How many troy ounces of fine gold are in 31.1035 grams of 9K gold?
⚡ Instant Calculated Answer
31.1035 grams of 9K gold contains exactly 11.664 grams (0.375 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $1,012.50 USD at $2,700/oz t.
0.375 oz t
Primary Metric
11.664 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: 31.1035 g × 0.3750 = 11.664 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 11.664 / 31.1034768 = 0.375 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 11.664 / 11.6638 = 1.0 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 15.55 g | 5.831 g | 0.1875 oz t | $506.25 |
| 24.88 g | 9.33 g | 0.3 oz t | $810.00 |
| 31.1 g | 11.663 g | 0.375 oz t | $1,012.50 |
| 37.32 g | 13.995 g | 0.4499 oz t | $1,214.73 |
| 46.66 g | 17.497 g | 0.5625 oz t | $1,518.75 |
| 62.21 g | 23.329 g | 0.75 oz t | $2,025.00 |
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