Question

Determine the empirical formula for a compound that is 70.79% carbon, 8.91% hydrogen, 4.59% nitrogen, and 15.72% oxygen. determine the empirical formula for a compound that is 70.79 carbon, 8.91 hydrogen, 4.59 nitrogen, and 15.72 oxygen. c18h27no2 c18h27no3 c17h27no3 c17h26no3

Answer 1

Hello!

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

C: 70.79% = 70,79 g 
H: 8.91% = 8.91 g
N: 4.59% = 4.59 g
O: 15.72% = 15.72 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:

$C: \dfrac{70.79\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 5.89\:mol$

$H: \dfrac{8.91\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 8.91\:mol$

$N: \dfrac{4.59\:\diagup\!\!\!\!\!g}{14\:\diagup\!\!\!\!\!g/mol} \approx 0.328\:mol$

$O: \dfrac{15.72\:\diagup\!\!\!\!\!g}{16\:\diagup\!\!\!\!\!g/mol} = 0.9825\:mol$

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:

$C: \dfrac{5.89}{0.328}\to\:\:\boxed{C\approx 18}$

$H: \dfrac{8.91}{0.328}\to\:\:\boxed{H\approx 27}$

$N: \dfrac{0.328}{0.328}\to\:\:\boxed{N\approx 1}$

$O: \dfrac{0.9825}{0.328}\to\:\:\boxed{O\approx 3}$

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

$\boxed{\boxed{C_{18}H_{27}N_1O_3\:\:or\:\:C_{18}H_{27}NO_3}}\end{array}}\qquad\checkmark$

I hope this helps. =)