Exam 1 approaches!!! DUN DUN DUUUUUUUHHHHHH.

I just finished up the homework due by Sunday and found myself feeling more confident than ever about this exam. I really feel that I understand this whole derivatives concept, and I love the new easy way to find derivatives. I don’t think I have a favorite, but it’s certainly faster to find the derivative applying the new rules we got this week in class.

I’ve actually been enjoying it so much, I decided that my Chemistry class should have such clear equations for math we have to do in there, and decided to write my own equations for the test we have coming up in Chemistry on Wednesday. I just think I will do so much better if I can plug numbers into an equation rather than second guessing myself. Before I go crazy on my Chemistry math… to make sure I help out my fellow classmates, I want to post the newest rules for Derivatives.

When f(x)=ab, f'(x)=(a)b'+(b)a' Or in other words… when F(x)=B(g)C(d), F'(x)=B(g)*C'(d)+C(d)*B'(g)

When F(x)=\dfrac{a}{b}, F'(x)=\dfrac{ba'-ab'}{b^2} Or in other words… when F(x)=\dfrac{B(g)}{C(d)}, F'(x)=\dfrac{C(d)*B'(g)-B(g)*C'(d)}{C(d)}

So a workable example of this would be:

F(x)=(8x-3)(3x^2) F'(x)=(8x-3)6x+(3x^2)8 if you are wondering how I got 6x and 8… refer to my last post.

F'(x)=48x^2 -18x + 24x^2 F'(x)=72x^2-18x

Example 2:

F(x)=\dfrac{8x-3}{3x^2} F'(x)=\dfrac{(3x^2)8-(8x-3)6x}{(3x^2)^2} F'(x) = \dfrac{24x^2-(48x^2-18x)}{(3x^2)^2} F'(x)=\dfrac{-24x^2+18x}{9x^4} which can be simplified into F'(x)= \dfrac{-8x+6}{3x^3}

On to Chemistry Math!

For Combustion Reactions:

A + O_2 \to is the symbol for any combustion reaction with A being any compound reacting with the diatom of O.

CH_4+2O_2 \to CO_2 + 2H_2O represents the combustion of hydrocarbons (compounds made only of hydrogen and carbon)

C_6H_{12}O_6 + 6O_2 \to 6CO_2 + 6H_2O represents a combustion of C,H, and O that also forms CO_2 and H_2O

Easy enough, but now we move on to more difficult concepts

The Mole:

1 Mole = the amount of matter that contains as many objects (atoms, molecules, ions, etc.) as the number of atoms in exactly 12 g of Carbon 12.

Avogadro’s Number helps us understand how many objects are in a mole.

6.022 x 10^{23} objects in a mole

To better understand what a mole is, (and yes, this totally applies to math right now) 1 mole of anything is 6.022 x 10^{23} objects. The mole helps us measure really really small things (i.e. atoms) in workable numbers.

Here’s how my chemistry notes put it:

(Molecular Mass) 1 molecule of H_2O = 18.02 amu

(Molar Mass) 1 mole of H_2O = 18.02 g

1 Mole of H2O = 6.022 x 10^{23} molecules of H2O

1 Mole of H2O contains (2)(6.022 x 10^{23})atoms of Hydrogen

I find that to be useful but not as useful as if I had an equation that laid it out for me. So here’s what I came up with:

1 mol x_ay_b has (a)(6.022x10^{23})atoms of x, and (b)(6.022x10^{23}) atoms of y.

To find mole using the number of atoms you can use the following equation:


To find grams in a mole:

Xgrams_{Y_a}=(\dfrac{1moleY_a}{1})(\dfrac{CgY_a}{1 moleY_a}) where Y stands for an element, A stands for the subscript and C stands for the number in grams of that element.

