Mass of crucible, lid, and Mg = 42.185 g Mass of crucible and lid = 40.760 g Mas
ID: 890230 • Letter: M
Question
Mass of crucible, lid, and Mg = 42.185 g Mass of crucible and lid = 40.760 g Mass of crucible, lid, and oxide of Mg First weighing = 43.115 g Second weighing = 43.169 g Third weighing = 43.171 gFind... 1. Mass of Mg 2. Mass of Oxide of Mg 3. Mass of oxygen in oxide of Mg
Calculate the percent by mass and the emperical formula for magnesium oxide.
If the reaction of oxygen and Mg were calculated as having formed Mg2O, what procedural error do you think you might have caused this incorrect chemical formula to be derived from the lab data?
If you started with 2.678 miles of Mg and an unlimited amount of O2, how much magnesium oxide could you form of the reaction went to completion? Mass of crucible, lid, and Mg = 42.185 g Mass of crucible and lid = 40.760 g Mass of crucible, lid, and oxide of Mg First weighing = 43.115 g Second weighing = 43.169 g Third weighing = 43.171 g
Find... 1. Mass of Mg 2. Mass of Oxide of Mg 3. Mass of oxygen in oxide of Mg
Calculate the percent by mass and the emperical formula for magnesium oxide.
If the reaction of oxygen and Mg were calculated as having formed Mg2O, what procedural error do you think you might have caused this incorrect chemical formula to be derived from the lab data?
If you started with 2.678 miles of Mg and an unlimited amount of O2, how much magnesium oxide could you form of the reaction went to completion? Mass of crucible, lid, and Mg = 42.185 g Mass of crucible and lid = 40.760 g Mass of crucible, lid, and oxide of Mg First weighing = 43.115 g Second weighing = 43.169 g Third weighing = 43.171 g
Find... 1. Mass of Mg 2. Mass of Oxide of Mg 3. Mass of oxygen in oxide of Mg
Calculate the percent by mass and the emperical formula for magnesium oxide.
If the reaction of oxygen and Mg were calculated as having formed Mg2O, what procedural error do you think you might have caused this incorrect chemical formula to be derived from the lab data?
If you started with 2.678 miles of Mg and an unlimited amount of O2, how much magnesium oxide could you form of the reaction went to completion? Mass of crucible and lid = 40.760 g Mass of crucible, lid, and oxide of Mg First weighing = 43.115 g Second weighing = 43.169 g Third weighing = 43.171 g
Find... 1. Mass of Mg 2. Mass of Oxide of Mg 3. Mass of oxygen in oxide of Mg
Calculate the percent by mass and the emperical formula for magnesium oxide.
If the reaction of oxygen and Mg were calculated as having formed Mg2O, what procedural error do you think you might have caused this incorrect chemical formula to be derived from the lab data?
If you started with 2.678 miles of Mg and an unlimited amount of O2, how much magnesium oxide could you form of the reaction went to completion?
Explanation / Answer
MASS of Mg = 42.185- 40.760 = 1.425 grams
mass of Mgo = 43.169 - 40.760 = 2.409 grams
mass of oxygen in oxide of Mg = 2.409-1.425 = 0.984 grams
percent by mass of oxygen = 0.984/2.409*100 = 40.85%
percent by mass of magnesium = 1.425/2.409*100 = 59.15%
emperical formula
Mg = 1.425/24.305 = 0.0586 O = 0.409/16 = 0.0255
simplest ratio
Mg = 0.0586/0.0255 = 2.3 rounded off = 2 O = 0.0255/0.0255 = 1
emperical formula = Mg2o
2Mg + o2 ---> 2MgO
No of moles of Mg = 2.678 mole
No of moles of Mgo = 2.678 mole
MOLARMASS OF MGO = 40.3044 g/mol
mass of MgO = 2.678*40.3044 = 107.935 grams