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A compound is found to contain 36.5% Na, 25.4% S, and 38.1% O. Find its empirical formula.What is the empirical mass of the said compound in g/mol?Use:Na = 23 amuS = 32 amuO = 16 amu

Question

A compound is found to contain 36.5% Na, 25.4% S, and 38.1% O. Find its empirical formula.What is the empirical mass of the said compound in g/mol?Use:Na = 23 amuS = 32 amuO = 16 amu

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Solution

Step 1: Assume a 100g sample. This means we have 36.5g of Na, 25.4g of S, and 38.1g of O.

Step 2: Convert the mass of each element to moles.

For Na: 36.5g / 23g/mol = 1.587 moles For S: 25.4g / 32g/mol = 0.793 moles For O: 38.1g / 16g/mol = 2.381 moles

Step 3: Divide each mole quantity by the smallest quantity to get the mole ratio.

For Na: 1.587 / 0.793 = 2 For S: 0.793 / 0.793 = 1 For O: 2.381 / 0.793 = 3

Step 4: The empirical formula is therefore Na2S1O3 or more simply, Na2SO3.

Step 5: To find the empirical mass, add up the atomic masses of all atoms in the empirical formula.

For Na2SO3: (2 * 23) + 32 + (3 * 16) = 46 + 32 + 48 = 126 g/mol

So, the empirical mass of the compound is 126 g/mol.

This problem has been solved

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