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

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so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer