Formulas and Nomenclature. Charge of Ions called OXIDATION NUMBER Related to the number of electrons...

Formulas and Nomenclature

Charge of Ions

• called OXIDATION NUMBER• Related to the number of electrons that are lost or gained when an atom becomes an ion Listen to this

Group2

Group 1

Group 13

Group 14

Group 15

Group 16

Group 17

Group 18

+1

+2

+3

+4

-3

-2

-1

0

Oxidation numbers • Same for all elements in a group or family

• (+) means that the atom loses electrons when forming a cation; formed by metals

• (-) means that the atom gains electrons when forming an anion ; formed by nonmetals

• The Noble Gases have oxidation number of zero because they are nonreactive and do not lose or gain electrons.

Transition Metals and ChargeSome transition metals form ions that have multiple charges. (MULTIVALENT)Ex. Fe2+ or Fe3+

Ex. Mn2+ or Mn5+ or Mn7+

These metals DO NOT form ions with more than one charge. (MONOVALENT)

Ag1+, Zn2+, Cd2+, Al3+ (Remember AZCA, they also make a stair step)

Classifying metals:

• Type 1 metal (MONOVALENT): Group 1,2 and Al, Zn, Cd, Ag

• These metals only have 1 charge• Al is +3• Zn and Cd are +2• Ag is +1

• Type 2 metal (MULTIVALENT): all metals that are not type 1 metals

• Can have 2 or more charges

POLYATOMIC ION (PAI)

• Two or more nonmetal atoms bonded together

• Act as a single unit with a net charge• LISTED ON YOUR STAAR CHART

• Chlorate, ClO3-1 is the same as (ClO3)-1 which

means that 1 chlorine atom is bonded to 3 oxygen atoms and as a group

• it has a charge of -1

Chemical Formula• A symbolic representation of a chemical compound.

Formula Anatomy of an Ionic Compound

2MgCl2

Elements

Coefficient- tells the number of formula units

Subscript- tells the ratio of atoms1Mg : 2 Cl

Chemistry definition of formula unit- the empirical formula of any ionic compound used as an independent entity for stoichiometric calculations.

Formula Unit – The smallest “piece” of an ionic compound.

Ratios- we use ratios because the charge (cation to anion) must be neutral.

Added InfoBinary Compound- compounds made of two TYPES of elements. Ex. Al2S3 Non ex. AlPO4

Ternary Compound – compounds made of more than two types of elements Ex. AlPO4

Contains Polyatomic Ion

Does not contain Polyatomic Ion

WRITING FORMULAS FOR IONIC COMPOUNDS : 1. Write the symbol for each ion. Positive ion goes first.

2. Write the oxidation number for each ion above the symbol.

• For type 2 metals, the Roman numeral after the metal tells you the oxidation number

3. Criss-cross the oxidation numbers (not the + or -). These numbers become the subscripts.

4. If a subscript is needed for the PAI, you must enclose it in parentheses first, then write the subscript outside of the parentheses.

5. Watch This

When writing formulas the more metallic element goes first, then the nonmetal

Example:

• Write the formula for the compound that contains aluminum and sulfur.

Example:

• Write the formula for the compound that forms when sodium and oxygen combine.

Example:

• Write the formula for the iron (II) chloride.

Example:

• Write the formula for the compound that contains calcium and phosphate.

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WRITING NAMES FOR WRITING NAMES FOR IONIC COMPOUNDS IONIC COMPOUNDS

1. Write the name of the metal first.2. IF THE METAL IS A TYPE 2 METAL, write the

oxidation number of the metal in parentheses. (You will have to work backwards from the subscripts to determine the oxidation number.)

3. Write the name of the nonmetal, changing the ending to “ide.”4. IF EITHER PART OF THE COMPOUND IS A

POLYATOMIC ION, write the name of the polyatomic nonmetal from the chart. DO NOT CHANGE THE ENDING OF THE PAI.

5. Watch This

Watch This

Example:

• Write the name for the compound that contains lithium and nitrogen.

Example:

• Name this: NiF2

Example:

• Name this: Ca(C2H3O2)2

Example:

Write the name for the compound that contains lithium and nitrogen.

Example:

Name this compound: NiF2

Example:

Name this compound: Ca(C2H3O2)2

Percent composition

Watch This

Percent Composition write allPercent Composition write all

Percent Composition (% mass): the percentage by mass of each element in a compoundWatch This

100compound of massmolar

element of mass Total Mass %

% Composition Practice

What is the percent of Li and O in Li2O?

What is the percent of each element in Na2SO3?

How much sodium is in 25 grams of Na2SO3 ?

