Chapter 10, Problem 3PEB

(a)

To determine

The mass percentage composition of K, C and O in potash, ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$

Answer to Problem 3PEB

Solution:

Percentage composition of K, C and Oin ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ are56.6%, 8.7% and 34.7% respectively.

Explanation of Solution

Given data:

Atomic mass ofK = 39.1 u.

Atomic mass ofC = 12 u.

Atomic mass of O = 16 u.

Formula used:

$\text{Percentage composition of an atom in compound= }\frac{\left(\begin{array}{l}\text{Atomic weight of }\\ \text{ the atom}\end{array}\right)\left(\begin{array}{l}\text{Number of atoms of }\\ \text{ the given element}\end{array}\right)}{\text{Formula weight of the given compoud}}$

Explanation:

Step 1: To calculate the formula weight of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$.

In ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$, there are 2 atoms of potassium, one atom of carbon and 3 atoms of oxygen.

$\begin{array}{c}{\text{Formula weight of K}}_2{\text{CO}}_3=(39.1\text{ u})\times (2)+12.0\text{ u + }(\text{16 u)}\times \text{(3})\\ \text{=138}\text{.2 u}\end{array}$

Step 2: To calculate the percentage composition.

$\begin{array}{c}\text{Percentage composition of K = }\frac{(39.1\text{ u of K})\times (2)}{138.2{\text{ u of K}}_{\text{2}}{\text{CO}}_{\text{3}}}\times 100\%{\text{ of K}}_{\text{2}}{\text{CO}}_{\text{3}}\\ =56.6\%\end{array}$

$\begin{array}{c}\text{Percentage composition of C = }\frac{(12\text{ u of C})\times (1)}{138.2{\text{ u of K}}_{\text{2}}{\text{CO}}_{\text{3}}}\times 100\%{\text{ of K}}_{\text{2}}{\text{CO}}_{\text{3}}\\ =8.7\%\end{array}$

$\begin{array}{c}\text{Percentage composition of O=}\frac{(16\text{ u of O})\times (3)}{138.2{\text{ u of K}}_{\text{2}}{\text{CO}}_{\text{3}}}\times 100\%{\text{ of K}}_{\text{2}}{\text{CO}}_{\text{3}}\\ =34.7\%\end{array}$

Conclusion:

Percentage composition of K, C and Oin ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ are56.6%, 8.7% and 34.7% respectively.

Potassium has the highest percentage composition and carbon has the least.

(b)

To determine

The mass percentage composition of Ca, S and O in gypsum, ${\text{CaSO}}_4$.

Answer to Problem 3PEB

Solution:

Percentage composition of Ca, S and Oin ${\text{CaSO}}_4$ are29.4%, 23.6% and 47% respectively.

Explanation of Solution

Given data:

Atomic mass ofCa = 40.1 u.

Atomic mass ofS = 32.1 u.

Atomic mass of O = 16 u.

Formula used:

$\text{Percentage composition of an atom in compound= }\frac{\left(\begin{array}{l}\text{Atomic weight of }\\ \text{ the atom}\end{array}\right)\left(\begin{array}{l}\text{Number of atoms of }\\ \text{ the given element}\end{array}\right)}{\text{Formula weight of the given compoud}}$

Explanation:

Step 1: To calculate the formula weight of ${\text{CaSO}}_4$.

In ${\text{CaSO}}_4$, there is 1 atom of calcium, one atom of sulphur and 4 atoms of oxygen.

$\begin{array}{c}{\text{Formula weight of CaSO}}_4=40.1\text{ u}+12.0\text{ u + }(\text{16 u)}\times \text{(4})\\ \text{=136}\text{.2 u}\end{array}$

Step 2: To calculate the percentage composition.

$\begin{array}{c}\text{Percentage composition of Ca = }\frac{(40.1\text{ u of Ca})\times (2)}{136.2{\text{ u of CaSO}}_4}\times 100\%{\text{ of CaSO}}_4\\ =\text{ 29}\text{.4}\%\end{array}$

$\begin{array}{c}\text{Percentage composition of S = }\frac{(32.1\text{ u of C})\times (1)}{136.2{\text{ u of CaSO}}_4}\times 100\%{\text{ of CaSO}}_4\\ =\text{ 23}\text{.6}\%\end{array}$

$\begin{array}{c}\text{Percentage composition of O=}\frac{(16\text{ u of O})\times (4)}{136.2{\text{ u of CaSO}}_4}\times 100\%{\text{ of CaSO}}_4\\ =47\%\end{array}$

Conclusion:

Percentage composition of Ca, S and Oin ${\text{CaSO}}_4$ are29.4%, 23.6% and 47% respectively.

Oxygen has the highest percentage composition and sulphur has the least.

