Chapter 2, Problem 2.112QA

Interpretation Introduction

To find:

To calculate:

a. The formula masses of CdS and CdSe

b. Atoms of Se, if a nanoparticle of CdSe contains 2.7 x 10⁷ atoms of Cd

c. Gram of Cd and S in 4.3 x 10^-15 g of CdS

Answer to Problem 2.112QA

Solution:

a) Formula mass of CdS is 144.475 g/mol and formula mass of CdSe is 191.37 g/mol

b) Atoms of Se in CdS = 2.7 x 10⁷

c) Mass of Cd in the CdS sample is 3.3 x 10^-15 g and mass of S in the nanoparticle = 9.5 10 ^-16 g.

Explanation of Solution

Formula and concept: The problem can be solved using a series of unit conversions.

A mole is the SI unit of amount chemical substance. When writing units, it is written as “mol”.

Molar mass of a compound is the sum of the atomic mass of each element.

Molar mass of the substance is treated as a conversion factor to interconvert grams and moles. Mole ratio within the compound is treated as ratio of atoms in that compound.

For example: 1 mole of CdS contains 1 mole of Cd and 1 mole of S

Similarly, 1 molecule of CdS contains 1 atom of Cd and 1 atom of S

A mole is the SI unit of amount chemical substance. When writing units, it is written as “mol”.

Given information:

We are given

a. Molecular formulae for CdS and CdSe

b. Atoms of CdS are given as 2.7 x 10⁷ in the sample

c. Mass of CdS is given as 4.3 x 10^-15 g

Calculations:

a. Atomic mass of Cadmium = 112.411 g per mol

Atomic mass of S = 32.065 g per mol

Molecular formula = CdS, the molar mass for CdS will be:

$Molar mass of CdS= Atomic weight of Cd + atomic weight of S$

$= 112.411\frac{g}{mol}+32.065\frac{g}{mol}=144.476\frac{g}{mol}$

Thus, the molar mass for CdS is 144.476 g per mol

Similarly, the molar mass for CdSe is calculated as:

$\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{C}\mathrm{d}\mathrm{S}\mathrm{e}=\mathrm{ }\mathrm{A}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{i}\mathrm{c}\mathrm{ }\mathrm{w}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{C}\mathrm{d}\mathrm{ }+\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{i}\mathrm{c}\mathrm{ }\mathrm{w}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{S}\mathrm{e}\mathrm{ }\mathrm{ }\mathrm{ }$

$=112.411\frac{g}{mol}+78.960\frac{g}{mol}=191.371\frac{g}{mol}$

Thus, the molar mass for CdSe is 191.371 g per mol

b. 1 molecule of CdSe = 1 atom of Cd and 1 atom of Se

This can be treated as a conversion factor to find atoms of Se from atoms of Cd as follows:

2.7 $\times {10}^7\mathrm{ }\mathrm{C}\mathrm{d}\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}\mathrm{ }\times \frac{\left(1\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{S}\mathrm{e}\right)}{1\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{C}\mathrm{d}}\mathrm{ }=2.7\mathrm{ }\times {10}^7\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{S}\mathrm{e}$

c. Molar mass of CdS is 144.475 g/mol that is 1 mole of CdS sample weighs 144.475 gram.

We know that 1 molecule of CdS contains 1 atom of Cd

Atomic weight for Cd is 112.41 g/mol that is 1 mole of Cd weighs 112.41 gram.

So, the conversion factors needed are:

$1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{C}\mathrm{d}\mathrm{S}=144.475\mathrm{ }\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{C}\mathrm{d}\mathrm{S}$

$1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{C}\mathrm{d}\mathrm{S}=1\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{C}\mathrm{d}$

$1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{C}\mathrm{d}=112.41\mathrm{ }\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{C}\mathrm{d}$

 Hence, the set up would be:

$4.3\mathrm{ }\times {10}^{-15}\mathrm{ }\mathrm{g}\mathrm{ }\mathrm{C}\mathrm{d}\mathrm{S}\mathrm{ }\mathrm{ }\times \mathrm{ }\frac{1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{C}\mathrm{d}\mathrm{S}}{144.475\mathrm{ }\mathrm{g}}\mathrm{ }\mathrm{ }\times \mathrm{ }\frac{1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{C}\mathrm{d}}{1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{C}\mathrm{d}\mathrm{S}}\mathrm{ }\mathrm{ }\times \mathrm{ }\frac{112.41\mathrm{ }\mathrm{g}\mathrm{ }\mathrm{C}\mathrm{d}}{1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{C}\mathrm{d}} =3.3392\mathrm{ }\times {10}^{-15}\mathrm{g}\mathrm{ }\mathrm{C}\mathrm{d}$

Thus, the mass of Cd atoms in the given sample of CdS is 3.3 x 10^-15 g.

On similar lines, we can calculate mass of S atoms as follows:

$4.3\mathrm{ }\times {10}^{-15}\mathrm{g}\mathrm{ }\mathrm{C}\mathrm{d}\mathrm{S}\mathrm{ }\mathrm{ }\times \mathrm{ }\frac{1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{C}\mathrm{d}\mathrm{S}}{144.475\mathrm{ }\mathrm{g}}\mathrm{ }\mathrm{ }\times \mathrm{ }\mathrm{ }\frac{1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{S}}{1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{C}\mathrm{d}\mathrm{S}}\mathrm{ }\mathrm{ }\times \mathrm{ }\frac{32.065\mathrm{ }\mathrm{g}\mathrm{ }\mathrm{S}}{1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{S}}\mathrm{ }=\mathrm{ }\mathrm{ }9.543\mathrm{ }\times {10}^{-16}\mathrm{g}\mathrm{ }\mathrm{S}$

Thus, mass of S atoms in the given sample of CdS is 9.5 x 10^-16 g.

Conclusion:

Molar mass of a substance is the sum of atomic weights of each element present in that substance. Just like mole-to-mole ratio, you can write the molecule-to-atoms ratio for a given compound. Molar mass of the compound is used to convert between moles and mass and vice a versa.