How many troy ounces of fine gold are in 10 grams of 18K gold?
⚡ Instant Calculated Answer
10 grams of 18K gold contains exactly 7.5 grams (0.2411 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $650.97 USD at $2,700/oz t.
0.2411 oz t
Primary Metric
7.5 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: 10 g × 0.7500 = 7.5 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 7.5 / 31.1034768 = 0.2411 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 7.5 / 11.6638 = 0.643 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 5 g | 3.75 g | 0.1206 oz t | $325.62 |
| 8 g | 6.0 g | 0.1929 oz t | $520.83 |
| 10 g | 7.5 g | 0.2411 oz t | $650.97 |
| 12 g | 9.0 g | 0.2894 oz t | $781.38 |
| 15 g | 11.25 g | 0.3617 oz t | $976.59 |
| 20 g | 15.0 g | 0.4823 oz t | $1,302.21 |
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