Remembering Some Basics:
Metric Units:
Mass – Grams (g)
Length – Meters (m)
Volume – Cubic centimeters (cm3) or milliliters (mL)
Density – g/mL
“Milli” – 1/1000 – there are 1000 milligrams in one gram
“Centi” – 1/100 – there are 100 centimeters in one meter
“Deci” - 1/10 – there are 10 decimeters in one meter
“Kilo” – 1000 – there are 1000 g in one kilogram
“Mega” – 1,000,000 – there are 1 million bytes in one megabyte
How many KB (kilobytes) are there in 1 MB (megabyte)?
Convert 3.45 grams to Kg:
Convert 0.0098 moles to millimoles:
Significant Figures: There’s nothing worse than a student doing a calculation and
rounding values with four decimal places to a single sig fig…
0.9515g divided by 140.1 g/mol = ?? moles
a. 0.0067915774
b. 6.79158 x 10-3
c. 0.007
d. 6.792 x 10-3 or 0.006792
Multiply 0.007 by 140.1 and you get 0.9807 – in the lab, that’s a significant difference
(pardon the pun)!
The Forgotten Rules:
1.
Non-Zero Digits are ALWAYS significant.
845 has ____ sig figs
1.234
has ____sig figs
2.
Zeros between Non-Zeros are ALWAYS significant.
606 has ____ sig figs
40306 has ____sig figs
3.
Zeros on LEFT of first Non-Zero are NOT significant (they are just place
holders for the decimal point position).
0.0000123 has ____ sig figs
4.
5.
For numbers greater than 1: all Zeros to the RIGHT are significant.
2.00 has ____ sig figs
4.00520 has ____ sig figs
For numbers less than 1: only Zeros between Non-Zeros and Zeros at the right
of a number are significant.
0.3050 has ____ sig figs
0.0035 has ____sig figs (see Rule #3)
When doing calculations:
1.
Adding and Subtracting Sig Figs:
The answer cannot have more digits to the right of the decimal point than either of the
original numbers
89.332 + 1.1 = 90.432 (without sig fig consideration)
89.332 has 3 sig figs after the decimal point
1.1 has 1 sig fig after the decimal point
The answer can only have 1 sig fig after the decimal point. 90.432 è 90.4 (ANSWER!
One sig fig to right of decimal point)
This brings up the issue of rounding:
When considering whether to round up or not, remember that if the NEXT number is
less than 5, you do not round up. If the next number is a 5 or higher, then go ahead and
round the number up by 1.
2.
Multiplying and Dividing Sig Figs:
The number of sig figs in the answer is determined by whichever number has the
smallest number of sig figs.
2.8 x 4.5036 = 12.61092 (without sig fig consideration)
2.8
has 2 total sig fig
4.5036 has 5 total sig fig
The answer cannot have more than 2 total sig fig.
12.61092 è 13 (after rounding)
2
When doing a series of calculations, remember to check sig figs as you move through
the calculations. Consider multiplying A x B to get an answer C, which must then be
multiplied by D to get to final answer E.
A = 3.66
B = 8.45
D = 2.11
Multiply A x B first, and determine your sig figs (C = 12.1) and then multiply that by D to
get 65.2. Check your sig figs as you progress through each calculation; don’t wait until
the end of a series to determine sig figs.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Conversion Problems (Unit Conversions):
Compound X has a molecular weight of 100.0 g/mole and a density of 2.0 g/mL.
Determine the number of moles in 30.0 g of Compound X:
Determine the number of milliliters (mL) in 30.0 g of Compound X:
Determine the number of grams in 3 moles of Compound X:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider the following BALANCED reaction equation:
Compound X
catalyst
Compound Y
If the equation is BALANCED as shown, how many moles of Y will form from 4 moles of
X?
One mole of Compound X will form one mole of Compound Y. So four moles of X will
form four moles of Y.
Consider the next equation – how many moles of Compound B are going to form from 2
moles of Compound A?
Compound A
catalyst
2 Compound B
If 1 mole of A forms 2 moles of B, then 2 moles of A form 4 moles of B.
Now reverse the thought process, if you need 6 moles of Compound B, how many moles
of Compound A do you need to start with?
3!
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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More questions for that BALANCED reaction equation:
Compound X
catalyst
Compound Y
Compound X has a molecular weight of 80.0 g/mole and Compound Y has a molecular
weight of 50.0 g/mole.
Solve the following:
How many moles of Compound X are in 5.00 grams?
? moles X= 5.00 g (1/80.0 g/mole) = 0.0625 moles X
How many moles of Compound Y will form from that 5.00 g Compound X?
