Moles Flashcards

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?

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
If 25g of calcium carbonate thermally decomposes, how much calcium oxide is produced? CaCO₃ → CaO + CO₂ *Ca = 40, O = 16
``` CaCO₃ CaO Mass 25g 14g RFM 100 56 Moles 0.25 0.25 Ratio 1 1 ``` 14g of calcium oxide is produced
26
Avogadro's Law
Equal volumes of gases at the same temperature and pressure contain the same # of molecules
27
How much space does one mole of a gas occupy at rtp?
24dm³
28
What is rtp?
Room temperature and pressure
29
How much is 1 litre in: 1. dm³? 2. cm³?
1. 1dm³ | 2. 1000cm³
30
Gas moles equation in: 1. dm³ 2. cm³
1. Volume = moles * 24 | 2. Volume = moles * 24,000
31
Calculate the volume of 0.01g of hydrogen gas at rtp in cm³
``` Moles = 0.01 / 2 Moles = 1/200 Volume = moles * 24,000 Volume = 1/200 * 24 = 120cm³ ```
32
What volume of CO₂ is produced in the thermal decomposition of 8g of CaCO₃? CaCO₃ → CO₂ + CaO *C = 12, O = 16, Ca = 40
``` CaCO₃ CO₂ Mass 8g RFM 100g Moles 0.08 0.08 Ratio 1 1 ``` ``` Volume = 0.08 * 24,000 Volume = 1920cm³ ```
33
Percentage yield
The efficiency of a chemical reaction
34
Why might some reactions not react fully?
1. Loss of solid or liquid when transferring container 2. Side reactions occurring 3. Reversible reactions
35
Theoretical yield
The quantity of a product that would be obtained from a complete reaction
36
Actual yield
The amount of product actually obtained in a chemical reaction
37
Equation to calculate percentage yield
Percentage yield = actual yield / theoretical yield * 100
38
In a reaction where the theoretical yield of magnesium oxide is 20g, only 14g is produced. What is the percentage yield of this reaction?
Percentage yield = 14/20 * 100 | Percentage yield = 70%
39
Atom economy
A measure of the amount of starting materials that become useful products
40
Equation for atom economy percentage
Atom economy % = mass of useful products / total mass of reactants * 100
41
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?
Atom economy % = 461/463 * 100 | Atom economy % = 99.6%
42
N₂ + 3H₂ → 2NH₃ Work out the atom economy for the above reaction, given that ammonia is the useful product.
Atom economy % = 2 * (14 + 3 * 1) / (2 * 14 + 3 * 2 * 1) * 100 Atom economy % = 17/17 * 100 Atom economy % = 100%