Question

Density of CO2 = 1.711 g/L
Temperature of water: 23 Celcius
Atmospheric Pressure: 738.8 torr

Trial	Mass of Unknown Carbonate Used	Volume of 6.00 M HCl Used	Volume of CO2 produced
I	1.00g	5.00 mL	211 mL
II	1.00g	5.00 mL	241 mL

5. Determine the average volume of CO2 produced in the 2 trials of the experiment, and, using the density of CO2, determine the average mass of CO2 produced from the unknown in those 2 trials.

6. Using your answer to Question 5, plus the average percent yield from the previous experiment (69%), determine the number of moles of unknown carbonate which must have been present to produce the CO2 which you collected.

7. Recalling that your average sample mass for the unknown was 1.00g, determine the formula weight of your unknown using your anser to question 6 (Remember, the formula weight is the mass in grams of 1 mole).

8. The actual unknown used was: CaCO3
Determine the actual formula weight of the unknown which you used and determine the % error in your experimental formula weight.

Answer 1

5)

average volume =( 211 + 241 ) / 2 = 226 mL

average mass CO2 produced = 0.226 * 1.711 = 0.39 g

6) experimetally moles of CO2 produced = 0.39 / 44 = 0.00886 moles

theorital mole of CO2 =  0.00886 / 0.69 = 0.0128 moles

since 1 mole of carbonate gives 1 moles of CO2 so

so theoretical moles of Carbonate present = 0.0128 moles

7) moles = weight taken / formula weight

formula weight = weight taken / moles

formula weight = 1 / 0.0128 = 78.125 g / mol

8)

actual formula weight = 100.09 g / mol

% error = ( 100.09 - 78.125 / 100.09 )*100

% error = 21.94 %