From the equation – 1 mole of Cmpd X forms 1 mole of Cmpd Y. With a 1:1 ratio, 0.0625
moles of X will form 0.0625 moles of Y.
Show in the conversion problem:
? moles Y= (5.00 g)(1/80.0 g/mole)(1 mole X/1 mole Y) = 0.0625 moles Y
How many grams of Compound Y will you form (a.k.a. what is the theoretical yield of
Compound Y) from 5.0 g Compound X?
Once you know moles of Y, you use the molecular weight of Y to convert back to grams:
? grams Y= 0.0625 moles Y (50.0 g Y/1 mole Y) = 3.125 g = 3.13 g
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
As a single conversion problem: If you use 10.0 g of Compound A, what will be the
theoretical yield of Compound B? (Show how to set up the conversion problem)
Again, Compound A has a molecular weight of 100.0 g/mole and Compound B has a
molecular weight of 200.0 g/mole.
Compound A
catalyst
2 Compound B
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Some Non-Mathematical Things You May or May not Recall…
All of organic chemistry revolves around molecules, and reactions of molecules with
reagents or other molecules, to build different or bigger molecules. All of these
molecules will have formulas for their structures.
Formulas are the mole ratios of the elements in a molecule. MOLE RATIOS. For
example, CH4 is the formula for methane and the formula tells you that the MOLE
RATIO between carbon and hydrogen is 1:4.
What is the empirical formula of a compound?
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It is the smallest whole number ratio (SMALLEST WHOLE NUMBER MOLE RATIO) of
the elements in the molecule.
How is the empirical formula determined for a compound?
Combustion analysis (also known as elemental analysis) is a common method for
determining the empirical formula for a molecule. A compound is burned in the presence
of oxygen and the amount of H2O and CO2 are measured and the elemental percentages
are then determined.
If the elemental analysis for Compound Q is 64.82% carbon, 13.60% hydrogen and
21.59% oxygen, what is its empirical formula?
How do you solve this question?
Assume that you were doing this process with 100 g of Compound Q. Your percentages
are then equivalent to gram amounts for each element.
Now what? Divide each of those gram values you are now considering by the atomic
weights of each element.
Why?
Because you are trying to determine the empirical formula... which is A MOLE RATIO
of the elements… and if you want to convert grams to moles, you need to divide by the
atomic weight of each element.
64.82g C ÷ (12.011g C/mole C) = 5.397 mole C
13.60g H ÷ (1.0079g H/mole H) = 13.49 mole H
21.59g O ÷ (15.9994g O/mole O) = 1.349 mole O
To get to the smallest whole number ratio, divide each by the smallest value (1.349):
5.397 mole C/1.349 = 4
13.49 mole H/1.349 = 10
1.349 mole O/1.349 = 1
C4H10O is the empirical formula.
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If you tackle a problem and find yourself not with all whole numbers but perhaps with a
half-factor (i.e. 2.5 or 4.5) then multiply all values by 2 to get rid of the half-factor.
Third-factors (4.333 or 2.666) will be adjusted by multiplying by 3.
Consider the following: 62.33% C, 6.54% H and 31.13% O.
62.33/12.011 = 5.189/1.946 = 2.666
6.54/1.0079 = 6.488/1.946 = 3.333
31.13/15.9994 = 1.946/1.946 = 1
0.333 = 1/3 x 3 = 1
0.666 = 2/3 x 3 = 2
62.33/12.011 = 5.189/1.946 = 2.666 x 3 = 8
6.54/1.0079 = 6.488/1.946 = 3.333 x 3 = 10
31.13/15.9994 = 1.946/1.946 = 1 x 3 = 3
C8H10O3 is the empirical formula.
So you can now calculate the empirical formula for any compound whose elemental
analysis (a.k.a. combustion analysis) you are given. And that is the smallest whole
number ratio… and that ratio is a mole ratio…
So - If the empirical formula is C4H8O, how many moles of carbon are in one mole of
C4H8O?
Four.
If you have 3 moles of C4H8O, how many moles of H are there?
24
By definition, what’s the molecular formula for a compound?
The molecular formula contains the actual number of each element in the molecule and
therefore contains the actual mole ratio of elements in a compound. There’s more
information in a molecular formula than an empirical formula.
CH2 tells you for every carbon atom there are 2 hydrogen atoms, but that’s just a ratio
and may not be all the atoms involved… CH2 is the empirical formula for C2H4, C3H6 or
C6H12.
A molecular formula like C6H12 tells you exactly how many carbon and hydrogen atoms
but tells you nothing about how they are connected…
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Each of the following fit the C6H12 formula:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Hidden inside the molecular formula is another piece of information called the
Unsaturation Number. Unsaturations cause the carbon skeleton to have less than the
maximum number of hydrogens possible. (Ever heard the phrase polyunsaturated fats??)
