What is the empirical formula of a compound that contains 43.38% sodium, 11.33% carbon, and 45.29% oxygen?

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Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of Na = 43.38 g

Mass of C = 11.33 g

Mass of O = 45.29 g

Step 1 : convert given masses into moles.

Moles of Na=

Moles of C =

Moles of O =

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Na =

For C =

For O =

The ratio of Na : C: O = 2: 1: 3

Hence the empirical formula is .

it would be a gas formation

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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