Question

A compound contain 21.6%of sodium 33.3%chlorine and 45%oxygen .derive it's empirical formula

Answer 1

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

$Na: \dfrac{21.6\:\diagup\!\!\!\!\!g}{23\:\diagup\!\!\!\!\!g/mol} \approx 0.94\:mol$

$Cl: \dfrac{33.3\:\diagup\!\!\!\!\!g}{35.4\:\diagup\!\!\!\!\!g/mol} \approx 0.94\:mol$

$O: \dfrac{45\:\diagup\!\!\!\!\!g}{16\:\diagup\!\!\!\!\!g/mol} = 2.8125\:mol$

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:

$Na: \dfrac{0.94}{0.94}\to\:\:\boxed{Na = 1}$

$Cl: \dfrac{0.94}{0.94}\to\:\:\boxed{Cl = 1}$

$O: \dfrac{2.8125}{0.94}\to\:\:\boxed{O \approx 3}$

Thus, the minimum or empirical formula found for the compound will be:

$\boxed{\boxed{Na_1Cl_1O_3\:\:\:or\:\:\:NaClO_3}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR!

Answer 2

Answer:

here is ur answer........