Respuesta :
Answer: Â The molecular formula will be [tex]C_6H_{15}O_9[/tex]
Explanation:
If percentage are given and we are given total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C= 31.17 g
Mass of H = 6.54 g
Mass of O = 62.29 g
Step 1 : convert given masses into moles.
Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{31.17g}{12g/mole}=2.6moles[/tex]
Moles of H=[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.54g}{1g/mole}=6.54moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{62.29g}{16g/mole}=3.9moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C= [tex]\frac{2.6}{2.6}=1[/tex]
For H = [tex]\frac{6.54}{2.6}=2.5[/tex]
For O =[tex]\frac{3.9}{2.6}=1.5[/tex]
The ratio of C: H: O= 1: 2.5: 1.5
Converting them into whole number ratios by multiplying by 2.
Hence the empirical formula is [tex]C_2H_5O_3[/tex]
The empirical weight of [tex]C_2H_5O_3[/tex] = 2(12)+5(1)+3(16)= 77 amu
The molecular weight = 231.2 amu
Now we have to calculate the molecular formula.
[tex]n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{231.2}{77}=3[/tex]
The molecular formula will be=[tex]3\times C_2H_5O_3=C_6H_{15}O_9[/tex]
The molecular formula will be [tex]C_6H_{15}O_9[/tex]
