A compound composed of 3.3% H, 19.3% C, and 77.4% O3.3% 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?

Molecular formula of the compound = H₂CO₃

Explanation:

First, the empirical formula of the compound is determined

Percentage by mass of each element is given as shown below:

H = 3.3% ; C = 19.9%; O = 77.4%

Mole ratio of the elements= percentage mass/ molar mass

H = 3.3/ 1 = 3.3

C = 19.3/12 = 1.6

O = 77.4/16 = 4.8

whole number ratio is obtained by dividing through with the smallest ratio

H = 3.3/1.6; C = 1.6/1.6; O = 4.8/1.6

H : C : O = 2 : 1 : 3

Empirical formula = H₂CO₃

Molecular formula/mass = n(empirical formula/mass)

60 = n(2*1 + 12*1 + 16*3)

60 = n(62)

n = 60/62 = 0.96

n is approximately = 1

Therefore, molecular formula of the compound = (H₂CO₃) * 1

Molecular formula of the compound = H₂CO₃

because u need two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei to form isotopes.

so since they all have the same shape? i mean maybe because m& m and reeses are both chocolates