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Arlecino [84]
3 years ago
7

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

Chemistry
2 answers:
dem82 [27]3 years ago
8 0

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

Maurinko [17]3 years ago
6 0
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
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