Chemistry help! Problem about mass/mole fractions?

(247 g hexane)(mol hexane / 86.175 g hexane)(kg benzene / 0.391 mol hexane)(1000 g/kg) = 7.33e3 g benzene

7.33e3 g + 247 g = 7.58e3 g solution

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(1.42 mol benzene / kg THF)(78.1118 g benzene / mol benzene) = 111 g benzene / kg THF

111 g + 1000 g = 1111 g solution

(111/1111)(1630 g) = 163 g solute

(1000/1111)(1630 g) = 1.47e3 g solvent

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l The Zn you be responsive to weighs l.641 grams, so the Fluorine interior the compound could weigh the form between 2.591 grams compound and l.641 gms Zn it particularly is .ninety 5 grams of F Now replace those plenty into moles by using dividing each and each mass by using mass according to mole of the element Moles Zn = l.641 gms over sixty 5 gms/mole = .0.5 moles Zn Moles F = .ninety 5 gms over 19 gms/mole = .05 moles F Now see in case you will get an entire huge form ratio between those 2 molar quantities by using dividing the better form of moles by using the smaller form of moles .05 moles F over .0.5 moles Zn = 2 so which you have 2 moles of F for each mole of Zn interior the compound, so its easiest formulation is ZnF2. 2. whilst given the breakdown of the compound in % basically assume you have l00 grams of the compound to artwork with and convert each and each % into grams. so which you have fifty seven.14 gms C, 6.sixteen gms H, 9.fifty two gms N, and 27.18 gms O Now divide each and all of the quantities by using the wt. according to mole of the element. fifty seven.14 gms C over l2 gms/mole = 4.seventy six moles C 6.sixteen gms H over l.008 gms/mole = 6.11 moles H 9.fifty two gms N over l4 gms/mole = .sixty 8 moles N 27.18 gms O over l6 gms/mole = l.7 moles of O Now back divide each and each of those molar quantities by using the smallest one. C 4.seventy six over .sixty 8 = 7 H 6.11 oveer .sixty 8 = 9 N .sixty 8 over .sixty 8 = l O l.7 over .sixty 8 = 2.5 you choose total numbers for subscripts so double basically about all those numbers to get the formulation Cl4Hl8N2O5 3. upload up the atomic weights interior the empirical formulation to get a finished of 242 Now divide this 242 into the mass of a mole it particularly is 483 and dad out with an answer of two. so which you be responsive to that the real "molecular" formala is double the empirical formulation or Fe2(NO3)6 This substance does not particularly form molecules, so the term formulation weight rather of molecular weight is greater suited. 4. back convert the percentages to grams and divide each and each by using its respective wt according to mole C sixty two.02 grams over l2 = 5.sixteen H l0.40 3 g over l.008 = 10.35 O 27.fifty 4 over l6 = l.seventy two Now divide all numbers by using the smallest I have been given C3H6O for the empirical formulation and its empirical formulation wt provides as much as fifty 8 and the real molecular wt is approximately double that so the molecular formulation could be double the empirical formulation or C6H12O2.