Chapter 11, Problem 72A

Interpretation Introduction

Interpretation: The mass of Na₃Ag(S₂O₃)₂is to be calculated from given mass AgBr from photographic film.

Concept introduction: 1 mol is defined as $6.02214076\times {10}^{23}$ particles of any substance.

Molecular mass is defined as the mass of one mole of a substance.

The formula to calculate moles of substance

  $\text{Moles=}\frac{\begin{array}{ccc}\text{Mass}& \text{of}& \text{substance}\end{array}}{\begin{array}{cccc}\text{Molar}& \text{mass}& \text{of}& \text{substance}\end{array}}$

Answer to Problem 72A

0.587 g of Na₃Ag(S₂O₃)₂ is produced

Explanation of Solution

The given compound is AgBr.

The formula to calculate formula mass is,

  $\text{Formula}\text{mass}=\left(N_1\times x_1\right)+\left(N_2\times x_2\right)+\mathrm{...}$

Where,

• $N_1$ represents the number of atoms of the first element in a chemical formula.
• ${\text{x}}_1$ represents the atomic mass of the first element.
• $N_2$ represents the number of atoms of the second element in a chemical formula.
• ${\text{x}}_1$ represents the atomic mass of the second element.

The atomic mass of silver is $107.87\text{g}/\text{mol}$ .

The atomic mass of bromine is $79.909\text{g}/\text{mol}$ .

Substitute the value of atomic mass in the above formula.

  $\begin{array}{c}\text{Formula}\text{mass}=\left(N_1\times x_1\right)+\left(N_2\times x_2\right)+\mathrm{...}\\ =\left(1\times 107.87\right)+\left(1\times 79.909\right)\\ =187.779\text{g/mol}\end{array}$

The formula mass or molar mass of AgBr is $187.779\text{g/mol}$ .

Given mass of AgBr is 0.275 g

  $\begin{array}{c}\text{Moles=}\frac{\begin{array}{ccc}\text{Mass}& \text{of}& \text{substance}\end{array}}{\begin{array}{cccc}\text{Molar}& \text{mass}& \text{of}& \text{substance}\end{array}}\\ \text{=}\frac{\text{0}\text{.275g}}{\text{187}\text{.78g/mol}}\\ \text{=0}\text{.001464moles}\end{array}$

The product produced after the reaction is Na₃Ag(S₂O₃)_2.

The formula to calculate formula mass is,

  $\text{Formula}\text{mass}=\left(N_1\times x_1\right)+\left(N_2\times x_2\right)+\mathrm{...}$

Where,

• $N_1$ represents the number of atoms of the first element in a chemical formula.
• ${\text{x}}_1$ represents the atomic mass of the first element.
• $N_2$ represents the number of atoms of the second element in a chemical formula.
• ${\text{x}}_1$ represents the atomic mass of the second element.

The atomic mass of sodium is $22.9898\text{g}/\text{mol}$ .

The atomic mass of silver is $107.87\text{g}/\text{mol}$ .

The atomic mass of oxygen is $15.99\text{g}/\text{mol}$ .

The atomic mass of sulfur is $32.064\text{g}/\text{mol}$ .

Substitute the value of atomic mass in the above formula.

  $\begin{array}{c}\text{Formula}\text{mass}=\left(N_1\times x_1\right)+\left(N_2\times x_2\right)+\mathrm{...}\\ =\left(3\times 22.98\right)+\left(1\times 107.87\right)+(4\times 32.06)+(6\times 15.99)\\ =401.09\text{g/mol}\end{array}$

From the stoichiometry of equation,

1 mole of AgBr produce = 1 mole of Na₃Ag(S₂O₃)₂

0.001464 moles of AgBr produce = 0.001464 moles of Na₃Ag(S₂O₃)₂

  $\begin{array}{c}Massofsubs\tan ce=moles\times molarmassofsubs\tan ce\\ =0.001464\times 401.09\\ =0.587g\end{array}$

Conclusion

0.587 g of Na₃Ag(S₂O₃)₂ is produced.