How many troy ounces of fine gold are in 10 grams of 22K gold?
⚡ Instant Calculated Answer
10 grams of 22K gold contains exactly 9.167 grams (0.2947 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $795.69 USD at $2,700/oz t.
0.2947 oz t
Primary Metric
9.167 g Fine Au
Secondary Metric
📐 Mathematical Formula & Step-by-Step Substitution
- Determine fineness ratio of 22K gold: 22 / 24 = 0.9167 (or 916.7‰ millesimal fineness).
- Multiply gross weight by fineness: 10 g × 0.9167 = 9.167 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 9.167 / 31.1034768 = 0.2947 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 9.167 / 11.6638 = 0.7859 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 5 g | 4.583 g | 0.1473 oz t | $397.71 |
| 8 g | 7.333 g | 0.2358 oz t | $636.66 |
| 10 g | 9.167 g | 0.2947 oz t | $795.69 |
| 12 g | 11.0 g | 0.3537 oz t | $954.99 |
| 15 g | 13.75 g | 0.4421 oz t | $1,193.67 |
| 20 g | 18.333 g | 0.5894 oz t | $1,591.38 |
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