Respuesta :
Answer : (1) The empirical formula of a compound is, [tex]C_3H_8CO[/tex]
(2) The molecular formula of a compound is, [tex]C_3H_8CO[/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 = 59.94 g
Mass of H = 13.44 g
Mass of O = 26.62 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{59.94g}{12g/mole}=4.995moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{13.44g}{1g/mole}=13.44moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{26.62g}{16g/mole}=1.664moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{4.995}{1.664}=3.00\approx 3[/tex]
For H = [tex]\frac{13.44}{1.664}=8.07\approx 8[/tex]
For O = [tex]\frac{1.664}{1.664}=1[/tex]
The ratio of C : H : O = 3 : 8 : 1
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]C_3H_8CO_1=C_3H_8CO[/tex]
The empirical formula weight = 3(12) + 8(1) + 1(16) = 60 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}}[/tex]
[tex]n=\frac{60.11}{60}=1[/tex]
Molecular formula = [tex](C_3H_8CO)_n=(C_3H_8CO)_1=C_3H_8CO[/tex]
Therefore, the molecular of the compound is, [tex]C_3H_8CO[/tex]
