Student Handout Determining M esor o sample 1oo maas % so,, - guam lost mass % H

Question

Student Handout Determining M esor o sample 1oo maas % so,, - guam lost mass % HO x 100 - mass % metal - 100 %-(mass % H2O + mass % so? ) Assume a 100 g sample: moles so moles 11,0 · HMH,0 moles metalMM meta x-moles water (x-integer) Assume a formula of MSO XH:O moles metal- moles SO, metalg MM metal moles So MM SO (% metal) g 7 X L's moles sulfate MM metal Assume a formula of M.SO XH O 06 metal moles metal -2(moles SO MMmtal MM metal 2moles So What metal would you report if MM-39.0 g/mol for MSO.XH:O and MM-19.5 g/mol for M:SO.XH:0?

Explanation / Answer

                                                          Trial 1                                                 Trial 2

MAss percent of sulfate:                     41.9%                                                 39.5%

average = 40.7%

Mass of crucible + unknown + lid         31.378                                              33.004

Mass of unknown                              31.378-(19.489+11.074)                      33.004-(21.269+11.012)

                                                              0.815g                                         0.723g

AFter first heating                           30.967                                               32.634

AFter second heating                     30.964                                                  32.633

Mass of unknown after heating         30.964-((19.489+11.074)                      32.633-(21.269+11.012)

                                                       0.401                                                  0.352

Mass of water lost 0.820 - 0.401 = 0.419g                                 0.723-0.352=0.371

Percentage of water =                    0.419 X 100 / 0.820 = 51.09%                       0.371 X 100 / 0.723 = 51.

Average = 51.09 + 51.3 = 51.2%

So the percentage of metal = 100- (51.2 + 40.7) = 8.1%  

[note: You have not taken the average of % of sulphate in your calcualtions]

Let there is 100 grams of the compound

Mass of water = 51.2 grams , therefore moles of water = Mass / Mol wt = 51.2 / 18 = 2.84

Mass of sulphate = 40.7 grams, therefore moles of sulphate = 40.7 / Mol wt = 40.7/ 96 = 0.424

Mass of metal = 8.1 grams

(i) assuming the formula: MSO4.xH2O

moles of metal = Moles of SO4-2

So moles of metal = 0.424

Molar mass of metal = Mass of metal / moles of metal = 8.1 / 0.424 = 19 (approx) , the metal must be potassium

(ii)assuming the formula: M2SO4.xH2O

moles of metal = 2 X Moles of SO4-2

So moles of metal = 2 X 0.424 = 0.848

Molar mass of metal = Mass of metal / moles of metal = 8.1 / 0.848 = 9.55 (approx) , the metal could be Be

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