Final answer:
To determine the empirical formula, convert the given masses of Na, S, and O to moles, calculate the mole ratios, and adjust them to obtain whole numbers. The compound's empirical formula is Na2S2O3.
Step-by-step explanation:
To find the empirical formula of a compound from a sample containing 1.358 g Na, 1.911 g S, and 1.429 g O, we first convert the masses of the elements to moles:
- For sodium (Na), atomic weight approximately 23 g/mol:
1.358 g ÷ 23 g/mol = 0.05905 mol Na - For sulfur (S), atomic weight approximately 32 g/mol:
1.911 g ÷ 32 g/mol = 0.05972 mol S - For oxygen (O), atomic weight approximately 16 g/mol:
1.429 g ÷ 16 g/mol = 0.08931 mol O
Next, we determine the mole ratio of these elements by dividing by the smallest number of moles among them:
- Mole ratio of Na: 0.05905 mol ÷ 0.05905 mol = 1
- Mole ratio of S: 0.05972 mol ÷ 0.05905 mol = 1.01 ≈ 1
- Mole ratio of O: 0.08931 mol ÷ 0.05905 mol = 1.51 ≈ 1.5
Since the oxygen ratio is approximately 1.5 and empirical formulas must have whole numbers, we can multiply all ratios by 2 to normalize them:
- Na: 1 × 2 = 2
- S: 1 × 2 = 2
- O: 1.5 × 2 = 3
Thus, the empirical formula of the compound is Na2S2O3.