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Dimensional Analysis

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Jonathan Verrano

on 3 March 2015

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Transcript of Dimensional Analysis

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
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