Question

elemental analysis of an unknown compound was found to give the following data 70.4% carbon, 8.08% hydrogen and 10.12 % of nitrogen. it was found to have a molecular weight of 137.18 Da. determine the empirical formula, molecular formula, and the degree of unsaturation for this compound   

Answer 1

1)

we have mass of each elements as:

C: 70.4 g

H: 8.08 g

N: 10.12 g

Divide by molar mass to get number of moles of each:

C: 70.4/12.01 = 5.8618

H: 8.08/1.008 = 8.0159

N: 10.12/14.01 = 0.7223

Divide by smallest to get simplest whole number ratio:

C: 5.8618/0.7223 = 8

H: 8.0159/0.7223 = 11

N: 0.7223/0.7223 = 1

  

So empirical formula is:C8H11N

Answer: C8H11N

2)

Molar mass of C8H11N,

MM = 8*MM(C) + 11*MM(H) + 1*MM(N)

= 8*12.01 + 11*1.008 + 1*14.01

= 121.178 g/mol

Now we have:

Molar mass = 137.18 g/mol

Empirical formula mass = 121.178 g/mol

Multiplying factor = molar mass / empirical formula mass

= 137.18/121.178

= 1

So molecular formula is:C8H11N

Answer: C8H11N

3)

Degree of unsaturation:

DOU = (2C + 2 + N - X - H)/2

X is number of halogens

C is number of carbon

N is number of Nitrogen

H is number of Hydrogen

DOU = (2C + 2 + N - X - H)/2

= (2*8 + 2 + 1 - 0 - 11)/2

= (19 - 11) / 2

= 8/2

= 4

Answer: 4