Respuesta :
Answer : The empirical of the compound is, [tex]C_1H_2O_3[/tex]
Solution : Given,
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 = 19.36 g
Mass of H = 3.25 g
Mass of O = 77.39 g
Molar mass of C = 12 g/mole
Molar mass of H = 1 g/mole
Molar mass of O = 16 g/mole
Step 1 : convert given masses into moles.
Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{19.36g}{12g/mole}=1.613moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{3.25g}{1g/mole}=3.25moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{77.39g}{16g/mole}=4.837moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{1.613}{1.613}=1[/tex]
For H = [tex]\frac{3.25}{1.613}=2.01\approx 2[/tex]
For o = [tex]\frac{4.837}{1.613}=2.99\approx 3[/tex]
The ratio of C : H : O = 1 : 2 : 3
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]C_1H_2O_3[/tex]
Therefore, the empirical of the compound is, [tex]C_1H_2O_3[/tex]
