Questions: A sample of a "suboxide" of cesium gives up 1.6907% of its mass as gaseous oxygen when gently heated, leaving pure cesium behind. Determine the empirical formula of this binary compound.

Solution Steps

Given that the sample releases \(1.6907\%\) of its mass as gaseous oxygen, we can assume a sample mass of 100 grams for simplicity. Therefore, the mass of oxygen in the sample is: \[ \text{Mass of oxygen} = 1.6907 \, \text{g} \]

The remaining mass after the oxygen is released is pure cesium. Thus, the mass of cesium is: \[ \text{Mass of cesium} = 100 \, \text{g} - 1.6907 \, \text{g} = 98.3093 \, \text{g} \]

To find the empirical formula, we need to convert the masses of cesium and oxygen to moles using their molar masses:

• Molar mass of cesium (Cs) = 132.91 g/mol
• Molar mass of oxygen (O) = 16.00 g/mol

Calculate moles of cesium: \[ \text{Moles of Cs} = \frac{98.3093 \, \text{g}}{132.91 \, \text{g/mol}} = 0.7397 \, \text{mol} \]

Calculate moles of oxygen: \[ \text{Moles of O} = \frac{1.6907 \, \text{g}}{16.00 \, \text{g/mol}} = 0.1057 \, \text{mol} \]

To find the empirical formula, divide the moles of each element by the smallest number of moles calculated: \[ \text{Ratio of Cs} = \frac{0.7397}{0.1057} \approx 7.000 \] \[ \text{Ratio of O} = \frac{0.1057}{0.1057} = 1.000 \]

Final Answer