Crime scene investigators keep a wide variety of compounds on hand to help with identifying unknown substances they find in the course of their duties. One such investigator, while reorganizing their shelves, has mixed up several small vials and is unsure about the identity of a certain powder. Elemental analysis of the compound reveals that it is 67.31 % carbon, 6.978 % hydrogen, 4.617 % nitrogen, and 21.10 % oxygen by mass. Which of the compounds could the powder be?

Respuesta :

The powder could be acetaminophen, analgesic  having chemical formula [tex]C_{17}H_{21}NO_{4}[/tex]

What is an empirical formula?

A chemical formula showing the simplest ratio of elements in a compound rather than the total number of atoms in the molecule.

We are given:

Percentage of C = 67.31 %

Percentage of H = 6.978 %

Percentage of N = 4.617 %

Percentage of O = 21.10 %

Let the mass of the compound be 100 g. So, the percentages given are taken as mass.

Mass of C = 67.31 g

Mass of H = 6.978 g

Mass of N = 4.617 g

Mass of O = 21.10 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of carbon = [tex]\frac{mass}{molar \;mass}[/tex]

Moles of carbon =[tex]\frac{67.31g}{12g/mole}[/tex]

=5.60 moles

Moles of hydrogen = [tex]\frac{mass}{molar \;mass}[/tex]

Moles of hydrogen = [tex]\frac{6.978 g}{1 g/mole}[/tex]

=6.978 moles

Moles of nitrogen =[tex]\frac{mass}{molar \;mass}[/tex]

Moles of nitrogen = [tex]\frac{4.617 g}{14 g/mole}[/tex]

=0.329 moles

Moles of oxygen =[tex]\frac{mass}{molar \;mass}[/tex]

Moles of oxygen =[tex]\frac{21.10 g}{16 g/mole}[/tex]

=1.31 moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.329 moles.

We get the ratio of C : H : N : O = 17 : 21 : 1 : 4

The empirical formula for the given compound is [tex]C_{17}H_{21}NO_{4}[/tex].

Learn more about the empirical formula here:

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