How to calculate the concentration of the elements after dissolution of iron(III) chloride?

Question

I was presented with the following problem:

In the laboratory you dissolve $\pu{24.7 g}$ of iron(III) chloride in a volumetric flask in water to a total volume of $\pu{375 ml}$.

• What is the molarity of the solution?
• What is the concentration of iron(III) cation in M?
• And what is the concentration of chloride anion in M?

I know how to do the first part of the problem, the answer is $\pu{0.406 M}\ \ce{FeCl3}$. But when I go to balance the equation, the computer system tells me the equation is wrong.

My equation: $$ 2\ce{FeCl3 -> 2 Fe + 3Cl2} $$

The computer system:

$$ \ce{FeCl3 -> Fe^3+ + 3Cl-} $$

I don't understand the computer systems' answer. Simply because when by itself, chlorine is a diatomic molecule therefore, should be $\ce{Cl2}$ not just $\ce{Cl}$. Am I wrong to think this? Am I correct or the computer system? If the computer system is correct, why?

Answer 1

When answering such problems, I recommend writing everything know down, then use the appropriate formula to solve for the unknown. For the first part you already did that correctly.

In the laboratory you dissolve $\pu{24.7 g}$ of iron(III) chloride in a volumetric flask in water to a total volume of $\pu{375 ml}$.

\begin{align} m(\ce{FeCl3 (s)}) &= \pu{24.7 g}\\ M(\ce{FeCl3}) &= \pu{162.3 g//mol}\\ V(\ce{FeCl3 (aq)}) &= \pu{375 mL}\\ \end{align}

• What is the molarity of the solution?

Molarity is another word for amount concentration, so you are looking for $c(\ce{FeCl3 (aq)})$. \begin{align} c &= \frac{n}{V}\\ n &= \frac{m}{M}\\ c(\ce{FeCl3 (aq)}) &= \frac{m(\ce{FeCl3 (s)})}{M(\ce{FeCl3})\cdot V(\ce{FeCl3 (aq)})}\\ c(\ce{FeCl3 (aq)}) &= \frac{\pu{24.7 g}}{\pu{162.3 g//mol}\cdot\pu{375E-3L}}\\ c(\ce{FeCl3 (aq)}) &= \pu{0.406 mol//L} \end{align}

• What is the concentration of iron(III) cation in M?
• And what is the concentration of chloride anion in M?

You are looking for the ion (amount) concentrations in your solution, i.e. $c(\ce{Fe^3+ (aq)})$ and $c(\ce{Cl- (aq)})$. For these two questions you need to know how the substance dissolves. The wording here is already very explicit, as it says iron(III) cation, i.e. $\ce{Fe^3+ (aq)}$, and chloride anion, i.e. $\ce{Cl-}$. Otherwise, with some more experience you will know that this compound is a salt and it dissociates into ions when dissolved in polar solvents like water. Therefore the reaction equation is $$\ce{FeCl3 (s) ->[H2O] Fe^3+ (aq) + 3 Cl- (aq)}.$$ From this you can derive the actual relationships between the ion concentrations. You can clearly see that there is one iron(III) ion and three chloride ions per formula of $\ce{FeCl3}$; therefore \begin{align} c(\ce{Fe^3+ (aq)})&= c(\ce{FeCl3 (aq)}) &&=\pu{0.406 mol//L},\\ c(\ce{Cl- (aq)}) &= 3\times c(\ce{FeCl3 (aq)}) &&=\pu{1.22 mol//L}. \end{align}

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The other equation would probably be appropriate when you (thermally) decompose the substance (in absence of everything) as it is the reverse equation which can be used to prepare the molecule: $$\ce{2 Fe (s) + 3 Cl2 (g) -> 2 FeCl3 (s)}.$$

Answer 2

you finish question 1 and you need to finish question 2 and 3 but why are you making a equation for that. it seem like that was what stopped you working in the first place.

but to make it balanced... you said dissolved in volumetric flask. if it dissolved its ionic. and that means there should be a "cation +" and a "anion -" hence the $$\ce{fe^3+ + 3cl- }$$ and the $\ce{fe^3+}$ is third oxidation state floating around in solution it could had been a $\ce{fe^2+}$ if you started with $\ce{fe(oh)2}$

so the pc is correct....

1 mole of $\ce{fecl3}$ will yield's 1,5 mole of cl2 therefore it has more $cl$ then $fe$ ions.

concentration of $fe$ ion in $M = 0.406$

concentration of $cl$ ion in $M = 1.218$

$final\ equation\ =$ $$molarity\ of\ \ce{fecl3} * number\ of\ fe\ ion's\ in\ the\ molecule$$ $$\ce{fe^3+}=\ 0.406M * 1 = 0.406M$$ $$molarity\ of\ \ce{fecl3} * number\ of\ cl\ ion's\ in\ the\ molecule$$ $$\ce{cl^-}=\ 0.406M * 3 = 1.218 M$$