# Original concentration calculation homework

A student performs the following gravimetric analysis of iron ions in a water system $$\ce{Fe^2+ (aq) + CO3^2- (aq) <=> FeCO3 (s)}$$ If this student used excess carbonate and obtained the following data, what was the original concentration in moles/L) of $\ce{Fe^2+}$ in the water sample?

Total Volume of Solution: 100.00 mL

Mass of $\ce{FeCO3}$ collected: 23.758 grams.

Change 100 ml to 0.1 liters $x + 115.583 = 23.758 \times 0.1 = 0.02~\mathrm{l}$ - would this be how you solve this problem?

• Hi, just a heads up: in order to make a new paragraph, you have to make two line breaks after a paragraph. Regarding your solution itself, the first thing I noticed is that you converted mL to L wrongly. Apart from that I'm afraid you'll have to make it a little clearer what you are trying to say. What does $x$ represent? Where does 11.583 come from? Oct 1 '15 at 11:41
• Trying to find the original concentration of the equation I used M1V1 = M2V2 Oct 1 '15 at 11:50
• liters should be .1 liters M1 is the molar mass of FeCO3 = 115.853g X V1 = 23.758g X .1 litter = .02 liters - would this be the original concentration Oct 1 '15 at 11:53

The mass of $\ce{FeCO3}$ is given as $m_{\ce{FeCO3}}=23.758\ \mathrm g$.

You already found that the molar mass of $\ce{FeCO3}$ is $M_{\ce{FeCO3}}=115.85\ \mathrm{g\ mol^{-1}}$.

For a pure sample, molar mass $M$ is defined as $$M=\frac mn$$ where $m$ is mass and $n$ is amount of substance.
Thus, the amount of $\ce{FeCO3}$ can be calculated as follows: \begin{align}n_{\ce{FeCO3}}&=\frac {m_{\ce{FeCO3}}}{M_{\ce{FeCO3}}}\\[6pt] &=\frac {23.758\ \mathrm g}{115.85\ \mathrm{g\ mol^{-1}}} \end{align}

The chemical equation $$\ce{Fe^2+ + CO3^2- <=> FeCO3}$$ indicates that $1\ \mathrm{mol}$ of $\ce{Fe^2+}$ yields $1\ \mathrm{mol}$ of $\ce{FeCO3}$. Therefore, $$n_{\ce{Fe^2+}}=n_{\ce{FeCO3}}$$

The volume of the solution is given as $V=100.00\ \mathrm{ml}=0.10000\ \mathrm l$.

The concentration $c$ is defined as $$c=\frac nV$$ where $n$ is amount of substance and $V$ is the volume of the solution. Thus, the concentration of $\ce{Fe^2+}$ can be calculated as follows: \begin{align}c_{\ce{Fe^2+}}&=\frac {n_{\ce{Fe^2+}}}V\\[6pt] &=\frac {n_{\ce{Fe^2+}}}{0.10000\ \mathrm l}\end{align}

• I consider this to be the blueprint of how to answer homework questions!
– Jan
Oct 21 '15 at 13:29