Chemistry Lecture Notes

Key Formula and Variables

• Formula: [ C = \frac{n}{V} ]
• Variables:
  • C: Concentration (measured in moles per decimeter cubed or mol·dm⁻³)
  • n: Moles
  • V: Volume (measured in decimeters)

Important Units

• Concentration (C): mol·dm⁻³
• Volume (V): decimeters
• Moles (n): dimensionless value

Usage of the Formula

• Solutions: This formula is used when dealing with solutions (liquids) not gases.

Gas Formula

• When dealing with gases, use an alternative formula specific to gas calculations.

Alternate Forms of the Formula

• Formula Derivation:
  • From [ n = \frac{M}{m} ] and [ C = \frac{n}{V} ]
  • Substituting gives [ C = \frac{M}{mV} ]
• Solving for Different Parameters:
  • n = CV (moles = concentration * volume)
  • V = \frac{n}{C} (volume = moles / concentration)

Concept of Dilution

• Example with Orange Juice (Oros):
  • Adding more water to a solution dilutes it.
  • Molecular Example: 7 Oros molecules in solution remain constant though volume increases, reducing concentration.
  • Concentration Decrease: As volume increases (more water), concentration (sweetness) decreases.

Important Relationships and Examples

• Moles in Solution I = Moles in Solution II:
  • Regardless of dilution, the number of moles remains constant.
  • Formula: [ C_1 V_1 = C_2 V_2 ] (initial concentration and volume = final concentration and volume)

Sample Problems Using the Formula

Determining Concentration Example

1. Problem: 6 moles in 2 dm³

• Solution: [ C = \frac{6}{2} = 3 ; mol·dm⁻³ ]

Including Mass in Calculation

2. Problem: 60 grams of NaOH in 2 dm³

• Convert Mass to Moles:
  • Molar Mass (NaOH): Na = 23, O = 16, H = 1, Total = 40 g/mol
  • [ n = \frac{60}{40} = 1.5 ; mol ]
• Solution: [ C = \frac{1.5}{2} = 0.75 ; mol·dm⁻³ ]

Using C1V1 = C2V2

3. Problem: New volume with added water

• Initial: C1 = 5 mol·dm⁻³, V1 = 20 dm³
• Added Water: 40 dm³
• Total Volume: V2 = 60 dm³
• Solution:
  • [ 5 \times 20 = C_2 \times 60 ]
  • [ 100 = 60C_2 \implies C_2 = \frac{100}{60} = 1.67 ; mol·dm⁻³ ]

Another Example with Volume Change

2. Problem: Changing volume to 15 dm³

• Initial: C1 = 3 mol·dm⁻³, V1 = 6 dm³
• Final Volume: V2 = 15 dm³
• Solution:
  • [ 3 \times 6 = C_2 \times 15 ]
  • [ 18 = 15C_2 \implies C_2 = \frac{18}{15} = 1.2 ; mol·dm⁻³ ]

Determining Mass From Concentration

4. Problem: HCl Mass Problem
   • Given: 73g HCl in 10 dm³
   • Molar Mass (HCl): H = 1, Cl = 35.5, Total = 36.5 g/mol
   • Using [ C = \frac{m}{MV} ]:
   • Solution: [ C = \frac{73}{36.5 \times 10} \approx 0.2 ; mol·dm⁻³ ]

Calculating Mass in a New Solution

5. Problem: CaCO3 with given concentration and volume

   • Given: C = 2 mol·dm⁻³, V = 4 dm³
   • Find Mass:

   Using [ n = CV ] :

   • [ n = 2 \times 4 = 8 ; mol ]
   • Molar Mass (CaCO3): Ca = 40, C = 12, O3 = 48 (3x16)
   • Total = 100 g/mol
   • Using [ m = nM ]:
   • Solution: [ m = 8 \times 100 = 800 ; g ]