Answer: The empirical formula will be=[tex]C_2H_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= (100-11) = 89 g
Mass of H = 11 g
Step 1 : convert given masses into moles.
Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{89g}{12g/mole}=7.4moles[/tex]
Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{11g}{1g/mole}=11moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{7.4}{7.4}=1[/tex]
For H = [tex]\frac{11}{7.4}=1.5[/tex]
The ratio of C : H: = 1: 1.5
Convertig them into whole number ratios
The ratio of C : H: = 2: 3
Hence the empirical formula is [tex]C_2H_3[/tex]
The empirical formula will be=[tex]C_2H_3[/tex]