0.914 g of CaCO3 heated CaCO3→CaO+ CO2. The gas was sealed 225 mL container over water at 25.0°C the pressure141.4 torr. The vapor pressure of water at 25.0°C is 23.8 torr.

Question

A 0.914 g sample of a mixture of calcium carbonate and an inert impurity is heated strongly until the calcium carbonate fully decomposes according to the reaction below:

CaCO₃ (s) → CaO (s) + CO₂ (g)

 

The gaseous product of the reaction is collected in sealed 225 mL container over water at 25.0°C, and the pressure inside the container is measured to be 141.4 torr. The vapor pressure of water at 25.0°C is 23.8 torr. What was the percent by mass of calcium carbonate in the original mixture?

Answer 1

First we will determine the number of moles of CO₂ that were formed and collected.

To do this, we will use the ideal gas law, PV = nRT, but before we can use the pressure (P) we must correct for the vapor pressure of water at 25ºC. So, we have the following values:

P = pressure of CO₂ = 141.4 torr - 23.8 torr = 117.6 torr x 1 atm/760 torr = 0.155 atm

V = volume in liters = 225 ml x 1 L/1000 ml = 0.225 L

n = moles = ?

R = gas constant = 0.0821 Latm/Kmol (since we are using atm in this value, we will change torr to atm)

T = temperature in K = 25.0ºC + 273.15 = 298.15K

Solving the ideal gas law for n (moles) we have..

n = PV/RT = (0.155 atm)(0.225L)/(0.0821 Latm/Kmol)(298.15K)

n = 0.00142 moles of CO₂

From this value and the balanced equation we can find the moles of CaCO₃ originally present:

0.00142 mol CO₂ x 1 mol CaCO₃/mol CO₂ = 0.00142 moles CaCO₃ originally present

Converting this to grams of CaCO₃ we have 0.00142 mol CaCO₃ x 100 g/mol = 0.142 g CaCO₃

Since the original sample weighed 0.914 g, the % CaCO₃ by mass = 0.142 g/0.914 g (x100) = 15.6%