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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

Respuesta :

Edufirst

Answer:

  • Percent deviation = 1.21%

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

        [tex]pV=nRT[/tex]

       [tex]n=\dfrac{pV}{RT}[/tex]

       [tex]n=\dfrac{1.1267atm\times 2.498liter}{0.08206\frac{atm\cdot liter}{K\cdot mol}\cdot 358.75K}[/tex]

        [tex]n=0.095605mol[/tex]

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%

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