Nice hit of nearly 16ozs of gold per ton.Let’s calculate the...

Nice hit of nearly 16ozs of gold per ton.

Let’s calculate the number of ounces of gold in an ore body with an intersection of 8.2 meters at a grade of 22.4 grams per tonne (g/t). As with the previous question, we need to estimate the tonnage, so I’ll use the same simplifying assumption: a 1-meter by 1-meter cross-section with a rock density of 2.7 tonnes per cubic meter (a typical value for gold-bearing rock).

Given:

• Intersection: 8.2 meters (m)
• Grade: 22.4 grams per tonne (g/t)

Step-by-Step Calculation:

1. Volume of the ore body:
   • Length = 8.2 m
   • Assume width = 1 m (simplification)
   • Assume height/thickness = 1 m (simplification)
   • Volume = 8.2 m × 1 m × 1 m = 8.2 cubic meters (m³)
2. Tonnage of the ore body:
   • Average density of rock = 2.7 tonnes/m³
   • Tonnage = Volume × Density = 8.2 m³ × 2.7 tonnes/m³ = 22.14 tonnes
3. Gold content in grams:
   • Grade = 22.4 g/t
   • Total gold = Tonnage × Grade = 22.14 tonnes × 22.4 g/t = 495.936 grams
4. Convert grams to ounces:
   • 1 troy ounce = 31.1035 grams (standard for precious metals)
   • Ounces of gold = 495.936 grams ÷ 31.1035 grams/ounce ≈ 15.94 ounces