Focus Question – Can the mass of magnesium combusted in excess oxygen be used to determine the empirical formula of magnesium oxide?
Hypothesis – Combustion of Magnesium will generate data which can be used to calculate the EF of Magnesium Oxide
Data Collection and Processing
* The Magnesium burnt with a very bright flame. (as seen in figure 2 below) * White smoke was formed and some escaped before the lid was positioned on the crucible. * The Magnesium needed to be reignited several times by lifting the lid to supply more oxygen. * At the completion of the experiment a grayish – white powder remained in the crucible and on the lid. (as seen in figure 3 below)
Figure 2: Magnesium burning
Figure 3: White magnesium oxide powder
Quantitative Raw Data
Table 1 - raw data for mass of magnesium and crucible values
mass of crucible and lid (g) ± 0.01| mass of crucible and lid + Mg (g) ± 0.01| mass of crucible and lid + Mg oxide (g) ± 0.01| 26.02| 26.53| 27.04|
28.03| 28.65| 29.04|
24.10| 24.51| 25.02|
27.09| 27.92| 28.46|
27.35| 28.25| 28.86|
26.78| 27.83| 28.53|
The mass of magnesium and oxygen were determined from the raw data and these values used to calculate the moles of magnesium and oxygen for each trial. The moles of magnesium and oxygen were then averaged (to reduce random error) and these values used to determine the experimental mole ratio for Mg : O. Uncertainties for each value were also calculated. Formula and worked examples for these are shown in table 2 below.
Excel was used to graph moles of Mg verses moles of O. The equation for the line of best fit was then used to determine the mole ratio of Mg:O and thus the empirical formula. This value was compared to the empirical formula calculated using the average moles of magnesium and oxygen.
Table 2 – calculations for processed data
Formula used| Worked example (trial 1 raw data used)|
mass Mg (g) = mass crucible and lid + Mg (± 0.01g) - mass of crucible and lid (± 0.01g)| Mass Mg = 26.53 ± 0.01g - 26.02 ± 0.01g = 0.51± 0.02g| % uncertainty mass Mg = absolute uncertainty/ mass Mg x 100/1| % uncertainty mass Mg = 0.02/ 0.51 x 100/1 = 3.9% = 4% (1sf)| mass O atoms = mass crucible and lid + MgO (± 0.01g) – mass of crucible and lid + Mg (± 0.001g)| Mass O = 27.04 ± 0.01g - 26.53 ± 0.01g = 0.51± 0.02g| % uncertainty mass O atoms = absolute uncertainty/ mass O x 100/1| % uncertainty mass O = 0.02/ 0.51 x 100/1 = 3.9% = 4% (1 sf)| Moles of Mg = mass Mg ± % uncertainty/ molar mass of Mg| Moles Mg = 0.51 ± 4%/ 24.31± 0% = 0.02 ± 4%| Moles of O atoms = mass O ± % uncertainty/ molar mass of O| Moles O atoms = 0.51 ± 4%/ 16.00 ± 0% = 0.03 ± 4%| Mole ratio of Mg to O = moles Mg / moles O| Moles ratio Mg : O = 0.02 / 0.03 = 0.67| % uncertainty mole ratio Mg : O = % uncertainty mole Mg + % uncertainty mole O atoms| % uncertainty on mole ratio Mg:O = 4% +4% = 8%| Absolute uncertainty of Mg : O mole ratio = 8/100 x mole ration Mg : O| Absolute uncertainty = 8/100 x 0.66 = ± 0.05 moles| Average mole ratio Mg : O = sum mole ratio Mg : O / total number of trials| Average mole ratio Mg : O = (0.66 + 1.08 +...