Final answer:
To determine the empirical formula of a compound with 21.65% Na, 33.3% Cl, and 45.1% O, you convert the mass percentages to moles, find the simplest whole number ratio, and ascertain that the compound's empirical formula is NaClO3.
Step-by-step explanation:
The question is asking to find the empirical formula of a compound with a specific composition by mass: 21.65% sodium (Na), 33.3% chlorine (Cl), and 45.1% oxygen (O). To find the empirical formula, we need to convert these percentages into moles and then find the simplest whole number ratio of the elements.
Firstly, assume you have 100 grams of the compound, making the mass of each component equal to the percent given. Therefore, you would have 21.65 grams of Na, 33.3 grams of Cl, and 45.1 grams of O. Next, you convert the mass of each element to moles by dividing by its atomic mass (Na = 22.99 g/mol, Cl = 35.45 g/mol, O = 16.00 g/mol):
- Na: 21.65 g ÷ 22.99 g/mol = 0.941 moles of Na
- Cl: 33.3 g ÷ 35.45 g/mol = 0.939 moles of Cl
- O: 45.1 g ÷ 16.00 g/mol = 2.819 moles of O
Find the ratio by dividing each mole value by the smallest of the mole values obtained:
- Na: 0.941 ÷ 0.939 = 1
- Cl: 0.939 ÷ 0.939 = 1
- O: 2.819 ÷ 0.939 = 3
The smallest whole number ratio of Na:Cl:O is 1:1:3. Therefore, the empirical formula for the compound is NaClO3.