Question

a red dye is found to be 59.99% carbon, 4.15 % hydrogen , 12.35% nitrogen , 9.42% sulfur, and 14.10% oxygen. the molecular mass ofthe dye is 68.0/ mol . determine the empirical formula and the molecular formula

Answer 1

we have mass of each elements as:

C: 59.99 g

H: 4.15 g

N: 12.35 g

S: 9.42 g

O: 14.1 g

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

C: 59.99/12.01 = 4.995

H: 4.15/1.008 = 4.1171

N: 12.35/14.01 = 0.8815

S: 9.42/32.07 = 0.2937

O: 14.1/16.0 = 0.8812

Divide by smallest to get simplest whole number ratio:

C: 4.995/0.2937 = 17

H: 4.1171/0.2937 = 14

N: 0.8815/0.2937 = 3

S: 0.2937/0.2937 = 1

O: 0.8812/0.2937 = 3

  

So empirical formula is:

C17H14N3SO3

Molar mass of C17H14N3SO3,

MM = 17*MM(C) + 14*MM(H) + 3*MM(N) + 1*MM(S) + 3*MM(O)

= 17*12.01 + 14*1.008 + 3*14.01 + 1*32.07 + 3*16.0

= 340.382 g/mol

Now we have:

Molar mass = 680.0 g/mol

Empirical formula mass = 340.382 g/mol

Multiplying factor = molar mass / empirical formula mass

= 680.0/340.382

= 2

Hence the molecular formula is : C34H28N6S2O6

Answer:

empirical formula is:C₁₇H₁₄N₃SO₃

molecular formula is : C₃₄H₂₈N₆S₂O₆