Question

Determine the Molecular formula of a compound that is 42.1% C, 6.49% H, and 51.4% O. The molecular weight is 342.5 g/mol. (Hint, do not round in the pseudoformula)

Answer 1

we have mass of each elements as:

C: 42.1 g

H: 6.49 g

O: 51.4 g

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

C: 42.1/12.01 = 3.5054

H: 6.49/1.008 = 6.4385

O: 51.4/16.0 = 3.2125

Divide by smallest to get simplest whole number ratio:

C: 3.5054/3.2125 = 1

H: 6.4385/3.2125 = 2

O: 3.2125/3.2125 = 1

  

So empirical formula is:CH2O

Molar mass of CH2O,

MM = 1*MM(C) + 2*MM(H) + 1*MM(O)

= 1*12.01 + 2*1.008 + 1*16.0

= 30.026 g/mol

Now we have:

Molar mass = 342.5 g/mol

Empirical formula mass = 30.026 g/mol

Multiplying factor = molar mass / empirical formula mass

= 342.5/30.026

= 11

So molecular formula is:C11H22O11

Answer: C11H22O11