Amolecule is found to contain 47.35% by mass c, 10.60% by mass h, and 42.05% by mass o. what is the empirical formula for this molecule?

The empirical formula of the compound is C₃H₈O₂

We'll begin by listing out what was given from the question. This includes:

Carbon (C) = 47.35%

Hydrogen (H) = 10.60%

Oxygen (O) = 42.05%

The empirical formula of the compound can be obtained as follow:

C = 47.35%

H = 10.60%

O = 42.05%

C = 47.35 / 12 = 3.946

H = 10.60 / 1 = 10.60

O = 42.05 / 16 = 2.628

C = 3.946 / 2.628 = 1.5

H = 10.60 / 2.628 = 4

O = 2.628 / 2.628 = 1

C = 1.5 × 2 = 3

H = 4 × 2 = 8

O = 1 × 2 = 2

Therefore, the empirical formula of the compound is C₃H₈O₂

Learn more: link

they it depends on if one believes about radioactivity, they make a geiger counter beep but maybe it was going to beep anyway electronics do wierd things all the time or it was programmed to beep near certain metals.

може ли да зареждате повече информация, моля