How many troy ounces of fine gold are in 5 grams of 9K gold?
⚡ Instant Calculated Answer
5 grams of 9K gold contains exactly 1.875 grams (0.0603 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $162.81 USD at $2,700/oz t.
0.0603 oz t
Primary Metric
1.875 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: 5 g × 0.3750 = 1.875 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 1.875 / 31.1034768 = 0.0603 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 1.875 / 11.6638 = 0.1608 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 2.5 g | 0.938 g | 0.0302 oz t | $81.54 |
| 4 g | 1.5 g | 0.0482 oz t | $130.14 |
| 5 g | 1.875 g | 0.0603 oz t | $162.81 |
| 6 g | 2.25 g | 0.0723 oz t | $195.21 |
| 7.5 g | 2.812 g | 0.0904 oz t | $244.08 |
| 10 g | 3.75 g | 0.1206 oz t | $325.62 |
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