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?
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Ответ:
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%
Empirical formula =?The empirical formula of the compound can be obtained as follow:
C = 47.35%
H = 10.60%
O = 42.05%
Divide by their molar massC = 47.35 / 12 = 3.946
H = 10.60 / 1 = 10.60
O = 42.05 / 16 = 2.628
Divide by the smallestC = 3.946 / 2.628 = 1.5
H = 10.60 / 2.628 = 4
O = 2.628 / 2.628 = 1
Multiply by 2 to express in whole numberC = 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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Ответ:
x + y = 25
x - y = 7
2x = 32
x = 16
y = 25 - 16 = 9