How many troy ounces of fine gold are in 15 grams of 24K gold?
⚡ Instant Calculated Answer
15 grams of 24K gold contains exactly 14.998 grams (0.4822 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $1,301.94 USD at $2,700/oz t.
0.4822 oz t
Primary Metric
14.998 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: 15 g × 0.9999 = 14.998 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 14.998 / 31.1034768 = 0.4822 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 14.998 / 11.6638 = 1.2859 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 7.5 g | 7.499 g | 0.2411 oz t | $650.97 |
| 12 g | 11.999 g | 0.3858 oz t | $1,041.66 |
| 15 g | 14.998 g | 0.4822 oz t | $1,301.94 |
| 18 g | 17.998 g | 0.5786 oz t | $1,562.22 |
| 22.5 g | 22.498 g | 0.7233 oz t | $1,952.91 |
| 30 g | 29.997 g | 0.9644 oz t | $2,603.88 |
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