Remembering some basics
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Remembering Some Basics. A little math for science…. A little “why ”… to get started….

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Remembering Some Basics

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Remembering some basics

Remembering Some Basics

A little math for science…


A little why to get started

A little “why”… to get started…

While organic chemistry isn’t filled with mathematics to the same extent as you probably found it in General Chemistry, there are still elements of mathematical calculations that you will be required to know how todo.


How far back should we go

How far back should we go?

Believe it or not, at some point, someone is going to break the rules of simple addition, subtraction, multiplication and division that you probably learned in first or second grade.

Remember that when you see a math equation, things in parentheses should be dealt with first. Multiplication and Division are a second high priority. Any time addition and subtraction are involved, you always start on the left and progress to the right.

Let’s try some basics…


How far back should we go1

How far back should we go?

Try these:

1. 3 + 4 – 2 + 1 = ??

2. 7 – 3.5 + 1.5 + 2 = ??

3. 8 – (18/3) + (6/2) +1 = ??

4. 10 – ½(21) + ½(1) +1 = ??


How far back should we go2

How far back should we go?

How did you do?

1. 3 + 4 – 2 + 1 = 3 + 4 = 7 - 2 = 5 + 1 = 6

2. 7 – 3.5 + 1.5 + 2 = 7 – 3.5 =3.5 + 1.5 = 5 + 2 =7

Be sure NOT to begin in the middle by combining 3.5 and 1.5. Addition and Subtraction rules require you to begin on the left, not in the middle…

3. 8 – (18/3) + (6/2) +1 =8 – 6 = 2 + 3 = 5 + 1 = 6

Do the multiplication and division first, then begin at the left…

4. 10 – ½(21) + ½(1) +1 = 10 – 10.5 = -0.5 + 0.5 = 0 +1 = 1


Basic units you should know

Basic Units you should know

These should be relatively straightforward for most everyone but conversion problems can be tricky…

Masses are commonly weighed in grams (g) in the lab

Length may be measured in centimeters (cm) or millimeters (mm).

Volume is most often measured in terms of milliliters (mL), which is the same as cubic centimeters.

Density (the one with the most confusing units of all) is always mass (g) divided by volume (mL): [g/mL]


Basic units you should know1

Basic Units you should know

While I’m sure you remember these, sometimes they are easy to confuse:

“Milli-”, as in milliliter means 1/1000 and there are 1000 milliliters (mL) in one liter (L)

“Centi”, as in centimeter means 1/100 and there are 100 centimeters (cm) in one meter (m)

“Deci-”, as in decimeter means 1/10 and there are 10 decimeters (dm) in one meter (m)

“Kilo-”, as in kilogram means 1000 and there are 1000 grams (g) in one kilogram (kg)


Basic units you should know2

Basic Units you should know

Convert the following values:

1.456 grams (g) to kilograms (kg)

1.456 kg to g

0.234 g to milligrams (mg)

0.0567 moles (mol) to millimoles (mmol)

1.4 mL Liquid X to g (if the density of Liquid X is 0.70 g/mL)

3.0 g Liquid X to mL (if the density of Liquid X is 0.70 g/mL)


Conversion factors

Conversion Factors

Convert the following values:

1.456 grams (g) to kilograms (kg)

? kg = 1.456 g ( 1 kg ) = 0.001456 kg

(1000 g)

1.456 kg to g

? g = 1.456 kg (1000 g) = 1456 g

( 1 kg )

0.234 g to milligrams (mg)

? mg = 0.234 g (1000 mg) = 234 mg

( 1 g )


Conversion factors1

Conversion Factors

Convert the following values:

0.0567 moles (mol) to millimoles (mmol)

? mmol = 0.0567 mol(1000 mmol) = 56.7 mmol

( 1 mol )

1.4 mL Liquid X to g (if the density of Liquid X is 0.70 g/mL)

? g = 1.4 mL (0.70 g) = 0.98 g

( 1 mL )

Remember that density is always some number of grams in 1 mL.


Conversion factors2

Conversion Factors

Convert the following values:

3.0 g Liquid X to mL (if the density of Liquid X is 0.70 g/mL)

? mL = 3.0 g ( 1 mL ) = 4.3 mL

(0.70 g )

Remember that density is always some number of grams in 1 mL.When converting volume to weight, you multiply by the density but when converting weight to volume, you have to divide and that means flipping the values over in the conversion problem, as shown above.


Calculations and the evil significant figures

Calculations and the evil Significant Figures

Now that we are doing calculations and not just conversion factors, we need to talk about how many significant figures you should use in your answer. It should contain all the digits that are known plus the next, which is an estimate…

More often than not, we don’t have a panic attack about sig figs when doing basic calculations in the lab but sometimes someone does something crazy and so a reminder would seem to be in order.

When doing a calculation like 1.23 x 0.954, it is just as wrong to write 1 as an answer, as it is to write 1.17342. The first answer (1) implies little to no precision (reliability of a measurement) and the second (1.17342) implies way more than it deserves.

