How many troy ounces of fine gold are in 50 grams of 9K gold?
⚡ Instant Calculated Answer
50 grams of 9K gold contains exactly 18.75 grams (0.6028 troy ounces) of pure fine gold, with an intrinsic scrap melt value of approximately $1,627.56 USD at $2,700/oz t.
0.6028 oz t
Primary Metric
18.75 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: 50 g × 0.3750 = 18.75 grams of fine elemental gold (Au).
- Convert grams to troy ounces (1 oz t = 31.1034768 g): 18.75 / 31.1034768 = 0.6028 troy ounces.
- Convert grams to tolas (1 tola = 11.6638 g): 18.75 / 11.6638 = 1.6075 tolas.
📊 Nearby Parameter Reference Table
| Gross Weight | Fine Gold (g) | Fine Gold (oz t) | Melt Value ($2,700/oz) |
|---|---|---|---|
| 25 g | 9.375 g | 0.3014 oz t | $813.78 |
| 40 g | 15.0 g | 0.4823 oz t | $1,302.21 |
| 50 g | 18.75 g | 0.6028 oz t | $1,627.56 |
| 60 g | 22.5 g | 0.7234 oz t | $1,953.18 |
| 75 g | 28.125 g | 0.9042 oz t | $2,441.34 |
| 100 g | 37.5 g | 1.2057 oz t | $3,255.39 |
Want to test custom numbers?
Launch the full interactive simulator pre-loaded with these exact calculation parameters.