Basic carbon dating problems I've seen before, but this seems to take more details into account, both in model and the questions, which I have not seen before, nor understand. Anyone that can help me?
Problem:
Carbon consist of the stable isotope ${}^{12}$C and the unstable isotope ${}^{14}$C that have a half-time of 5730 years. Any other isotopes are assumed non-existing. The relationship between ${}^{14}$C and ${}^{12}$C in living material is $1.5\cdot10^{-12}$. 1 mol is $6\cdot10^{23}$ particles and ${}^{12}$C weighs 12 grams/mol and ${}^{14}$C about 14 grams/mol.
A. An item was said to be from year 400. ${}^{14}$C measurement in 1977 showed 6.08 decays/gram/year. From what year is this item?
B. An piece of char cole had 4.09 decays/gram/year. Living material had 6.68. When was the char cole made?
C. In skeleton residue the proportion ${}^{14}$C/${}^{12}$C was found to be 6.24% of what it is in living material. When did the person live.
My work (so far):
A. Let $N_{14}(t)=N_{14}(0)\cdot2^{-t/5730}$ denote the number of ${}^{14}$C particles before decay starts. Assume 1 gram material from the item. It consists of $N_{12}(0)$ ${}^{12}$C particles and $N_{14}(0)$ ${}^{14}$C particles. We have a total of $$ N_{12}(0)+N_{14}(0) $$ ${}^{12}$C and ${}^{14}$C particles.
Divide by $6\cdot10^{23}$ particles/mol and we get $$ \frac{N_{12}(0)}{6\cdot10^{23}}+\frac{N_{14}(0)}{6\cdot10^{23}} $$ mol and multiply by 12 and 14 g/mol we get $$ 12\cdot\frac{N_{12}(0)}{6\cdot10^{23}}+14\cdot\frac{N_{14}(0)}{6\cdot10^{23}} $$ grams which should equal 1, i.e. $$ 12\cdot\frac{N_{12}(0)}{6\cdot10^{23}}+14\cdot\frac{N_{14}(0)}{6\cdot10^{23}}=1 \quad\Leftrightarrow\quad 12\cdot\frac{\frac{N_{14}(0)}{1.5\cdot10^{-12}}}{6\cdot10^{23}}+14\cdot\frac{N_{14}(0)}{6\cdot10^{23}}=1 $$ since $$ \frac{N_{14}(0)}{N_{12}(0)}=1.5\cdot10^{-12} \quad\Leftrightarrow\quad N_{12}(0)=\frac{N_{14}(0)}{1.5\cdot10^{-12}}. $$
This gives $$ N_{14}(0)=7.5\cdot10^{10} $$ ${}^{14}$C particles. Hence $$ N_{14}(t) =N_{14}(0)\cdot2^{-t/5730} =7.5\cdot10^{10}\cdot2^{-t/5730} $$
If this is even correct calculations, I don't know how to continue. Any thoughts?
Edit: you are using the currently understood value of 5730 for the half-life. Apparently it was once accepted that the half-life is 5568 years. This resource is probably an older one that is using that value. I will edit my answer below to reflect this.
Assume there is one gram of carbon in front of you, almost all of which is $^{12}\mathrm{C}$. Back when the material was alive, it was also one gram of $^{12}\mathrm{C}$. The amounts of $^{14}\mathrm{C}$ are so small to not yet be relevant to the calculation.
So with one gram of $^{12}\mathrm{C}$, there was $\frac{1}{12}$ of a mole of $^{12}\mathrm{C}$. So there were about $\frac{1}{12}\times6\mathrm{E}23=5\mathrm{E}22$ atoms of $^{12}\mathrm{C}$.
The $1.5\mathrm{E}{-12}$ ratio is a ratio of atoms of $^{14}\mathrm{C}$ to atoms of $^{12}\mathrm{C}$. Therefore when the material was alive, there were $5\mathrm{E}22\times 1.5\mathrm{E}{-12}=7.5\mathrm{E}10$ atoms of $^{14}\mathrm{C}$.
So the $^{14}\mathrm{C}$ atom count function over time is $7.5\mathrm{E}10\left(\frac12\right)^{-t/5568}$. Its derivative is about $-9.337\mathrm{E}6\left(\frac12\right)^{-t/5568}$, in lost atoms per year. But part (A) says you are currently losing $6.08$ atoms per year (from your one gram sample). So $$\begin{align} -9.337\mathrm{E}6\left(\frac12\right)^{-t/5568}&\approx-6.08\\ \left(\frac12\right)^{-t/5568}&\approx6.512\mathrm{E}{-7}\\ \frac{t}{5568}&\approx20.55\\ t&\approx114424 \end{align}$$
This says the material is over 114 thousand years old. That's pretty far off from the suggested age in part (A). And I think carbon dating is not supposed to be usable/reliable beyond something like 50,000 years into the past.
In a comment, you say the age should really come out to 702 years (from 1275 to 1977). If that were the case, the current absolute loss rate of atoms would work out to be $1.019\mathrm{E}7$ atoms per gram per year, far more than $6.08$. I suspect something is misprinted somewhere.
Parts (B) and (C) seem a lot easier. For (B), solve $\left(\frac12\right)^{-t/5568}=\frac{4.09}{6.68}$. For part (C), solve $\left(\frac12\right)^{-t/5568}=0.0624$