A student carried out a dehydration experiment using MgSO_4. nH_2O. The student

Question

A student carried out a dehydration experiment using MgSO_4. nH_2O. The student placed an evaporating dish which contained 3.8457 g of MgSO_4. nH_2O in an oven at 110 degree C. At this temperature, some but not all of the water in this hydrate was lost. When a constant mass was achieved, the salt weighed 2.4402 g. The student then placed the evaporating dish and contents back in the oven at 160 degree C. The residue came to a constant mass of 18779 g. Stronger beating produced no further loss in so assume all the evaporated. Assume that this final residue is the completely compound (see 1st Page for definition). The in molecular steps and the formulas must be written with integers and fractions: What is the formula of the 1.8779 g sample? What is the formula of the 2.4402 g sample? What is the formula of the 3.8457 g sample? A hydrate of sodium carbonate loses 63 % of its mass when heated to 110 degree C. How many molecules of water are attached to each molecule of sodium carbonate?

Explanation / Answer

(a) Formula of 1.8779 mass is MgSO4

(b) Mass loss =3.8457 g - 2.4402g = 1.4055 g

This is due to water. Moles of water = 1.4055g/18g/mol = 0.078 mol

0.1093 mol corresponds to 7 molecules of H2O (see Part C). So, 0.078 mol corresponds to 7 *0.078/0.1093 = 4.99~5 molecules

Formula of 2.4402 g mass = MgSO4.2H2O

(c)

mass loss = 3.8457 g-1.8779 g = 1.9678 g

This mass corresponds to H2O.

moles of H2O = 1.9678 g/18g/mol = 0.1093 mol

Mass of anhydrous MgSO4 = 1.8779g

Moles of anhydrous MgSO4 = 1.8779g/120.37g/mol = 0.0156 mol

So, 0.0156 mol MgSO4 contains 0.1093 moles water.

moles of water in 1 mol MgSO4 = 7.007 ~7

Formula of 3.8457g sample = MgSO4.7H2O

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.