Mole Concept & Stoichiometry

Question 1

Relative atomic mass

Answer 1

The relative atomic mass, A_r, of an element is denied as the ratio of the average mass of one atom of the element to 1/12 of the mass of an atom of ¹²C isotope.

Question 2

Relative isotopic mass

Answer 2

The relative isotopic mass, A_r, of a particular isotope is defined as the ratio of the mass of one atom of the isotope to 1/12 of the mass of an atom of ¹²C isotope.

Question 3

Relative molecular mass

Answer 3

The relative molecular mass, M_r, of a substance is defined as the ratio of the average mass of one molecule to 1/12 of the mass of an atom of ¹²C isotope.

Question 4

Isotopes

Answer 4

Atoms of the same element that have the same number of protons but different number of neutrons

Question 5

Relative isotopic abundance (of each isotope in a sample)

Answer 5

Proportion of each isotope in a sample
Relative abundance = Percentage of the isotope of the sample by mass

Question 6

Mole

Answer 6

The amount of substance contain a number of particles equal to the Avogadro’s constant (6.02 x 10²³ mol^-1)

Question 7

Avogadro’s constant L

Answer 7

• 6.02 x 10²³ mol^-1
• Therefore number of particles = mol x L

Question 8

Molar mass M

Answer 8

• The mass of one mole of a substance
• Units: g mol^-1
• Therefore mass of substance = mol x M

Question 9

Molar gas volume V_m

Answer 9

• The volume that one mole of gas occupies at a particular set of
  temperature and pressure
• At stp 273 K and 10⁵ Pa/1 bar
  1 mol of any gas occupies 22.7 dm³
• At rtp 293 K and 101325 Pa/1 atm
  1 mol of any gas occupies 24.0 dm³
• Therefore volume of gas = mol x V_m

Question 10

Concentration

Answer 10

• Amount of solute (mol or g), dissolved per unit volume of solution
• concentration = (mass or mol) / volume
• can also expressed as parts per million or parts per billion

Question 11

Empirical formula

Answer 11

The simplest ratio of the atoms of the different elements in a compound

Question 12

Molecular formula

Answer 12

The actual ratio of the atoms of the different elements in a compound

Question 13

Limiting reagents

Answer 13

• Reactant that is not in excess
• Reactant that is completely used up in the reaction

Question 14

Complete combustion of hydrocarbons

Answer 14

C_xH_y + (x + y/4) O₂ → xCO₂ +y/2H₂O

Question 15

Avogadro’s Hypothesis

Answer 15

Two equal volumes of gases , under the same T and P, contain the same number of molecules

Question 16

Percentage Purity

Answer 16

(mass of pure sample/mass of total impure sample) x 100%

Question 17

Percentage yield

Answer 17

(actual yield/theoretical yield) x 100%

Question 18

Acid-base reaction

Answer 18

acid + base = salt + water

n_acid/n_base = [acid]/[base]

Question 19

Dilution

Answer 19

• Addition of more solvent to a given solution
• The amount of solute present does not change
• Concentration decreases; the solute is now dispersed in larger final
  volume
• Amount of solute before dilution = Amount of solute after dilution
• C₁V₁ = C₁V₂

Question 20

Double indicator titration

Answer 20

• When a weak diprotic acid, carbonate, etc is neutralised, it
  undergoes 2 stages of reaction
• Hence when titrating such reagents, each stage of neutralization can be detected by using different indicators

Question 20

Sampling

Answer 20

• Collection of a portion from a given solution
• The concentration of sample is the same as the original solution
• The amount of solute present changes
• Concentration of original solution = Concentration of sampled
  solution
• n₁/V₁ = n₁/V₁
• Scaling factor = V2/V1

Question 21

Continuous double indicator titration

Answer 21

• Uses two indicators in the same solution to detect the end point of a
  titration process
• Involves performing two titrations of the same analyte back-to-back,
  with the first titration using one indicator and the second using
  another
• The titration is considered continuous because the second indicator
  is added after the first titration is complete

Question 22

Discontinuous double indicator titration

Answer 22

• Uses two separate solutions with a different indicator each
• Conduct two separate titrations for each stage of neutralisation
• The titration is considered discontinuous because the two stages of
  neutralisation occur in two separate titrations