Find the Relative Error in the Mass of the Product

I have the following question,



> 
> A student was given the assignment of $4.00$ $\pu{mL}$ of $0.50$ $\pu{M}$ $\ce{KI}$ with $16.00$ $\pu{mL}$
>  of $0.50$ $\pu{M}$ $\ce{Pb(NO3)2}$ in a stoichiometry study using the method of
>  continuous variations. If the student used only $3.50$ $\pu{mL}$ of $0.50$ $\pu{M}$ $\ce{KI}$,
>  what would be the relative error in the mass of the product?
> 
> 
> 


I need to find the relative error, I know that
Relative Error = Absolute Error/Actual Value. I also know that Absolute Error = Expected Value - Actual Value. 


I believe that the way to solve this is to find the moles of $\ce{KI}$ at $4.00$ $\pu{mL}$ and add that with the moles of $\ce{Pb(NO3)2}$ at $16.00$ $\pu{mL}$. Both at $0.50$ $\ce{M}$. This would be the Expected Value?


Then I need to find the moles of $\ce{KI}$ at $3.50$ $\pu{mL}$ and then add this with the moles of $\ce{Pb(NO3)2}$ at $16.00$ $\pu{mL}$. This would be the Actual Value?




More of a long comment than answer.


Well you are all wrong.


Absolute Error = Actual Value - Expected Value 


That way if you have too little product the error is negative, and if you have too much product the value is positive. 


Relative Error would typically be expressed as a percentage. 


$ \text{Relative Error} = \frac{\text{Absolute Error}}{\text{Expected Value}} \times 100\%$.


First when you mix $\ce{KI}$ and $\ce{Pb(NO\_3)\_2}$ what happens? 



> 
> Hint - What is the formula for the chemical reaction?
> 
> 
> 


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