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GCSE Chemistry > Moles - 4th > Flashcards

Moles - 4th Flashcards

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1
Q

what is relative formula mass?

A
  • the mass of a formula unit of a substance
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2
Q

how is RFM calculated?

A
  • by adding the relative atomic masses (RAMs) of the atoms or ions.
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3
Q

calculate the RFM for H20.

A

RFM = (2*1) + 16 = 18

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4
Q

calculate the RFM for Ba(OH)2.

A

RFM = 137 + (2*(16+1)) = 171

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5
Q

calculate the RFM for CuSO4 dot 5H20.

A

RFM = 63.5 + 32 + (416) + (518) = 249.5

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6
Q

what is percentage by mass?

A
  • shows the mass of a substance that comes from a particular element.
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7
Q

how is percentage by mass calculated?

A
  • from the formula for the substance, using the RFM
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8
Q

calculate the percentage of C in C3H8.

A

%C = (C3/C3H8) * 100
= ((312) / ((312)+8) )* 100
= 81.8%

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9
Q

calculate the percentage of N in NH4NO3.

A

%N = ((N+N)/NH4NO3) * 100
= ((214) / 80) 100
= 35%

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10
Q

what is a mole?

A
  • the unit of amount of substance in chemistry
  • it represents how many particles of a substance there are, taking into account that particles of different substances have different masses
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11
Q

how is the amount of substance in moles calculated?

A
  • from its mass and RFM
  • moles = mass (g) / RFM
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12
Q

calculate the amount in moles in 11g of CO2.

A

= mass / RFM
= 11/44 = 0.25 mol

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13
Q

calculate the mass of 2.5 moles of H2O.

A

= moles * RFM
= 2.5 * 18
= 45g

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14
Q

what does the empirical formula of a substance contain?

A
  • the simplest mole ratio of its elements
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15
Q

how can empirical formula be calculated?

A
  • from the masses (or percentages) of each element
    1. convert the masses (or percentages) into moles
    2. cancel down the mole ratio
    3. write down the simplest mole ratio into a formula
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16
Q

calculate the empirical formula of a substance containing 6.2g P and 4.8g O .

A
  • draw a ‘mass, moles, ratio’ table and stick the numbers in:
    P O
    mass (g) 6.2 4.8
    moles 0.2 0.3
    ratio 2 3
  • whole number ratio = 2:3
  • P203
17
Q

calculate the empirical formula of a substance containing 83.7% C and 16.3% H.

A
  • use the percentages as masses (as if we had 100g of the substance
  • draw a ‘mass, moles, ratio’ table and stick the numbers in:
    C H
    mass (g) 83.7 16.3
    moles 6.975 16.3
    ratio 3 7
  • whole number ratio = 3:7
  • C3H7
18
Q

what leads to errors in empirical formula calculations?

A
  • rounding moles is v.dangerous and can lead to errors
  • you can get away with rounding moles to 2dp, but never round to 1dp or whole numbers
19
Q

what does the molecular formula of a substance contain?

A
  • the actual numbers of atoms in a molecule - like empirical formula, but uncancelled
20
Q

how can the molecular formula be calculated?

A
  • from the empirical formula and RFM
  • this works by finding the uncancelled multiple of the empirical formula with the correct RFM
21
Q

calculate the molecular formula if the empirical formula is CH2 and the RFM is 70.

A
  • RFM of CH2 = 14
  • multiple = 70/14 = 5
  • molecular formula = 5 * CH2 = C5H10
22
Q

what do hydrated salt crystals contain?

A
  • water trapped inside them in a fixed mole ratio
  • this water is called water of crystallisation and it is written into a formula like so:
    CuSO4 “DOT” 5H2O
    salt “dot” water of crystallisation
23
Q

what happens to hydrated salt crystals when they are heated?

A

they lose their water as vapour
- e.g.
CuSO4”dot”5H2O -> CuSO4 + 5H2O

24
Q

how is water of crystallisation calculated?

A
  • from the masses of salt and water produced on heating
25
Q

calculate ‘x’ if 16.7g of MgCl2”dot”xH20 crystals decreased in mass to 9.5 on heating.

A
  1. calculate the mass of water lost - 16.7g - 9.5g = 7.2g
  2. draw a ‘mass, moles, ratio’ table and stick the numbers in. be careful to use the final mass of solid for the salt, not the original mass of crystals
    MgCl2 H2O
    mass (g) 9.5 7.2
    moles 0.1 0.4
    ratio 1 4
    - whole number ratio = 2:3
    - x = 4
    - MgCl2 “dot” 4H2O
26
Q

what term do chemists use when there isn’t enough substance to do a reaction?

A
  • the substance is limiting
27
Q

what term do chemists use when there is more than enough substance to do a reaction?

A
  • the substance is in excess
28
Q

what happens when we mix 0.9 moles of Zn with 1.1 moles of CuO?

A
  • Zn + CuO -> ZnO + Cu
  • the equation tells us the moles of both should be the same, so it must be that:
    Zn is limiting, because there isn’t enough of it and CuO is in excess, because there’s too much of it.
29
Q

what does the limiting substance determine?

A
  • how many moles are actually used and made
30
Q

how can you work out which substance is in limiting/excess?

A
  • because there are multiple substances in a reaction, each with a mass and potentially different balancing numbers in the reaction equation, it can be difficult to work out which is in excess
  • pick one at random, then calculate how much of the other would be needed to react with it, and then compare that to how much of the other there actually is.
31
Q

what happens if we mix 10g of Mg and 10g of HCl?

A
  • Mg + 2HCl -> MgCl2 + H2
  • start with Mg and calculate how much HCl would be needed to react with that amount of Mg, and then compare that to how much HCl there actually is.
  • 1Mg : 2HCl
  • 0.416 moles : 0.833…
  • mass of HCl - 30.4g
  • 10g of Mg would need to react with 30.4g of HCl, but we only have 10g of HCl!
  • HCl is limiting - there isn’t enough - Mg is in excess
32
Q

how is percentage yield of a reaction calculated?

A

% yield = (experimental (from experiment) / theoretical (from calc)) * 100

33
Q

how is experimental mass and theoretical mass found?

A
  • EM - measured in an experiment (or given to you in question)
  • TM - calculated based on the mass of a reactant
34
Q

calculate the percentage yield if 10g of Al produces 0.96g of H2 in an experiment.

A
  1. find/invent a balanced equation for the reaction :
    2Al + 6HCl -> 2AlCl3 + 3H2
  2. calculate the theoretical mass of H2 that 10g of Al should have produced
    - 2Al : 3H2
    mole ratio - 2 : 3
    actual moles - 0.37 : 0.55
    - mass of H2 = 1.11g
    - % yield = (0.96/1.11) * 100 = 86.4%

GCSE Chemistry (15 decks)

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