100

54
Q

High atom economy means…

A

A large proportion of products are desired (few waste)

Good sustainability because atoms well utilised and good use of natural resources

55
Q

Atom economy problem

A

ASSUMES 100% YIELD - ALL REACTANTS ARE CONVERTED TO PRODUCTS

56
Q

Improving atom economy

A

Helps in preserving finite resources and prevents disposal of waste - improves efficiency and in an ideal chemical process a use would be found for all products (thus atom economy would be 100%)

57
Q

What else should be taken into account when judging sustainability?

A

The availability of reactants (starting materials may be low or not - affects prices)
Efficiency depends on percentage yield too

A-Level Chemistry (28 decks)

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