Question

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

Here we have to get the correct molecular formula of the compound.

The molecular formula of the compound is C₆H₆O₆ i.e. option C.

Let assume, the empirical formula of the compound is $(CHO)_{n}$.

The given molar mass of the compound is 176.124 g/mole.

The percent of carbon in the compound is $\frac{283.4}{692.5}$×100 = 40.924.

The percent of hydrogen in the compound is $\frac{31.7}{692.5}$×100 = 4.577.

The percent of oxygen in the compound is $\frac{377.4}{692.5}$×100 = 54.498.

Now the ratio of the atomic number in the compound for carbon is $\frac{40.924}{12}$ = 3.410

Now the ratio of the atomic number in the compound for hydrogen is $\frac{4.577}{1}$ = 4.577

Now the ratio of the atomic number in the compound for oxygen is $\frac{54.498}{16}$ = 3.406

So, the C, H and O lowest ratio is $\frac{3.410}{3.406}$ = 1, $\frac{4.577}{3.406}$ = 1 and $\frac{3.406}{3.406}$ = 1

Thus the empirical formula of the compound is $[CH_{1}O] _{n}$ (where n = integer.

12n + 1n + 16n = 176.24

29n = 176.24

n = 6 (approx)

Thus the molecular formula of the compound is C₆H₆O₆.