# Answer to Question #11910 in Other Chemistry for mod

Question #11910

1.2380 g sample composed of lead, dithoionate and water, of unknown proportions[Pb(S2O6)n.mH2O] is analysed to determine the proportions. It is heated in a crucible and on loss of water, the mass is reduced to 1.0331g. The remaining lead dithionate is taken up in solution, processed and percipitate as PbS. 680.4mg of PbS was obtained gravimetrically. From the given information, determine the unknown n and m in the molecular formula for the compound.

Expert's answer

We won't use equation but scheme of reaction

1.2380 0.6804 1.0331

[Pb(S

(207 +(32*2+16*6)*n)+ m*18 207+32*2n 18*m

[1.2380* (207+32*2n)=0.6804*((207 +(32*2+16*6)*n)+ m*18)

{

[1.2380*18*m=1.0331*( (207 +(32*2+16*6)*n)+ m*18)

[256,266+79.232n =140.8428+108.864n+12.2472m{

[22,284m=213.8517+165.296n +18.5958m

[22,284m=213.8517+165.296n +18.5958m

[22,284m - 18.5958m =213.8517+165.296n

[3.6882m==213.8517+165.296n

[m==(213.8517+165.296n)/3.6882m=57.98+44.82n

256,266+79.232n =140.8428+108.864n+12.2472m

256,266+79.232n =140.8428+108.864n+12.2472* (57.98+44.82n)

256,266+79.232n =140.8428+108.864n -709.833 + 548.92n

79.232n -108.864n - 548.92n =140.8428 - 709.833 +

-256.266

-578,552n=-825,3

n=1.45

m=m=57.98+44.82*1.45=122.505

1.2380 0.6804 1.0331

[Pb(S

_{2}O_{6})n.mH_{2}O] -----> PbS(2n) + mH_{2}O +......(207 +(32*2+16*6)*n)+ m*18 207+32*2n 18*m

[1.2380* (207+32*2n)=0.6804*((207 +(32*2+16*6)*n)+ m*18)

{

[1.2380*18*m=1.0331*( (207 +(32*2+16*6)*n)+ m*18)

[256,266+79.232n =140.8428+108.864n+12.2472m{

[22,284m=213.8517+165.296n +18.5958m

[22,284m=213.8517+165.296n +18.5958m

[22,284m - 18.5958m =213.8517+165.296n

[3.6882m==213.8517+165.296n

[m==(213.8517+165.296n)/3.6882m=57.98+44.82n

256,266+79.232n =140.8428+108.864n+12.2472m

256,266+79.232n =140.8428+108.864n+12.2472* (57.98+44.82n)

256,266+79.232n =140.8428+108.864n -709.833 + 548.92n

79.232n -108.864n - 548.92n =140.8428 - 709.833 +

-256.266

-578,552n=-825,3

n=1.45

m=m=57.98+44.82*1.45=122.505

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