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

The emperical formula of MgO

Extracts from this document...


The Empirical Formula of Magnesium Oxide Data collection: This table shows the mass of different chemicals: Chemicals Mass/g + 0.01 crucible lid 38.12 crucible lid+ Magnesium 38.30 crucible lid + contents after heating 38.40 Table (1): Raw data Uncertainties: 1. balance +/-0.01 Data processing and presentation: Quantity Mass+ the uncertainty Magnesium 38.30- 38.12=0.18+0.1 oxygen 38.40- 38.30=0.10+0.1 number of moles of Mg 0.18/24.31=0.00699 number of moles of oxygen 0.10/16=0.00625 Table (2) Mass of Mg= (the mass of crucible lid +Mg) - (the mass of crucible lid) = 38.30- 38.12= 0.18 Mass of Oxygen= (the mass of crucible lid + contents after heating) - (the Mass of crucible lid+ Magnesium) =38.40- 38.30=0.10 -3 Number of moles of Mg=0.18/24.31= 6.7 X10 moles Number of moles of oxygen= 0.10/16.00 = 6.3 X10 moles Ratio Mg/O =0.00699/ 0.00625 =1.11+0.1 The actual value of Mg/O=1.00 The experimental empirical formula Mg O 6.7 X10 6.3 X10 6.3 X10 6.3 X10 Mg O 1.1 1 Magnesium oxide is made up of a ratio by mass of approximately 1:1 Magnesium to Oxygen, giving up an empirical formula of MgO. ...read more.


The theoretical empirical formula for Magnesium Oxide was Mg2.4O while the experimental empirical formula was MgO. Because the theoretical mole ratio differed from the experimental mole ratio, there is a mole ratio percent error. To find the mole ratio percent error, the Oxide as 2 magnesium atoms per 2 oxygen atoms. This theoretical ratio is 2:2which equals 1:1. From the Experimental empirical formula, the experimental percent composition was determined to be 65.2% magnesium in the magnesium oxide and 34.5%oxygen in the magnesium oxide. These numbers differ from the theoretical percent composition, as the theoretical percent composition was 60.3% magnesium and 39.7%oxgyen. The percent error in the experimental percent compositions for both magnesium and oxygen were calculated using the theoretical percent compositions for magnesium and oxygen, and the experimental percent compositions for these elements, and then plugging these numbers into the percent error formula. The percent error for magnesium was 8.6%, with the percent error of the oxygen being 13.1 %.( Refer to percent error under results for complete calculations) ...read more.


Also, although the crucible was taken off the flame only when the magnesium inside looked fully oxidized, and therefore altered the ratio of magnesium to oxygen in the resulting magnesium oxide, because of the amount of magnesium which was not chemically combined with oxygen. Improvements may be made to the lab to lessen the margin for error. The shape of the metal could be standardized to lessen the possibility for leftover unoxidized magnesium. If the magnesium was in smaller particles, the greater surface area might allow a more complete oxidation process since there would be more surface area for the oxygen to chemically react with. Also a way to ensure complete oxidation of the magnesium would be beneficial. Although stirring the magnesium oxide to check for any unoxidized magnesium would probably remove some magnesium oxide on the stirring instrument, maybe the crucible could simply be left on the flame for an additional period of time to lessen the chance for unoxidized magnesium. ...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. Experiment - The Empirical Formula of Magnesium Oxide

    =16.00g/mol, the masses has already been calculated above. Thus, substitute the values into the formula: n(Mg)=0.086/24.31=3.5*10^-3mol n(O)=0.057/16.00=3.6*10^-3mol The results gained were rounded up to 3 decimals so that they are consistent with the absolute uncertainties. Percentage uncertainties associated with masses and number of moles Due to division was employed; the absolute uncertainties of the mass of each element

  2. Enthalpy of Combustion of Alcohols Lab

    Step 1: Calculating the number of moles of Ethanol combusted n = m/M C2H5OH M = (2 x 12)

  1. Empirical Formula of Magnesium Oxide

    mass of Mg Therefore, the average amount of magnesium obtained throughout the trials was 0.2g�0.02g or 0.2g�10% This table summarizes the differences calculated above Average Experimental mass of O2 (g) Average Experimental mass Mg (g) 0.08g�0.02g or 0.08g�25% 0.2g�0.02g or 0.2g�10% Determining the empirical formula of magnesium oxide Find the

  2. Lab Experiment : The change in mass when magnesium burns. (Finding the empirical formula ...

    to get whole number: 1.2 x 5 = 6 1 x 5 = 5 The ratio of Mg to O is therefore 6:5 The empirical formula of Magnesium Oxide found for this trial is therefore Mg6O5 Trail 1. Observations: (Observations are made by lifting the lid after every 2 minutes)

  1. Chemistry Investigation to find the Empirical Formula of Magnesium Oxide

    Nevertheless, when displayed in Table 6 the average ratio was 1 : 1 matching the theoretical of 1 : 1 which supports the hypothesis. Evaluation of Data In Graph 1, it was noticed that there were many random errors with the experiment.

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

    In both cases, the same electronic balance was used to take the mass readings for all trials. An electronic balance was preferred in this experiment because it had lesser uncertainty than an analogue weighing scale. 1 × electronic stopwatch (0.01s)

  1. Discovering the formula of MgO

    Magnesium ?ribbon? will be used in all cases. 3. The experiment is performed in the same laboratory on the same day i.e., the atmospheric conditions and oxygen concentration will be the same. 4. The same Bunsen burner was used every time, with the air hole completely open. Apparatus and chemicals: 1.

  2. The purpose of this lab was to calculate the heat of formation for magnesium ...

    âTHCl QHCl = 100g×4.18J/g°C×5.3°C QHCl= 418J/°C × 5.3°C QHCl= 2215.4 J QH20= 2.2154 kJ 1. Calculate the moles of the Mg burned. Solution: nMg = nMg = nMg = 0.005154 ± 0.0798% mol nMg = 5.154x10-3 ± 0.0798% mol 1.

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