Question

A compound has carbon 48% hydrogen 8% nitrogen 28% and oxygen 60% calculate the empirical formula of the compound

Answer 1

Hello!

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100g solution)

C: 48% = 48 g
H: 8% = 8 g
N: 28% = 28 g
O: 60% = 60 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:

$C: \dfrac{48\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} = 4\:mol$

$H: \dfrac{8\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 8\:mol$

$N: \dfrac{28\:\diagup\!\!\!\!\!g}{14\:\diagup\!\!\!\!\!g/mol} = 2\:mol$

$O: \dfrac{60\:\diagup\!\!\!\!\!g}{16\:\diagup\!\!\!\!\!g/mol} = 3.75\:mol$

We note that the values ​​found above, only the oxygen is not integer, so let's divide these values ​​by the smallest of them so that the ratio is not changed, let's see:

$C: \dfrac{4}{2}\to\:\:\boxed{C = 2}$

$H: \dfrac{8}{2}\to\:\:\boxed{H = 4}$

$N: \dfrac{2}{2}\to\:\:\boxed{N = 1}$

$O: \dfrac{3,75}{2}\:\to\:O = 1.875\to\:\:\boxed{O \approx 2}$

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

$\boxed{\boxed{C_2H_4N_1O_2\:\:or\:\:C_2H_4NO_2}}\end{array}}\qquad\checkmark$


I Hope this helps, greetings ... DexteR!