Question

Compute the molecular formula for a compound of molecular weight 120g/mole which isshown to contain 40% C, 6.67% H and 53.33% oxygen?Group of answer choicesC1H2O1C2H1O2C2H4O2C3H6O3C4H8O4

Answer 1

The first step to solve this question is to assume that the percent of each element is the mass of the corresponding element. Then divide it by the corresponding molecular mass.

`\begin{gathered} (40gC)/(12gC)=3.33 \\ (6.67gC)/(1gC)=6.67 \\ (53.33gC)/(16gC)=3.33 \end{gathered}`

Divide all the results by the minimum result, it means, divide all the results by 3.33:

`\begin{gathered} (3.33C)/(3.33)=1 \\ (6.67H)/(3.33)=2 \\ (3.33O)/(3.33)=1 \end{gathered}`

These results will be the subscripts of each of the compounds of the empiric formula. It means that the empiric formula is:

`C_1H_2O_1`

To find the molecular formula, find the molecular weight of the empiric formula:

`1\cdot12g/mol+2\cdot1g/mol+1\cdot16g/mol=30g/mol`

Now, divide the molecular weight of the molecular formula by the molecular weight of the empiric formula:

`n=(120g/mol)/(30g/mol)=4`

Multiply the subscripts of the empiric formula by this value to find the molecular formula. It means that the subscript of C is 4, of H is 8 and of O is 4. Then the molecular formula is:

`C_4H_8O_4`

It means that the correct answer is the last choice C4H8O4.