The empirical formula of the compound is [tex]HClO_3[/tex].
Given:
The compound that is 1.2% hydrogen, 42.0% chlorine, and 56.8% oxygen.
To find;
The empirical formula of the compound.
Solution:
Let the empirical formula of the compound be [tex]H_xCl_yO_z[/tex]
consider 100 grams of compound.
Mass of hydrogen = [tex]\frac{1.2}{100}\times 100g = 1.2 g[/tex]
Moles of hydrogen :
=[tex]\frac{1.2g}{1.00 g/mol}=1.20 mol[/tex]
Mass of chlorine = [tex]\frac{42.0}{100}\times 100g = 42.0 g[/tex]
Moles of chlorine:
=[tex]\frac{42.0g}{35.45g/mol}=1.18 mol[/tex]
Mass of oxygen = [tex]\frac{56.8}{100}\times 100g = 56.8 g[/tex]
Moles of oxygen:
=[tex]\frac{56.8 g}{15.99 g/mol}=3.55 mol[/tex]
Now, divide the least moles of an element with moles of elements present in the compound.
[tex]x=\frac{1.20 mol}{1.18mol}=1\\\\y=\frac{1.18mol}{1.18 mol}=1\\\\z=\frac{3.55mol}{1.18 mol}=3\\\\[/tex]
The empirical formula of the compound :
[tex]H_1Cl_1O_3=HClO_3[/tex]
The empirical formula of the compound is [tex]HClO_3[/tex].
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