So – here goes… one more time, as a reminder…


The forgotten rules of significant figures

The Forgotten Rules of Significant Figures

1. Non-Zero Digits are ALWAYS significant.

845 has three sig figs

1.234 has four sig figs.

2. Zeros BETWEEN Non-Zeros are also ALWAYS significant.

606 has three sig figs

40306 has five sig figs

3. Zeros on the LEFT of the first Non-Zero number is not significant (and they are considered only placeholders for the decimal point position)

0.0000123 has only three sig figs.


The forgotten rules

The Forgotten Rules

4. For numbers greater than 1: All Zeros to the RIGHT (Rule 4) are significant

2.00 has three sig figs

4.00520 has six sig figs (two zeros between non-Zeros and one zero at the right counts also)

5. For numbers less than 1: only Zeros between Non-Zeros (Rule 2) and Zeros to the right (Rule 4) are significant

0.3050 has four sig figs (ignoring zero on left)

0.00305 has threesig figs (ignoring all zeros on left)

3.05000 x 10-2 has six sig figs (all zeros count!)


The forgotten rules1

The Forgotten Rules

So how do you use this mathematical genius knowledge of sig figs in your calculations?

When doing calculations:

1. Addition and Subtraction: The answer cannot have more digits to the right of the decimal point than either original number.

Example: 89.332 + 1.1 = 90.432, without considering sig figs.

89.332 has three sig figs after the decimal place. 1.1 has one sig fig after the decimal place. The answer should only have one sig fig after the decimal place in it.

Thus 90.432 should actually be 90.4.


The forgotten rules2

The Forgotten Rules

90.432 was easy to “chop off” for sig figs. Essentially you just ROUNDED…

Remember when rounding: if the NEXT number is less than 5, you do not round up, just chop off the remaining digits. If the next number is a 5 or higher, then round the digit up by 1.

90.432 becomes 90.4 with three sig figs (because the fourth sig fig is a 3, which is less than 5).

How about this one?

Round 90.487 to three sig figs.

Three sig figs means 90.487 should become 90.5 (since the fourth sig fig is 8, which is greater than 5).


The forgotten rules3

The Forgotten Rules

When doing a series of calculations, remember to check sig figs as you move through the calculations. Consider multiplying A x B x C.

You technically first multiply A x B first, adjusting your sig figs in youranswer, prior to multiplying by C.

Try it:

Multiply 3.66 x 8.45 x 2.1

What would be your answer?


The forgotten rules4

The Forgotten Rules

Multiply 3.66 x 8.45 x 2.1

First multiply 3.66 x 8.45 = 30.927 (without sig figs). Since both numbers have three sig figs, 30.927 must be corrected to three sig figs also – 30.9 – prior to multiplying by 2.1 (only two sig figs).

Final Answer? 30.9 x 2.1 = 64.89 which should also only have two sig figs so 65 is the final answer.

One final point – conversion factors between metric units don’t count towards determining the correct number of sig figs. For example, 1 kg = 1000 g has no sig figs to worry about…


Some random calculations

Some random calculations…

Always in a reaction, organic chemists are calculating the amounts of reagents required, the limiting reagent, theoretical yield and percent yield (because we never get 100%!) so these would be useful for you to be able to do.

Consider the following: Compound X has a molecular weight of 100.0 g/mol and a density of 2.0 g/mL. Determine the number of moles in 30.0 g of Compound X.

Then determine the number of milliliters in 30.0 g of Compound X.

And to be sure you have it all under control, determine the number of grams in 3.5 moles of Compound X.


Calculations

Calculations

To determine moles, divide by MW:

? mol X = 30.0 g (1 mol ) = 0.300 mole X

(100.0 g)

To determine milliliters, divide by density:

? mL X = 30.0 g (1 mL) = 15 mL X

( 2.0 g)

To determine grams, multiply by MW:

? gX = 3.5 moles (100.0 g) = 350 g X or 3.5x102g X

(1 mole)


Equation calculations

Equation Calculations

Consider the following balanced equation:

How many moles of Y will form from 1 mole of X?

If the equation is balanced, the coefficients (being 1 unless another number is shown) tell you the ratio. One mole of Compound X will form one mole of Compound Y.

How many moles of Y will form from 4 moles of X?

If the ratio is 1:1, then four moles of X form four moles of Y.


Equation calculations1

Equation Calculations

Consider the next equation:

How many moles of Compound B are going to form from 2 moles of Compound A?

Since 1 mole of A forms 2 moles of B, then 2 moles of A form 4 moles of B.

? mol B = 2 mol A (2 mole B) = 4 mole B

(1 mole A)

Reverse the thinking – If you need to make 6 moles of B, how many moles of A do you begin with?


