Chapter 2, Problem 1PE

How many atoms of each element are expressed by the following formulas? (Hint: Pay special attention to counting elements within parentheses.)

(a) ${\text{SF}}_{\text{6}}$

(b) ${{\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{)}}_{\text{2}}{\text{N}}_{\text{2}}{\text{H}}_{\text{2}}$

(c) ${\text{Ca}}_3{(P{O}_4)}_2$

(d) ${\text{Co(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{•6H}}_{\text{2}}\text{O}$

Interpretation Introduction

Interpretation: Number of atoms of each element in given formula is to be determined.

Concept Introduction:

Atoms are composed of tiny particles and these tiny particles are known as protons, electrons, and neutrons.

The substances which are unable to be degraded into simpler substances are known as atoms.

Elements are made up of one type of atoms. Elements join together to make molecules.

Atoms of different elements join together to form compound. All compounds are molecules but all molecules are not compounds.

Answer to Problem 1PE

Solution:

(a) $1\text{ S, 6F}$

(b) $\text{4 C, 12}\text{}\text{ H, 2 N}$

(c) $\text{3 Ca, 2 P, 8 O}$

(d) $1\text{ Co, 2 N, 12 O, 12 H}$

Explanation of Solution

a) ${\text{SF}}_6$

The formula-${\text{SF}}_6$ consists of two elements S and F. In compound ${\text{SF}}_6$, there is one atom of element S, and six atoms of element F are present. Therefore, the formula ${\text{SF}}_6$ represents:

$1\text{ S 6F}$

Hence, number of atoms in each element in ${\text{SF}}_6$ are $1\text{ S, 6F}$.

b) ${\left({\text{C}}_2{\text{H}}_5\right)}_2{\text{N}}_2{\text{H}}_2$

The formula ${\left({\text{C}}_2{\text{H}}_5\right)}_2{\text{N}}_2{\text{H}}_2$ consists of three elements C, H, and N. In ${\left({\text{C}}_2{\text{H}}_5\right)}_2{\text{N}}_2{\text{H}}_2$, each ${\text{C}}_2{\text{H}}_5$ consists of two C and five H atoms so, two units of ${\text{C}}_2{\text{H}}_5$ consist of four C and ten H atoms. In the ${\text{N}}_2{\text{H}}_2$ unit, there are two additional H atoms and two N atoms. Therefore, the formula ${\left({\text{C}}_2{\text{H}}_5\right)}_2{\text{N}}_2{\text{H}}_2$ represents:

$\text{4 C 12}\text{}\text{ H 2 N}$.

Hence, number of atoms in each element in ${\left({\text{C}}_2{\text{H}}_5\right)}_2{\text{N}}_2{\text{H}}_2$ are $\text{4 C, 12}\text{}\text{ H, 2 N}$.

c) ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$

The formula ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ consists of three elements Ca, P and O. In ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$, each ${\text{PO}}_4$ consists of one P atom and four O atoms so, two ${\text{PO}}_4$ units consist of two P atoms and eight O atoms. The molecule also consists of three Ca atoms. Therefore, the formula ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ represents:

$\text{3 Ca 2 P 8 O}$

Hence, number of atoms in each element in ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ are $\text{3 Ca, 2 P, 8 O}$.

d) $\text{Co}{\left({\text{NO}}_3\right)}_2.6{\text{H}}_2\text{O}$

$\text{Co}{\left({\text{NO}}_3\right)}_2.6{\text{H}}_2\text{O}$ is the formula used for a hydrate which is indicated by the raised dot. For every $\text{Co}{\left({\text{NO}}_3\right)}_2.6{\text{H}}_2\text{O}$, there are six water molecules which consist of twelve H and six O atoms. The molecule also consists of one Co. In ${\left({\text{NO}}_3\right)}_2$ there are two ${\text{NO}}_3$ units, each unit contains one N and three O atoms, so, both units contain total two N and six O atoms. Thus, the formula $\text{Co}{\left({\text{NO}}_3\right)}_2.6{\text{H}}_2\text{O}$ represents:

$1\text{ Co 2 N 12 O 12 H}$

Hence, number of atoms in each element in $\text{Co}{\left({\text{NO}}_3\right)}_2.6{\text{H}}_2\text{O}$ are $1\text{ Co, 2 N, 12 O, 12 H}$.