# The Chemical Equation

*law of multiple proportions*very precisely and correctly concluded that this was significant. Because the creation or destruction of nuclei is very unlikely without a nuclear reactor, the same number of atoms of each type (element) must exist both before and after

*any*chemical transformation. This means that molecules must react in

**simple, whole-number ratios**(i.e. the law of multiple proportions), as can be seen by the following example:

`X`_{1} * CH_{4} + X_{2} * O_{2} = X_{3} * CO_{2} + X_{4} * H_{2}O

**stoichiometric**coefficients. The process of 'balancing' a chemical equation is simply determining a set of coefficients {X

_{i}}, which represent the proportions of whole molecules in the reaction that balance the number of atoms of each element on both sides of the equal sign. Note: an equal sign, as well as single and double headed arrows, is sometimes used to separate reactants and products in a chemical equation. Nomatter what is used as a separator, a chemical reaction must be written as an

**equation**, which means that the same number of each type of atom exist each side. (Sometimes unscrupulous chemistry intructors provide unbalanced chemical equations (which are really not equations at all) in problems given to students, but they argue that these are the cards that life deals us. Nonetheless, the equations always must be

__checked__before proceeding further with the use of said equation.) How do you balance a chemical equation?

- Identify all the different elements in the chemical equation.
- Count the number of atoms on both sides of the equal sign of the element that appears in the fewest different molecules in the equation. Make sure to use the correct molecular formula (structure) in this arithmetic
- Adjust the Stoichiometric Coefficents of the species that contain this element to balance the count of this element in the equation.
- Repeat the last two steps for each element identified in step 1. You must maintain the ratio of coefficients of every molecule that has been balanced for a previous element.
- When all the elements have been balanced, multiply the coefficients in the entire equation by any number you wish. This is usually done to obtain the smallest whole number (integer) coefficients, {X
_{i}}

`Carbon: X`_{1} = X_{3}Hydrogen: X_{4} = (2)*X_{1}Oxygen: X_{2} = X_{3} + (1/2)*X_{4}

If one chooses X_{1}=1, then X

_{3}= 1, then X

_{4}= 2, then X

_{2}= 2, and the equation is balanced. Practice some more equation balancing on your own.

**Counting Atoms by Weight**

Hardware store owners solved a problem long ago with a method that can be applied to almost everything, including chemistry. People need nails. Store owners have nails, but want money for them. People need lots of nails, but counting lots of nails so that you can charge for them is a drag, because they are small and pointy. Shopkeepers decided on an easier method than counting each individual nail, they simply sold nails *by weight*. But how many nails do you get when you buy a pound of nails? It obviously depends on the weight of an individual nail. Even now, the weight of an individual nail is still found on in Hardware Stores; It is listed as Penny Weight. A 10 penny nail weighs 1/2 oz (1 penny = 1/20 oz). So, how many 10 penny nails are in 1 lb of nails? answer.

**Isotopes and the 'Average' Mass of an Atom**

In order to count molecules, we need to weigh them, because they are too small to count individually. Molecules also come in different sizes, just like nails, but there are actually an infinite number of possible molecular weights! We are saved by the fact that all molecules are made up of atoms, and there are only about 100 of these, so all we need to know is the atomic weights of the elements.

**average atomic weight**s of the elements in mole calculations and in the determination of average molecular weights. For example, for every 100,000 Hydrogen (Z=1) atoms(ions) we count in our mass spectrometer, 99,985 atoms weigh 1.007825 grams/mole and 15 weigh 2.0140 grams/mole. The minor isotope of hydrogen is so important it is called deuterium, but it is not a different element. The presence of naturally occuring deuterium makes the average atomic mass of the element with Z=1 as follows:

**Atomic Mass of Hydrogen = (0.99985)*(1.007825) + (0.000015)*(2.0140) = 1.00797 g/mol**

The natural abundance of the isotopes of Neon are as follows:

`Abundance(`^{20}Ne) = 90.92 %

Abundance(^{21}Ne) = 0.257 %

Abundance(^{22}Ne) = 8.82 %

and the atomic masses of these isotopes are: `Atomic Mass(`^{20}Ne) = 19.99244 g/mol

Atomic Mass(^{21}Ne) = 20.99395 g/mol

Atomic Mass(^{22}Ne) = 21.99138 g/mol

What is the average atomic weight of Neon from these data? answer Once we know the average weight (mass) of all the elements (these are usually listed in the Periodic Table), we can calculate the molar mass of all the molecules just by knowing their molecular formula. For example, the molecular weight of Methane, CH

_{4}, is simply

MW{Methane} =

(1 carbon atom per molecule)*(12.011 grams per mole for carbon atoms)

+ (4 atoms of hydrogen per molecule)*(1.00797 grams per mole for hydrogen atoms)

= 16.0429 grams per mole methane

Limiting Reagents

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