How many troy ounces of fine gold are in 1000 grams of 18K gold?
⚡ Instant Calculated Answer
1000 grams of 18K gold contains exactly 750.0 grams (24.1131 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $65,105.37 USD at $2,700/oz t.
24.1131 oz t
Primary Metric
750.0 g Fine Au
Secondary Metric
📐 Mathematical Formula & Step-by-Step Substitution
- Determine fineness ratio of 18K gold: 18 / 24 = 0.7500 (or 750.0‰ millesimal fineness).
- Multiply gross weight by fineness: 1000 g × 0.7500 = 750.0 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 750.0 / 31.1034768 = 24.1131 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 750.0 / 11.6638 = 64.3015 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 500 g | 375.0 g | 12.0565 oz t | $32,552.55 |
| 800 g | 600.0 g | 19.2904 oz t | $52,084.08 |
| 1000 g | 750.0 g | 24.1131 oz t | $65,105.37 |
| 1200 g | 900.0 g | 28.9357 oz t | $78,126.39 |
| 1500 g | 1125.0 g | 36.1696 oz t | $97,657.92 |
| 2000 g | 1500.0 g | 48.2261 oz t | $130,210.47 |
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