How many troy ounces of fine gold are in 31.1035 grams of 18K gold?
⚡ Instant Calculated Answer
31.1035 grams of 18K gold contains exactly 23.328 grams (0.75 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $2,025.00 USD at $2,700/oz t.
0.75 oz t
Primary Metric
23.328 g Fine Au
Secondary Metric
📐 Mathematical Formula & Step-by-Step Substitution
- Determine fineness ratio of 18K gold: 18 / 24 = 0.7500 (or 750.0‰ millesimal fineness).
- Multiply gross weight by fineness: 31.1035 g × 0.7500 = 23.328 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 23.328 / 31.1034768 = 0.75 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 23.328 / 11.6638 = 2.0 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 15.55 g | 11.663 g | 0.375 oz t | $1,012.50 |
| 24.88 g | 18.66 g | 0.5999 oz t | $1,619.73 |
| 31.1 g | 23.325 g | 0.7499 oz t | $2,024.73 |
| 37.32 g | 27.99 g | 0.8999 oz t | $2,429.73 |
| 46.66 g | 34.995 g | 1.1251 oz t | $3,037.77 |
| 62.21 g | 46.657 g | 1.5001 oz t | $4,050.27 |
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