Question

Chloe t. followed the procedure of this experiment to determine the empirical formula of a compound of iron (fe) and chlorine (cl). to do so, she added 2.15 g of zn to a solution containing 1.750 g of fe(x)cl(y). after the reaction was complete, she isolated 0.771 g of fe. the mass of cl in the fe(x)cl(y) solution is

Answer 1

`m(\text{Cl}) = 0.979 \; \text{g}`

Step-by-step explanation

The
`1.750 \; \text{g}` sample contains only the atoms of

• Iron,
  `\text{Fe}`, and
• Chlorine,
  `\text{Cl}`

Thus no matter what natural numbers
`x` and
`y` are,
`m(\text{Fe}_x\text{Cl}_y) = m(\text{Fe}) + m(\text{Cl})` shall holds.

Recall that zinc
`\text{Zn}` is more reactive than iron as seen in the reactivity series; Adding zinc to the solution would reduce all iron atoms (regardless of the oxidation state,
`+2` or
`+3`) to their elementary form, hence displacing them out of the solution.
`0.771 \; \text{g}` of iron was obtained and therefore
`m(\text{Fe}) = 0.771 \; \text{g}`.

Thus

`\begin{array}{lll}m(\text{Cl})& =& m(\text{Fe}_x\text{Cl}_y) - m(\text{Fe})\\ & = &1.750 \; \text{g}- 0.771\; \text{g}\\ &=&0.979\; \text{g}\end{array}`