Question

A sample of a substance with the empirical formula XBr2 weighs 0.5210 g. When it is dissolved in water and all its bromine is converted to insoluble AgBr by addition of an excess of silver nitrate, the mass of the resulting AgBr is found to be 0.6586 g. The chemical reaction is XBr2 + 2 AgNO32AgBr + X(NO3)2

(a) Calculate the formula mass of XBr2. Formula mass XBr2 = g mol-1

(b) Calculate the atomic mass of X. Atomic mass X = g mol-1

Answer 1

The balanced chemical reaction is as follows:

XBr₂ + 2AgNO₃ 2 AgBr + X(NO₃)₂

Mass of sample of XBr₂ = 0.5210 g

Mass of AgBr precipitate = 0.6586 g

Determine the number of moles of AgBr as follows:

= 0.6586 g AgBr x (1 mol AgBr / 187.77 g AgBr)

= 0.003507 mol AgBr

Use the moles of AgBr and the mole ratio from the balanced chemical reaction and determine the moles of XBr₂ as follows:

= 0.003507 mol AgBr x ( 1 mol XBr₂ / 2 mol AgBr)

= 0.001754 mol XBr₂

(a) Determine the formula mass of XBr₂ as follows:

Mass of sample of XBr₂ = 0.5210 g

Moles of XBr₂ = 0.001754 mol

Molar mass = Mass(g) / Number of moles

Molar mass = 0.5210 g / 0.001754 mol

Molar mass of XBr₂ = 297.1 g/mol

(b) Determine the atomic mass of X as follows:

Atomic mass of X + (2 x Atomic mass of Br) = Molar mass of XBr₂

Rearrange the formula for atomic mass of X as follows:

Atomic mass of X = Molar mass of XBr₂ -(2 x Atomic mass of Br)

Atomic mass of X = 297.1 g/mol - (2 x 79.904 g/mol)

Atomic mass of X = 137.3 g/mol