Chapter 7 Compounds Names and Formulas Chemistry Notes

Importance of writing and

naming chemical formulas

It is how

chemists

communicate

with each

Chemical formulas & compounds

Chemists that speak

different languages can communicate with each other using the common language of chemical

How to write the correct formula for a

binary ionic compound:

1. Determine the oxidation number for each element in the compound:

Example: Li and Cl

Li+1 _{and Cl}-1 _{= LiCl}

2. To write the correct formula, make sure the oxidation numbers add up to zero by writing

subscripts for each element – The cation is written first and the anion is written second!

What if the charges don’t add up to zero?

• _{If the charges don’t add up to zero, then} subscripts will be needed in the formula. • _Examples:

Li+1 and O-2 = Li

2O

Ca+2 _{and Br}-1 = _CaBr

2

Be+2 _{and P}-3 = _Be

Naming Binary Ionic Compounds

• _{Simply write the name of the metal}

• _{Change the ending of the non-metal to ide}

Examples:

LiCl is lithium chloride Li₂O is lithium oxide

CaCl₂ is calcium chloride

Be₃P₂ is beryllium phosphide

What about metals that have more than

one oxidation number?

• _{Most transition metals have more than 1}

oxidation number; for example copper can be +1 or +2 or iron can be either +2 or +3

• _{Important exceptions within the transition}

elements, Zn+2 _{and Ag}+1 _{do not have multiple}

oxidation numbers

Naming compounds that contain metals with multiple oxidation numbers:

• _Cu+1 _{& Cl}-1 _{= CuCl}

• Cu+2 _{& Cl}-1 _{= CuCl}

2

• _{How do we name these 2 compounds?}

Naming compounds with these type of

metals

• _{The rules are the same for ordinary binary} ionic compounds with one important

exception!

• _{Write the name of the binary ionic}

Roman numerals:

1 = I

5 = V

2 = II

6 = VI

3 = III

7= VII

Therefore:

(Remember, copper in this compound is +1)

• CuCl₂ Copper (II) chloride

Practice:

• _{Write the formula and name the following:} • _{Na & F}

NaF sodium flouride

• _{Ba & Br}

BaBr₂ Barium bromide

• _Co+2 _{& O}

CoO Cobalt (II) oxide

• _{Zn & N}

Ternary Ionic Compounds

compounds that contain a polyatomic ion. Examples:

MgSO₄

NaNO₃ Li₂CO₃

Consult the list of common polyatomic

ions

•

The names, formulas, and

names for common

polyatomic ions are listed

so you don’t need to

Writing and Naming Ternary Ionic

Compounds

The rules for writing and naming ternary ionic compounds are very similar to those for

Practice:

Na2SO4 sodium sulfate Mg & OH

Mg(OH)₂ Magnesium hydroxide Fe+3 _{& NO}

3

Writing and Naming Binary Molecular

Compounds

•

Remember, molecular compounds only

contain non-metals!

Rules for naming binary molecular

compounds

:

• _{When writing these types of formulas, we} use prefixes to designate how many of

each type of element are in the compound.

Prefixes:

1 = mono 2 = di 3 = tri

Naming Molecular Compounds

1. For the first element, if there is only one

atom, just write the name of the element. If there is more than one, a prefix to indicate how many atoms must be used.

Practice:

CO

= Carbon dioxide

NH

= Nitrogen trihydride (Ammonia)

CO = Carbon monoxide

H

O = Dihydrogen monoxide (Water)

CF

= Carbon tetrafluoride

Using Chemical Formulas

• _{Find the formula mass of potassium chlorate,} KClO₃

– _{Sum the masses of the following one K atom, one Cl}

atom and 3 oxygen atoms. Find the mass on the periodic table

• _{1 K = 39.10 amu} • _{1 Cl = 35.45 amu}

• _{3 O = (16.0 x 3) = 48.0 amu}

• _{Add up the totals 39.10+ 35.45 + 48.0 = 122.55} amu

Molar Mass

• _{Is numerically equal to the formula mass} • _{Difference is formula mass is measured in}

amu, molar mass is g/mol, so just find the

formula mass and just give the units to g/mol • _{If you have 1 mole of a compound you have}

• _{What is the molar mass of barium nitrate,}

Ba(NO₃)₂?

1 mol Ba x 137.3 g Ba = 137.3 g Ba

2 mol N x 14.01 g N = 28.02 g N

6 mol O x 16.0 g O = 96.0 g O

Using Molar Mass of Compound

the molar mass of a compound can be used to convert the number of moles of a

compound to grams and grams of a compound to moles.

Molar mass as a Conversion Factor

a conversion factor to relate an amount in moles to a mass in grams for a given

substance.

• _{How do we do this?????}

Example:

• _{What is the mass in grams of 2.50 mol of oxygen}

gas?

– _{Given: 2.50 mol O2}

– _{Unknown mass of O2 in grams}

• _{First step:}

– _{Calculate the molar mass of O2}

– _{2 mol O x 16.00g = 32.00g (mass of O2)}

• _{Second step:}

– _{Multiple the amount of O2 given in moles x molar mass (g/mol)}

= mass in grams

Example: Convert 3.5 moles of NaOH to grams:

Molar mass of NaOH = 39.9969 g/mol

Converting from grams to moles

Here the process would be reversed

.

•

To convert from grams of a

Example

• _{Convert 50 g of phosphorus trichloride to}

moles:

Molar mass of PCl₃ = 137.333 g/mole

Percentage Composition by Mass

a particular element in a chemical compound.

• _{To find the mass percentage of an element in}

Example

What is the percentage of each element in water?

Total formula mass of water = 18.016 amu H: (2 x 1.008) x 100 = 11.2% H

18.016

Empirical Formulas

The empirical formula of a compound is the

lowest whole number ratio of elements in that compound.

Examples:

C6H12O6 Empirical formula: CH2O

H2O2 Empirical Formula: HO

Calculating Empirical Formulas

A compound has 31.75 g silver, 4.125 g of nitrogen and 14.125 g oxygen. What is the empirical formula of the compound?

To solve, convert each element’s mass into

Like this:

31.75 g Ag ÷107.86 g/mol = 0.29 moles Ag

4.124 g N ÷ 14.007 g/mol = 0.29 moles N

Divide all by the lowest number of moles to get the ratio of elements in the compound

0.29 moles Ag ÷ 0.29 moles = 1 Ag

0.29 moles N ÷ 0.29 moles = 1 N

0.88 moles O ÷ 0.29 moles = 3 O

Empirical formulas with

percentages

This is done the same as before, just

convert the percentages of each element into grams and then proceed to solve the problem. Example:

Determine the empirical formula of a

Converting grams to moles

52.11 g C ÷ 12.011 g/mol = 4.33 mole C

13.14 g H ÷ 1.0079 g/mol = 13.03 mole H

Now find mole ratios

4.33 mole C ÷ 2.17 = 2 moles C

13.03 mole H ÷ 2.17 = 6 moles H

2.17 mole O ÷ 2.17 = 1 mole O