Question

A compound composed of 3. 3 % h, 19. 3 % c, and 77. 4 % o has a molar mass of approximately 60 g/mol. What is the molecular formula of the compound?.

Answer 1

The molecular formula of the given compound is $\mathrm{H}_{2} \mathrm{CO}_{3}$ also known as  Carbonic acid.

What is empirical formula and molecular formula?

The simplest whole-number ratio of the various atoms in a compound is represented by an empirical formula.

The precise number of various atom types present in a compound's molecule is indicated by the molecular formula.

Given that,

H = 3.3%

C = 19.3%

O = 77.4%

No. of moles of H = 3.3/1

No. of moles of H = 3.3

No. of moles of C = 19.3 / 12

No. of moles of C = 1.60

No. of moles of O = 77.4/16

No. of moles of O = 4.83

Therefore, the ratio of the atoms of C, H and O = 3.3 : 1.60 : 4.83

Divide by smallest value which you get =3.3 / 1.60 : 1.60 / 1.60 : 4.83 / 1.60

The ratio of the atoms of C, H and O = 2 : 1 : 3

So, the empirical formula is $\mathrm{H}_{2} \mathrm{CO}_{2}$

Let the molecular formula is $\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right) \mathrm{n}$

Then, molar mass $=(2 \times 1+1 \times 12+3 \times 16) n$

Molar mass = 62n

As the question, 62 n = 60

n = 0.96 or n = 1 (rounded off to nearest ones)

So, the molecular formula is $\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right) 1=\mathrm{H}_{2} \mathrm{CO}_{3}$ i.e., the compound is Carbonic acid.

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