How many troy ounces of fine gold are in 5 grams of 22K gold?
⚡ Instant Calculated Answer
5 grams of 22K gold contains exactly 4.583 grams (0.1473 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $397.71 USD at $2,700/oz t.
0.1473 oz t
Primary Metric
4.583 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: 5 g × 0.9167 = 4.583 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 4.583 / 31.1034768 = 0.1473 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 4.583 / 11.6638 = 0.3929 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 2.5 g | 2.292 g | 0.0737 oz t | $198.99 |
| 4 g | 3.667 g | 0.1179 oz t | $318.33 |
| 5 g | 4.583 g | 0.1473 oz t | $397.71 |
| 6 g | 5.5 g | 0.1768 oz t | $477.36 |
| 7.5 g | 6.875 g | 0.221 oz t | $596.70 |
| 10 g | 9.167 g | 0.2947 oz t | $795.69 |
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