How many grams of chloride ions are in 8.3 g of copper (II) chloride

Question

asked Jun 14, 2020 126k views

1 Answer

Andy Finkenstadt · Answer 1 · 2020-06-17T15:09:50+0000

$4.4\; {\rm g}$ .

Refer to a modern periodic table for relative atomic mass:

Formula of copper (II) chloride: CuCl₂.

Formula mass of copper (II) chloride:

63.546 + 2 × 35.45 =
$134.446\; {\rm g \cdot mol^(-1)}$ .

How many
${\rm mol}$ of formula units in 8.3 grams of CuCl₂?

$(8.3\; {\rm g}) / (134.446\; {\rm g \cdot mol^(-1)}) \approx 0.0617\; {\rm mol}$

Note the subscript "2" in the formula CuCl₂. There are
$2\; {\rm mol}$ of chloride ions in every
$1\; {\rm mol}$ formula unit of CuCl₂.

There would be
$2 * 0.0617\; {\rm mol} \approx 0.123\; {\rm mol}$ of
${\rm Cl^(-)}$ ions in
$0.0617\; {\rm mol}$ formula units of CuCl₂.

The mass of
$1\; {\rm mol}$ of chloride ion is 35.45 grams.

The mass of
$0.123\; {\rm mol}$ of
${\rm Cl^(-)}$ ions will be
$35.45 * 0.123 \approx 4.4\; {\rm g}$ .

Round the final value to two sig. fig. since the mass in the question is also 2 sig. fig.