Methanol+Vs+Ethanol+Fuel+Lab

Methanol or Ethanol: Which one would be the best fuel and release the most heat?


 * Introduction**

Right now, one of the major fuels is a non-biodegradable hydrocarbon that comes from petroleum. Once the world's supply of petroleum has been depleted, an alternative fuel will be quintessential to energize many of the major world technologies, like cars. Two potential alternative fuels, methanol and ethanol, are currently being tested to see which releases the most heat and would be the best replacement for petroleum.

Methanol's Balanced Equation - CH3OH(l) + O2(g) → CO2(g) + 2H20(g) + heat energy

Ethanol's Balanced Equation - C2H5OH(l) + 3O2(g) → 3H2O(l) + 2CO2(g) + heat energy


 * Independent Variable**

The independent variable in this experiment is the different liquid used - either Methanol or Ethanol. The important thing to note about the methanol and ethanol is how many moles of the liquid it takes to raise the


 * Dependent Variable(s)**

The dependent variable in this experiment is the amount of heat released when the liquid is burned, calculated in kJ. This can be measured using the equation ΔH = mcΔT. In this equation, m is the mass of the water getting heated, c is the specific heat value of water (4.187 J/g*K), and ΔT is the change in temperature, measured in degrees Kelvin. The liquid that uses the least amount of moles to release a certain amount of heat will be deemed the better fuel.


 * Controlled Variables**

The controlled variables in this experiment are the experimental environment, which includes room temperature, materials, procedure, time of burning.


 * Hypothesis**

If methanol and ethanol are both burned, then ethanol will release more heat. This hypothesis is based on the currently provided statistics, which state that methanol will release approximately 725 kJ of heat energy per mole and ethanol will release approximately 1,364 kJ of heat energy per mole.


 * Materials**

Electronic Triple Beam Balance Spirit Burner with Methanol Spirit Burner with Ethanol Cap for spirit burner 2 125 mL conical flasks 100 grams (100 mL) of distilled water Clamp, Stand, Boss Wire Gauze Thermometer Matches Safety Eye Wear Lap Apron 100 mL Graduated Cylinder Conical Funnel Pipette Ruler


 * Procedure**

1. Set up experiment station. Prepare the clamp, stand, and boss. Ensure that all participants have proper safety eye wear and a lab apron on. 2. Fill up the 100 mL Graduated Cylinder with 100 mL of distilled water, then pour it into the 250 mL conical flask. 3. Using the thermometer, take an initial temperature of the distilled water. Record this data. 4. Place the conical flask on the wire gauze, then place both on top of the clamp. Make sure that the clamp is tight. 5. Mass the spirit burner with methanol, along with the cap. Record this data. 6. Place the spirit burner underneath the conical flask, leaving approximately 2-5 cm. Measure this with a ruler. 7. Carefully light the wick on top of the spirit burner. 8. Using the thermometer, carefully stir the water slowly until the temperature has risen by 20°K. 9. Extinguish the flame on the spirit burner. 10. Mass the spirit burner, noting the change in mass from the initial. Record this data as the mass of methanol burned. 11. Repeat steps 2-10 two more times, recording the data each time. 12. Repeat steps 2-11, substituting the methanol spirit burner with the ethanol spirit burner. Record all relevant data. 13. Clean up.


 * Data Collection**


 * Methanol || Initial Temp || Final Temp || Initial Mass || Final Mass || ∆ Temp (°C) || ∆ Mass grams) ||
 * Trial 1 || 17.6°C || 37.6°C || 159.91 grams || 158.06 grams || 20.0 || 1.85 ||
 * Trial 2 || 18.8°C || 38.8°C || 158.06 grams || 156.47 grams || 20.0 || 1.59 ||
 * Trial 3 || 19.2°C || 39.2°C || 188.39 grams || 185.88 grams || 20.0 || 2.51 ||


 * Ethanol || Initial Temp || Final Temp || Initial Mass || Final Mass || ∆ Temp (°C) ||  || ∆ Mass grams) ||
 * Trial 1 || 18.8°C || 38.8°C || 176.06 grams || 174.35 grams || 20.0 ||  || 1.71 ||
 * Trial 2 || 18.8°C || 39.1°C || 174.35 grams || 172.66 grams || 20.3 ||  || 1.69 ||
 * Trial 3 || 19.6°C || 39.7°C || 172.66 grams || 171.00 grams || 20.1 ||  || 1.66 ||

