Respuesta :
Answer: The molecular formula of the compound is [tex]C_3H_{12}O_3[/tex]
Explanation:
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 = 37.5 g
Mass of H = 12.5 g
Mass of O = 50.0 g
Step 1 : convert given masses into moles.
Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{37.5g}{12g/mole}=3.125moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{12.5g}{1g/mole}=12.5moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{50.0g}{16g/mole}=3.125moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =[tex]\frac{3.125}{3.125}=1[/tex]
For H= [tex]\frac{12.5}{3.125}=4[/tex]
For O =[tex]\frac{3.125}{3.125}=1[/tex]
The ratio of C: H: O = 1: 4: 1
Hence the empirical formula is [tex]CH_4O[/tex].
The empirical weight of [tex]CH_4O[/tex] = 1(12)+4(1)+1(16)= 32g.
The molecular weight = 93.0 g/mole
Now we have to calculate the molecular formula.
[tex]n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{93}{32}=3[/tex]
The molecular formula will be=[tex]3\times CH_4O=C_3H_{12}O_3[/tex]
