Question

Chemistry by Tro 3rd Edition, For More Practice 3.19  

This is a question from the 3rd Edition, That is exactly the name of

those questions as they are posted. It follows Example 3.19

The question: A compound with the percentage composition show next has a molar mass of 60.10 g/mol. Determine its molecular formula.   C, 39.97% H, 13.41% N, 46.62%

Answer 1

Case study:-

(a)Molar mass of composition=60.10 g/molf

(b)Percentage weight of composition C=39.97%,H=13.41% & N=46.62%

Answer:-

Assumption:-suppose we have 100 gm of composition

so that C=39.97 g,H=13.41 g & N=46.62 g

Molecular weight of C=12,H=1,N=14

Moles of C=39.97/12=3.28 , H=13.41/1=13.41 , N=46.62/14=3.33

Now dividing by lowest one and get smallest number ration

C=3.28/3.28=1

H=13.41/3.28=4.08

N=3.33/3.28=1.01

Thus Empirical formula of compound is C₁H_4.08N_1.01

Now Molecular formula is C₁H_4.08N_1.01

_{=12+4.08(1)+14(1.01)=30.22}

_{Molar mass of compound is 60.10 g/mol so that amount present in compound =60.10/30.22=1.99=2}

Chemical Formula =2(C₁H₄N₁)=C₂H₈N₂