A compound contains 21.6% sodium, 33.3% chlorine, 45.1% oxygen. Derive its empirical formula.
Answers
Explanation:
Hello!
We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100g solution)
Na (sodium): 21.6% = 21.6 g
Cl (chlorine): 33.3% = 33.3 g
O (oxygen): 45% = 45 g
Let us use the above mentioned data (in g) and values will be converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see: datos: Na = 23 u; Cl = 35,4 u; O = 16 u
Na: \dfrac{21.6\:\diagup\!\!\!\!\!g}{23\:\diagup\!\!\!\!\!g/mol} \approx 0.94\:molNa:
23╱g/mol
21.6╱g
≈0.94mol
Cl: \dfrac{33.3\:\diagup\!\!\!\!\!g}{35.4\:\diagup\!\!\!\!\!g/mol} \approx 0.94\:molCl:
35.4╱g/mol
33.3╱g
≈0.94mol
O: \dfrac{45\:\diagup\!\!\!\!\!g}{16\:\diagup\!\!\!\!\!g/mol} = 2.8125\:molO:
16╱g/mol
45╱g
=2.8125mol
We note that the values found above are not integers, so let's divide these values by the smallest of them, so that the proportion is not changed, let's see:
Na: \dfrac{0.94}{0.94}\to\:\:\boxed{Na = 1}Na:
0.94
0.94
→
Na=1
Cl: \dfrac{0.94}{0.94}\to\:\:\boxed{Cl = 1}Cl:
0.94
0.94
→
Cl=1
O: \dfrac{2.8125}{0.94}\to\:\:\boxed{O \approx 3}O:
0.94
2.8125
→
O≈3
Thus, the minimum or empirical formula found for the compound will be: