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

The molecular formula will be

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 =

Moles of O =

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

For Cs =

For O =

The ratio of Cs : O= 1:1

Hence the empirical formula is

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

The molecular weight = 298 g/mole

Now we have to calculate the molecular formula.

The molecular formula will be=

1. 1 mole of anything contains 6.022x10^23 pieces  

d  

2. a  

3. c, number  

5. 1.24x10^24atoms / 6.022x10^23atoms/mole = 2moles h2 atoms  

6. fe2br3 mass = 2moles x mass fe + 3 moles x mass br  

molar mass = grams / number of moles = g/mole  

to find moles = g / molar mass  

to find mass = g = molar mass / moles