To find molecules of a compound with a given number of grams:

For our purposes, we’re going to use an element of E_a.

xmoleculesE_a=Ygrams of E_a(\dfrac{1 E_a molecule}{g of E_a in 1 molecule})(\dfrac{6.022x10^{23}}{1 mole})

To find atoms of a specific element in a compound in a specified amount in grams:

x atomsE=(\dfrac{H grams E_a}{1})(\dfrac{1 mole of E_a}{B grams E_a})(\dfrac{a moles E}{1 mole E_a})(\dfrac{6.022x10^{23} atoms}{1 mole E}) where E=element, a=subscript (number of elements in the compound), H=grams of the compound present, B=grams of the compound in 1 mole.

I now see how confusing this could be. Especially if you aren’t in Chemistry… but it makes a lot more sense to me now that I’ve written it all out, and I’m going to conclude with my final math equations for Chemistry.

Determining Empirical Formulas from Molecular Formulas and Vice Versa

This is pretty simple. It’s kind of like factoring actually. In order to determine an empirical formula from a molecular one, all you have to do is get the simplest whole number ratio.

For example, the molecular equation for Glucose is this: C_6H_{12}O_6 the subscripts are all divisible by 6, so the empirical formula would look like this: CH_2O

It’s a little harder going from the Empirical formula to the Molecular one… but here’s how it works.

First you need the molecular weight. Then you need the Formula Weight.

For example, Formaldehyde has a molecular weight of 30.03. It’s molecular formula is CH_2O look familiar? That’s because it’s molecular formula is also its empirical formula, which happens to also be the empirical formula of glucose… awkward, right? Cause I’d much rather eat some glucose than some formaldehyde… so it’s pretty important that we recognize that an empirical formula doesn’t usually give us enough info.

Next we have Acetic Acid. It’s Empirical formula is also CH_2O, however its molecular weight is 60.06. In order to find out what it’s Molecular Formula is, we can do this simple equation:

Where MW= Molecular Weight, FW=Formula Weight \dfrac{MW}{FW}=x Once you solve for x, you can use it to create the new Molecular Formula which would look like this C_xH_2xO_x

To show exactly what I mean, we can plug-in the info for Acetic Acid and get its molecular formula since we know it’s MW=60.06 and it’s FW = 30.03 (weight of formaldehyde) (you can also get this by adding the molecular weight of 1 carbon, 2 hydrogens, and 1 oxygen using the periodic table as a reference).  Let’s plug it in:

\dfrac{60.06}{30.03}=2 C_2H_2(2)O_2=C_2H_4O_2

And we have our Molecular Formula!

We can also find it using the Molecular Mass, for example:

Cortisol’s Molecular Mass = 362.47 g. It’s made of 69.6% C, 8.34% H, and 22.1% O by mass. What is in its molecular formula?

I created an equation to help solve this and it goes a little something like this:




C=252g H=30.23g O=80.11g

Now that I have those numbers I can take the g/mole of each element and get the number of each element in the molecule.




Now we know that the molecular formula of Cortisol is C_{21}H_{30}O_5

But what if we didn’t have the Molecular Mass? If we had just the percentages, we could solve for 100g. It could look something like this:

x moles c = \dfrac{69.6g}{12.011g/mol}=5.80mol

x moles h = \dfrac{8.34g}{1.01g/mol}=8.26mol

x moles o = \dfrac{22.1g}{16g/mol}=1.38mol

But we need to get to a whole number so first we have to divide by the lowest number:

\frac{1.38}{1.38}=1moloxygen \frac{8.26}{1.38}=6molhydrogen \frac{5.8}{1.38}=4.2

That won’t work, because we still have a fraction, and we need a whole number, so we have to find a multiple that will make it whole, which in this case happens to be 5.

Multiply each by 5, and you get what we got before: C_{21}H_{30}O_5


So I realize this was a lot of Chemistry, but I felt that the math really was connected to my calculus class, and I am certainly more adept at doing the algebra now that I’m working with more complex figures like Derivatives.

Good Luck to you ALL on the 1st Exam! In celebration of making it to our first Exam, I am going to change up my blog background.

Oh… and happy early Valentines to you all!