Question

Naphthalic acid, a large organic molecule, is found to have the following composition: C, 66.67%;H, 3.73%;O, 29.60% by mass.

Determine the empirical formula of this compound.

B. Given that the molar mass of naphthalic acid is 216g/mol, what is the compound's molecular formula?

Answer 1

Explanation:

The steps for determining the empirical formula of a compound are as follows:

Step 1: Obtain the mass of each element present in grams

Element % = mass ( in g)

Step 2: Determine the number of moles of each type of atom present

mass / atomic mass = Mole (M)

Step 3: Divide the number of moles of each element by the smallest number of moles

Mole / least moles value = Atomic Ratio (R)

Step 4: Convert numbers to whole numbers. This set of whole numbers are the subscripts in the empirical formula.

R * whole number = Empirical Formula

Calculations:

Step 1: Obtain the mass of each element present in grams

Element % of H = 3.73 % = 3.73 g

Element % of C = 66.67 % = 66.67 g

Element % of O = 29.60 % = 29.60 g

Step 2: Determine the number of moles of each type of atom present

We know

mass / atomic mass = Mole (M)

for H atomic mass = 1 g/mol hence, moles= 3.73 g / 1 g/mol = 3.73 mol

for C atomic mass = 12 g/mol hence, moles= 66.67 g / 12 g/mof = 5.556 mol

for O atomic mass = 16 g/mol hence, moles= 29.60 g / 16 g/mol = 1.85 mol

Step 3: Divide the number of moles of each element by the smallest number of moles

Here list mole is 1.85

Atomic ratio for H = 3.73 / 1.85 = 2

Atomic ratio for C = 5.556 / 1.85  = 3

Atomic ratio for O = 1.85 / 1.85 = 1

it is already in whole number hence step 4 not required

Answer (A) : Empirical formula = C3H2O

Step 4: Calculation of molecular formula

first determine the empirical formula mass

here, atomic mass of C = 12  g/mol and O = 16 g/mol and H =1 g/mol

Hence, empirical formula mass = C3H2O = ( 3 × 12 + 2 × 1 + 16 ) g/mol = 54 g/mol

Now divide molecular mass by empirical mass to get the number(x)

(x) = 216 g/mol / 54 g/mol = 4

Hence, Molecular formula = (Empirical formula ) × 4 = (C3H2O) × 4 = C12H8O4

Answer :(b)  Molecular formula = C12H8O4