How many troy ounces of fine gold are in 10 grams of 21K gold?
⚡ Instant Calculated Answer
10 grams of 21K gold contains exactly 8.75 grams (0.2813 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $759.51 USD at $2,700/oz t.
0.2813 oz t
Primary Metric
8.75 g Fine Au
Secondary Metric
📐 Mathematical Formula & Step-by-Step Substitution
- Determine fineness ratio of 21K gold: 21 / 24 = 0.8750 (or 875.0‰ millesimal fineness).
- Multiply gross weight by fineness: 10 g × 0.8750 = 8.75 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 8.75 / 31.1034768 = 0.2813 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 8.75 / 11.6638 = 0.7502 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 5 g | 4.375 g | 0.1407 oz t | $379.89 |
| 8 g | 7.0 g | 0.2251 oz t | $607.77 |
| 10 g | 8.75 g | 0.2813 oz t | $759.51 |
| 12 g | 10.5 g | 0.3376 oz t | $911.52 |
| 15 g | 13.125 g | 0.422 oz t | $1,139.40 |
| 20 g | 17.5 g | 0.5626 oz t | $1,519.02 |
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