How many troy ounces of fine gold are in 31.1035 grams of 24K gold?
⚡ Instant Calculated Answer
31.1035 grams of 24K gold contains exactly 31.1 grams (0.9999 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $2,699.73 USD at $2,700/oz t.
0.9999 oz t
Primary Metric
31.1 g Fine Au
Secondary Metric
📐 Mathematical Formula & Step-by-Step Substitution
- Determine fineness ratio of 24K gold: 24 / 24 = 0.9999 (or 999.9‰ millesimal fineness).
- Multiply gross weight by fineness: 31.1035 g × 0.9999 = 31.1 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 31.1 / 31.1034768 = 0.9999 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 31.1 / 11.6638 = 2.6664 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 15.55 g | 15.548 g | 0.4999 oz t | $1,349.73 |
| 24.88 g | 24.878 g | 0.7998 oz t | $2,159.46 |
| 31.1 g | 31.097 g | 0.9998 oz t | $2,699.46 |
| 37.32 g | 37.316 g | 1.1997 oz t | $3,239.19 |
| 46.66 g | 46.655 g | 1.5 oz t | $4,050.00 |
| 62.21 g | 62.204 g | 1.9999 oz t | $5,399.73 |
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