Question f4c23

May 13, 2015

The purity of the copper (II) carbonate will be equal to 79.3%.

$C u C {O}_{3 \left(s\right)} \stackrel{\textcolor{red}{\text{heat}}}{\to} C u {O}_{\left(s\right)} + C {O}_{2 \left(g\right)}$

Notice that you have a $1 : 1$ mole ratio between copper (II) carbonate and carbon dioxide, which means, regardless of how many moles of the former react, you'll produce the same number of moles of the latter.

Since you didn't provide the pressure and temperature conditions, I'll assume you're at STP (Standard Pressure and Temperature).

At STP , which implies a pressure of 100 kPa and a temperature of 273.15 K, 1 mole of any ideal gas occupies exactly 22.7 L. This is known as the molar volume of a gas at STP.

You can use the molar volume of a gas to determine how many moles of carbon dioxide were produced by the reaction

${n}_{C {O}_{2}} = {V}_{C {O}_{2}} / {V}_{\text{molar" = (3.600 * cancel("L"))/(22.7cancel("L")) = "0.1586 moles}}$ $C {O}_{2}$

According to the aforementioned mole ratio, if this many moles of $C {O}_{2}$ were produced, the number of moles of $C u C {O}_{3}$ that reacted was

0.1586cancel("moles "CO_2) * ("1 mole "CuCO_3)/(1cancel("mole "CO_2)) = "0.1586 moles" $C u C {O}_{3}$

Use the compound's molar mass to determine how many grams would contain this many moles

0.1586cancel("moles "CuCO_3) * "124 g"/(1cancel("mole "CuCO_3)) = "19.7 g"

Since the total sample weighs 24.8 g, its purity will be

$\text{% purity" = m_(CuCO_3)/m_"sample} \cdot 100$

"% purity" = (19.7cancel("g"))/(24.8cancel("g")) * 100 = color(green)("79.3%")#

SIDE NOTE Many problems still require you to use the old definition of STP, which specify a pressure of 1 atm and a temperature of 273.15 K. At that pressure and temperature, the molar volume of a gas is 22.4 L.

If your teacher or instructor wants you to use those values for pressure and temperature, simply redo the calculations using 22.4 L instead of 22.7 L.