# Percentage error. How to find percentage error?

The Question

Calculate the percentage dissociation of $\ce{H2S(g)}$ if $0.1\ \mathrm{mol}$ of $\ce{H2S}$ is kept in a $0.5\ \mathrm L$ flask at $1000\ \mathrm K$. The value for $K_c$ for the reaction $\ce{2H2S(g) <=> 2H2(g) + S2(g)}$ is $1\times10^{-7}$.

MY ATTEMPT Initially there are $0.1\ \mathrm{mol}$ of $\ce{H2S}$ and nothing for $\ce{H2}$ and $\ce{S2}$

Now at equilibrium let us assume $x$ moles dissociated so

$0.1 - x\ \mathrm{mol}$ of $\ce{H2S}$, $x\ \mathrm{mol}$ of $\ce{H2}$, and $x/2\ \mathrm{mol}$ of $\ce{S2}$

Now I know that degree of dissociation = $\alpha$ = amount dissociated/ initial amount in our case $x/0.1=\alpha$ so $x = 0.1\alpha$. So I substitute.

now we have $0.1-0.1\alpha$ of $\ce{H2S}$, $0.1\alpha$ of $\ce{H2}$, and $0.1\alpha/2$ of $\ce{S2}$

Thus $$K_c = \frac{\left[0.1\alpha/2V\right](\ce{S2})\left[0.1\alpha/V\right]^2(\ce{H2})}{[0.1-0.1\alpha/V]^2(\ce{H2S})}$$ Substituting value of $V$ and solving I reach here

$$\frac{0.1\alpha^3\alpha^3}{( 0.1 - 0.1\alpha)^2}$$

Now my doubt. I assumed $0.1\alpha$ very small in comparison to $0.1$ so I neglected it and assumed the whole $0.1 - 0.1\alpha$ to be $0.1$ only. Now my doubt is that what is the percentage error in my assumption.

On further solving I get percentage of degree of dissociation as $1\,\%$ and that is correct. My method is correct, my only doubt is how to calculate percentage error in my assumption.

$$\frac{0.1\alpha^3\alpha^3}{( 0.1 - 0.1\alpha)^2}$$
So the molarity of the solution is 0.1 +/- 0.05. If $\alpha < 0.1$ then the assumption that $0.1\alpha$ is negligible is reasonable.
(2) You can solve the equation for alpha exactly by trial and error. Solve as you did for $\alpha$, then use a factor 10 times larger and one 10 times smaller. One number should be too big and the other too small as compared to Kc. Calculate the average of the two values and test that number. That average will be too big or two small. Discard one side of the bracket. So you're creating an ever shrinking bracket about the "true" value. 6-8 iterations should get you very close.
Now with the "true value" and the estimate from the assumption that $0.1\alpha$ was negligible you can calculate the "real" percentage error for $\alpha$.