# Percent yield - 1 | Chemical Calculations

15.0 g of copper are reacted with an excess of chlorine gas. If the percent yield is 87.5%, how many grams of copper(II) chloride are obtained?

Chemical equation:

Cu + Cl2 → CuCl2

From the chemical equation: 1 mole of CuCl2 is produced for each mole of Cu.
nCuCl2 = nCu
nCuCl2 = $\frac{{\mathrm{m}}_{\mathrm{Cu}}}{{\mathrm{M}}_{\mathrm{Cu}}}$ = $\frac{15.0}{63.5}$ =2.36 x 10-1 mol
mCuCl2 = nCuCl2 x MCuCl2 = 2.36 x 10-1 x 134.45 = 31.8 g

% yield = mobtained / mtheoretical x 100 = mobtained / mCuCl2 x 100

⇒ mobtained = 0.875 x 31.8 = 27.8 g