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

Answer:

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

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

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

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

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 $\rm C_{(4 \times 3)} H_{(6 \times 3)}O_{3}$, which simplifies to $\rm C_{12} H_{18} O_{3}$.

Answer 2

Answer:

The molecular formula is C12H18O3

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