Equation calculations2

Equation Calculations

Since 1 mole of A forms 2 moles of B, then to form 6 moles of B you must start with 3 moles of A:

? mol A = 6 molB (1 molA) = 3 molA

(2 molB)

The equation must be balanced first before doing these types of conversion problems…

On to more types of calculations you need to know…


Equation calculations3

Equation Calculations

Consider the following balanced equation:

Compound X has a molecular weight (MW) of 80.0 g/mol. Compound Y has a MW of 50.0 g/mol.

How many moles of Compound X are in 5.00 g of Compound X?


Equation calculations4

Equation Calculations

Conversion Problem: To convert grams to moles, you must use the molecular weight

? molX = 5.00 g X (1 mole X) = 0.0625 molX

(80.0 g X)

Since Compound X is the only reagent in the reaction, it must be the limiting reagent. How many moles of Compound Y can possibly be formed from 5.00 g of Compound X?

Need to consider that balanced equation again…


Equation calculations5

Equation Calculations

The ratio of this equation is 1:1 for Compounds X and Y.

Conversion Problem: To convert moles to moles, use the mole ratio:

? molY= 0.0625 molX(1 molY) = 0.0625 molY

(1 molX)

Now that you know the number of possible moles of product Compound Y, determine the number of possible grams (theoretical yield) of Compound Y that would form.


Equation calculations6

Equation Calculations

You’ll need to find that MW for Compound Y (50.0 g/mol)

Conversion Problem: To convert moles to grams, you must use the molecular weight

? g Y = 0.0625 molY(50.0 g Y) = 3.125 g Y (1 molY)

With only 3 allowed sig figs, 3.125 g should be rounded to 3.13 g of Y.


Equation calculations7

Equation Calculations

One more 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)

In this problem, Compound A has a molecular weight of 100.0 g/moland Compound B has a molecular weight of 200.0 g/mol.

Balanced Equation:


Equation calculations8

Equation Calculations

Solution: Keep in mind the 1 : 2 ratio between A and B.

? g B = 10.0 gA(1 mol A)(2 mole B) (200 g B) = 40.0 g B

(100 g A) (1mole A)

From here it’s a short hop to determining a percent yield for a reaction. Recall:

Percent Yield = Actual Yield/Theoretical Yield x100

So if the reaction above, starting with 10.0 g A ended with on 25.5 g of B, what would be the percent yield for the reaction?

Percent Yield = 25.5/40.0 x 100 = 63.75 or 63.8 %


Mole ratios

Mole Ratios

So we’ve reviewed calculations for reactions. The importance of a balance chemical equation should be obvious, since that relationship between compounds is a mole ratio that you need to know to complete these calculations.

Just because its not obviously stated (or so it would seem after all these years), its probably not a bad time to point out that even a chemical formula is a mole ratio, only this time it’s a mole ratio of elements.

There are two types of formulas this applies to: empirical formulas and molecular formulas.

What’s the difference between the two?


Mole ratios1

Mole Ratios

An Empirical Formula is the smallest whole number ratio of elements in a compound (molecule). That smallest whole number ratio is actually a MOLE ratio.

Empirical formulas are determined through a process called combustion analysis (or elemental analysis) where a compound is burned in the presence of oxygen and the amount of H2O and CO2 that form are measured and percentages of elemental amounts are determined. We’ll review that later in the semester when we need to.

For now, just answer this:

If the empirical formula is C4H8O, how many moles of carbon atoms are in one mole of C4H8O?


Mole ratios2

Mole Ratios

If the empirical formula is C4H8O, how many moles of carbon atoms are in one mole of C4H8O?

If you simply assume you have one mole (remember – no coefficient means 1 mole), then the formula is telling you that in this one mole of C4H8O, there are four moles of carbon, eight moles of hydrogen and (look – no number again!) one mole of oxygen.

Take it to the next level – What if you have 3 moles of C4H8O? How many moles of hydrogen would you have?

You can even set up a conversion problem, if it makes more sense…


Mole ratios3

Mole Ratios

What if you have 3 moles of C4H8O? How many moles of hydrogen would you have?

? Mol H = 3 mol C4H8O ( 8 mol H ) = 24 mol H

(1 molC4H8O)

Again, its all about the mole ratios…

The other kind of formula is the molecular formula which is similar to the empirical formula because it also describes the mole ratio of elements in a compound but the molecular formula contains the actual mole ratio of elements, not just the smallest ratio.

An empirical formula like CH2 tells you that for every single carbon atom there are two hydrogen atoms. C6H12 has the same general ratio but is more specific about what is actually in the moleculeinformation.


Mole ratios4

Mole Ratios

Unfortunately, you are still limited when you only know a molecular formula. The compound C6H12 has many possible shapes:

Notice the ring features and double bonds in these possible molecules. Without some other sort of identification tool, you wouldn’t know which possible structure is the correct structure.

Well, that should cover the basic information that you need to remember for calculations… at least to get things started… Good luck with those sig figs!


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