A0.999-g sample of a metal chloride, mcl2, is dissolved in water and treated with excess aqueous silver nitrate. the silver chloride that formed weighed 1.286 g. calculate the atomic mass of m.

Mass of silver chloride formed = 1.286 g

Molar mass of silver chloride = 143.32 g/mol AgCl

Converting mass to moles of AgCl using the molar mass:

The balanced chemical equation between metal chloride and silver nitrate can be represented as,

Calculating the moles of MCl2 from moles of AgCl that form:

The given mass of =0.999 g

Molar mass of =

Molar mass of M= Molar mass of - 2 (Molar mass of Cl)

                          =151.6g/mol

Atomic mass of M is .

Further Explanation:

Given reaction occurs as follows:

The formula to calculate moles of AgCl is as follows:

               ...... (1)

Substitute for mass of AgCl and for molar mass of AgCl in equation (1).

According to the balanced chemical reaction, one mole of reacts with two moles of to form two moles of AgCl and one mole of , So stoichiometric ratio between  and AgCl is 1:2.

Since two moles of AgCl are produced by one mole of , number of moles of that can produce 0.00897 moles of AgCl can be calculated as follows:

The formula to calculate number of moles of is as follows:

                             ...... (2)

Rearrange equation (2) to calculate molar mass of .                                ...... (3)

Substitute 0.999 g for mass of and 0.00449 mol for moles of  in equation (3).

The formula to calculate molar mass of  is as follows:

 ....... (4)

Rearrange equation (4) to calculate atomic mass of M.

  ...... (5)

Substitute 222.5 g/mol for molar mass of and 35.453 g for atomic mass of Cl in equation (5).

Therefore atomic mass of M is 151.6 g.

Learn more:

1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride:

2. Calculate the moles of ions in the solution:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: AgCl, MCl2, 151.6 g, atomic mass, molar mass, 222.5 g/mol, 35.453 g, 1.286 g, 0.999 g, stoichiometric ratio, 1:2.