Q. 6.123 - An Introduction to General, Organic, and Biological Chemistry

Question

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

a. tin(II) carbonate

b. lithium phosphide

c. silicon tetrachloride

d. manganese(III) oxide

e. tetraphosphorus triselenide 

f. calcium bromide 

Step-by-Step Solution

Answer

(a) tin(II) carbonate  has the formula $SnC{O}_3$.

(b) lithium phosphide has the formula $L{i}_{3}P$.

(c) silicon tetrachloride has the formula $SiC{l}_4$.

(d) manganese(III) oxide has the formula $M{n}_2{O}_3$.

(e) tetraphosphorus triselenide  has the formula $P_{4}S{e}_3$.

(f) calcium bromide has the formula $CaB{r}_2$.

Part(a) Step 1: Given information

We have been given tin(II) carbonate 

Sn, C and O symbols used in this formula.

Part(a) Step 2: Explanation

Now, We must locate valency and exchange valency. 

So, the valency Sn$=2$ and $C{O}_3=-2$,

Interchange valency in the subscript $S{n}_1$ and $C{O}_3$

Then the compound has formula $SnC{O}_3$.

Part(b) Step 1: Given information

We have been given lithium phosphide  ,

Li and P symbols are used in this formula.

Part(b) step 2: Explanation

Now, We must locate valency and exchange valency. 

So, the valency Li$=+1$ and P$=3$

Interchange valency in the subscript $L{i}_3$and $P_1$,

Hence the formula is $L{i}_{3}P$.

Part(c) Step 1: Given information

We have been given silicon tetrachloride  ,

Si and Cl symbols are used in this formula.

Part(c) Step 2: Explanation

Now, We must locate valency and exchange valency. 

The valency Si$=4$ and Cl$=-1$,

Interchange valency in the subscript $S{i}_1$and $C{l}_4$,

Then formula is $SiC{l}_4$.

Part(d) Step 1: Given information

We have been given manganese(III) oxide ,

Mn and O symbols are used in this formula.

Part(d) Step 2: Explanation

Now, We must locate valency and exchange valency. 

The valency Mn$=3$ and O$=2$,

Interchange valency in the subscript $M{n}_2$and $O_3$

hence the formula is $M{n}_2{O}_3$.

Part(e) Step 1: Given information

We have been given tetraphosphorus triselenide ,

P and Se symbols are used in  this formula.

Part(e) Step 2: Explanation

Now, We must locate valency and exchange valency. 

So, the valency P$=3$ and Se$=2$,

As there are $4$molecule of phosphorus and $3$molecule of Selenium,

So, the formula of given compound is $P_{4}S{e}_3$.

Part(f) Step 1: Given information

We have been given calcium bromide   ,

Ca and Br symbols are used in this formula.

Part(f) Step 2: Explanation

Now, We must locate valency and exchange valency. 

So, the valency Ca$=2$ and Br$=1$,

Interchange valency in the subscript $C{a}_1$and $B{r}_2$

Hench the formula is $CaB{r}_2$.