Question

learning: molecules and compounds calculationsquestion ⓘthe empirical formula is a chemical formula that provides the smallest, whole-number ratio of elements in a compound.the molecular formula is a whole-number multiple of the empirical formula.the multiple is determined from the ratio of the molar mass to the empirical formula mass.whole-number multiple = $\frac{molar mass}{empirical formula mass}$what is the molecular formula if the molar mass is 318.3 g/mol and the empirical formula is $c_5h_4o$?answer$c_{20}h_{16}o_4$$c_5h_4o$$c_{15}h_{12}o_3$$c_{16}h_{20}o_4$i dont know yet

Explanation:

Step1: Calculate empirical formula mass

Calculate the molar mass of $\text{C}_5\text{H}_4\text{O}$ using atomic masses: $M(\text{C})=12.01\ \text{g/mol}$, $M(\text{H})=1.008\ \text{g/mol}$, $M(\text{O})=16.00\ \text{g/mol}$.

$$\begin{align*} \text{Empirical mass} &= (5\times12.01) + (4\times1.008) + 16.00 \\ &= 60.05 + 4.032 + 16.00 \\ &= 80.082\ \text{g/mol} \end{align*}$$

Step2: Find whole-number multiple

Divide given molar mass by empirical mass.

$$ \text{Multiple} = \frac{318.3}{80.082} \approx 4 $$

Step3: Scale empirical formula

Multiply each subscript in $\text{C}_5\text{H}_4\text{O}$ by 4.

$$ \text{Molecular formula} = \text{C}_{5\times4}\text{H}_{4\times4}\text{O}_{1\times4} = \text{C}_{20}\text{H}_{16}\text{O}_4 $$

Answer:

$\text{C}_{20}\text{H}_{16}\text{O}_4$