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2.2g of a compound of phosphorous and sulphur has 1.24g of P in it.Determineits empirical formula. 2.2g of a compound of phosphorous and sulphur has 1.24g of P in it.Determineits empirical formula.
Elements
Given mass
%
Atomic mass
Moles
Mole ratio
Whole no. ratio
P
1.24g
56.4
30.974
1.8
1.3
13
S
0.96
43.6
32.065
1.4
1
10
So the empirical formula is P13S10
atomic weight of P : 31 and S :32 wt of P:1.24g and wt pf : 0.96g ratio of moles of P:S = (1.24/31) : (0.96/32) = 4:3 empirical formula is : nP4S3 -- Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Naga Ramesh IIT Kgp - 2005 batch
atomic weight of P : 31 and S :32
wt of P:1.24g and wt pf : 0.96g
ratio of moles of P:S = (1.24/31) : (0.96/32)
= 4:3
empirical formula is : nP4S3
--
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Naga Ramesh IIT Kgp - 2005 batch
Ratio of P and the compound is 1.24 to 2,2 is 0.564In compound P4S3 , mass of P is 124g and total mass of P4S3 is 220. Ratio is 124/220 =0.564So, answer is option3, P4S3.
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