Respuesta :
Answer: Molecular formula will be [tex]C_3H_6O_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= 40 g
Mass of H = 6.71 g
Mass of O = 100 - (40+6.71 ) = 53.29 g
Step 1 : convert given masses into moles.
Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40g}{12g/mole}=3.33moles[/tex]
Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.71g}{1g/mole}=6.71moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.29g}{16g/mole}=3.33moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{3.33}{3.33}=1[/tex]
For H =[tex]\frac{6.71}{3.33}=2[/tex]
For O = [tex]\frac{3.33}{3.33}=1[/tex]
The ratio of C: H: O= 1 : 2: 1
Hence the empirical formula is [tex]CH_2O[/tex]
The empirical weight of [tex]CH_2O[/tex] = 1(12)+2(1)+1(16)= 30g.
The molecular weight = 90.08 g/mole
Now we have to calculate the molecular formula:
[tex]n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{90.08}{30}=3[/tex]
The molecular formula will be=[tex]3\times CH_2O=C_3H_6O_3[/tex]
Molecular formula will be [tex]C_3H_6O_3[/tex]
