How many troy ounces of fine gold are in 25 grams of 9K gold?
⚡ Instant Calculated Answer
25 grams of 9K gold contains exactly 9.375 grams (0.3014 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $813.78 USD at $2,700/oz t.
0.3014 oz t
Primary Metric
9.375 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: 25 g × 0.3750 = 9.375 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 9.375 / 31.1034768 = 0.3014 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 9.375 / 11.6638 = 0.8038 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 12.5 g | 4.688 g | 0.1507 oz t | $406.89 |
| 20 g | 7.5 g | 0.2411 oz t | $650.97 |
| 25 g | 9.375 g | 0.3014 oz t | $813.78 |
| 30 g | 11.25 g | 0.3617 oz t | $976.59 |
| 37.5 g | 14.062 g | 0.4521 oz t | $1,220.67 |
| 50 g | 18.75 g | 0.6028 oz t | $1,627.56 |
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