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Archeologists removed some charcoal from a Native American campfire, burned them in O2, and bubbled the CO2 formed into Ca(OH)2 solution (limewater). The CaCO3 that precipitated was filtered and dried. If 4.80 g of the CaCO3 had a radioactivity of 3.3 d/min, how long ago was the campfire? (t1/2 of 14C = 5730 years, and the ratio of 12C/14C in living organisms results in a specific activity of 15.3 d/min g.)
1 Answer
- Roger the MoleLv 75 years ago
(4.80 g CaCO3) / (100.0869 g CaCO3/mol) x (1 mol C / 1 mol CaCO3) x (12.01070 g C/mol) = 0.57601 g C
(3.3 d/min) / (0.57601 g C) = 5.7291 d/min g
Let z be the age of the campfire in years:
(5.7291 d/min g) / (15.3 d/min g) = (1/2) ^ (z / 5730 y)
Solve for z algebraically:
0.37445 = (1/2) ^ (z / 5730 y)
log 0.37445 = (z / 5730 y) log (1/2)
log 0.37445 / log (1/2) = z / 5730 y
1.41715500522 = z / 5730 y
z = 1.41715500522 (5730 y) = 8120 y
