How many troy ounces of fine gold are in 500 grams of 9K gold?
⚡ Instant Calculated Answer
500 grams of 9K gold contains exactly 187.5 grams (6.0283 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $16,276.41 USD at $2,700/oz t.
6.0283 oz t
Primary Metric
187.5 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: 500 g × 0.3750 = 187.5 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 187.5 / 31.1034768 = 6.0283 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 187.5 / 11.6638 = 16.0754 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 250 g | 93.75 g | 3.0141 oz t | $8,138.07 |
| 400 g | 150.0 g | 4.8226 oz t | $13,021.02 |
| 500 g | 187.5 g | 6.0283 oz t | $16,276.41 |
| 600 g | 225.0 g | 7.2339 oz t | $19,531.53 |
| 750 g | 281.25 g | 9.0424 oz t | $24,414.48 |
| 1000 g | 375.0 g | 12.0565 oz t | $32,552.55 |
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