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 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?

Answer 1

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