How many troy ounces of fine gold are in 100 grams of 9K gold?
⚡ Instant Calculated Answer
100 grams of 9K gold contains exactly 37.5 grams (1.2057 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $3,255.39 USD at $2,700/oz t.
1.2057 oz t
Primary Metric
37.5 g Fine Au
Secondary Metric
📐 Mathematical Formula & Step-by-Step Substitution
- Determine fineness ratio of 9K gold: 9 / 24 = 0.3750 (or 375.0‰ millesimal fineness).
- Multiply gross weight by fineness: 100 g × 0.3750 = 37.5 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 37.5 / 31.1034768 = 1.2057 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 37.5 / 11.6638 = 3.2151 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 50 g | 18.75 g | 0.6028 oz t | $1,627.56 |
| 80 g | 30.0 g | 0.9645 oz t | $2,604.15 |
| 100 g | 37.5 g | 1.2057 oz t | $3,255.39 |
| 120 g | 45.0 g | 1.4468 oz t | $3,906.36 |
| 150 g | 56.25 g | 1.8085 oz t | $4,882.95 |
| 200 g | 75.0 g | 2.4113 oz t | $6,510.51 |
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