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Calhoun Community College - CHM 111 - Class Notes - Week 9

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Lesson 11

Wednesday, June 15, 2016 5:29 PM

• Atomic Weight / atomic mass

○ Decimal number on the periodic table

○ Weighted average of all the isotopes

○ We use it as grams

○ Weight and/or mass of one atom

• Mole

○ Describes an amount of substance/compound

○ Mole is the SI unit used to represent 6.022 x 10^23 particles

○ 6.022 x 10^23 = Avogadro's number

○ 1 mole of an atom - the weight (atomic mass) in grams

○ 1 mole of Cu = 63.54 grams

▪ 1 mole of Cu = 6.022 x 10^23 atoms of Cu

○ 1 mole of O = 15.9994 grams

▪ 1 mole of O = 6.022 x 10^23 atoms of O

○ Molecular weight is used for molecules

○ Atomic weight is used for atoms

• Formula / molecular weights

○ Formula weight (FW) = ionic compounds

○ Molecular Weight (MW) = covalent compounds

○ Calculation

▪ The atomic weights of all the atoms present in the compound

• Calculations of grams, moles, molecules, and atoms

○ If you start with moles:

▪ You will ALWAYS multiply Don't forget about the age old question of What is the resonance condition in nmr spectroscopy?

▪ To get to grams:

□ Multiply by the molecular weight of the compound or molecule

▪ To get to the number of molecules or atoms:

□ Multiply by Avogadro's number (6.022 x 10^23)

○ If you start with grams:

▪ To get to moles:

□ DIVIDE by the molecular weight

▪ To get to the number of molecules or atoms:

11 Pae 10

▪ To get to the number of molecules or atoms:

□ Divide by the molecular weight

□ Then multiply by Avogadro's number

○ If you start with the number of molecules or atoms

▪ To get to moles:

□ Divide by Avogadro's number

▪ To get to grams:

□ Divide by Avogadro's number

□ Then multiply by the molecular weight We also discuss several other topics like What family patterns do we see across cultures and specifically in the us?

○ To find the number of atoms in a molecule

▪ Multiply the number of molecules by the subscript of the atom you are trying to find

• Percent Composition

○ (Mass of the element / mass of the compound ) x100

11 Pae 11

College Chemistry 1 – Mona Chaudhary

6/14/2016

Important Words or

Definitions Important People Important Concepts Lesson 12

∙ Formulas from element composition

o Empirical formula

∙ Simplest formula

∙ Relative ration of atoms

∙ Smallest whole number ratio of atoms

o Molecular (true) Formula

∙ True ratio of atoms

∙ Multiple of empirical formula ratios

o Molecular formula = (empirical formula x n)

∙

Element Relative mass of atoms

Relative # of atoms

Divide by the

smallest #

Whole #

ratio

Symbol

Change %

sign to Grams

Divide the relative mass by the atomic weight of the element

Divide by the

smallest relative value

Whole

number

ratio

We also discuss several other topics like What are the 3 branches of government?

∙ Example:

o A compound contains the following: 40% carbon, 6.7% hydrogen, 53.3% oxygen. What is the empirical formula?

o What is the molecular formula if the compound has a molecular mass of 60.0 g/mol?

o

Element

Relative mass of atoms

Relative # of

atoms

Divide by the

smallest #

Whole #

ratio

C

40 g

40/12.01

3.33/3.33

1

H

6.7g

6.7/1.01

6.63/3.33

2

O

53.3g

53.3/16

3.33/3.33

1

We also discuss several other topics like What is the common use of mythic etiologies?

We also discuss several other topics like What is the difference between theory and law?

Empirical formula = CH2O

Molecular formula:

Molecular weight of empirical formula = 12.01+2.02+16 = 30.03 If you want to learn more check out What is the recipe for science?

60/30.03 = 1.998 (almost 2)

Multiply all the subscripts by 2

C2H4O2