A compound contains 74.2 g Na and 25.8 g O. Determine the Empirical Formula for this compound
please show your work I need an explanation on how to do it

The empirical formula for the compound is Na2O

Explanation:

Data obtained from the question include:

Sodium (Na) = 74.2g

Oxygen (O) = 25.8g

We can obtain the empirical formula for the compound as follow:

First, divide the above by their individual molar mass as shown below:

Na = 74.2/23 = 3.226

O = 25.8/16 = 1.613

Next, divide the above by the smallest number

Na = 3.226/1.613 = 2

O = 1.613/1.613 = 1

Therefore, the empirical formula is:

Na2O

The empirical formula is Na2O

Explanation:

Step 1: Data given

Mass of Na = 74.2 grams

Molar mass of Na = 22.99 g/mol

Mass of O = 25.8 grams

Molar mass of O = 16.0 g/mol

Step 2: Calculate moles Na

Moles Na = mass Na / molar mass Na

Moles Na = 74.2 grams / 22.99 g/mol

Moles Na = 3.23 moles

Step 3: Calculate moles O

Moles O = 25.8 grams / 16.0 g/mol

Moles O = 1.61 moles

Step 4: Calculate the mol ratio

We divide by the smallest amount of moles

Na: 3.23 moles / 1.61 moles = 2

O: 1.61 moles / 1.61 moles = 1

This means for every O atom we have 2 Na atoms

The empirical formula is Na2O

c. is the answer

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