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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

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dexteright02
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:

[tex]C: \dfrac{70.79\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 5.89\:mol[/tex]

[tex]H: \dfrac{8.91\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 8.91\:mol[/tex]

[tex]N: \dfrac{4.59\:\diagup\!\!\!\!\!g}{14\:\diagup\!\!\!\!\!g/mol} \approx 0.328\:mol[/tex]

[tex]O: \dfrac{15.72\:\diagup\!\!\!\!\!g}{16\:\diagup\!\!\!\!\!g/mol} = 0.9825\:mol[/tex]

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:

[tex]C: \dfrac{5.89}{0.328}\to\:\:\boxed{C\approx 18}[/tex]

[tex]H: \dfrac{8.91}{0.328}\to\:\:\boxed{H\approx 27}[/tex]

[tex]N: \dfrac{0.328}{0.328}\to\:\:\boxed{N\approx 1}[/tex]

[tex]O: \dfrac{0.9825}{0.328}\to\:\:\boxed{O\approx 3}[/tex]

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

[tex]\boxed{\boxed{C_{18}H_{27}N_1O_3\:\:or\:\:C_{18}H_{27}NO_3}}\end{array}}\qquad\checkmark[/tex]

I hope this helps. =)
Eduard22sly

The empirical formula of the compound is C₁₈H₂₇NO₃

From the question given above, the following data were obtained:

Carbon (C) = 70.79%

Hydrogen (H) = 8.91%

Nitrogen (N) = 4.59%

Oxygen (O) = 15.72%

Empirical formula =?

The empirical formula of the compound can be obtained as follow:

C = 70.79%

H = 8.91%

N = 4.59%

O = 15.72%

Divide by their molar mass

C = 70.79 / 12 = 5.899

H = 8.91 / 1 = 8.91

N = 4.59 / 14 = 0.328

O = 15.72 / 16 = 0.9825

Divide by the smallest

C = 5.899 / 0.328 = 18

H = 8.91 / 0.328 = 27

N = 0.328 / 0.328 = 1

O = 0.9825 / 0.328 = 3

Therefore, the empirical formula of the compound is C₁₈H₂₇NO₃

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