Quantity1. Mass of crucible + cover2. Mass of crucible + cover +Magnesium3. Mass of Magnesium4. Mass of crucible + cover +magnesium oxide5. Mass of magnesium oxide6. Mass of oxygen that combined withthe magnesium7. Atomic mass of magnesium8. Atomic mass of oxygen9. Moles of magnesium10.Moles of oxygenSubtract # 2 - #1Subtract #4 - #1Subtract #4 - #2From the periodic tableFrom the periodic tableDivide #3/#7(In other words, divide the grams ofmagnesium by 24.3 to get the molesof magnesium)Your valueFind the simplest whole number ratio11. Empirical formula of magnesium between #9 and # 10. Those wholeoxide based on this datanumbers become of the subscripts ofthe formula*20.00 g24.86 g28.06 g24.316.0moles of MgDivide #6/#8(In other words, divide the moles of moles of Ooxygen by 16.0 to get the moles ofoxygen*For example, if the moles of Mg were found to be 2.0 moles, and the moles of O were found to be 6.0 moles,then you would write the following: Mg₂O6Next, divide each of the subscripts by 2 to generate whole numbers.Mg2O6 This gives us Mg₁03 which is the empirical formula based on the data in this example.Another student repeats the experiment and gets the following data. Complete the table and show the empiricalformula for magnesium oxide based on this student's data:Quantity1. Mass of crucible + cover2. Mass of crucible + cover +Magnesium3. Mass of Magnesium4. Mass of crucible + cover +magnesium oxide5. Mass of magnesium oxide7. Atomic mass of magnesium6. Mass of oxygen that combined with Subtract #4 - #2the magnesium8.Atomic mass of oxygen9. Moles of magnesiumSubtract #2 - #110. Moles of oxygenSubtract # 4 - #1From the periodic tableFrom the periodic tableDivide #3/#7(In other words, divide the grams ofmagnesium by24.3 to get the molesof magnesium)Divide #6/#8(In other words, divide the moles ofoxygen by 16.0 to get the moles ofoxygen11. Empirical formula of magnesium Find the simplest whole number ratiooxide based on this databetween #9 and #10.Your value20.00 g24.86 g31.26 g24.316.0

Empirical formula: The formula in which the simplest whole number ratio of constituent atoms ( that…

Quantity 1. Mass of crucible + cover 2. Mass of crucible + cover. Magnesium 3. Mass of Magnesium 4. Mass of crucible + cover+ magnesium oxide 5. Mass of magnesium oxide 7. Atomic mass of magnesium 6. Mass of oxygen that combined with subtract 84-82 the magnesium 8. Atomic mass of oxygen 9. Moles of magnesium 10. Subtract #2 #1 Moles of oxygen Subtract #4 #1 From the periodic table From the periodic table Divide #3/87 (In other words, divide the grams of magnesium by 24.3 to get the moles of magnesium) Your value Find the simplest whole number ratio 11. Empirical formula of magnesium between #9 and #10. Those whole oxide based on this data numbers become of the subscripts of the formula 20.00 g 24.86 g 28.06 g 24.3 16.0 moles of Mg Divide #6/88 (In other words, divide the moles of moles of O oxygen by 16.0 to get the moles of oxygen *For example, if the moles of Mg were found to be 2.0 moles, and the moles of O were found to be 6.0 moles, then you would write the following: Mg₂06 Next, divide each of the subscripts by 2 to generate whole numbers. Mg₂O This gives us Mg₁O3 which is the empirical formula based on the data in this example. Another student repeats the experiment and gets the following data. Complete the table and show the empi formula for magnesium oxide based on this student's data: Quantity 1. Mass of crucible + cover 2. Mass of crucible cover+ Magnesium 3. Mass of Magnesium 4. Mass of crucible +cover+ magnesium oxide 5. Mass of magnesium oxide 7. Atomic mass of magnesium 6. Mass of oxygen that combined with subtract #4-82 the magnesium 8. Atomic mass of oxygen 9. Moles of magnesium Subtract #2 #1 10. Moles of oxygen Subtract #4 #1 From the periodic table From the periodic table 20.00 g 11. Empirical formula of magnesium Find the simplest whole number ratio oxide based on this data between #9 and #10 Your value 24.86 g Divide #3/87 (In other words, divide the grams of magnesium by 24.3 to get the moles of magnesium) Divide #6/88 (In other words, divide the moles of oxygen by 16.0 to get the moles of oxygen 24.3 31.26 g 16.0