Covalent MoleculesWhat are the common names for these molecules:

Dihydrogen monoxide H2O Water

Carbon tetrahydrideCH4 methane

Nitrogen TrihydrideNH3 Ammonia

Naming Covalent Binary Molecules1. Write the name of the

first element. The less electronegative element goes first.

2. Change the nonmetal’s ending to –ide.

3. Use prefixes to indicate the number of each type of atom.

*Exception-the first element will never have the prefix mono

• 1-mono• 2-di• 3-tri• 4-tetra• 5-penta• 6-hexa• 7-hepta• 8-octa• 9-nona• 10-deca

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Practice

• Write the name for the following molecules:

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Writing formulas for molecules• The prefixes tell you the subscript for each atom.

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Practice

• Write the formulas for the following molecules:

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Empirical formula

• Empirical - To be derived from observation, experiment, or data.

• Empirical formula - the simplest whole number ratio between two (or more) elements

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Watch This

Watch this first

Steps to determine the empirical formula of a compound:

1. Determine the mass of each element in the sample.2. Divide the mass of each element by the molar mass

(from the PT) to determine the number of moles of each element. Round to the thousandths (._ _ _ )!

3. Divide the # of moles of each element by the smallest # of moles. This is the mole ratio for each element in the compound.

4. If your answers to step 3 are whole numbers, these are written as the subscripts.

5. If your answers to step 3 are NOT whole numbers, multiply by 2, 3, or 5 to obtain a whole number if increments of 0.5, 0.3 or 0.2 are given, respectively.

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exampleWhat is the empirical formula for a sulfur oxide compound containing 50% sulfur and 50% oxygen?

Step 1: Since % means “parts per hundred”, assume we are working with a 100 g sample. That means we have 50 g of sulfur and 50 g of oxygen.

Step 2: Use dimensional analysis to convert grams to moles

50 g S __1 mol_ = 1.558 moles S 32.1 g

50 g O _1 mol_ = 3.125 moles O 16.0 g

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LabelProperly!!

Round to thousandths (._ _ _)

Step 3: Divide by the smallest number of moles to obtain a mole ratio.

1.558 moles S

1.558 moles S = 1 S

LabelProperly!!

3.125 moles O

1.558 moles S = 2 O

So, we have 1 S for every 2 O. These numbers become the subscripts and the formula is SO2

In our example, we did not need the 4th step since the ratio came out to a whole number. 38

A compound contains 54.1 g of Mg and 45.9 g of P. Determine the compound’s empirical formula.

Note: This time, we already have the number of grams so we can skip to step 2.

Another example

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Step 2: Use Dimensional Analysis to convert grams to moles.54.1 g Mg___1 mol_ = 2.226 moles Mg 24.3 g45.9 g P __1 mol_ = 1.481 moles P 31.0 g

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LabelProperly!!

Round to thousandths place (._ _ _ )

Step 3: Divide by the smallest number of moles to obtain a mole ratio.

1.482 moles P

1.482 moles P = 1 P

LabelProperly!!

2.226 moles Mg

1.482 moles P = 1.5 Mg

Notice, the bottom answer did not come out to a whole number this time.

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Skip to step 5 since answers are not whole numbers.

Step 5: Multiply answers from step 3 so that you get whole numbers.

We had 1.5 Mg and 1 P

0.5 = ½ flip it and you have your scale factor, 2.

1.5 x 2 = 3 Mg and 1 x 2 = 2 P. The ratio did not change, it is just a whole number ratio now.

So, we have 3 Mg for every 2 P or the formula Mg3P2

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Now that you know the steps, here is a jingle to make them easier to remember:

Percent to mass step 1Mass to mole step 2Divide by small step 3Multiply ‘til whole steps 4 and 5

Note: You may not need all of the steps.

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Molecular formula

A formula that is reducible. It is a multiple of an empirical formula.

Ex. Can C8H12 be reduced?Of course, it’s divisible by 4. So, dividing by 4

reduces the formula to C2H3. C8H12 is the molecular formula.C2H3 is the empirical formula.

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Empirical formula Mass of empirical formula

molecular formula Mass of molecular formula

This template can help you organize your information and find what you are missing.

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Ex. The molar mass of a molecular formula is 283.88 g/mole and it’s empirical formula is P2O5. Determine the molecular formula.

Draw your chart and fill in the info from the problem.

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P2O5

283.88 g/mol?

141.943 g/moleNow, divide the molar mass of the MF by the molar mass of the EF. (283.88 g/mole)/(141.943 g/mole) = 2. Scale factor is 2. Multiply the subscripts in the EF by 2 and the MF is… P4O10

from

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use PT to

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practiceA compound is made from 2.00 g carbon, 0.335 g hydrogen, and 2.66 g oxygen. Its molar mass is 90.0 g/mole. Determine the molecular formula.

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