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Your T. A.'s Name Lab meeting time Your Name Pre-Laboratory Assignment (Part 3) 1. A student determined percent of Br in a sample of an unknown bromide gravimetrically by dissolving 1.691 of solid unknown in water, adding excess of AgNO, and precipitating AgBr. He obtained 3.450 g of AgBr. Using these data, calculate % bromide in the unknown. (Give set-up and numerical answer. If you use more than one step in your calculations, show all work.) 2. The only other component of the unknown mentioned in iis known to be Magnesium. a) What is the per cent by weight of Magnesium in the unknown? b) Using your answers to questions #1 and 2a, determine the empirical formula of the unknown bromide. SHOW ALL WORK!

Answer 1

Solution 0 given, mass of unknown solid = 1-6919 Mass of product obtained (AgBx) = 3.450 g Molarmass of AgBr = 108 + 80 = 188 g/mol so we can say that In 188 g Agbr , bromide is = 80 g so 3.450g Agor, bromide is = 80g x 3.450g - 1889 = 1.47g It is given that AgNO3 is in excess so all bromide present in solid unknown will be convert into AgBr so after reaction ! bromide present in Ag Br = bromide present in solid unknown (before reaction) so we can say that the post en 1.691 g solid untenown, bromide present - 1-479 so in loog solid unknown, bromic present = 1-479 X loog 1.691g - 86.939 or 86.93 % Scanned with Anc- percent cbromide in the unknown is 86-93% ·

@@ given, The only other component present in unknown solid is Magnesium. So, Mass of Magnesium = Mass of unknown salid - Mass of bromide = 1.691g - 1.479 = 0.2219 So In 16919 unknown solid magnesium present = 0.2219 so In loog unknown solid magnesium present = 0.221g Xloog 1 1.691g = 13.079 or 13.07% or We can calculate it by % by weight of (Mg), magnesium = Total percent - percent of bromick s 100 – 86.93 = 13.07% Ans:- 13.07 % magnesium present in unknown solid. o samoles of bromide (Bi) in unknown solid = " mass of bromide melarmass of bromide 1:479 - Bogimal = 0.018 mel CS Scanned with CamScanner

moles of magnesium in unknown solid - mass of magnesium Atomic mass of magnesium 0.2219 24-3g/mol = 0.009 mal and step Divide each male with smallest number of males Mg = 0.0og mal ng = O.oogmat = ! o.ol&mal Bromide = 0.009 mal Empirical formula = Mg Bra Ans: Mg Brz cs Scanned with CamScanner