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

Chemitry Lab - Molar Volume of a Gas

Extracts from this document...


Introduction: In this lab I am going to find out the volume of one mole of hydrogen gas at room temperature and atmospheric pressure. The room pressure only slightly differs from the standard, but can be taken into consideration when calculating the results. The molar volume is 22.41 liters per mole at STP (Standard pressure), in other words, at zero degrees centigrade. Figure 3.1 (the experiment set up) Procedure: 1. Set up all equipment. 2. Cut a piece of Magnesium ribbon about 20cm in length. 3. Calculate the weight of the ribbon from the weight of a 1 m long ribbon. 4. Measure 1.0 M Hydrochloric acid to a volume of 25-30ml. 5. Pour the HCl to the reaction flask. 6. Add the Mg ribbon to the reaction flask and secure the mouth of the flask as fast as possible with a hose. Make sure that the hydrogen gas cannot escape from the flask. 7. Follow the temperature 8. Collect the gas until no further reaction is observed in the reaction flask. 9. Carefully remove the gas collection flask so that no gas escapes from the flask. 10. Light the gas. 11. Determine the volume of the gas. ...read more.


* After the reaction, when lighting the hydrogen gas, collection beaker made a popping sound but the reaction flask actually burned and formed a thin flame. * The flame from the reaction flask gave out a lot of heat, which was not noticed when lighting the gas in the collection flask. Figure 3.2 Time (min) Temperature (C) 0 22 1 27 2 26 3 25 4 25 5 25.5 6 26 7 26 8 26 9 26 10 26 11 26 12 26 13 25.5 14 25.5 15 25 Figure 3.3 Calculations: 1.000m � 1mm of Mg ribbon weights 1.06 g �0.005g 25.8cm �1mm Mg ribbon used weight = 0.27348g 1.06g ? 25.8cm = 0.27348 ? 0.96% (0.003g) 100 30ml � 1ml (3.3%) of HCl m(Mg)=0.273g M(Mg)=24.31 n=0.273 = 0.011 n(H2) 24.31 Molar volume at the conditions in the room? molar volume= liters mol V(m)= 0.292l = 26.545 l/mol 0.011moles Correction for temperature 26.545 l at 294 K ? l at 273 K pV=nRT V1=V2 T1 T2 24.545 l/mol = X X = 22.79 l/mol at room temperature 294 K 273 K Correction for pressure P1 =750 TORR P2= 760 TORR 22.79 = X X= 22.5 l/mol 760 750 22.49 - 22.41 = 0.36% error 22.49 Theoretical = 22.5 l/mol Vm= 22.41 0.36 % error The molar volume was 0.36% too large. ...read more.


The experimental value turned out unexpectably very small. 0.36% error in the experiment seems very small, unless there has been some unnoticed mistakes that have influenced the experimental value. The theoretical value is 22.5 l/mol. Temperature changes during the experiment turned out some interesting results, for the temperature seems to start falling soon after the chemical reaction has ended, yet it begins to rise a little after a few minutes and stays constant for a long time before starting to fall (figure 3.2). From the information gained during this experiment, it is difficult to state why this happened; therefore, some extra research should take place if performing the experiment again. Improved investigation: For further investigation, temperature should be measured for longer than 15 minutes in order to find out the rate the temperature is going to fall in a closed flask. The distance between water surface in the container and the surface in the flask should also be measured. For more accurate results, factors such as air and water vapour in the flask should be taken into consideration when calculating the final values. Advisable would be to do some research on why the temperature changed the way it did in this experiment. ?? ?? ?? ?? The molar volume of hydrogen gas 20.10.2009 ...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. Rates of Reaction Lab

    There may be several reasons behind this, such as; the apparatus used may not have been completely leak-proof and airtight. Another reason behind this may have been because the syringe piston was not lubricated sufficiently, so that a certain volume of gas had to be collected before the piston was pressured to move outwards.

  2. Acids/Bases Design Lab. How does a change in the pH value of a solution ...

    The 500.0 cm3 volumetric flask now contained approximately 257.4 cm3 of hydrochloric acid solution. The 500.0 cm3 volumetric flask was then filled completely with distilled water, so that the final volume read 500.0cm3. 6. The hydrochloric acid solution in the 500.0 cm3 volumetric flask was emptied into a 500cm3 beaker.

  1. Change of Potential Difference in Voltaic Cells Lab Report

    However, there were few major weaknesses for the method used. The voltmeter used for this experiment didn't represent accurate readings as a trial sometimes would be repeated more than once to be certain of one reading for a certain concentration.

  2. Airbag design lab. Is it possible to use baking soda, NaHCO3(s), and 2.00 ...

    not homogenous Water * transparent * colourless * odourless * liquid Carbon Dioxide * clear transparent gas * present every where Quantitative Observations - Data Table II Variable Value Volume of Ziploc bag 750mL Room Temperature 20 Atmospheric Pressure =76.15cm/Hg =101.52kPa Pressure exerted in the experiment (atmospheric pressure-water pressure)

  1. Aim: To find the molar mass of butane, by finding the number of moles ...

    + (5*10-5/0.024 * 100) + (0.01/301.2 * 100)] = �[0.05208 + 0.208+ 0.033] = 0.264% (3.d.p) Therefore Let Absoulute Uncertaintly of n be: �0.00001 (5 d.p.), because 0.264 /100 = 0.002 /100 = 0.00264 (5 d.p.) 6. Abesolute Uncertainty of RMM (?RMM)= ?n + ?mass = � [0.00001 + 0.01g] = �0.01 Qualitative Results

  2. Molar Heat combustion chemistry - investigate the effect of molar mass on the molar ...

    4 19.0 5 19.0 1-Butanol 1 23.5 2 20.5 3 24.0 4 22.0 5 23.0 1-Pentanol 1 21.0 2 21.0 3 20.0 4 22.0 5 23.0 Sample Calculations 3. Finding the heat of combustion To find heat of combustion, use the following equation: 1. Convert all masses to kg 2.

  1. Calculating the Molar Volume of a Gas Experiment.

    Volume 2 = 0.489+0.171+1.35 = 2.01% % uncertainty to absolute uncertainty for Volume 2 = 2.01*0.0346/100 = 0.000696 ∴ Absolute uncertainty for Volume = 0.034600 dm3 ±0.000696 Molar Volume Moles = Volume/Molar Volume 0.00146 = 0.0346/Molar

  2. Chemistry lab reort-molar volume of hydrogen

    0.039 Temperature of water () 25.0 Temperature of air () 20.0 Initial volume of HCl () 49.70 Final volume of HCl () 10.45 Height of HCl column () 12.9 Reference Data Table Atmospheric pressure () 102.44 Vapor pressure of water (kPa)

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