Uncertainties Calculation

Recently I did an experiment on isolating casein and now that I am processing the results, I feel my method of uncertainty calculation is not feasible. The casein was isolated 3 times and the results are as follows: 2.386, 2.628 and 2.164 grams. The scale had an uncertainty of 3 decimal places. Furthermore, the casein was isolated from 100 mL of milk, measured using a measuring cylinder with an uncertainty of 1 mL. The mean uncertainty was found using the following formula:


$$Δx\_\mathrm{avg} = \frac{Δx}{\sqrt{N}} = \frac{R}{2\sqrt{N}}$$


Hence, my calculations were:


$$\mathrm{Range} = 2.628 - 2.164 = 0.464$$


Thus the uncertainty for the average would be:


$$\frac{0.464}{2\sqrt{3}} = 0.133945262 = 0.2$$


This would result in a final value for the average amount of casein in the isolated milk of $\pu{2.4 ± 0.2 g}$.


However, this error seems unreasonable considering that my measurements were to 3 decimal places. Furthermore, how would I add the uncertainty of the 100 mL of milk to this value? Would I simply add the uncertainties? 




If we have 2 variables 
with absolute uncertainties(+)
$A \pm \sigma\_A$ and $B \pm \sigma\_B$


(+) expressed as standard deviations, but applicable to confidence intervals as well.


Then $C = A \pm B$ has uncertainty 


$$\sigma\_C = \sqrt{{\sigma\_A}^2 + {\sigma\_B}^2}$$


For $C=A/B$ or $C=A\cdot B$,


$$\sigma\_C = C \cdot \sqrt{\left(\frac {\sigma\_A}{A}\right)^2+ \left(\frac {\sigma\_B}{B}\right)^2 }$$


Let $A$ is content of cassein in 100 ml of milk.   

Let $B$ is the actual volume of used milk.   

Let $C=A\cdot B$ is the mass of cassein in the volume of milk used in analysis.   



Then
$\sigma\_C$
is the estimation of the error of the amount of cassein (known as the result error)  

$\sigma\_A$
is the estimation of the error of the cassein mass in 100 ml of milk (unknown)  

$\sigma\_B$
is the estimation of the error of the milk volume (1 ml )  



The error introduced by scales can be neglected, or introduced as additive variable $D=0$ with absolute error estimation $\sigma\_D=0.001~ \text{g}$.


Another, the biggest factor is the used analytical method with its bias $E$ and error $\sigma\_E$.


$$\begin{align}
C &= D + E + A\cdot B \\
\sigma\_{AB} &= A\cdot B \cdot \sqrt{\left(\frac {\sigma\_A}{A}\right)^2+ \left(\frac {\sigma\_B}{B}\right)^2 }\\
\sigma\_C &= \sqrt{{\sigma\_D}^2 + {\sigma\_E}^2 + {\sigma\_{AB}}^2}\\
\sigma\_{AB} &= 2.4 \cdot \sqrt{\left(\frac {\sigma\_A}{2.4}\right)^2+ \left(\frac {1}{100}\right)^2 }\\
0.13 &= \sqrt{{0.001}^2 +{\sigma\_E}^2 + {\sigma\_{AB}}^2}\\
\end{align}$$


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