How many troy ounces of fine gold are in 10 grams of 9K gold?
⚡ Instant Calculated Answer
10 grams of 9K gold contains exactly 3.75 grams (0.1206 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $325.62 USD at $2,700/oz t.
0.1206 oz t
Primary Metric
3.75 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: 10 g × 0.3750 = 3.75 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 3.75 / 31.1034768 = 0.1206 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 3.75 / 11.6638 = 0.3215 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 5 g | 1.875 g | 0.0603 oz t | $162.81 |
| 8 g | 3.0 g | 0.0965 oz t | $260.55 |
| 10 g | 3.75 g | 0.1206 oz t | $325.62 |
| 12 g | 4.5 g | 0.1447 oz t | $390.69 |
| 15 g | 5.625 g | 0.1808 oz t | $488.16 |
| 20 g | 7.5 g | 0.2411 oz t | $650.97 |
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