(c)

To determine

The mass percentage composition of K, N and O in gypsum, ${\text{KNO}}_3$.

Answer to Problem 3PEB

Solution:

Percentage composition of K, N and Oin ${\text{KNO}}_3$ are38.7%, 13.8% and 47.5% respectively.

Explanation of Solution

Given data:

Atomic mass ofK = 39.1 u.

Atomic mass ofN = 14 u.

Atomic mass of O = 16 u.

Formula used:

$\text{Percentage composition of an atom in compound= }\frac{\left(\begin{array}{l}\text{Atomic weight of }\\ \text{ the atom}\end{array}\right)\left(\begin{array}{l}\text{Number of atoms of }\\ \text{ the given element}\end{array}\right)}{\text{Formula weight of the given compoud}}$

Explanation:

Step 1: To calculate the formula weight of ${\text{KNO}}_3$.

In ${\text{KNO}}_3$, there is one atom of potassium, one atom of nitrogen and 3 atoms of oxygen.

$\begin{array}{c}{\text{Formula weight of KNO}}_3=39.1\text{ u}+14.0\text{ u + }(\text{16 u)}\times \text{(3})\\ \text{=101}\text{.1 u}\end{array}$

Step 2: To calculate the percentage composition.

$\begin{array}{c}\text{Percentage composition of K = }\frac{(39.1\text{ u of K})\times (2)}{101.1{\text{ u of KNO}}_3}\times 100\%{\text{ of KNO}}_3\\ =\text{ 38}\text{.7}\%\end{array}$

$\begin{array}{c}\text{Percentage composition of N = }\frac{(14.0\text{ u of N})\times (1)}{101.1{\text{ u of KNO}}_3}\times 100\%{\text{ of KNO}}_3\\ =\text{ 13}\text{.8}\%\end{array}$

$\begin{array}{c}\text{Percentage composition of O =}\frac{(16\text{ u of O})\times (3)}{101.1{\text{ u of KNO}}_3}\times 100\%{\text{ of KNO}}_3\\ =47.5\%\end{array}$

Conclusion:

Percentage composition of K, N and Oin ${\text{KNO}}_3$ are38.7%, 13.8% and 47.5% respectively.

Oxygen has the highest percentage composition and potassium has the least.

(d)

To determine

The mass percentage composition of C, H, N and O in gypsum, ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$.

Answer to Problem 3PEB

Solution:

Percentage composition of C, H, N and Oin ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$ are49.5%, 5.2%, 28.9% and 16.5% respectively.

Explanation of Solution

Given data:

Atomic mass ofC = 12 u.

Atomic mass of H = 1 u.

Atomic mass of N = 14 u.

Atomic mass of O = 16 u.

Formula used:

$\text{Percentage composition of an atom in compound= }\frac{\left(\begin{array}{l}\text{Atomic weight of }\\ \text{ the atom}\end{array}\right)\left(\begin{array}{l}\text{Number of atoms of }\\ \text{ the given element}\end{array}\right)}{\text{Formula weight of the given compoud}}$

Explanation:

Step 1: To calculate the formula weight of ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$.

In ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$, there is 8 atoms of carbon, 10 atoms of hydrogen, 4 atoms of nitrogen and 2 atoms of oxygen.

$\begin{array}{c}{\text{Formula weight of C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}=(12\text{ u)}\times \text{(8)}+(1\text{ u})\times (8)+(14.0\text{ u )}\times \text{(4)+ }(\text{16 u)}\times \text{(2})\\ \text{=194}\text{.0 u}\end{array}$

Step 2: To calculate the percentage composition.

$\begin{array}{c}\text{Percentage composition of C = }\frac{(12\text{ u of C})\times (8)}{194{\text{ u of C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}}\times 100\%{\text{ of C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}\\ =\text{ 49}\text{.5}\%\end{array}$

$\begin{array}{c}\text{Percentage composition of H = }\frac{(1\text{ u of H})\times (10)}{194{\text{ u of C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}}\times 100\%{\text{ of C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}\\ =\text{ 5}\text{.2}\%\end{array}$

$\begin{array}{c}\text{Percentage composition of N = }\frac{(14\text{ u of N})\times (4)}{194{\text{ u of C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}}\times 100\%{\text{ of C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}\\ =\text{ 28}\text{.9}\%\end{array}$

$\begin{array}{c}\text{Percentage composition of O = }\frac{(16\text{ u of O})\times (2)}{194{\text{ u of C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}}\times 100\%{\text{ of C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}\\ =\text{ 16}\text{.5}\%\end{array}$

Conclusion:

Percentage composition of C, H, N and O in ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$ are49.5%, 5.2%, 28.9% and 16.5% respectively. Carbon has the highest percentage composition and hydrogen has the least.