Moles and Conversion of Mass

Question 1

Describe a mole

Answer 1

A measure of the amount of any substance

1 mole is exactly 6.02214076 × 10²³ particles or ‘things’

Question 2

Recall Avogadro’s number

Answer 2

6.02214076 × 10²³

The number of units or ‘things’ in one mole of any substance

Question 3

Describe the relationship between mass, moles and molar mass

Answer 3

n = m/M

• n = number of moles
• m = mass (g)
• M = molar mass (gmol⁻¹)

Question 4

Describe molar mass

Answer 4

The mass of 1 mole of a substance in gmol⁻¹

For example:
M(C₅H₁₂)
= (5 × C) + (12 × H)
= (5 × 12.01) + (12 × 1.01)
= 72.12 g/mol

Question 5

Define stoichiometry

Answer 5

The calculation of products and reactants in a chemical reaction

Question 6

Describe limiting reagent

Answer 6

The reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed

Question 7

Describe excess reactant

Answer 7

The reactant in excess after combining with all of the limiting reactant

Question 8

Describe theoretical and experimental yield

Answer 8

The maximum amount of product that can be produced by a chemical reaction assuming perfect completion of reaction

The experimental yield may be less than the theoretical yield

Question 9

Describe empirical formula

Answer 9

The simplest whole-number ratio of elements in a compound

Question 10

Describe percentage composition

Answer 10

The percentage by mass of each of the different elements in the compound

Used to identify unknown compounds by comparing the percentage composition to one calculated from a known formula and to determine the empirical formula or the molecular formula [if given the relative molecular mass]

Question 11

Recall how the mass of products and reactants are calculated

Answer 11

The law of conservation of mass states that in a chemical reaction, the total mass of reactants is equal to the total mass of products

For example:
2.43g = Mg
1.28g = O₂
Balance the equation
2Mg + O₂ → 2MgO
Find the number of moles
N[Mg] = 2.43/24.31 ≈ 0.10 moles of Mg
N[O₂] = 1.28/(16+16) ≈ 0.04 moles of O₂
Find the mole ratio
2Mg:O₂
2:1 ratio ∴ Oxygen is limiting and Magnesium is excess
O₂:2MgO
1:2 ratio
0.04 moles of O₂ = (0.04 × 2) moles of MgO
Find the mass
Mass of MgO = 0.08 × (24.31 + 16) = 3.2g

Question 12

Recall how the percentage composition is calculated

Answer 12

[mass of element]/[molecular mass] × 100

Question 13

Recall how the percentage yield is calculated

Answer 13

percentage (%) yield = [experimental yield]/[theoretical yield] × 100

Question 14

Recall how the molecular formula of a compound it calculated from its empirical formula and molar mass

Answer 14

• Assume 100g of the compound, and determine the mass of each element.
• Determine the number of moles of each element.
• Find the simplest ratio (divide the number of moles by the smallest number then round)
• Calculate the empirical formula mass
• Determine how many empirical units there are in the molecule
• Multiply the empirical formula by the number of empirical units to find the molecular formula

For example:
49.4% = C, 5.2% = H, 16.6% = O, 28.8% = N, molar mass = 198.18g/mol
Mass in 100g
49.4 = C, 5.2 = H, 16.6 = O, 28.8 = N
Moles
C = 49.4/12.0 = 4.12
H = 5.2/1.01 = 5.15
O = 16.6/19.0 = 1.04 [smallest number]
N = 28.8/14.0 = 2.06
Simplest ratio
C = 4.12/1.04 = 3.9 = 4
H = 5.15/1.04 = 5
O = 1.04/1.04 = 1
N = 2.06/1.04 = 1.9 = 2
Empirical formula
C₄H₅ON₂
Empirical formula mass
= (4xC) + (5xH) + (1xO) + (2xN) = (4x12.0) + (5x1.01) + (1x16.0) + (2x14.01) = 97.07
Empirical units
[relative molecular mass]/[relative empirical mass] = 198.18/97.07 = 2.04 = 2 [round]
Molecular formula
= 2 x C₄H₅ON₂ = C₈H₁₀O₂N₄