Respuesta :
Answer : The molecular of the compound is, [tex]C_8H_{16}O_2[/tex]
Solution : Given,
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 = 54.53 g
Mass of H = 9.15 g
Mass of O = 36.32 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{54.53g}{12g/mole}=4.54moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{9.15g}{1g/mole}=9.15moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{36.32g}{16g/mole}=1.14moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{4.54}{1.14}=3.98\approx 4[/tex]
For H = [tex]\frac{9.15}{1.14}=8.02\approx 8[/tex]
For O = [tex]\frac{1.14}{1.14}=1[/tex]
The ratio of C : H : O = 4 : 8 : 1
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]C_4H_8O_1[/tex]
The empirical formula weight = 12(4) + 8(1) + 1(16) = 72 gram/eq
Now we have to calculate the molecular formula of the compound.
Formula used :
[tex]n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}=\frac{132}{72}=1.83\approx 2[/tex]
Molecular formula = [tex](C_4H_8O_1)_n=(C_4H_8O_1)_2=C_8H_{16}O_2[/tex]
Therefore, the molecular of the compound is, [tex]C_8H_{16}O_2[/tex]