Temperature uncertainty: ±.1°C Mass uncertainty: ±.01 grams Average change in mass for methanol: 1.98 grams Average change in mass for ethanol: 1.69 grams (END OF GROUP WORK) >
 * __~OTHER OBSERVATIONS~__**
 * The two fuels produced a different color flame during the tests. The methanol was burning with a pure orange color while the ethanol had mixtures of blue and white flames as well as the orange flames.Ethanol burns at a much fast pace than the Methanol fuel. Main reason why we were able to squeeze 6 trials into one class period

__**~CALCULATIONS~**__ To determine which fuel is more efficient, our target find is the amount of heat energy produced in a mole. Ratio: Heat Produced/Amount of Moles in Experiment = Unkown Heat/1 Mole HEAT ENERGY PRODUCED FORMULA: Heat Energy = Mass of Substance X Change in Temperature X Specific Heat Capacity of Substance

VARIABLES: Substance = 100ML of Water Mass = 100 Grams Specific Heat Capacity = 4.187 Joules/GramChange in Temperature = Refer to Chart Above Methanol Mass Change = 1.98 grams Ethanol Mass Change = 1.69 grams WORKED OUT SOLUTIONS: Methanol Trials: Heat Produced = 100 ML X 20.0 X (4.187 J/G) Heat Produced = 8374 J of Heat Energy Produced Significant Figures = 8370 J Ethanol Trials: Heat Produced = 100 ML X 20.1 X (4.187 J/G) Heat Produced = 8415.87 J of Heat Energy Produced Significant Figures = 8420 J MOLAR MASS EQUATION:

Methanol Molar Mass: 32.05 Experiment Mass: 1.98 Ratio = 1.98/32.05 = .062 Moles

Ethanol Molar Mass: 46.08 Experiment Mass: 1.69 Ratio = 1.69/46.08 = .037 Moles RATIO SOLUTIONS:

Methanol: 8370 J / .062 Mole = X / 1 Mole 8370 J / .062 = Amount of Heat Produced in One Mole = 135000 J = 135 KJ

Ethanol: 8420 J / .037 Mole = X / 1 Mole 8420 J / .037 = Amount of Heat Produced in One Mole = 228000 J = 228 KJ Methanol: % Error = (|135-725| / 725) X 100 = 81.38 % Ethanol: % Error = (|228-1364| / 1364) X 100 = 83.28 % Methanol came out with a ratio of 135 KJ of Heat Energy produced per every Mole of Methanol Fuel burned. Ethanol came out witha ratio of 228 KJ of Heat Energy produced per every Mole of Ethanol Fuel burned. Given these results, it is possible to conclude that Ethanol is a much more efficient fuel than Methanol. Producing approximately 70% more Heat Energy. Proving that the hypothesis is correct. Ethanol proves to be more efficient. There is a direct relationship between the amount of hydrogen and carbon within the fuel. The more of either, will result in the fuel releasing more Heat Energy when burned. Since there are more Carbon and Hydrogen atoms in ethanol than methanol, this supports the relationship found. A check that can be used to also support the end result, is calculating the heat energy from the breaking and reforming of the bonds in the chemical equations. The methanol equation had 2061 kj/mol energy to break the bonds and 3466 kj/mol to put together the bonds. 2061 - 3466 = -1405, an exothermic reaction.The ethanol equation had 4728 kj/mol energy to break the bonds and 6004 kj/mol to put together the bonds. 4728 - 6004 = -1726, an exothermic reaction.The ethanol fuel is giving off more heat energy when breaking and connecting the bonds, therefor proving the hypothesis that the ethanol fuel gives off more energy. During our tests, we had a temperature uncertainty of ±.1°C. This may have been a factor since we were using temperature as a stop mark instead of using time. The plan was to stop when the temperature difference reaches 20°C.Also, when taking the temperature of a mass of water, it is very difficult to ensure that the temperature is the same in all areas. The temperature probe may not have been stirred enough or put in the wrong place, thus creating an incorrect reading. So with the possible incorrect readings from the probe, and the temperature uncertainty, these could very possibly be the reasons why we had a very high percentage error. Despite the large errors, the errors were very close. So the percentage error wouldn't have influenced the test results in too much of a way. The numbers would be different but the results will remain the same with Ethanol being more efficient. So besides the large percentage errors, our lab in the end completed its task to prove how ethanol fuel is more efficient than methanol fuel.
 * Percentage Error:**
 * __~CONCLUSION~__**
 * __~EVALUATING~__**