Final answer:
The compound's empirical formula is CsO, with cesium and oxygen in a 1:1 ratio. The molecular formula is determined to be Cs₂O because the molar mass is twice that of the empirical formula. The correct answer is (d) Cs₂O.
Step-by-step explanation:
To determine the molecular formula of a compound containing 89% cesium (Cs) and 11% oxygen (O) with a molar mass of 298 g/mol, we first need to find the empirical formula by comparing the moles of each element in the compound.
- Cesium (Cs) has a molar mass of approximately 132.9 g/mol.
- Oxygen (O) has a molar mass of approximately 16.00 g/mol.
For cesium: (89/132.9) = 0.670 moles
For oxygen: (11/16.00) = 0.688 moles
To find the simplest whole number ratio, we divide by the smallest number of moles, which is 0.670:
Cs: 0.670/0.670 = 1
O: 0.688/0.670 = 1.027
Approximating to the nearest whole number, we get a ratio of Cs to O as 1:1.
Thus, the empirical formula is CsO.
To find the molecular formula, we compare the empirical formula mass (CsO = 132.9 + 16.00 = 148.9 g/mol) to the given molar mass (298 g/mol).
298 g/mol ÷ 148.9 g/mol = 2, which means the molecular formula has twice as many atoms as the empirical formula.
Therefore, the molecular formula is Cs₂O. Answer option (d) is correct.