How many troy ounces of fine gold are in 31.1035 grams of 14K gold?
⚡ Instant Calculated Answer
31.1035 grams of 14K gold contains exactly 18.144 grams (0.5833 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $1,574.91 USD at $2,700/oz t.
0.5833 oz t
Primary Metric
18.144 g Fine Au
Secondary Metric
📐 Mathematical Formula & Step-by-Step Substitution
- Determine fineness ratio of 14K gold: 14 / 24 = 0.5833 (or 583.3‰ millesimal fineness).
- Multiply gross weight by fineness: 31.1035 g × 0.5833 = 18.144 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 18.144 / 31.1034768 = 0.5833 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 18.144 / 11.6638 = 1.5556 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 15.55 g | 9.071 g | 0.2916 oz t | $787.32 |
| 24.88 g | 14.513 g | 0.4666 oz t | $1,259.82 |
| 31.1 g | 18.142 g | 0.5833 oz t | $1,574.91 |
| 37.32 g | 21.77 g | 0.6999 oz t | $1,889.73 |
| 46.66 g | 27.218 g | 0.8751 oz t | $2,362.77 |
| 62.21 g | 36.289 g | 1.1667 oz t | $3,150.09 |
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