Question

A carbohydrate residue sample recovered at an archaeological site at Queen Charlotte Island, British Columbia is believed to be a food residue of the early indigenous people who lived in this area. Carbon analyses of the sample show: C-12-3.50 x 102 mole/g sample C-13-3.85 x 10 mole/g sample C-14 -2.10 x 10 mole/g sample Carbon analyses of a Pee Dee Belemnite sample are shown to be exactly the same as a modem plant sample, with the following information: C-12 = 3.50 x 102 mole/g sample C-13-3.90 x 10 mole/g sample C-14-3.50 x 10"mole/g sample Other necessary information: Half-life of CH4 = 5720 years Isotopic fractionation-{[(C-13/C-12 of sample) - (C-13/C-12 of standard)]/(C- 13/C-12 of standard)) x 1000 Isotopic fractionation (8 C-13) of C3 plants i.e. rice or wheat) --27 to -21 Isotopic fractionation (8 C-13) of C4 plants (i.e. maize or sugar cane).17 to .9 Based on the above information, (i) calculate how old this carbohydrate residue is, and (ii) predict the possible food source of the early indigenous people who lived in this area.

Answer 1

Based on the above information,

1. calculate how old this carbohydrate residue is

First, Calculate isotopic fractionation

[(C13/C-12 of sample) - (C-13/C-12 of standard)]/(C- 13/C-12 of standard)) x 1000

Susbtitute the values:

[(3.85 x 10 mole/g sample /3.50 x 102 mole/g sample) - (3.90 x 10 mole/g sample / 3.50 x 102 mole/g sample)]/(3.90 x 10 mole/g sample / 3.50 x 102 mole/g sample) x 1000

Isotopic fractionation is 11.89 % (or 0.1189)

calculate the C14 percentage decay:

If modern sample have C-14 3.50 x 10 mole/g sample this will be considered your 100%

And the food residue sample had C-14 2.10 x 10 mole/g sample. Then, calculate the corresponding percentage:

2.10 x 10 mole/g X 100% / 3.50 x 10 mole/g

The corresponding percentage of C14 is 60%

Second, use the formula to calculate how old a sample is by knowing the percentage of C14 decay:

t = [ln (N_f/N_o) / (isotopic fractionation) x t_1/2

Where N_f/N_o is the percentage od C14 decay, and t_1/2 is the half-life of CH4 = 5720 years and using the isotopic fractionation calculated before:

t = [ln (0.6) / (-0.1189) x 5720 years

t = 24,574 years

ii. Predict the possible food source of the early indigenous people who lived in this area.

Based on the above information, where isotopic fractionation was 0.1189, then it must have been an C4 plants (i.e. maize or sugar cane) whose isotopic fractionation is known to be between 0.17 to 0.9.