How many troy ounces of fine gold are in 150 grams of 18K gold?
⚡ Instant Calculated Answer
150 grams of 18K gold contains exactly 112.5 grams (3.617 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $9,765.90 USD at $2,700/oz t.
3.617 oz t
Primary Metric
112.5 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: 150 g × 0.7500 = 112.5 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 112.5 / 31.1034768 = 3.617 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 112.5 / 11.6638 = 9.6452 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 75 g | 56.25 g | 1.8085 oz t | $4,882.95 |
| 120 g | 90.0 g | 2.8936 oz t | $7,812.72 |
| 150 g | 112.5 g | 3.617 oz t | $9,765.90 |
| 180 g | 135.0 g | 4.3404 oz t | $11,719.08 |
| 225 g | 168.75 g | 5.4254 oz t | $14,648.58 |
| 300 g | 225.0 g | 7.2339 oz t | $19,531.53 |
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