How many troy ounces of fine gold are in 15 grams of 9K gold?
⚡ Instant Calculated Answer
15 grams of 9K gold contains exactly 5.625 grams (0.1808 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $488.16 USD at $2,700/oz t.
0.1808 oz t
Primary Metric
5.625 g Fine Au
Secondary Metric
📐 Mathematical Formula & Step-by-Step Substitution
- Determine fineness ratio of 9K gold: 9 / 24 = 0.3750 (or 375.0‰ millesimal fineness).
- Multiply gross weight by fineness: 15 g × 0.3750 = 5.625 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 5.625 / 31.1034768 = 0.1808 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 5.625 / 11.6638 = 0.4823 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 7.5 g | 2.812 g | 0.0904 oz t | $244.08 |
| 12 g | 4.5 g | 0.1447 oz t | $390.69 |
| 15 g | 5.625 g | 0.1808 oz t | $488.16 |
| 18 g | 6.75 g | 0.217 oz t | $585.90 |
| 22.5 g | 8.438 g | 0.2713 oz t | $732.51 |
| 30 g | 11.25 g | 0.3617 oz t | $976.59 |
Want to test custom numbers?
Launch the full interactive simulator pre-loaded with these exact calculation parameters.