A small sample of wood from an archaeological site in Clo-vis, New Mexico, was burned in O2 and the CO2 produced was bubbled through a solution of Ba(OH)2 to produce a precipitation of BaCO3. When the BaCO3 was collected by filtration, a 1.000 g sample was found to have a radioactivity of 4.0 x 10^-5 Bq. The half-life of 14C is 5715 y, and living organisms have a radioactivity due to 14C if 15.3 disintegrations/min per gram of carbon. If the analysis was carried out in 1960, what is the date of the Clovis site?

Video transcript

Welcome back, everyone. We need to determine the age of an ancient fabric sample. You burn a small piece of fabric and excess oxygen and pass the produced carbon dioxide through slaked lime, calcium to hydroxide to produce a precipitate of calcium carbonate. The radio activity of a 0.545 g sample of calcium carbonate is 2.15 times 10 to the negative fourth power becquerel given the radio activity of the sample is due to carbon 14 Which has a half life of 5,715 years. And the radio activity of the recent samples is 15.3 disintegration per minute program of carbon. How old is the fabric sample? Alright. So we're going to begin with initially recalling that whenever we have radioactive decay, we're always going to follow first order kinetics. And so we want to recall the following rate law in which we would take the natural log Ln of N where N represents our radio activity present in our precipitate calcium carbonate. And then we have this divided by N0 which represents the radio activity in the carbon 14 sample. This is all set equal to negative one times our decay constant K, which is then multiplied by time T. So again, noting down that N is our radio activity in our precipitate calcium carbonate sample. And, and zero again, it's our radio activity in our carbon 14 sample, which is our initial conditions. Now, we are actually given N in the prompt where it tells us that we have The sample due to carbon With a radio activity of the recent samples being 15.3 disintegration per minute per gram of carbon. And so that is actually our value for N. So we can say N is equal to 15. disintegration is per minute per gram of carbon. And we don't know, we're sorry, that is our initial, so sorry about that mistake. This is the value we know now, but we don't know the radio activity of calcium carbonate because they're given to us in Becquerels. So we have it in Becquerels, but we need it in disintegration per minute program of carbon just like our initial sample. And so let's begin by finding our Mueller mass of calcium carbonate using our periodic tables Where for our mass of calcium, we find it in group two a and see that it has a mass of 40.07, 8 g per mole. Adding to this, we have our massive carbon which we see in group four A on the periodic table which has a mass of point oh one g per mole. And adding to this, we have three atoms of oxygen. So we multiply by three and an oxygen in group six A has a mass of 16.0 g per mole. And so we would result in a molar mass for calcium carbonate or precipitate equal to 100.08, 8 g per mole. Now, with the smaller mass, we want to figure out the mass of carbon in our calcium carbonate sample. And so referring back to the prompt, we see that they tell us the mass of calcium carbonate is 0.545 g for the sample. So we're going to begin with that. And we're gonna utilize the molar mass that we just calculated in the denominator to go from grams of calcium carbonate, two moles of calcium carbonate. And so we would plug in in the denominator 100.0 88 g for one more. So canceling out grams, we're going to move from our next step in molds of calcium carbonate, two moles of carbon where we're going to utilize the Chemical formula of calcium carbonate. And we see that we just have one mole of carbon. So we have for one mole of our calcium carbonate, one mole of carbon, that's a 1-1 ratio. And so canceling out moles of calcium carbonate. We now want to figure out again the mass of carbon. So we're going to multiply by our last conversion factor where we plug in the molar mass of carbon in the numerator And then moles of carbon in the denominator. And so we see that we have on the periodic table 12.01 g of carbon equivalent to one mole of carbon. So canceling out moles of carbon, we're left with grams of carbon, which is what we want. And this calculates for a mass as 0. 9695 g of carbon in our calcium carbonate sample. Now, with this information, we can figure out the radio activity of our calcium carbonate sample in the proper units. But we need to recall that one becquerel is equivalent to one disintegration per second. So we're going to have to do so, Akiyama tree. So again, calculating the calcium carbonate radio activity in the proper units, since we're given Becquerels, we're finding our term end here. And so we're going to begin with the radio activity that they give us in Becquerels as 2. times 10 to the negative fourth power. But instead of Becquerels, we're going to interpret it as disintegration per second. And then we need to divide by our massive carbon in our calcium carbonate, which we just determined above as 0.6539695 g of carbon. And so we need to get rid of the second units. So we're going to multiply by the next conversion factor between 60 seconds in the numerator and one minute in the denominator. And so canceling out 60 seconds with seconds in the denominator. Here, we're left with disintegration per minute per gram of carbon, which is what we want. This is going to result in a radio activity equal to a value of 0.1 97 to 569. And then our units as we want our dis integrations per minute per gram of carbon. So now that we have each of our radio activity of our samples and the matching units, we can go ahead and get into the rate law to calculate for time to get the age of the fat burke. But recall that we were not given our DK constant. So recall that we can calculate our decay constant K by taking the Sorry, the Ln of two. This is going to be divided by the half life given to us in the prompt. And so to calculate this, we'll get that the Ln of two in our numerator is equal to a value of 0.69, 3.5 life given to us in the prompt for carbon is 5715 years. So in our calculators were going to results in a decay constant of 1.2125984 inverse years. And it's important to have this many significant figures because we want an accurate answer. And by that I mean significant figures after the decimal. So let's go ahead and get into our rate law to calculate for a time. So age of fabric is the final step here where we would begin with our formula where as we stated, we would take the natural log of N where we determined N for our activity of carbon in the calcium carbonate sample being 0.1972569 disintegration per minute per gram of carbon, which is divided by and zero, which is our radio activity of carbon in the carbon. 14 recent samples given in the prompt as 15. disintegration per minute per gram of carbon. This is all set equal to negative one times are decay constant which we just solved above as 1.2125984 inverse years which are multiplied by the time or the age of our fabric in this case. So we need to begin by simplifying the left hand side as well as multiplying our right hand side. And so what we would do is for the Ln of our quotient on the left, we would result in a value of negative 4.3511012, which is equal to the right hand side, which is the product as negative 1.2125984 inverse years. And so we need to isolate T, so we're going to divide both sides by negative 1.2125984 inverse years. This cancels out on the right hand side and on the left hand side, we do the same H sorry about that. Alright. And we will now have T isolated where we would find that T is equal to a value of 35882 . years. Now notice that when we divided by inverse years here, we now have years in the numerator. So we no longer have inverse years. But recall that in the prompt and sorry, this is a t here. Recall that in the prompt, we have a minimum number of 366. So we need to round this 2, 366, which will result in a value of 35, years. And this would be the age of our fabric sample which corresponds to choice C in the multiple choice. So it's highlighted in yellow is our final answer. I hope that everything I explained was clear. If you have any questions, please leave them down below. And I'll see everyone in the next practice video.