Respuesta :
Answer: The empirical formula is [tex]C_3H_6O[/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 = 62.04 g
Mass of H = 10.43
Mass of O = (100-(62.04+10.43)) g = 27.53 g
Step 1 : convert given masses into moles.
Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{62.04g}{12g/mole}=5.17[/tex]moles
Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{10.43g}{1g/mole}=10.43moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{27.53g}{16g/mole}=1.72moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{5.17}{1.72}=3[/tex]
For H = [tex]\frac{10.43}{1.72}=6[/tex]
For O =[tex]\frac{1.72}{1.72}=1[/tex]
The ratio of C :H : O= 3: 6 : 1
Hence the empirical formula is [tex]C_3H_6O[/tex]
