Hydrates Percent Composition Empirical Formula Naming Chapter 8

What is a hydrate? An ionic compound that contains water inside its crystal structure. These compounds do not look or feel wet. The water molecules are loosely attached and can be removed by heating. Generic Formula MxNmy • nH2O MxNmy • nH2O hydrate MxNmy • nH2O anhydrate MxNmy • nH2O water

How do we analyze a hydrate? The water molecules are loosely attached and can be removed by heating. MxNmy • nH2O → MxNmy + nH2O The remaining anhydrate (anhydrous crystals) may look the same, may have a different texture or may even be a different color Mx – Metal Ion; Nmy – Nonmetal Ion; n = prefix # ∆

Law of Constant Composition Determining the mass ratio of unknown “SG” SG was a hydrate, G was the water S was the anhydrate Look at this Lab Data SG was actually Potassium aluminum sulfate dodecahydrate KAl(SO4)2 • 12H2O Calculate the theoretical % of water in this compound water hydrate 45.6% water Per 5 SG 14.505 S 7.773 G (water) 6.732 % water 46.4 12(18.02) [216.2] x100 = 39.1+27+2(32.07)+8(16)+12(18) [474.4]

Lets test a copper(II) sulfate hydrate Measure the mass of the hydrate 8.986 g hydrate to start with Heat the compound to drive off the water Measure the mass of the remaining anhydrate 5.746 g anhydrate Calculate the mass of water removed, H2O 8.986 - 5.746 = 3.243 g water removed Calculate moles of water 3.243 * 1mol/18g = 0.180 mol H2O Calculate moles of anhydrate, formula 5.764 * 1mol/159.6 = 0.036 mol CuSO4 Determine the mole ratio water / anhydrate 0.180 / 0.036 = 4.999/1 Thus CuSO4 • 5 H2O

Lets test barium chloride hydrate Measure the mass of the hydrate 2.11 g hydrate to start with Heat the compound to drive off the water Measure the mass of the remaining anhydrate 1.78 g anhydrate Calculate the mass of water removed, H2O 2.11 – 1.78 = 0.33 g water removed Calculate moles of water 0.33 * 1mol/18g = 0.0183 mol H2O Calculate moles of anhydrate, formula 1.78 * 1mol/208.2 = 0.00855 mol BaCl2 Determine the mole ratio water / anhydrate 0.0183mol H2O/ 0.00855 = 2.1 Thus BaCl2 • 2 H2O

A 344 gram sample of hydrated calcium sulfate and the sample is heated to evaporate the water. The dry sample of calcium sulfate has a mass of 272 grams. What is the mole ratio between the calcium sulfate, CaSO4 and water, H2O? What is the formula of the hydrate? Step 1: Calculate the difference between the hydrated sample and the dry sample. 344 grams (hydrate) - 272 grams (dry sample) = 72.0 grams of water Step 2: Convert mass to moles for each sample. 272g CaSO4(1 mol/136.14g) = 1.9979 mol CaSO4 72.0g H2O(1mol/18.02g) = 3.996 mol H2O Step 3: Divide by lowest molar value 1.99/1.99 = 1 mol CaSO4 3.99/1.99 = 2 mol H2O CaSO4 • 2 H2O

A 87.2 gram sample of a hydrate of MgI2 was heated thoroughly in a porcelain crucible, until its weight remained constant. After heating, 57.4 grams of the anhydrous compound remained. What is the formula of the hydrate? Step 1: Calculate the difference between the hydrated sample and the dry sample. 87.2 grams (hydrate) – 57.4 grams (dry sample) = 29.8 grams of water Step 2: Convert mass to moles for each sample. 57.4g MgI2(1 mol/278.11g) = 0.107 mol MgI2 29.8g H2O(1mol/18.02g) = 3.996 mol H2O Step 3: Divide by lowest molar value 1.99/1.99 = 1 mol MgI2 3.99/1.99 = 2 mol H2O MgI2 • 2 H2O 8

A hydrate was analyzed and determined to be 34% water A hydrate was analyzed and determined to be 34% water. Further the anhydrate was analyzed to be 23.96% nickel, 17.16% nitrogen, and 58.88% oxygen. Determine the formula for this hydrate

A hydrate was analyzed and determined to be 34% water A hydrate was analyzed and determined to be 34% water. Further the anhydrate was analyzed to be 23.96% nickel, 17.16% nitrogen, and 58.88% oxygen. Determine the formula for this hydrate