A certain anesthetic contains 64.9 percent C, 13.5 percent H, and 21.6 percent O

Question

A certain anesthetic contains 64.9 percent C, 13.5 percent H, and 21.6 percent O by mass. At 120°C and 0.987 atm, 1.00 L of the gaseous compound weighs 2.30 g. What is the molecular formula of the compound?
A certain anesthetic contains 64.9 percent C, 13.5 percent H, and 21.6 percent O by mass. At 120°C and 0.987 atm, 1.00 L of the gaseous compound weighs 2.30 g. What is the molecular formula of the compound?
A certain anesthetic contains 64.9 percent C, 13.5 percent H, and 21.6 percent O by mass. At 120°C and 0.987 atm, 1.00 L of the gaseous compound weighs 2.30 g. What is the molecular formula of the compound?

Explanation / Answer

Given that, at 120°C and 0.987 atm, 1.00 L of the gaseous compound weighs 2.30 g.

Hence, the number of moles of gasses compound are

n = PV/RT

= (0.987 atm)* 1.00 L/ (8.314 atm L/ mole K)*(120+273 K)

= 0.987/ 3267.402

= 0.000302

= 3.02x10-4

Now, the molecular weight of the compound is

Mol. weight = weight/ no. of moles

= 2.30/ 3.02x10-4

= 7614.007

= 7614.007 g/mole

Given that, compound contains 64.9 % C, 13.5 % H, and 21.6 % O by mass

Hence,

Weight of C = (64.9/100)*molecular weight of the compound

= 0.649*7614.007

= 4941.49 g

ie the no of C atoms per molecule are = 4941.49/ atomic weight of C

= 4941.49/ 12

= 411.79

Similarly, weight of H = (13.5/100)*7614.007

= 1027.8914

ie the no of H atoms per molecule are = 1027.8914/ 1

= 1027.8914

Similarly, weight of O = (21.6/100)*7614.007

= 1644.817

ie the no of O atoms per molecule are = 1644.817/ 16

= 102.8011

Hence the molecular formula is C411.79H1027.8914O102.8011

We can write it as C412H1028O103

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