**DIMENSIONAL ANALYSIS**

Dimensional Analysis, also known as the unit factor method is an equivalence statement between units used for converting from one unit to another.

Each quantity includes two things:

(1) A Unit

(2) A Chemical Substance

Always write both of these down

**MOLECULAR WEIGHTS**

The sum of the atomic weights of atoms in a molecule's molecular formula.

**PERCENTAGE COMPOSITION**

The percent composition of a component in a compound is the percent of the total mass of the compound that is due to that component.

To calculate the percent composition of a component in a compound:

Find the molar mass of the compound by adding up the masses of each atom in the compound using the periodic table or a molecular mass calculator.

Calculate the mass due to the component in the compound you are for which you are solving by adding up the mass of these atoms.

EMPIRICAL FORMULA

CONCEPT OF MOLE

Chemical reactions usually happen at levels where using grams, or even absolute numbers of atoms/molecules/ions as units, would be confusing and complicated.

The solution is the mole- a unit used for counting particles in elements, compounds, or ionic substances

Just like there are about 110 elements on the periodic table, there are 6.022x10^23 atoms, formula units, or molecules, in a single mole6.022x10^23 is known as Avogadro's number.

**JONATHAN W. VERRANO**

**CHEMISTRY PROJECT**

Avogadro's Number

Avogadro's number is 602 sextillion particles, an inconceivably large amount

To help you understand how big Avogadro's number really is, here are some examples...

If you covered the earth in a mole of

doughnuts, the depth would be

5 miles

If 1 mole of pennies could be distributed to the world's population, every person could spend a million dollars per hour, every hour, for the rest of their lives.

MOLE CONVERSIONS

Converting moles to grams can be a confusing concept for students to grasp, but referring to a chart can be helpful

Grams Moles Particles

Molar Mass equals the atomic mass of the element. Molar mass is used as a conversion factor from grams to moles and grams to particles and visa versa

Use Molar Mass Use Avogadro's Number

For this conversion, Avogadro's number

is used as a conversion factor

6.022x10^23

Molar mass is not needed because

we are NOT going from, or to, grams

Moles Atoms/molecules/formula units

How many grams of nitrogen are in 35 moles of nitrogen?

35 moles of nitrogen x 14.007 nitrogen grams

------------------------- ----------------------------- =

1 1 mole

Explanation:

1. To convert moles to grams, use the molar mass of Nitrogen

14.007g Nitrogen = 1 mole Nitrogen

2. Use it as a conversion factor to cancel out moles, so only grams remain

3. Don't forget to use significant figures

EXAMPLES

490.245

grams of Nitrogen

How many atoms of vanadium are in 13.5 moles of vanadium?

---------------------- x -------------------------------- =

1 1 mole of Vanadium

Explanation:

1. 1 mole= 6.022x10^23 atoms, formula units, or molecules

2. Moles have to cancel out in order to achieve atoms, so multiplication is used

EXAMPLES OF MOLES TO ATOM

13.5 moles of Vanadium 6.022 x 10^23 atoms of Vanadium

8.13 x 10^24

Atoms of

Vanadium

Formula Units

When a substance is ionic

Molecules

When a substance is a compound

Atoms

When it is a single element

Labels

How many molecules of Magnesium Sulfate are in 4.1 moles of Magnesium Sulfate?

Formula: MgSO4

4.1 moles of MgSO4 6.022 x 10^23 molecules of MgSO4

------------------------ x ------------------------------------------- = 2.5 MgSO4 molecules

1 1 mole of MgSO4

CALCULATING MOLES TO MOLECULES

Explanation

1. There are 6.022x10^23 molecules of MgSO4 in 1 mole MgSO4

2. Cross moles out to obtain the number of MgSO4 molecules

EXAMPLE : H2O

The sum of the atomic weights of the atoms in a molecule's empirical formula.

(Almost identical to the Molecular Weight, only with the empirical formula)

FORMULA WEIGHT

Element Atomic Mass Total Mass

C 12.011 12.011

H 1.00794 x 4 4.03176

------------------------------------

CH4 = 16.043 (Formula Weight)

EXAMPLE : CH4

See, it's calculated the same as Molecular Weight, it just uses the empirical formula and has a different name!

The mass of a substance divided by its amount of substance (kg/mol).

Again, it is very similar to Molecular Weight and Formula Weight

(And calculated in the same manner.)

MOLAR MASS

Percent Composition = (Mass due to specific component) / (Total molar mass of compound) x 100

Percent Composition =

Mass Of Element In Compound

Total Molar Mass Of Compound

x

100

Find the percent of hydrogen and oxygen in water.

1)

find the molar mass of the compound by adding the masses of each element.

molar mass of H O = (2)(1.01g H) + 16.00g O = 18.02g H O

2)

calculate the total molar mass of the specific elements in the compound.

mass of H in H O = 1.01g H 2 = 2.02g H

mass of O in H O = 16.00g O 1=16.00g O

3)

divide the total mass of the specific element by the total mass of the compound

and multiply by 100.

EXAMPLES:

A formula that gives the simplest whole-number

ratio of atoms in a compound

Step 1

Convert the grams of each element to moles.

If given percentages, assume that the total mass of the compound is 100g so that

percent given = mass of element in compound

ex. 57.14% C, 6.16% H, 9.52% N, and 27.18% O

moles of C =

57.14g C

1 mol C

12.01g C

= 4.757mol C

moles of H =

6.16g H

1 mol H

1.01g H

= 6.10mol H

moles of N =

9.52g N

1 mol N

14.00g N

= .680mol N

moles of O =

27.18g O

1 mol O

16.00g O

= 1.698mol O

Step 2

Divide each mole value by the smallest number of moles calculated

4.757mol C, 6.10mol H, 0.680mol N, 1.698mol O

C :

4.757mol C

0.680mol N

= 6.995

7

H :

6.10mol CH

0.680mol N

= 8.97

9

N :

0.680mol N

0.680mol N

= 1

O :

1.698mol O

0.680mol N

= 2.497

2.5

These are the mole ratios, or subscripts of each element in the compound

However, if the mole ratio is too far to round to a whole number, then there is one more step necessary to find the empirical formula.

Step 3

if the mole ratio of one of the elements cannot be rounded to a whole number, then you must multiply

ALL

ratios by a constant to make all mole ratios whole numbers.

C :

4.757mol C

0.680mol N

= 6.995

7

H :

6.10mol CH

0.680mol N

= 8.97

9

N :

0.680mol N

0.680mol N

= 1

O :

1.698mol O

0.680mol N

= 2.497

(2) = 14

(2) = 2

(2) = 18

(2) = 10

Empirical Formula: C H N O

14

18

2

10

Step 4

If the molar mass of the molecular compound is given then...

(1) find the molar mass of the empirical formula

14(12.01g C) + 18(1.01g H) + 2(14.00g N) + 5(16.00g O)

= 294.32g C H N O

(2) divide the molar mass of the empirical formula by the molar mass of the molecular formula

294.30g C H N O

14

18

2

10

w

x

y

z

1

3) multiply the subscripts of each element by the whole number calculated in (2)

C H N O

14

18

2

10

molecular formula =

294.32g C H N O

14

18

2

10