How many troy ounces of fine gold are in 100 grams of 21K gold?
⚡ Instant Calculated Answer
100 grams of 21K gold contains exactly 87.5 grams (2.8132 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $7,595.64 USD at $2,700/oz t.
2.8132 oz t
Primary Metric
87.5 g Fine Au
Secondary Metric
📐 Mathematical Formula & Step-by-Step Substitution
- Determine fineness ratio of 21K gold: 21 / 24 = 0.8750 (or 875.0‰ millesimal fineness).
- Multiply gross weight by fineness: 100 g × 0.8750 = 87.5 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 87.5 / 31.1034768 = 2.8132 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 87.5 / 11.6638 = 7.5018 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 50 g | 43.75 g | 1.4066 oz t | $3,797.82 |
| 80 g | 70.0 g | 2.2506 oz t | $6,076.62 |
| 100 g | 87.5 g | 2.8132 oz t | $7,595.64 |
| 120 g | 105.0 g | 3.3758 oz t | $9,114.66 |
| 150 g | 131.25 g | 4.2198 oz t | $11,393.46 |
| 200 g | 175.0 g | 5.6264 oz t | $15,191.28 |
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