How many troy ounces of fine gold are in 1000 grams of 9K gold?
⚡ Instant Calculated Answer
1000 grams of 9K gold contains exactly 375.0 grams (12.0565 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $32,552.55 USD at $2,700/oz t.
12.0565 oz t
Primary Metric
375.0 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: 1000 g × 0.3750 = 375.0 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 375.0 / 31.1034768 = 12.0565 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 375.0 / 11.6638 = 32.1508 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 500 g | 187.5 g | 6.0283 oz t | $16,276.41 |
| 800 g | 300.0 g | 9.6452 oz t | $26,042.04 |
| 1000 g | 375.0 g | 12.0565 oz t | $32,552.55 |
| 1200 g | 450.0 g | 14.4678 oz t | $39,063.06 |
| 1500 g | 562.5 g | 18.0848 oz t | $48,828.96 |
| 2000 g | 750.0 g | 24.1131 oz t | $65,105.37 |
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