Question

What is tbe mass of ca present in 15g of cesium acetate, formula csch3coo?

Answer 1

The molecular formula of cesium acetate it s

CsCH3COO

Thus it means each mole of cesium acetate will have one mole of Cesium

The moles of cesium acetate present = MAss / Molar mass = 15 / 191.949

Moles of Cesium acetate = 0.0781

Moles of Cs present = 0.0781

Mass of Cs = Moles X atomic mass = 0.0781 X 132.9 = 10.38 g

Answer 2

10.386 grams is the mass of Cs present in 15 grams of cesium acetate.

Step-by-step explanation:

Atomic mass of cesium = 132.905 g/mol

Molecular mass of the
`CH_3COOCs` = 191.949 g/mol

Percentage of an element in a compound:

`\frac{\text{Number of atoms of element}* \text{Atomic mass of element}}{\text{molecular mass of element}}* 100`

Mass percentage of cesium in 1 molecule of
`CH_3COOCs`:

`(1* 132.905 g/mol)/(191.949 g/mol)* 100=69.24\%`

Mass of cesium in 15 grams of cesium acetate be x

`69.24\%=(x)/(15 g)* 100`

x = 10.386 g

10.386 grams is the mass of Cs present in 15 grams of cesium acetate.