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 =$\frac{\text{ given mass of Cs}}{\text{ molar mass of Cs}}= \frac{89g}{133g/mole}=0.67moles$

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

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

For Cs = $\frac{0.67}{0.67}=1$

For O =$\frac{0.69}{0.67}=1$

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.

$n=\frac{\text{Molecular weight of metal}}{\text{Equivalent of metal}}=\frac{298}{149}=2$

The molecular formula will be=$2\times CsO=Cs_2O_2$