Question

A molecule has the empirical formula C4H6O. If its molecular weight is determined to be about 212 g/mol, what is the most likely molecular formula?

a. C12H18O3
b. C8H12O3
c. C11H16O4
d. C4H6O
e. C2H3O

Answer 1

The molecular formula is most likely to be first option,
`\rm C_(12) H_(18) O_(3)`.

Explanation:

Both the molecular formula and the empirical formula of a molecule give the types of atoms in that molecule. However, the molecular formula of a molecule gives the exact number of atoms each molecule. On the other hand, the empirical formula gives only a simplified ratio.

For example, if the empirical formula of the molecule is
`\rm C_4H_6O`, then the molecular formula would be
`(\rm C_4H_6O)_k`, or equivalently
`\mathrm{C}_(4\, k)\mathrm{H}_(6\, k) \mathrm{O}_(k)`, where
`k` is a positive whole number (
`1,\, 2,\, \dots`, etc.) The goal here is to find the value of

Look up the relative atomic mass data on a modern periodic table:

• C: 12.011.
• H: 1.008.
• O: 15.999.

The formula mass of
`\rm C_4 H_6 O` would be

`\begin{aligned}&M(\rm C_4 H_6 O) \cr &\approx 4 * 12.011 + 6 * 1.008 + 15.999 \cr &= 70.091\;\rm g \cdot mol^(-1)\end{aligned}`.

The formula mass of
`\mathrm{C}_(4\, k)\mathrm{H}_(6\, k) \mathrm{O}_(k)` would be

`\begin{aligned}&M(\mathrm{C}_(4\, k)\mathrm{H}_(6\, k) \mathrm{O}_(k)) \cr &\approx (4\, k) * 12.011 + (6\, k) * 1.008 + k * 15.999 \cr &= ( 4 * 12.011 + 6 * 1.008 + 15.999)\, k \\ &= (70.091\, k)\;\rm g \cdot mol^(-1)\end{aligned}`.

On the other hand, since the molecule should have a molecular mass of
`\rm 212\; g \cdot mol^(-1)`,

`70.091\, k \approx 212`.

`k \approx 3`. (Round to the nearest whole number.)

Hence, the molecular formula would be
`\rm C_((4 * 3)) H_((6 * 3))O_(3)`, which simplifies to
`\rm C_(12) H_(18) O_(3)`.

Answer 2

The molecular formula is C12H18O3

Step-by-step explanation:

Step 1: Data given

The empirical formula is C4H6O

Molecular weight is 212 g/mol

atomic mass of C = 12 g/mol

atomic mass of H = 1 g/mol

atomic mass of O = 16 g/mol

Step 2: Calculate the molar mass of the empirical formula

Molar mass = 4* 12 + 6*1 +16

Molar mass = 70 g/mol

Step 3: Calculate the molecular formula

We have to multiply the empirical formula by n

n = the molecular weight of the empirical formula / the molecular weight of the molecular formula

n = 70 /212 ≈ 3

We have to multiply the empirical formula by 3

3*(C4H6O- = C12H18O3

The molecular formula is C12H18O3