American Gold Buffalo (1 oz) melt value
United States, 2006–present. Contains 1 troy oz of pure gold. The value below updates with the live spot price.
Specifications
| Country | United States |
|---|---|
| Years | 2006–present |
| Composition | .9999 fine gold |
| Total weight | 31.108 g |
| Purity | .9999 (99.99% gold) |
| Pure gold | 31.103 g (1 troy oz) |
| Diameter | 32.7 mm |
| Edge | Reeded |
| Obverse design | James Earle Fraser |
| Reverse design | James Earle Fraser |
| Face value | $50.00 |
Specifications: U.S. Mint specifications and U.S. coinage law.
Common questions
How much gold is in a American Gold Buffalo (1 oz)?
A American Gold Buffalo (1 oz) contains 1 troy ounces (31.103 grams) of pure gold. It weighs 31.108 grams in total and is .9999 fine.
How is the melt value of a American Gold Buffalo (1 oz) calculated?
Multiply the pure gold content (1 troy oz) by the current gold spot price. The calculator on this page does this automatically with live prices.
What will a dealer pay for a American Gold Buffalo (1 oz)?
Bullion coins and bars are usually bought close to the spot value of their metal content, while dealers sell them at a premium over spot. Ask any buyer what percent of melt they are paying and compare.
Related melt values
Melt value is an estimate of metal content at spot price. It is not an offer, appraisal or measure of collector value.