Respuesta :
Answer: The empirical formula for the given compound is [tex]NaClO_3[/tex] and the name of compound is sodium chlorate.
Explanation : Given,
Percentage of Na = 21.6 %
Percentage of Cl = 33.3 %
Percentage of O = 45.1 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of Na = 21.6 g
Mass of Cl = 33.3 g
Mass of O = 45.1 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Na =[tex]\frac{\text{Given mass of Na}}{\text{Molar mass of Na}}=\frac{21.6g}{23g/mole}=0.939moles[/tex]
Moles of Cl = [tex]\frac{\text{Given mass of Cl}}{\text{Molar mass of Cl}}=\frac{33.3g}{35.5g/mole}=0.938moles[/tex]
Moles of O = [tex]\frac{\text{Given mass of O}}{\text{Molar mass of O}}=\frac{45.1g}{16g/mole}=2.82moles[/tex]
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.938 moles.
For Na = [tex]\frac{0.939}{0.938}=1.00\approx 1[/tex]
For Cl = [tex]\frac{0.938}{0.938}=1[/tex]
For O = [tex]\frac{2.82}{0.938}=3.00\approx 3[/tex]
Step 3: Taking the mole ratio as their subscripts.
The ratio of Na : Cl : O = 1 : 1 : 3
Hence, the empirical formula for the given compound is [tex]Na_1Cl_1O_3=NaClO_3[/tex] and the name of compound is sodium chlorate.
The empirical formula is NaClO3.
Based on the given information,
• A compound comprises 21.6% of Na, 33.3% of Cl and 45.1% of O.
Let us assume that the weight of the compound is 100 grams.
Now the mass of the components present within the compound is,
Mass of Na = 21.6% * 100 g = 21.6 grams
Mass of Cl = 33.3% * 100 g = 33.3 grams
Mass of O = 45.1% * 100 g = 45.1 grams
Now the molecular weight of Na is 22.99 g/mol, Cl is 35.45 g/mol, and O is 16 g/mol.
The number of moles of the components will be,
Moles of Na = 21.6 g/22.99 g/mol = 0.940 moles
Moles of Cl = 33.3 g/35.45 g/mol = 0.939 moles
Moles of O = 45.1 g/16 g/mol = 2.82 moles
Now the mole ratio of the components is Na:Cl:O as 0.940:0.939:2.82
When we divide with the smallest number to get the lowest ratio, we get 1:1:3.
Thus, the empirical formula of the compound will be NaClO3.
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