Q. 6.124 - An Introduction to General, Organic, and Biological Chemistry

Question

Write the formula for each of the following: $(6.3,6.4,6.5)$

a. sodium carbonate

b. nitrogen dioxide

c. aluminum nitrate

d. copper(I) nitride

e. potassium phosphate

f. cobalt(III) sulfate

Step-by-Step Solution

Answer

(a) sodium carbonate has the formula $N{a}_{2}C{O}_3$.

(b) nitrogen dioxide has the formula $N{O}_2$.

(c) aluminum nitrate has the formula $Al{\left(N{O}_3\right)}_3$.

(d) copper(I) nitride has the formula $CuN{O}_3$.

(e) potassium phosphate has the formula $K_{3}P{O}_4$.

(f) cobalt(III) sulfate has the formula $C{o}_2{\left(S{O}_4\right)}_3$.

part(a) Step 1: Given information

We have been given sodium carbonate  ,

Na, C and O symbols are used in this formula.

Part(a) Step 2: Explanation

Now, We must locate valency and exchange valency. 

The valency Na$=1$, C$=4$ and O$=2$,

Interchange valency in the subscript $N{a}_2$ and $C{O}_3$,

Then formula is $N{a}_{2}C{O}_3$.

Part(b) Step 1: Given information

We have been given nitrogen dioxide ,

N and O symbols are used in this formula.

Part(b) Step 2: Explanation

Now, We must locate valency and exchange valency. 

The valency N$=3$ and O$=2$,

As there are $1$molecule of nitrogen and $2$molecule of oxygen ,

So, final formula is $N{O}_2$.

Part(c) Step 1: Given information

We have been given aluminum nitrate  ,

Al, N and O symbols are used in this formula.

Part(c) Step 2: Explanation

Now, We must locate valency and exchange valency. 

The valency Al$=3$ and $N{O}_3=-1$,

Interchange valency in the subscript $A{l}_1$and ${\left(N{O}_3\right)}_3$,

Then the final formula $Al{\left(N{O}_3\right)}_3$.

Part(d) Step 1: Given information

We have been given  copper(I) nitride  ,

Cu, N and O symbols are used in this formula.

Part(d) Step 2: Explanation

Now, We must locate valency and exchange valency. 

The valency Cu$=2$ and $N{O}_3=-2$,

Interchanged valency in the subscript $C{u}_1$and $N{O}_{3_1}$

Then final formula is $CuN{O}_3$.

Part(e) Step 1: Given information

We have been given potassium phosphate ,

K, P and O symbols are used in this formula.

Part(e) Step 2: Explanation

Now, We must locate valency and exchange valency. 

The valency K$=1$ and $P{O}_4=-2$,

Interchange valency in the subscript $K_2$and $P{O}_{4_1}$

Hence finally formula is $K_{2}P{O}_4$.

Part(f) Step 1: Given information

We have been given cobalt(III) sulfate ,

Co, S and O symbols are used in this formula.

Part(f) Step 2: Explanation

Now, We must locate valency and exchange valency. 

The valency Co$=3$ and $S{O}_4=2$,

Interchanged valency in the subscript $C{o}_2$and $S{O}_{4_3}$

Hence formula is $C{o}_2{\left(S{O}_4\right)}_3$.