How many troy ounces of fine gold are in 1000 grams of 24K gold?
⚡ Instant Calculated Answer
1000 grams of 24K gold contains exactly 999.9 grams (32.1475 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $86,798.25 USD at $2,700/oz t.
32.1475 oz t
Primary Metric
999.9 g Fine Au
Secondary Metric
📐 Mathematical Formula & Step-by-Step Substitution
- Determine fineness ratio of 24K gold: 24 / 24 = 0.9999 (or 999.9‰ millesimal fineness).
- Multiply gross weight by fineness: 1000 g × 0.9999 = 999.9 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 999.9 / 31.1034768 = 32.1475 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 999.9 / 11.6638 = 85.7268 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 500 g | 499.95 g | 16.0738 oz t | $43,399.26 |
| 800 g | 799.92 g | 25.718 oz t | $69,438.60 |
| 1000 g | 999.9 g | 32.1475 oz t | $86,798.25 |
| 1200 g | 1199.88 g | 38.577 oz t | $104,157.90 |
| 1500 g | 1499.85 g | 48.2213 oz t | $130,197.51 |
| 2000 g | 1999.8 g | 64.2951 oz t | $173,596.77 |
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