Compound X has a molar mass of 316.25 g*mol^-1 and the following composition: element & mass %
phosphorus & 39.18%
sulfur & 60.82%
Write the molecular formula of X.

Compound X has a molar mass of 316.25 g*mol^-1 and the following composition:

element & mass %

phosphorus & 39.18%

sulfur & 60.82%

Write the molecular formula of X.

Explanation:

The given molecule of phosphorus and sulfur has molar mass --- 316.25 g.

Empirical formula calculation:                      

element:              phosphorus                       sulfur

co9mposition:      39.185%                            60.82%

divide with

atomic mass:          39.185/31.0 g/mol           60.82/32.0g/mol

                              =1.26mol                           1.90mol

smallest mole ratio:   1.26mol/1.26mol =1      1.90mol/1.26 mol =1.50

multiply with 2:          2                                         3

Hence, the empirical formula is:

P2S3.

Mass of empirical formula is:

158.0g/mol

Given, molecule has molar mass --- 316.25 g/mol

Hence, the ratio is:

316.25g/mol/158.0 =2

Hence, the molecular formula of the compound is :

2 x (P2S3)

=

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

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