Experiment 13 Report Sheet A Carbonate Analysis: Molar Volume of Carbon Dioxide

Question

Experiment 13 Report Sheet A Carbonate Analysis: Molar Volume of Carbon Dioxide Lab Sec Name Desk No. Date A. Sample Preparation and Setup Apparatas Calculatioe of mass of CaCO, sample for analysis (from Prelaboratory Assignment, question 2) Unknown sample no. Trial I Trial 2 1. Mass of sample (g) -45.6559- 2. Mass of generator+ sample before reaction (g) 4- 3. Instructor's approval of apparatus C. Determination of Volume, Temperature, and Pressure of the Carbon Dioxide Gas 1. Initial reading of volume of water in CO.-collecting graduated cylinder (mL) 2. Final reading of volume of water in Co,-collecting 34 omL graduated cylinder (mL) 3. Volume of CO.(g) collected (L) 4. Temperature of water (c) 5. Barometric pressure (rorr) 6. Vapor pressure of H,0 at"C (torr) 7. Pressure of dry CO(g) (forr) D. Amount of Carbon Dioxide Gas Evolved 1. Mass of generator +sample after reaction (g) 2. Mass loss of generator mass CO2 evolved (g) 3. Moles of CO, evolved (mol 210 20.0 6. 04 0.001w Experiment 13 187

Explanation / Answer

Trial 2

Vapor pressure of H2O at 20 C = 17.5 torr

Pressure of dry CO2 = Barometric pressure - Vapor pressure of water

= 760.9 - 17.5 = 743.4 torr

Mass loss of generator = before reaction - after reaction

= 45.655 - 45.607

= 0.048 g

Mass of CO2 evolved = Mass loss of generator = 0.048 g

Moles of CO2 evolved = mass/molecular weight

= 0.048 / 44g/mol

= 0.001091 mol

Pressure of dry CO2 in atm = 743.4 torr x 1atm/760torr

= 0.9782 atm

Volume of CO2 at STP = volume of CO2 (exp) x (PCO2(exp)/760) x (273/TCO2)

= [0.001091 mol x 0.0821 L-atm/mol-K x (20+273)K]/[0.9782 atm] x [743.4/760] x [273/(20+273)]

= 0.02445 L

Molar Volume of CO2 at STP = 0.02445 L/0.001091 mol

= 22.41 L/mol

Moles of CaCO3 in sample = moles of CO2 = 0.001091 mol

Mass of CaCO3 = moles x molecular weight

= 0.001091 mol x 100.08 g/mol

= 0.1091 g

Mass of original sample = 0.18 g

% CaCO3 in sample

= Mass of CaCO3 x 100 / Mass of original sample

= 0.1091 x 100 / 0.180

= 60.66%

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