IB Chemistry on Ideal Gas Equation, RMM determination of volatile liquid or gas

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IB Chemistry on Ideal Gas Equation, RMM determination of volatile liquid or gas

  1. 1. Candidate Name: Jaeyoon Yi Candidate Number: Date: October 9, 2011 IB Chemistry HL Y1 Internal Assessment: Investigation on determining molecular mass of butane using ideal gas equation Data Collection Quantitative Data Trials Variables Average Average Uncertainty (Standard deviation) 1 2 3 Mass of flask + air 0.002/g 69.334 69.330 69.329 69.331 69.331 0.003 Mass of flask 0.002/g 69.203 69.199 69.198 69.200 69.200 0.003 Mass of flask + gas 0.002/g 69.432 69.428 69.427 69.429 69.429 0.003 Volume of flask 0.002/cm³ 112.697 112.720 112.688 112.702 112.7020 Mass of butane 0.229 0.229 0.004/g Temperature 290.4 290.4 ±1/K Pressure 101.3 101.3 ±0/kPa Table. 1 shows the data collected from experiment 0.229 0.229 290.3 290.4 290.400 101.3 101.3 101.3 0.229 0.0165 0 0.058 0 Qualitative Data When butane gas is released in the flask, after about 30seconds, it smelled like gas and phenomenon like heat shimmer was seen on the entry of the flask.
  2. 2. Data processing • • • Calculation for mass of flask o Mass of flask = (Mass of flask + air) – mass of air o Mass of air = D*V  D is density. The density of air at 293K is about 0.001166g cm-3, so it can be assumed that the density of air at 290.3 or 290.4 is about 0.001166g cm-3.  V is volume of air in flask. The unit is cm3 o Mass of flask in 1st trial = (69.334) – (0.001166*112.697) = 69.203g Calculation for mass of butane o Mass of butane = (flask + gas) – flask o Mass of butane in 1st trial = 69.432 – 69.203 = 0.229g Calculation for the average mass of flask o o Average = Trial1+ Trial2 + Trial3 3 Average mass of flask = 69.203+ 69.199 + 69.198 3  • = 69.200 Calculation for standard deviation of mass of flask o Standard deviation = "( X Trial !X ) 2 3 o 2 2 2 Standard deviation of mass of flask = (69.203! 69.200) + (69.199 ! 69.200) + (69.198 ! 69.200) 3  • = 0.003 Calculation for RMM of butane o RMM =      o mRT PV m = mass of butane in g R = ideal gas constant, 8.314 unit is kPa dm3 mol-1 K-1 T = temperature in K P = pressure in kPa V = volume in dm3 RMM of 1st trial = 0.229 ! 8.314 ! 290.4 = 48.511 g mol-1 101.3! 0.112697 For other calculation, refer to this calculation to find RMM of each trial Trial RMM/ g mol-1 1 48.511 2 48.512 3 48.501 Average 48.508 Table. 2 shows the RMM derived from each trial.
  3. 3. Error Analysis • The percentage error of RMM for 1st trial o The variables that have uncertainty are temperature, volume, and mass of butane o The percentage error of temperature = Uncerta int y !100 Temperature 1 !100 " 0.344% 290.4 Uncerta int y !100 o The percentage error of volume = Volume   o = = The percentage error of mass of butane =  o o o 0.002 !100 " 0.002% 112.697 = Uncerta int y Mass of Bu tan e !100 0.004 !100 " 1.747% 0.229 Add the two percentage errors = 0.344% + 0.002% +1.747= 2.093% Therefore, the uncertainty of RMM of 1st trial is (48.511 2.093%) g mol-1 When converting it into absolute error, (48.511 1.015) g mol-1 For other calculation, refer to this equation to find the uncertainty of each RMM Trial RMM/ g mol-1 1 48.511 (48.511 1.015) g mol-1 2 48.512 (48.512 1.015) g mol-1 3 48.501 (48.501 1.015) g mol-1 Average 48.508 RMM (48.508 Uncertainty 0.006) g mol-1 (standard deviation is used) Table. 3 shows uncertainty of RMM For average, Standard deviation can be used • Standard deviation = o • = "( X Trial !X ) 2 3 (48.511! 48.508)2 + (48.512 ! 48.508)2 + (48.501! 48.508)2 3 o = 0.006 o Therefore, the uncertainty of average using standard deviation is (48.508 0.006) g mol-1. Percentage error # & o Percentage error = % Literature Value ! Average of RMM "100 ( % % ( Literature Value $ ' o The original RMM of butane is 58g/mol, so literature value is 58
  4. 4. o • Percentage error = # 58.00 ! 48.508 & "100 ( % ) 16.37% % $ ' 58.00 Total error is 16.37% o Calculation for Random Error   o " Average RMM Percentage for Random Error = $ Uncerta int y of $ Average RMM $ # " 0.006 % !100 '% ( 0.012% # 48.508 & =$ Calculation for Systematic Error Total  Systematic Error =  =16.37 – 0.012 = 16.36% Error Percentage of Error/% % Error 16.37 Random Error 0.012 Systematic Error 16.36 Table. 5 shows final error in percentage. Error ! Random Error % ' !100 '% ' &
  5. 5. Conclusion In theory, the relative molecular mass of butane is 58g/mol, but in this experiment, it turned out that the molecular mass of gas in the flask was about 48.508g/mol. The data suggests that the result is not valid since the final molecular mass found is 48.508g/mol, which is much lighter than expected, and it can be concluded that this experiment has errors in procedure. Evaluation This experiment is not valid even though three results are obtained from three trials. The results are all consistent, but they are not accurate, for they are not close to 58g/mol at all. This is also can said through the uncertainty of the data. The random error, which is an error caused by the limitation of equipment, is 0.012%, which is very close to zero while the systematic error is 16.36%, which is over 90% of total error. This shows that random error does not affect the result very much, but systematic error affects the result to a great extent. The most crucial error is that nobody knows that whether the flask is fully filled with the butane gas or not just by seeing, and as the flask is about to be capped with cap, even during that short instant, other air components can trespass through the flask. Moreover, the exact mass of flask is impossible, for the flask is filled with air in normal state, so in fact the mass of flask found by a mere weight is mass of flask and the mass of air inside, so it was required to subtract the mass of air, but it is temperature dependent, so this also causes error of the data.
  6. 6. Limitation and Improvements Limitation Invisibility of butane It cannot be seen that the flask is fully filled or not, for butane is invisible. Due to this fact, nobody knows whether there are molecules other than butane inside the flask. This affects the whole experiment consequently. Finding mass of butane When one weighs a flask, it is same as weighing both flask and air in the flask, so to get the mass of butane inside flask, mass of flask alone should be found. However, mass of air is temperature dependent and the composition of air is not same everywhere. Therefore, the mass of flask alone cannot be determined easily. Finding volume of flask To get the volume of flask, I filled the flask with the water till it gets to the top (the bottom of cap). And weighs the mass of water, for the density of water is 1g/mL, but this density cannot be sure. Only if the condition of environment is STP, then density of water is 1g/mL, but the environment of the lab is not STP. Improvements Put butane for a longer time. Even though it smells gas, keep putting the gas inside the flask so that one can minimize the space of other substances. Find the temperature of the lab using thermometer or temperature prob, then use mass of air against temperature chart to find the mass of air. Try this several times and take the average mass of flask to minimize random error. Find the temperature of the lab using thermometer or temperature prob, then use density of water against temperature chart to find the density of water at specific temperature. After finding the finding the mass of water, then divide the mass by density to find volume. Then take average of the volume of flask to minimize random error.

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