Answer : The molecular formula of the compound is [tex]C_3H_8O_1[/tex].
Solution : Given,
Molecular weight = 60.10 amu
Empirical formula = [tex]C_3H_8O[/tex]
Molar mass of carbon = 12.01 g/mole
Molar mass of hydrogen = 1.01 g/mole
Molar mass of oxygen = 15.99 g/mole
First we have to calculate the Empirical mass of [tex]C_3H_8O_1[/tex].
Empirical mass of [tex]C_3H_8O_1[/tex] = [tex](3\times 12.01g/mole)+(1\times 1.01g/mole)+(1\times 15.99g/mole)=53.03g/mole[/tex]
Now we have to calculate the molecular formula.
Divide the molecular weight by the empirical mass and then multiply the subscripts by this number.
[tex]\text{ Molecular formula}=\frac{\text{ Molecular weight}}{\text{ Empirical mass}}= \frac{60.10g/mole}{53.03g/mole}=1.133\approx 1[/tex]
Now multiply the subscripts by 1, we get the molecular formula of the compound.
Therefore, the molecular formula of the compound is [tex]C_3H_8O_1[/tex].
