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 pae_???? Desk No. e Preparation and Setup Apparatus of mass of CaCO, sample for analysis (from Prelaboratory Assignment, question 2) A. Sample Unknown sample no. 1. Mass of sample (g) 2. Mass of generator + sample before reaction (8) 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 Trial I Trial 2 graduated cylinder (mL) 34 onL31 oml 2. Final reading of volume of water in CO-collecting graduated cylinder (mL) 3 Volume of CO.(g) collected (L) 4. Temperature of water (c) 5. Barometric pressure (rorr) Vapor pressure of H,O at C (torr) 7. Pressure of dry CO(g) (torr) D. Amount of Carbon Dioxide Gas Evolved 1. Mass of generator + sample after reaction (g) 2 Mass loss of generator mass CO, evolved (g) 3. Moles of CO, evolved (mol) 4.51?46e1- Experiment 13 187

Explanation / Answer

Trial 1

Vapor pressure of H2O at 21 C = 18.7 torr

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

= 760.9 - 18.7 = 742.2 torr

Mass loss of generator = before reaction - after reaction

= 45.559 - 45.514

= 0.045 g

Mass of CO2 evolved = Mass loss of generator = 0.045 g

Moles of CO2 evolved = mass/molecular weight

= 0.045g / 44g/mol

= 0.001023 mol

Pressure of dry CO2 in atm = 742.2 torr x 1atm/760torr

= 0.9766 atm

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

= [0.001023 mol x 0.0821 L-atm/mol-K x (21+273)K]/[0.9766 atm] x [742.2/760] x [273/(21+273)]

= 0.02553 L

Molar Volume of CO2 at STP = 0.02553 L/0.001023 mol

= 24.96 L/mol

Moles of CaCO3 in sample = moles of CO2 = 0.001023 mol

Mass of CaCO3 = moles x molecular weight

= 0.001023 mol x 100.08 g/mol

= 0.1023 g

Mass of original sample = 0.18 g

% CaCO3 in sample

= Mass of CaCO3 x 100 / Mass of original sample

= 0.1023 x 100 / 0.180

= 56.83%

Similarly trial 2 can be done.

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