Answer: The empirical formula is [tex]C_7H_{12}O_2[/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= 65.60 g
Mas of H = 9.44
Mass of O = (100-(65.6+9.44)= 24.96 g
Step 1 : convert given masses into moles.
Mass of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{65.5g}{12g/mole}=5.47moles[/tex]
Moles of H =[tex]\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{9.44g}{1g/mole}=9.44moles[/tex]
Moles of O=[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{24.96g}{16g/mole}=1.56moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{5.47}{1.56}=3.5[/tex]
For H = [tex]\frac{9.44}{1.56}=6[/tex]
For O =[tex]\frac{1.56}{1.56}=1[/tex]
The ratio of C:H:O= 3.5: 6: 1
Converting them into whole numbers, the ratio will be 7: 12: 2
Hence the empirical formula is [tex]C_7H_{12}O_2[/tex]
