Question

2. In a combustion reaction, a sample of carbon dioxide was collected. The following data were reported for properties of this carbon dioxide sample.

GIVEN FIND

Mass = 4.157 g Rexp
P = 856.3 torr
V = 2.498 L
T = 85.6 C

Answer 1

• Percent deviation = 1.21%

Step-by-step explanation:

The statement gives some properties of a sample of carbon dioxide collected during the experiment of a combustion reaction and asks to determine the percent deviation of the experiment.

Then, what you must calculate is the percent error on the mass.

That is:

% error = |measured mass - theoretical mass| / (theoretical mass) × 100

The measured mass was 4.157g

The theoretical mass can be calculated from the data of pressure, volume, and temperature, using the ideal gas equation:

• p = 856.3 torr × 1 atm/ 760.0 torr = 1.1267 atm
• V = 2.498 liter
• T = 85.6 + 273.15 = 358.75K

`pV=nRT`

`n=(pV)/(RT)`

`n=(1.1267atm* 2.498liter)/(0.08206(atm\cdot liter)/(K\cdot mol)\cdot 358.75K)`

`n=0.095605mol`

Now use the molar mass of CO₂ to calculate the theoretical mass:

• mass = number of moles × molar mass

• mass = 0.095605mol × 44.01g/mol = 4.20758g ≈ 4.208g

Finally, you can calculate the percent deviation or error:

• % error = |4.157g - 4.208g| / 4.208g × 100 = 1.21%