Question

A compound made of two elements, iridium (Ir) and oxygen (O), was produced in a lab by heating iridium while exposed to air. The following data was collected: Mass of crucible: 38.26 g Mass of crucible and iridium: 39.52 g Mass of crucible and iridium oxide: 39.73 g Show, or explain, your calculations as you determine the empirical formula of the compound.

Answer 1

the mass of crucible = 38.26

Mass of crucible + Ir = 39.52 g

So mass of Ir = 39.52 - 38.26 = 1.26 g

atomic mass of Ir = 192 g / mole

Moles of Ir = mass /  atomic mass = 1.26/ 192 = 0.0066

mass of crucible + Iridium oxide = 39.73 g

Mass of oxygen = total mass - (mass of Ir + mass of crucible )

Mass of oxygen = 39.73 - (39.52) = 0.21g

moles of O = mass / atomic mass = 0.21 / 16 = 0.0131

let us divide the moles of Ir and O with moles of Ir

moles of Ir = 0.0066 / 0.0066 = 1

Moles of O = 0.0131 /0.0066 = 1.98 = 2

the empirical formula = IrO2

Answer 2

1) Compund Ir (x) O(y)

2) Mass of iridium = mass of crucible and iridium - mass of crucible = 39.52 g - 38.26 g = 1.26 g

3) Mass of iridium oxide = mass of crucible and iridium oxide - mass of crucible = 39.73g - 38.26g = 1.47g

4) Mass of oxygen = mass of iridum oxide - mass of iridium = 1.47g - 1.26g = 0.21g

5) Convert grams to moles

moles of iridium = mass of iridium / molar mass of iridium = 1.26 g / 192.17 g/mol = 0.00656 moles

moles of oxygen = mass of oxygen / molar mass of oxygen = 0.21 g / 15.999 g/mol = 0.0131

6) Find the proportion of moles

Divide by the least of the number of moles, i.e. 0.00656

Ir: 0.00656 / 0.00656 = 1

O: 0.0131 / 0.00656 = 2

=> Empirical formula = Ir O2 (where 2 is the superscript for O)

Answer: Ir O2