Question

Select the correct answer. A compound is used as a food additive. The compound has a molar mass of 176. 124 grams/mole. A 692. 5-gram sample undergoes decomposition, producing 283. 4 grams of carbon, 31. 7 grams of hydrogen, and 377. 4 grams of oxygen. What is the molecular formula of the compound? A. C3H4O3 B. C4H8O4 C. C6H6O6 D. C6H8O6 E. C6H8O8.

Answer 1

The molecular formula is the expression of the number of atoms of the elements and the chemical symbols of the atoms. The correct molecular formula of the compound is $\rm C_{6}H_{6}O_{6}.$

What is the molecular formula?

A molecular formula is a depiction of the chemical symbols of the elements with the number of atoms in the subscripts present in the compound.

Let the empirical formula for the compound be, $\rm (CHO)_{n}$ and the molar mass of the compound is 176.124 g/mole.

The percentage of the carbon in the compound is calculated as,

$\begin{aligned} &=\dfrac{\text{mass of the carbon}}{\text{mass of the sample}} \times 100\\\\&= \dfrac{283.4}{692.5}\times 100\\\\&= 40.92\end{aligned}$

The percentage of the hydrogen in the compound is calculated as,

$\begin{aligned} &=\dfrac{\text{mass of the hydrogen}}{\text{mass of the sample}} \times 100\\\\&= \dfrac{31.7}{692.5}\times 100\\\\&= 4.57\end{aligned}$

The percentage of the oxygen in the compound is calculated as,

$\begin{aligned} &=\dfrac{\text{mass of the oxygen}}{\text{mass of the sample}} \times 100\\\\&= \dfrac{377.4}{692.5}\times 100\\\\&= 54.49\end{aligned}$

The ratio of the atomic number of the carbon in the compound is = 3.410

The ratio of the atomic number of the hydrogen in the compound is = 4.577

The ratio of the atomic number of the oxygen in the compound is = 3.406

Now dividing all the three ratios with the smallest ratio we get C: H: O = 1:1:1

Hence, the empirical formula of the compound is given as, $\rm [CHO]_{n}$ and the molecular formula as,

$\begin{aligned}\rm 12n + 1n + 16n & = 176.24\\\\29 \rm n &= 176.24\\\\\rm n &= 6 \end{aligned}$

Hence, the molecular formula of the compound is option C. $\rm C_{6}H_{6}O_{6}.$

Learn more about molecular formulas here:

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