How many troy ounces of fine gold are in 100 grams of 18K gold?
⚡ Instant Calculated Answer
100 grams of 18K gold contains exactly 75.0 grams (2.4113 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $6,510.51 USD at $2,700/oz t.
2.4113 oz t
Primary Metric
75.0 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: 100 g × 0.7500 = 75.0 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 75.0 / 31.1034768 = 2.4113 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 75.0 / 11.6638 = 6.4302 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 50 g | 37.5 g | 1.2057 oz t | $3,255.39 |
| 80 g | 60.0 g | 1.929 oz t | $5,208.30 |
| 100 g | 75.0 g | 2.4113 oz t | $6,510.51 |
| 120 g | 90.0 g | 2.8936 oz t | $7,812.72 |
| 150 g | 112.5 g | 3.617 oz t | $9,765.90 |
| 200 g | 150.0 g | 4.8226 oz t | $13,021.02 |
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