Question

A mixture of PCI3 and PCI5 weighing 10,00 g contains 81.12 % CI by mass.

1. how many grams of PCI3 is present in the mixture?

2. how many grams of PCI5 is present in the mixture?

express the answer to four significant figures and include the appropriate units.

Answer 1

First we need to calculate the molecular weight of the PCl3 and PCl5

P = 31 g/mol

Cl = 35.4 g/mol

PCl3 = 137.1 g/mol

PCl5 = 207.9 g/mol

Assuming that the If mix both of them in 1:1 mol ratio molar ratio, then the total mixture

wt = 137.1 + 207.9 = 345

If we add mol wt of chlorine from both the products, this molar mixture will have

total 8 chlorine atoms (8*35.4) = 283.2 g/mol

Then calculate % of chlorine in this mixture as follows:

= 283.2 x 100 / 345 = 82.1 %

It is very close to the mentioned chlorine mass, i.e. 81.12 %

From the above it is very much indicative that the mixture is having the PCl3 and PCl5 in their molar masses.

Claculate the weight of individual PCl3 and PCl5 in the mixture as follows based on the above facts:

A 345 g molar mixture is having 137.1 g of PC13

Hence, 1000 g mixture will have = (137.1 / 345) * 1000 = 397.4 g of  PCl3

Similarly, PCl5

Hence, 1000 g mixture will have = (207.9 / 345) * 1000 = 602.6 g of  PCl5

To verify the concept,

Total wt of PCl3 + PCl5 = 602.6 + 397.4 =1000 g