Question

a 1.537g sample of an unknown chloride compound was treated with excess AgNO3 and the AgCl product was collected and dried. The mass of recovered AgCl was 4.081g. What is the identity of the unknown chloride compound?
explain

Answer 1

Answer on Question #56087 - Chemistry - General chemistry

Question:

a 1.537g sample of an unknown chloride compound was treated with excess AgNO₃ and the AgCl product was collected and dried. The mass of recovered AgCl was 4.081g. What is the identity of the unknown chloride compound?

explain

Solution:

With the addition of silver nitrate $AgNO_{3}$ to unknown metal salt $MCl_{x}$, the precipitation of silver chloride $AgCl$ occurs. The model reaction equation is:

$$xAgNO_{3} + MCl_{x} \rightarrow xAgCl + M(NO_{3})_{x}$$

Then, let's calculate the number of the moles of $AgCl$ precipitated:

$$n(AgCl) = \frac{m(AgCl)}{M(AgCl)} = \frac{4.081\,g}{143.32\,g/mol} = 0.02848\,mol$$

As it is evident from $AgCl$ formula, the number of the moles of $AgCl$ is equal to the number of the moles of $Cl^{-}$ reacted and the number of the moles of unknown chloride will be $n(AgCl)/x$.

Let's calculate the molar mass of the unknown chloride:

$$M(MCl_{x}) = \frac{m(MCl_{x})}{n(MCl_{x})} \cdot x = \frac{1.537\,g}{0.02848\,mol} \cdot x = 53.98x$$

Then, molar mass of the metal is $M(M) = M(MCl_{x}) - xM(Cl)$:

$$M(M) = x(53.98 - 35.45) = 18.53x$$

Let's presume, that the $x$ value is 1, or 2, or 3. The molar mass of the metal, calculated with $x = 1$ and $x = 2$ don't correspond any elements in the periodic table (18.5 and 37.1 g/mol, respectively). If we use an assumption that $x = 3$, we will get 55.6 g/mol value, we can find out that the unknown metal is iron. Hence, the unknown compound is $FeCl_{3}$.

Answer: $FeCl_{3}$

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