Michael S. answered • 13d
B.S. in Chemistry, Indiana University; Organic Chem Teaching Intern
342 g of NaAu(CN)2
4 Au + 8 NaCN + O2 + 2 H2O → 4 NaAu(CN)2 + 4 NaOH
Step 1 — mole ratio from oxygen to product
O2 has an implied coefficient of 1, and NaAu(CN)2 has 4, so the ratio is 4:1:
n = 0.314 mol O2 × (4 mol NaAu(CN)2 / 1 mol O2) = 1.256 mol NaAu(CN)2
The invisible 1 in front of O2 is easy to overlook — but it is what makes this ratio 4, not 1.
Step 2 — molar mass of NaAu(CN)2
Na: 22.99
Au: 196.97
C: 2 × 12.01 = 24.02
N: 2 × 14.01 = 28.01
Total = 272.0 g/mol
Read the formula carefully: the subscript 2 sits outside the parentheses, so it multiplies both the carbon and the nitrogen. Two cyanides, not one — counting a single CN gives 245.98 g/mol and an answer about 10% low.
Step 3 — mass
m = (1.256 mol)(272.0 g/mol) = 342 g
Three significant figures, matching the 0.314 mol you were given.
Sanity check: gold is heavy, so a little over a mole of a gold compound landing in the hundreds of grams is entirely reasonable. If your answer came out near 85 g, you used a 1:1 ratio and skipped the coefficient.
About the chemistry: this is the cyanidation (MacArthur–Forrest) process, still the dominant method of extracting gold worldwide. It works because gold is otherwise almost unreactive — cyanide is one of the few ligands that will pull Au into solution, forming the very stable dicyanoaurate(I) complex, with oxygen serving as the oxidant that takes Au(0) to Au(I). Notice the balanced equation confirms this: oxygen is consumed, and the gold in the product is +1.
Worth knowing that the process is also why cyanide handling dominates mine safety and tailings regulation — the same stability that makes the extraction work makes the waste streams hazardous.
