Analysis of Alum Objectives To determine the percentage of water in alum hydrate and an unknown hydrate To calculate the water of crystallization of an unknown hydrate To develop the lab skills for analyzing a hydrate

Empirical Formulas of Hydrates A hydrate is an ionic solid AX that has some water trapped in it It is written as AX.yH2O AX is some known ionic compound like say Na2SO4 or BaCl2 etc Here is an example In the experiment we know what AX is, but we will not know y this we must find CoCl2 Anhydrous Colbalt (II) Chloride CoCl2.6H2O Colbalt(II)Chloride hexahydrate

Hydrates Prefix # water of hydration Hemi 1/2 Mono 1 Sesqui 3/2 Di 2 Tri 3 Tetra 4 Penta 5 Hexa 6 Hepta 7 Octa 8 Nona 9 Deca 10 Undeca 11 Dodeca 12 “A solid compound having a fixed number of water molecules H2O in its composition” The water in the hydrate is called the water of crystallization BaCl2.2H2O 2 waters of crystallization Barium Chloride dihydrate KAl(SO4)2.12H2O 12 waters of crystallization Alum hydrate or potassium aluminum sulfate dodecahydrate

How much water in my hydrate? We can measure the mass percent of water in a hydrate quite easily The mass percent of water in a hydrate is defined as If we already know the chemical formula for the hydrate we can calculate it theoretically from the molar masses assuming 1 mole of the hydrate

Calculate the theoretical % of H2O in alum hydrate (KAl(SO4)2.12H2O) To do this we figure out how many of each element are present in the formula 1K, 1Al, 2S, (8 + 12 = 20 O), and 24H We then calculate the mass of this many moles of each element and add them together 1 x 39.10 g = 39.10 g 1 x 26.98 g = 26.98 g 2 x 32.066 g = 64.132 g 20 x 16.00 g = 320.00 g 24 x 1.00 g = 24.00 g 474.46 g We the apply the formula from the last slide

Hydrate(s) + heat  anhydrous salt(s) + water of hydration(g) Experimental mass % H2O All hydrates can be heated to remove the water, leaving behind the anhydrous material Hydrate(s) + heat  anhydrous salt(s) + water of hydration(g) To get the experimental value we measure the mass before heating mbefore Heat the crystal to boil off the water Reweigh mafter The mass % H2O can be calculated from the formula

1. 150g Alum hydrate when dehydrated leaves a mass of 0 1.150g Alum hydrate when dehydrated leaves a mass of 0.627 g of the anhydrous product: what is the mass % water? Given mafter = 0.627 g mbefore = 1.150 g Use 𝑚𝑎𝑠𝑠 % 𝐻 2 𝑂= 1.150𝑔 −0.627𝑔 1.150𝑔 ×100% = 45.5%

Experimental Technique The proficiency of your technique can be inferred by comparing the experimental result (45.5%) with the theoretical value (45.58%) If your value is too small you haven’t removed all the water so heat some more If your value is too large you heated it too strongly and as the water came out of the crystal it blew some of the crystal out too! So next time heat it more gently

Getting the number of waters of hydration y We have just calculated % H2O, what if we also knew what the ionic solid was? Let’s say we knew we had CuSO4 but we didn’t know how many waters were in our hydrate In other words we want to find y in the formula CuSO4.yH2O We can use the following formula to get y 𝑦= 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑙𝑜𝑠𝑡 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑎𝑛ℎ𝑦𝑑𝑟𝑜𝑢𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡

Getting the number of waters of hydration y 2.200g of Copper(II)sulfate hydrate after heating has a mass of 1.4070 g What is the mass % water? What is y in the formula of our hydrate if it is written CuSO4.yH2O? The number of moles of Copper (II) Sulfate The number of moles of water lost 𝑚𝑜𝑙𝑒𝑠 𝑤𝑎𝑡𝑒𝑟 𝑙𝑜𝑠𝑡= 2.200−1.407 𝑔 18.0𝑔/𝑚𝑜𝑙 =4.40× 10 −2 𝑚𝑜𝑙𝑒𝑠 𝑤𝑎𝑡𝑒𝑟 What is the formula for the hydrate? CuSO4.5H2O 𝑚𝑎𝑠𝑠 % 𝐻 2 𝑂= 2.200𝑔 −1.407𝑔 2.200𝑔 ×100% = 36.05% 𝑚𝑜𝑙𝑒𝑠 𝐶𝑢𝑆 𝑂 4 = 𝑚𝑎𝑠𝑠 𝑎𝑛ℎ𝑦𝑑𝑟𝑜𝑢𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 = 1.4070𝑔 159.609𝑔/𝑚𝑜𝑙 =8.81× 10 −3 𝑚𝑜𝑙𝑒𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑦= 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑙𝑜𝑠𝑡 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑎𝑛ℎ𝑦𝑑𝑟𝑜𝑢𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 = 4.40× 10 −2 𝑚𝑜𝑙𝑒𝑠 𝑤𝑎𝑡𝑒𝑟 8.81× 10 −3 𝑚𝑜𝑙𝑒𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 =5.00

Getting the number of waters of hydration y You will be given a mystery compound but we will tell the molar mass of the part of the compound that doesn’t have the water. Since this is the formula of what is left one you heat it and remove the water we call the part without the water the ANHYDROUS COMPOUND (AC) Your formula is therefore AC.yH2O and you must find y We will tell you the molar mass of AC Then following what we did for the copper(II) sulfate 𝑦= 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑙𝑜𝑠𝑡 𝑚𝑜𝑙𝑒𝑠 𝐴𝐶

Procedure Weigh a clean dry 250-mL beaker covered with a watch glass m1 = mbeaker + mwatchglass Add about 1g of your unknown to the beaker and reweigh (with the watch glass on) m2 = m1 + mhydrate Determine the mass of the unknown mbefore = mhydrate = m2 – m1 Support beaker and watchglass on a ring stand with a wire gauze. Heat gently (burner 3 inches below beaker) – till the moisture has gone (Crystal should turn to powder at this point) Turn off burner, let it cool for 10 mins then reweigh, the beaker, its contents, and the watchglass together m3 mafter = manhydrous = m3 – m1 mwater= mbefore - mafter Discard powder in trash, clean beaker and repeat a second trial Calculate the mass % water for each trial Calculate y for each trial