45.62mg of toluene is combusted and we get 152.5mg of carbon dioxide and 35.67mg of water vapor. what is the empirical formula of toluene?

Empirical formula is  C₅H₆O₄₀₇.

Explanation:

Given data:

Mass of carbondioxide = 152.5 mg

Mass of water vapors = 36.67 mg

Mass of toluene = 45.62 mg

Empirical formula = ?

Solution:

First of all we will convert the mg into g.

Mass of carbondioxide = 152.5/1000 = 0.1525 g

Mass of water vapors = 36.67/ 1000 = 0.0367 g

Mass of toluene = 45.62 /1000 =0.0456 g

Percentage of each atoms:

C = 0.1525/0.0456 × 12/ 44 = 0.91

H = 0.0367/0.0456 × 2/ 18 = 0.089

O = 100 - ( 0.91+0.089)

O = 100- 0.999 = 99

Number of gram atoms:

Number of gram atoms of C = O.91/12 = 0.076

Number of gram atoms of H = 0.089/1 = 0.089

Number of gram atoms of O = 99 / 16 = 6.188

Atomic ratio:

C                       :            H                 :    O

0.076/0.076     :       0.089/0.076     :   6.188/0.076

     1                   :            1.17                    :         81.4

C : H : O = 5 ( 1: 1.7 : 81.4)

C : H : O = 5 : 6 : 407

Empirical formula is  C₅H₆O₄₀₇.

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.

16

explanation:

the marks go by 4s