Spirit Burner Experiment Aim: To perform an experiment in the labs to determine the heats of combustion of selected alcohols and then to use this information and the gathered data from chemical data sources to determine trends in the longer chain alcohols. The Variables that MUST be controlled throughout the experiment are: • The height from spirit burner nozzle to the base of the water filled beaker • Air drafts around the room must be kept to minimum so as to keep the heat on the base of the
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DETAILED LESSON PLAN IN ADVANCED CHEMISTRY IV- Euler LEARNING OBJECTIVES At the end of the lesson‚ students are expected to: Write the IUPAC name of certain alcohol compounds; SUBJECT MATTER Topic: Nomenclature of Carboxylic Acid References: General‚ Organic and Biochemistry by Denniston‚ Topping and Caret. Page 283-289. Fundamentals of General‚ Organic and Biochemistry by John R. Holum. Page 418-425. Materials: Ball and stick model of compounds Overhead projector Textbooks Pictures of certain
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Molar Heat of Combustion Aim: To find the molar heat of combustion for four different alkanols: 1. Methanol 2. Ethanol 3. 1-Propanol 4. 1-Butanol - And to compare the experimental value with the theoretical. Background: The Molar Heat of Combustion of a substance is the heat liberated when 1 mole of the substance undergoes complete combustion with oxygen at standard atmospheric pressure‚ with the final products being carbon dioxide gas and liquid water. (Ref. “Conquering Chemistry‚ Roland
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Results: Figure [ 1 ] (Temp1: Ethanol; Temp2: n-Propanol) Figure [ 2 ] (Temp1: 1-Butanol; Temp2: n-Pentane) Figure [ 3 ] (Temp1: Methanol; Temp2: n-Hexane) Chart [ 1 ] Chart [ 2 ] Chart [ 3 ] Table [ 1 ] Alcohols | Molecular Weight (g) | ∆T (°C) | Ethanol | 46.068 | 9.17 | 1-Propanol | 60.09 | 6.72 | 1-Butanol | 74.12 | 3.35 | Methanol | 32.04 | 15.615 | | | | | Calculations: 1. Calculating Molecular Weight a. Substance: Ethanol b. Formula: C2H5OH
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Responding variable : Heat of combustion Fixed variables : Volume of water‚ copper can‚ thermometer [ K1PP1(ii) - Able to write the hypothesis or variables correctly] Materials ethanol‚ propanol‚ butanol‚ water Apparatus Copper can‚ tripod stand‚ thermometer‚ 100 cm3 measuring cylinder‚ spirit lamp‚ weighing balance‚ wooden block‚ wind shield [ K1PP1(iii) - Able to list all the materials and apparatus correctly]
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James Matthew Jocson*‚ Gianvittorio Lanta‚ Chiqui Ann Llamado‚ Jeron Manaig College of Science Department of Biology University of Santo Tomas‚ Manila‚ Philippines Abstract Five oxygen-bearing organic compounds were given namely Methanol‚ 2-propanol‚ tert - butanol‚ formalin‚ and acetone. Different tests were done to each sample to differentiate their characteristics. These test were Dichromate test‚ Tollens Test‚ DNPH test‚ Iodoform Test‚ and Lucas Test. This was conducted to classify the samples from
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combustion for the five alcohols were determined; methanol‚ ethanol‚ propanol‚ butanol‚ and pentanol. As the line of best fit in the graph suggests‚ the enthalpy of combustion increased as the sizes of the molecules increased. This was predicted in the hypothesis and proves it to be correct. As seen on the graph‚ the enthalpy of combustion increases from 140kJ/mol for methanol‚ which has the smallest molecular mass‚ to 530kJ/mol for propanol‚ which has the largest molecular mass. The enthalpy of combustion
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stronger its attraction to other polar molecules will be. The next two alcohols were ethanol and propanol which followed the predicted pattern at 13.2 ºC and 5.3 ºC respectively. Butanol had the smallest change in temperature‚ 2.5 ºC which indicates much less evaporation. This leads to believe that butanol has the strongest intermolecular forces of the 4 alcohols tested. This is supported by the fact that butanol is the largest of the four‚ and has the most electrons‚ and therefore is the most polarizable
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they are polar or nonpolar‚ as well as which molecule has the strongest intermolecular forces. We then determine that either Butanol or Pentane would evaporate the quickest because they were similar in structure but they differed due to Butanol having five carbon molecules and also an oxygen molecule where Pentane has only four carbons and no oxygen molecules. Meaning Butanol has more hydrogen bonding to break apart. The next week‚ we then were given seven different types of substance to start this
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molecular size and mass. When the equations for combustion of these alcohols are listed in the order of increasing number of carbon atoms‚ Methanol 1 CH3OH(l) + 3/2 O2(g) ⋄ 1 CO2(g) + 2 H2O(l) Ethanol 1 C2H5OH(l) + 3 O2(g) ⋄ 2 CO2(g) + 3 H2O(l) Propanol 1 C3H7OH(l) + 9/2 O2(g) ⋄ 3 CO2(g) + 4 H2O(l) Butanol 1 C4H9OH(l) + 6 O2(g) ⋄ 4 CO2(g) + 5 H2O(l) Pentanol 1 C5H11OH(l) + 15/2 O2(g) ⋄ 5 CO2(g) + 6 H2O(l) The number of CO2 molecule increases in a linear fashion‚ as well as H2O. In the formation
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