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

1) average volume of CO₂ collected = (211 + 241)ml/2 = 226 ml = 0.226 lts

Density of CO₂ = 1.711 g/lt

So average mass of CO₂ collected = density × vol = 1.711 g/lt × 0.226 lts = 0.387 gms

Average mass of carbon dioxide = 0.387 gms

2) molar mass of CO₂ = 44.01 g/mol

Actual yield = 0.387 gms

Average percent yield = 69% = (actual yield/theoritical yield) ×100

Theoritical yield = actual yield × 100/69 = 0.387 gms ×100/69 = 0.561 gms

No. Of theortical moles of CO₂ produced = wt/molar mass = 0.561 g/44.01 g/mol = 0.0127 moles

XCO₃ -----> CO₂

1 mole of carbonate produce 1 mole of Carbon dioxide

So no. Of theortical moles of unknown carbonate would be 0.0127 moles

3) Average sample of unknown carbonate = 1gm

Formula wt = wt/no. Of moles = 1g /0.0127 mole = 78.74 g/mol

4) Actual unknown carbonate is CaCO₃

True Molar mass of CaCO₃ = 100.08 g/mol

Ca = 1× 40.078 g/mol = 40.078 g/mol

C = 1× 12.011 g/mol = 12.011 g/mol

O = 3 × 15.99 g/mol = 47.997 g/mol

CaCO₃ = 100.086 g/mol (40.098+12.011+47.997)

But observed/experimental molar mass of carbonate = 78.74 g/mol

% error = |exp value-true value|/true value)×100 =

= | (78.74 g/mol-100.086g/mol)|/100.086 g/mol)×100

=( 21.346/100.086)x100 = 21.33 %