Concentration of electrons in silicon sample with a phosphorus dopant

I'm studying electrical and computer engineering and all I remember from chemistry is from high school. We have a $\ce{Si}$ sample and we want to replace one in every million atoms with a phosphorus atom. 


I want to determine the concentration of conduction electrons after this process and I'm given the molar mass $\pu{0.028 kg/mol}$ and the density $\pu{2300 kg/m^3}$.


It might be very simple, but what can I get from these two? I checked molar mass since I haven't heard about it ever, but I'm getting more confused.




Unless there is something I'm missing entirely, the answer is quite straightforward. Lets assume that each phosphorus atom contributes one free electron. Then the concentration of electrons is equal to the concentration of $\ce{P}$ atoms doped to the silicon crystal:


$$c\_\mathrm{e} = c(\ce{P})\label{eq:1}\tag{1}$$


Concentration of phosphorus $c(\ce{P})$ can be found from its amount $n(\ce{P})$ and volume $V$:


$$c(\ce{P}) = \frac{n(\ce{P})}{V}\label{eq:2}\tag{2}$$


On the other hand, volume can also be determined from the mass $m$, density $\rho$ and molar mass $M$ of silicon:


$$V = \frac{m(\ce{Si})}{\rho(\ce{Si})} = \frac{n(\ce{Si})\cdot M(\ce{Si})}{\rho(\ce{Si})}\label{eq:3}\tag{3}$$


Now substituting unknown parameters in \eqref{eq:1} and \eqref{eq:2} with \eqref{eq:3}, and knowing that $n(\ce{P}) : n(\ce{Si}) = 1 : 10^6$, concentration of electrons can be found using the values you provided:


$$c\_\mathrm{e} = \frac{n(\ce{P})}{n(\ce{Si})} \cdot \frac{\rho(\ce{Si})}{M(\ce{Si})} = 10^{-6}\cdot\frac{\pu{2300 kg m-3}}{\pu{0.028 kg mol-1}} = \pu{8.21e-2 mol m-3}\tag{4}$$


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