Answer: The empirical formula is [tex]C_{3}H_{3}O_1[/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 = 65.5 g
Mass of H = 5.5 g
Mass of O = 29.0 g
Step 1 : convert given masses into moles.
Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{65.5g}{12g/mole}=5.5moles[/tex]
Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.5g}{1g/mole}=5.5moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{29g}{16g/mole}=1.8moles\approx 1moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{5.5}{1.8}=3[/tex]
For H = [tex]\frac{5.5}{1.8}=3[/tex]
For O =[tex]\frac{1.8}{1.8}=1[/tex]
The ratio of C : H : O= 3: 3: 1
Hence the empirical formula is [tex]C_{3}H_{3}O_1[/tex]
