Respuesta :
Answer : The empirical formula of a compound is, [tex]C_3H_8O_2[/tex]
Solution :
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C = 47.35 g
Mass of H = 10.60 g
Mass of O = 42.05 g
Molar mass of C = 12 g/mole
Molar mass of H = 1 g/mole
Molar mass of O = 16 g/mole
Step 1 : convert given masses into moles.
Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{47.35g}{12g/mole}=3.946moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{10.60g}{1g/mole}=10.60moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{42.05g}{16g/mole}=2.628moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{3.946}{2.628}=1.5[/tex]
For H = [tex]\frac{10.60}{2.628}=4.03\approx 4[/tex]
For O = [tex]\frac{2.628}{2.628}=1[/tex]
The ratio of C : H : O = 1.5 : 4 : 1
To make the ratio in whole number we are multiplying the ratio by 2, we get:
The ratio of C : H : O = 3 : 8 : 2
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]C_3H_8O_2[/tex]
Therefore, the empirical of the compound is, [tex]C_3H_8O_2[/tex]
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 mass
C = 47.35 / 12 = 3.946
H = 10.60 / 1 = 10.60
O = 42.05 / 16 = 2.628
Divide by the smallest
C = 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 number
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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