A compound has a molar mass of 100 g/mol and the percent composition (by mass) of 65.45% C, 5.45% H, and 29.09% 0. Determine the empirical formula and the molecular formula. A) CHO and C6H6O6

B) C3H3O and C6H6O2

C) C3HO and C6H2O2

D) CH2O and C4H8O4

E) CH4O and C3H12O3

B) C3H3O and C6H6O2

Explanation:

Given data:

Molar mass of compound = 100 g/mol

Percentage of hydrogen = 5.45%

Percentage of carbon = 65.45%

Percentage of oxygen = 29.09%

Empirical formula = ?

Molecular formula = ?

Solution:

Number of gram atoms of H = 5.45 / 1.01 = 5.4

Number of gram atoms of O = 29.09/ 16 = 1.8

Number of gram atoms of C = 65.45 / 12 = 5.5

Atomic ratio:

            C                      :      H            :         O

           5.5/1.8              :     5.4/1.8     :        1.8/1.8

            3                      :        3          :        1

C : H : O = 3 : 3 : 1

Empirical formula is C₃H₃O.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass = 12×3 + 1.01 ×3 + 16×1 = 55.03  

n = 100 / 5503

n = 2

Molecular formula = n (empirical formula)

Molecular formula = 2 (C₃H₃O)

Molecular formula = C₆H₆O₂

c-14 is carbon-14, or radiocarbon.   it is a radioactive isotope of carbon with an atomic nucleus containing 6 protons and 8 neutrons.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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