• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Enthalpy of paraffin wax Lab

Extracts from this document...


Chemistry Enthalpy of Paraffin Wax lab Aim: Calculate an experimental value for the enthalpy of Combustion of paraffin wax Variables Independent variable: Amount of heat released by the combustion of paraffin wax Dependant variable: Enthalpy of combustion of paraffin wax Controls: The candle used to release energy must be the same throughout the experiment, to ensure that the wax is the same. This will be done by only using one wax candle The aluminium can used to heat up the water must remain the same so that it's mass does not change due to the carbon molecules on the bottom from the heat and so that no water escapes the system. ...read more.


Measure the temperature of the water in the can and note it down 4. Place the candle below the can (which is suspended above it using a clamp attached to a pole) 5. light the candle 6. Wait around 5 minutes before noting down the temperature of the water 7. blow out the candle and push it aside carefully and swiftly 8. Take the mass of the candle and note it down in your datatable. 9. Weigh the can with the water, take the final mass down and jot down how much water was lost during the procedure 10. Repeat steps 1 to 9 for the second trial Datatable Trial 1 Trial 2 Initial temperature of water (c) ...read more.


Second Experimental Value for enthalpy of combustion (Kj/mol) Theoretical Value for enthalpy of combustion (Kj/mol) -10613.63 -10027.5 -13774 Reasons for difference : Improvements More trials could be done We could time the amount of time the paraffin wax candle burns to give us more even values Water was lost when removing and placing the thermometer, this could be avoided by having the thermometer in the water the whole time, but then the thermometer would absorb the heat as well. Sources of error : Some water and paraffin wax was lost to then environment during the experiment, this can not be changed. Some of the heat was escaping to the atmosphere from the candle, this could not be improved because the candle needs oxygen to burn so we could not place it in an insulated environment. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our International Baccalaureate Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Chemistry essays

  1. Enthalpy of Combustion Lab Report

    (1.24 ? 46.1) = 0.00269 (4.18 ? 8 ? 78.08) = 2611 2611 (2611 ? 0.00269) = 971 - 971 3. (1.50 ? 46.1) = 0.00325 (4.18 ? 9.7 ? 82.73) = 3354 3354 (3354 ? 0.00325) = 1032 - 1032 4. (1.75 ? 46.1) = 0.00380 (4.18 ? 10 ? 82.55) = 3451 3451 (3451 ?

  2. Hesss Law Lab, use Hesss law to find the enthalpy change of combustion of ...

    extent and my results are fairly accurate because they are lying just 4.5% below the table values. These points towards the fact that there are some small problems with the system for my results not being more accurate and I am going to discuss it in the paragraph below.

  1. IB chemistry revision notes

    E.g. , ethylamine, amines. o In Water * Strong base --> * Weak base --> * With the same concentrations, the pH of a strong base is higher than that of a weak base. o Conductivity of a strong base is higher than that of a weak base.

  2. Thermodynamics: Enthalpy of Neutralization and Calorimetry

    (42.170 �.0014 g)(41.4 �.5 C�-21.2 �.5 C�) + HC (41.4 �.5 C�-21.2 �.5 C�) (65.4 C�-41.4 C�(�.5)) (41.4 C�-21.2 C�(�.5)) 24.0 (�.7) C� 20.2 (�.7) C� ( 1.00 cal/gC�)(45.977 �.0014 g)(24.0 �.71 C�) = ( 1.00 cal/gC�)(42.170 �?.0014 g)(20.2 �.71 C�) + HC (20.2 �.71 C�) 45.977 �.0014 g � 24.0 �.71 C� 42.170 �.0014 g �

  1. Change of Potential Difference in Voltaic Cells Lab Report

    From the molar mass of copper sulfate pentahydrate, CuSO4.5H2O and the number of moles we can find the mass in grams of solute needed. Mass of copper sulfate pentahydrate = 1 249.71 = 249.71 g This mass is required to prepare 1L of 1M copper sulfate solution.

  2. The aim of this experiment is to examine the enthalpy of combustion of the ...

    = ρ (H2O) * V (H2O) = 1.0 g cm-3 * 40.0 cm3 = 40.0 g The change in temperature is found in the following way : ΔT (H2O) = Tmin – Tmax = 24.0 °C – 50.0 °C = 26.0°C ΔT (H2O)

  1. To determine the standard enthalpy of formation of Magnesium Oxide using Hess Law.

    As evident from reaction 2, MgO reacts with HCl much less vigorously as Mg itself and therefore, any presence of oxide on Mg is bound to slow down the reaction. In this experiment, due to presence of oxide on Magnesium, the mass of Mg that reacted would no longer be 0.05g.

  2. Analysis of the Standard Enthalpy of Combustion for Alcohols

    ?Store the latest run? in loggerpro. 8. Re-weigh the alcohol lamp (including cap) as soon as possible after extinguishing the lamp. 9. Repeat steps 2 ? 8 with the same alcohol to obtain trail 2, and trial 3 results. 10. Repeat steps 2 ? 9 for 4 other consecutive alcohols.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work