Measurements show that unknown compound has the following composition: element mass 62.1 % carbon, 10.5 % hydrogen and 27.6 % oxygen. Write the empirical chemical formula of this compound?

C3H6O

Explanation:

The percentage composition of the elements in the compound are given as follows:

62.1 % carbon = 62.1g of C

10.5 % hydrogen = 10.5g of H

27.6 % oxygen = 27.6g of O

Next, we convert each mass to mole by dividing by their molar/atomic mass

C = 62.1/12 = 5.175mol

H = 10.5/1 = 10.5mol

O = 27.6/16 = 1.725mol

Next, we divide each mole value by the smallest mole value (1.725)

C = 5.175mol ÷ 1.725 = 3

H = 10.5mol ÷ 1.725 = 6.086

O = 1.725mol ÷ 1.725 = 1

The empirical ratio approximately of C:H:O is 3:6:1, hence, the empirical formula is C3H6O

answer; talc and gypsum;

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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