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

Answer:

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

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:

$\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of element}}\times 100$

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

$\frac{1\times 132.905 g/mol}{191.949 g/mol}\times 100=69.24\%$

Mass of cesium in 15 grams of cesium acetate  be x

$69.24\%=\frac{x}{15 g}\times 100$

x = 10.386 g

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