a. Suppose you assumed that the atmospheric pressure was one atmosphere instead of the pressure that you recorded. Describe the error that would occur in reporting the molar mass. b. Calculate the percent error in your molar mass had this occurred.
The correct answer and explanation is:
a. Error in Reporting Molar Mass
If the atmospheric pressure used in the calculation is assumed to be one atmosphere (1 atm) instead of the actual recorded pressure, the volume of the gas would be calculated incorrectly. The Ideal Gas Law equation, PV=nRTPV = nRT, depends on both pressure and volume. Assuming standard atmospheric pressure while using a different actual pressure would result in a miscalculation of the gas volume. Since volume and pressure are inversely related in the Ideal Gas Law (Boyle’s Law), assuming a higher pressure than the actual value would lead to an overestimation of the gas volume. In turn, this would cause an underestimation of the molar mass of the gas.
The molar mass (MM) is calculated as: M=mRTPVM = \frac{mRT}{PV}
Where:
- mm is the mass of the gas
- RR is the gas constant
- TT is the temperature in Kelvin
- PP is the pressure
- VV is the volume
If pressure is assumed to be one atmosphere instead of the recorded value, the volume VV will be calculated as larger, which would decrease the calculated molar mass MM.
b. Calculating Percent Error in Molar Mass
The percent error in the molar mass can be calculated using the formula: Percent Error=(Actual Molar Mass−Calculated Molar MassActual Molar Mass)×100\text{Percent Error} = \left( \frac{\text{Actual Molar Mass} – \text{Calculated Molar Mass}}{\text{Actual Molar Mass}} \right) \times 100
Let’s assume that the actual molar mass of the gas is known to be 32.0 g/mol (e.g., oxygen). If the calculated molar mass, assuming one atmosphere of pressure, is 30.0 g/mol, the percent error would be: Percent Error=(32.0−30.032.0)×100=2.032.0×100=6.25%\text{Percent Error} = \left( \frac{32.0 – 30.0}{32.0} \right) \times 100 = \frac{2.0}{32.0} \times 100 = 6.25\%
Thus, the error in molar mass due to assuming a standard atmospheric pressure would result in a 6.25% underestimation of the molar mass.