i) What is the ratio of Magnesium/oxygenj) What is the percent error in the experiment? 111(a) Weight of empty crucible16-4409(b) Weight of crucible + magnesium16-7521 g(c) Weight of magnesium in crucible0-312 9(d) Weight of crucible after heating16-954 д0.517 g Mg0Imol Mar0.312 g Mg xX40.3049 mg 0Imol Mg Imal Mgoма(e) Weight of magnesium oxide present1 mol MgX24-3059 Ma(f) Weight of oxygen absorbed by magnesium0-5 mol O₂x 1 mol MgX24-3059 Mg(g) Moles of magnesium used0-312 g Mg x 1 mol Mg.24-305gMgх.32902-мама0.20590=0·20590mol Mg1 mol 0₂0.01284= 0-01284 mol Mg0-01284XImolImol= 0·01284 mol Mg(h) Moles of oxygen absorbed0-312 g Mg x 1 mol Mg24-305g Mg(i) Ratio of magnesium/oxygenRatio Mg=0=mol MgMol 0(j) % error in experiment(1 mol - 0.9651 mol) /0.96516 mol x100) = 3.617.Ths.6116.752 g -16.440g=0=312 g0-312 g Mg x.=0.517 g MgO

There are some errors in your already performed calculations. So, I will working on all of that.

(a) Weight of empty crucible 16-440 (b) Weight of crucible + magnesium 16-75215 (c) Weight of magnesium in crucible 0-312 9 (d) Weight of crucible after heating 16-954 9 0.517 9 И90 0.312 g Mg x I mol MaD. x (e) Weight of magnesium oxide present , ты ма 40.3049 mgo ·x 24-305 ама Imol Mg Imal Mgo absorbed by magnesium 0-5 mỗi bê ма ма (f) Weight of oxygen I mol Mg 0.2059 X X 1 -X.32 9 0₂- 1 mol 0₂ = 0·20590 Mg 24-305g (g) Moles of magnesium used 0.01284 0312 g Mg x 1 molimg. 24-305 = 0.01284 mol Mg gMg (h) Moles of oxygen 0-01284 Imol 0-312 9 mg x (i) Ratio of magnesium/oxygen Ratio Mg=0= (j) % error in experiment 16.7529-16.4409=0-312 9 0-312g Mg absorbed I mol Mg 24-3059 Mg mol Ma Mol 0 у Imol (1 mol 9 -0.517 g MgO mol Mg = 0·01284 mol Mg 3-417 0-9651 mol) /0.96516 mol xivo)=3-617-

(a) Weight of empty crucible 16-440 (b) Weight of crucible + magnesium 16-75215 (c) Weight of magnesium in crucible 0-312 9 (d) Weight of crucible after heating 16-954 9 0.517 9 И90 0.312 g Mg x I mol MaD. x (e) Weight of magnesium oxide present , ты ма 40.3049 mgo ·x 24-305 ама Imol Mg Imal Mgo absorbed by magnesium 0-5 mỗi bê ма ма (f) Weight of oxygen I mol Mg 0.2059 X X 1 -X.32 9 0₂- 1 mol 0₂ = 0·20590 Mg 24-305g (g) Moles of magnesium used 0.01284 0312 g Mg x 1 molimg. 24-305 = 0.01284 mol Mg gMg (h) Moles of oxygen 0-01284 Imol 0-312 9 mg x (i) Ratio of magnesium/oxygen Ratio Mg=0= (j) % error in experiment 16.7529-16.4409=0-312 9 0-312g Mg absorbed I mol Mg 24-3059 Mg mol Ma Mol 0 у Imol (1 mol 9 -0.517 g MgO mol Mg = 0·01284 mol Mg 3-417 0-9651 mol) /0.96516 mol xivo)=3-617-

Chemistry & Chemical Reactivity

10th Edition

ISBN:9781337399074

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter2: Atoms Molecules And Ions

Section: Chapter Questions

Problem 165SCQ

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DATA: weight of the crucible + lid (Worucibletlid) =30.4265 weight of the crucible + lid + magnesium (Worucible-tlid+Mg) =30.6831 weight of the product + crucible + id (Werucible-tid+Mg 0) =30.7168 mass of the product (mMgo) = ? CALCULATIONS: Mass of magnesium: mMg = Mass of oxygen: mo = Number of moles of Mg: PMg = Number of moles of O: no = Mole ratio of Mg to O in the compound: rMg / no = Empirical (simplest) formula of the compound: Molecular formula of the compound (Note that the compound is an ionic compound) |