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Chemistry GCSE > Moles > Flashcards

Moles Flashcards

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1
Q

RAM

A

Relative Atomic Mass: the average mass of one atom of an element compared to the mass of a carbon-12 atom

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2
Q

How do you calculate the RAM?

A

Calculate the weighted average of the isotopes:

  1. Multiply each isotope’s mass by its abundance
  2. Add all the totals together
  3. Divide this total by the total abundance number (may not be 100)
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3
Q

Isotopes

A

Atoms of the same element, with the same number of protons but a different number of neutrons

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4
Q

Calculate the RAM of boron, given that a sample of born is found to be composed:
20% has an isotopic mass of 10.0
80% has an isotopic mass of 11.0

A 
RAM = 10 * 0.2 + 11 * 0.8
RAM = 10.8 g/mol
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5
Q

RFM

A

Relative Formula Mass: the sum of all the RAMs for each of the elements in the formula

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6
Q

Calculate the RFM of water

*H = 1, O = 16

A 
RFM = 2 * 1 + 1 * 16
RFM = 18 g/mol
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7
Q

Calculate the RFM of ammonia

*N = 14

A 
RFM = 1 * 14 + 3 * 1
RFM = 17 g/mol
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8
Q

How would you calculate the percentage composition of element A in compound B?

A

% A = (RAM of A * number of A atoms in B) / RFM of B * 100

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9
Q

Mole

A

The amount of a substance that contains Avogadro’s Number’s worth of particles (atoms, molecules, ions, electrons, etc.)

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10
Q

Avogadro’s Number

A

6.02 * 10^23

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11
Q

How many particles are there in?

  1. 1 mole
  2. 10 moles
  3. 5 moles
A
  1. 6.02 * 10^23
  2. 6.02 * 10^24
  3. 3.01 * 10^24
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12
Q

What is the mass of one mole of a substance?

A

The substance’s RFM, measured in grams

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13
Q

What mass would 5 moles of oxygen (RFM = 32) have?

A 
Mass = 32 * 5
Mass = 160g
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14
Q

What equation links mass, RFM and moles?

A

Mass = RFM * # of moles

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15
Q

How many moles are there in 54g of water?

A

RFM = 18 g/mol

54 / 18 = 3 moles

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16
Q

0.02 moles of a compound is found to have a mass of 1.64g. Find the formula mass of the compound.

A 
RFM = mass / # of moles
RFM = 1.64 / 0.02
RFM = 82 g/mol
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17
Q

Steps of experiment to measure the change in mass when magnesium burns

A
  1. Clean the magnesium ribbon and loosely coil
  2. Weigh a clean crucible and lid. Place the magnesium inside and reweigh
  3. Heat the crucible for 5–10 minutes, lifting the lid a bit from time to time using tongs. Ensure that as little product escapes as possible
  4. Continue heating until glowing ceases (reaction complete)
  5. Cool the crucible and reweigh
18
Q

How would you find the masses needed to calculate the empirical formula of magnesium oxide, using the following values from the change in mass experiment?

  1. Mass of crucible
  2. Mass of crucible and magnesium
  3. Mass of crucible and magnesium oxide
A
  1. Find mass of magnesium by taking value #1 away from #2

2. Find mass of oxygen by taking #2 away from #3

19
Q

Empirical formula

A

The simplest ratio of atoms in a compound

20
Q

What is the empirical formula of H₂O₂?

A

HO

21
Q

Steps to calculate the empirical formula

A
  1. Determine mass or percentage of elements
  2. Divide the given masses or percentages by mass by their respective RAMs
  3. Divide the results by the smallest value
  4. Convert into the simplest whole number ratio
22
Q

What is the empirical formula of a compound made up of 20% calcium and 80% bromine by mass?

*Ca = 40, Br = 80

A

Ca: 20 / 40 = 1/2 Br: 80 / 80 = 1
Ca: 0.5 / 0.5 = 1 Br: 1 / 0.5 = 2
(Both already integers)
1 : 2 OR CaBr₂

23
Q

The empirical formula of sodium chloride is NaCl. It does not have a molecular formula. Why is this the case?

A

Sodium chloride does not exist as a molecule. It is an example of a giant ionic lattice

24
Q

Steps to find mass of A where the mass of B is known

A
  1. Calculate RAM of B (the chemical we have the most info about)
  2. Calculate # of moles of B
  3. Calculate # of moles of A using the molar ratio between A and B
  4. Calculate RAM of A
  5. Calculate mass of A
25
Q

If 25g of calcium carbonate thermally decomposes, how much calcium oxide is produced?

CaCO₃ → CaO + CO₂

*Ca = 40, O = 16

A 
CaCO₃         CaO
Mass       25g             14g
RFM         100              56
Moles     0.25            0.25
Ratio          1                  1

14g of calcium oxide is produced

26
Q

Avogadro’s Law

A

Equal volumes of gases at the same temperature and pressure contain the same # of molecules

27
Q

How much space does one mole of a gas occupy at rtp?

A

24dm³

28
Q

What is rtp?

A

Room temperature and pressure

29
Q

How much is 1 litre in:

  1. dm³?
  2. cm³?
A
  1. 1dm³

2. 1000cm³

30
Q

Gas moles equation in:

  1. dm³
  2. cm³
A
  1. Volume = moles * 24

2. Volume = moles * 24,000

31
Q

Calculate the volume of 0.01g of hydrogen gas at rtp in cm³

A 
Moles = 0.01 / 2
Moles = 1/200
Volume = moles * 24,000
Volume = 1/200 * 24 = 120cm³
32
Q

What volume of CO₂ is produced in the thermal decomposition of 8g of CaCO₃?

CaCO₃ → CO₂ + CaO

*C = 12, O = 16, Ca = 40

A 
CaCO₃         CO₂
Mass          8g
RFM         100g
Moles       0.08           0.08
Ratio           1                 1
Volume = 0.08 * 24,000
Volume = 1920cm³
33
Q

Percentage yield

A

The efficiency of a chemical reaction

34
Q

Why might some reactions not react fully?

A
  1. Loss of solid or liquid when transferring container
  2. Side reactions occurring
  3. Reversible reactions
35
Q

Theoretical yield

A

The quantity of a product that would be obtained from a complete reaction

36
Q

Actual yield

A

The amount of product actually obtained in a chemical reaction

37
Q

Equation to calculate percentage yield

A

Percentage yield = actual yield / theoretical yield * 100

38
Q

In a reaction where the theoretical yield of magnesium oxide is 20g, only 14g is produced. What is the percentage yield of this reaction?

A

Percentage yield = 14/20 * 100

Percentage yield = 70%

39
Q

Atom economy

A

A measure of the amount of starting materials that become useful products

40
Q

Equation for atom economy percentage

A

Atom economy % = mass of useful products / total mass of reactants * 100

41
Q

One mole of lead (II) iodide is formed in a reaction. Given that the RFM of lead (II) iodide is 461 g/mol and the total mass of reactants is 463g, what is the atom economy of the reaction?

A

Atom economy % = 461/463 * 100

Atom economy % = 99.6%

42
Q

N₂ + 3H₂ → 2NH₃

Work out the atom economy for the above reaction, given that ammonia is the useful product.

A

Atom economy % = 2 * (14 + 3 * 1) / (2 * 14 + 3 * 2 * 1) * 100
Atom economy % = 17/17 * 100
Atom economy % = 100%

Chemistry GCSE (4 decks)

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