Chem – Representation of Compounds and Molecules with Subscripts

 

Before we discuss the formation of any compounds, we have to make sure we have an understanding of how those compounds are represented in chemistry. In any type of chemistry class, the main focus of how many elements you have in a compound is what they call the subscripts. Subscripts are numbers that are placed to the right of and down towards the bottom of an element. The subscripts are a number that multiplies the element they are next to. If there is no subscript present then that means it is counted as a 1.

 

Examples: Identify what the subscripts of each element in the compounds below.

K2O K is 2 and O is 1
SrF2 Sr is 1 and F is 2
Mg3P2 Mg is 3 and P is 2

 

Some chemical representations also have parenthesis in them. These parenthesis are much like the ones you use in math class. That is parenthesis are used to multiply everything inside them by a number that is put outside of them.

 

Examples: How many atoms of each element are present in the following compounds.

(NH4)2S 2 N and 8 H and 1 S
Be(NO3)2 1 Be and 2 N and 6 O
Fe2(S2O3)3 2 Fe and 6 S and 9 O

 

 

VIDEO Counting Compound Subscripts Demonstrated Example 1: How many atoms of each element are present in the compound below.

Ca3(PO4)2

 

Step 1:

What is the first element?

Answer: Ca

 

Step 2:

What is the subscript of the first element?

Answer: 3

 

Step 3:

Is Ca in parenthesis?

Answer: No

 

Step 4:

How many Ca atoms do we have?

Answer: 3 Ca

 

Step 5:

What is the next element?

Answer: P

 

Step 6:

What is the subscript of P?

Answer: 1

 

Step 7:

Is P in parenthesis?

Answer: Yes

 

Step 8:

What is the subscript of the parenthesis?

Answer: 2

 

Step 9:

How many P atoms do we have?

Answer: 1 * 2 = 2 P

 

Step 10:

What is the last element?

Answer: O

 

Step 11:

What is the subscript of O?

Answer: 4

 

Step 12:

Is O in parenthesis?

Answer: Yes

 

Step 13:

What is the subscript of the parenthesis?

Answer: 2

 

Step 14:

How many O atoms do we have?

Answer: 4 * 2 = 8 O

 

Step 15:

COMPLETE ANSWER: 3 Ca and 2 P and 8 O

 

PRACTICE PROBLEMS: How much of each element are present in the following compounds.

LiCl 1 Li and 1 Cl
Rb2S 2 Rb and 1 S
Al2Se3 2 Al and 3 Se
CH4 1 C and 4 H
Cs2O 2 Cs and 1 O
Cr3P2 3 Cr and 2 P
Sr(OH)2 1 Sr and 2 O and 2 H
Cr2(SO3)3 2 Cr and 3 S and 9 O
Pb3(PO4)4 3 Pb and 4 P and 16 O

 

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