Question

What is the molecular formula of a compound containing 89% cesium (Cs) and 11% oxygen (O) with a molar mass = 298 g/mol?

Answer 1

M(Cs)=133 g/mol
M(O)=16 g/mol
M(CsxOy)=298 g/mol
w(Cs)=0.89
w(O)=0.11

CsxOy

x=M(CsxOy)w(Cs)/M(Cs)
x=298*0.89/133=2

y=M(CsxOy)w(O)/M(O)
y=298*0.11/16=2

Cs₂O₂ cesium peroxide

Answer 2

Answer: The molecular formula will be
`Cs_2O_2`

Explanation:-

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of Cs= 89 g

Mass of O = 11 g

Step 1 : convert given masses into moles.

Moles of Cs =
`\frac{\text{ given mass of Cs}}{\text{ molar mass of Cs}}= (89g)/(133g/mole)=0.67moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (11g)/(16g/mole)=0.69moles`

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Cs =
`(0.67)/(0.67)=1`

For O =
`(0.69)/(0.67)=1`

The ratio of Cs : O= 1:1

Hence the empirical formula is
`CsO`

The empirical weight of
`CsO` = 1(133)+1(16)= 149g.

The molecular weight = 298 g/mole

Now we have to calculate the molecular formula.

`n=\frac{\text{Molecular weight of metal}}{\text{Equivalent of metal}}=(298)/(149)=2`

The molecular formula will be=
`2* CsO=Cs_2O_2`