Respuesta :
Asnwer : Empirical formula of a compound is : [tex]C_{3}H_{4}O[/tex]
Given information : C = 64.3 % , H = 7.2 % , O = 28.5 %
Step 1 : Convert the given percentage (%) to grams.
Explanation : Let the total mass of the compound be 100 grams.
Mass of C = 64.3 g
[tex](100g)\times \frac{(64.3percent)}{(100percent)} = 64.3g[/tex]
Mass of H = 7.2 g
[tex](100g)\times \frac{(7.2percent)}{(100percent)} = 7.2g[/tex]
Mass of O = 28.5 g
[tex](100g)\times \frac{(28.5percent)}{(100percent)} = 28.5g[/tex]
Step 2 : Convert the grams of each compound to moles.
[tex]Moles = \frac{Grams}{Molar mass}[/tex]
Molar mass of C = 12.0g/mol
Molar mass of H = 1.0 g/mol
Molar mass of O = 16.0g/mol
[tex]Moles of C = \frac{64.3g}{12.0\frac{g}{mol}}[/tex]
Moles of C = 5.36 mol
[tex]Moles of H = \frac{7.2g}{1.0\frac{g}{mol}}[/tex]
Moles of H = 7.2 mol
[tex]Moles of O = \frac{28.5g}{16.0\frac{g}{mol}}[/tex]
Moles of O = 1.78 mol
Step 3 : Find the mole ratio of C , H and O
Mole ratio is calculated by dividing the mole values by the smallest value.
Mole of C = 5.36 mol , Mole of H = 7.2 mol , Mol of O = 1.78 mol
Out of the three mole values , mole value of O that is 1.78 mol is less , so we divide all the mole values by 1.78 mol.
[tex]Mole of C = \frac{5.36mol}{1.78mol} = 3.0[/tex]
[tex]Mole of H = \frac{7.2mol}{1.78mol} = 4.0[/tex]
[tex]Mole of O = \frac{1.78mol}{1.78mol} = 1.0[/tex]
C : H : O = 3 : 4 : 1
So empirical formula of the compound is [tex]C_{3}H_{4}O_{1} [/tex] or [tex]C_{3}H_{4}O[/tex]