An unsaturation might be a double bond or a ring (1 unsaturation each) or a triple bond
(2 unsaturations).
If a molecule contains 3 double bonds and one ring, how many unsaturations is that?
4
The number of unsaturations in a molecule is quickly calculated from its molecular
formula by using the following equation:
#C – ½ (#H + #X) + ½ (#N) +1
where:
H = Hydrogens
X = Halides (Group VII – F, Cl, Br, I) and
N = nitrogen (or technically any Group V element, like P)
So C6H12 would be 6 – ½ (12) + ½ (0) +1 = 1
Just remember when using this equation to start on the left and work across to the
right.
Try: C7H13Br
C7H13N
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
How do you determine the molecular formula of a compound once you’ve figured out the
empirical formula? C2H4, C4H8 and C6H12 all have the same empirical formula – but do
not have the same molecular formula. How would you know which one was the right one?
The molecular weight.
The empirical formula, CH2, weighs 14. The molecular formula is a multiple of that
number. For instance, C2H4 weighs 28 (14 x 2 or CH2 x 2). C4H8 weighs 56 (CH2 x 4).
C6H12 weighs 84. The molecular weight is 6X that of CH2.
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SDBS-Mass
MS-NW-0014
cyclohexane
How
do you figure
C6H12
SDBS NO.
897
out the molecular weight of a compound?
(Mass of molecular ion:
84)
Something called a mass spectrum:
Experimentally,
mass spectroscopy
measures the weights of atoms in a compound. The
Source Temperature:
200 °C
Sample weight
Temperature:
65 the
°C
molecular
is one of
heaviest weights possible for a compound because it
RESERVOIR, 75 eV
includes ALL of the atoms. On the mass spectrum, this particular peak is called the
molecular
ion. We will discuss this at greater length for Expt 7. For now, just know
peak data
that the x-axis gives you masses. This is the mass spectrum for C6H12.
MS-NW-0014 SDBS NO. 897
The molecular ion for this compound (the molecular weight) is always found on the right
National Metrology Institute of Japan(NMIJ)
hand side, because the molecular weight is one of the heaviest peaks at 84.
(c) National Institute of Advanced Industrial Science and Technology (AIST)
Notice that there’s a peak at 85. What might cause the molecule to have a weight
higher than all of the actual atoms??
1 of 1
Isotopes.
1/5/13 4:26 PM
What is an isotope?
Variations of atoms caused by differing numbers of neutrons in the nucleus. The most
prominent isotope of carbon is Carbon-12 and of hydrogen is Hydrogen-1. With this in
mind, what does C6H12 weigh?
8
(6 x 12) + (12 x 1) = 84.
What if the compound had an isotope in it – like C13. Then the formula (technically
C5C13H12) would weigh what?
85!
Notice that we AREN’T using 12.011 and 1.0079, respectively, for Carbon and Hydrogen.
Why not?
Those values on the periodic table are NOT the weights of individual atoms of carbon
and hydrogen. They are the average weights of carbons and hydrogen atoms, when all
possible percentages of all isotopes are calculated. What is the atomic mass for a single
carbon atom? It depends on what ISOTOPE you are referring to.
Isotopes…
AIST:RIO-DB
Spectral Database for Organic Compounds,SDBS
http://sdbs.riodb.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi
So, why a weight of 85 for C6H12? Perhaps C6H12 actually has a bit of C6H11D in it… or
Spectral Database for
Japanese
Introduction
Contact
maybe instead
of 6 SDBS
C-12 atoms,
there
are 5Disclaimer
C-12 andHELP
1 C-13!
Organic
Compounds
What's New
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SDBS-Mass
Try this question:
MS-NW-0960
SDBS NO.
704
3-heptanone
The
empirical formula for a compound is C7H14O. The mass spectrum is shown below:
C7H14O
(Mass of molecular ion:
Source Temperature: 280 °C
Sample Temperature: 150 °C
RESERVOIR, 75 eV
peak data
MS-NW-0960 SDBS NO. 704
114)
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The molecular ion appears at 114 so you know the MW = 114 g/mole. What is the
molecular formula?
How many unsaturations are in this molecule based on its molecular formula?
If you know there aren’t any multiple bonds in the molecule, what might be a valid
structure?
What if you knew it did not have any rings, what might be a valid structure?
We’ll talk more about formulas and structures in class. Later on in the semester, we’ll
be looking at how to figure out structures based on provided spectra and information.
Puzzles, puzzles, puzzles… I hope you all like doing puzzles….
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