## Atomic Mass

Atomic mass is represented by atomic mass unit (amu), and is defined by $$1amu=\frac{m\left(C-12\right)}{12}=1.660539\times {10}^{-24}g$$

where *m(C-12)* is the mass of carbon-12 isotope.

Using the above Xe example, hence, the average mass of Xe is 131.30 amu.

** Example:** How many atoms are there in 10.0 g of Na?

As we've seen in Chapter 1, we can use dimensional analysis to solve this problem. As you read the question, you can write the equation down as,

*?*

_{Na}= 10.0g_{Na}
should suffice. The conversion of 1 *amu* is 1.660539 x 10^{-24}
*g*.

$${?}_{Na}=10.0{g}_{Na}\left(\frac{1amu}{1.660539\times {10}^{-24}g}\right)\left(\frac{{1}_{Na}}{22.99amu}\right)=0.2619\times {10}^{24}Na$$

That's a huge number!

## Moles

Remember the first example in this chapter?
In a 17 g of ammonia, there are always
14 g of nitrogen and 3 g of hydrogen? Can we standardize the mass in terms of
gram quantity because gram is the quatity we can directly measure?
As *amu* is defined using C-12 as fundamental unit
(for one atom) of mass, we can express atomic mass in grams in the
macroscopic scale. For this we need to know the quantity (or number),
for example, of nitrogen in 14 g.

We define *Avogadro's number* to be the number of C-12 atoms present
in an exactly 12.0 g of C-12, and is 6.022 x 10^{23}. This comes
straight from the 1 amu = 1.660539 x 10^{-24}g (Take the recipical of
the number associated with g).

It is nuissance to write such a large number all the time: we must express
12.0g/6.022x10^{23} C-12. So we make this number
into another unit; we call it a *mole*, abbreviated as *mol*.
Therefore,
1 *mol* of substance always contains 6.022 x 10^{23} substances.

So, C-12 contains 6.022 x 10^{23} atoms, which is a mole of C-12.

** Example:** How many atoms are there in 12.0 g of N-15 and in 15.0 g of N-15?

We can solve this by dimensional analysis. One of the key conversion factors
is relating C-12 to N-14.
$${?}_{N-15}=12.0{g}_{N}\left(\frac{1amu}{1.660539\times {10}^{-24}g}\right)\left(\frac{{1}_{N-15}}{15amu}\right)=0.4818\times {10}^{24}=4.818\times {10}_{N-15}^{23}$$
$${?}_{N-15}=15.0{g}_{N}\left(\frac{1amu}{1.660539\times {10}^{-24}g}\right)\left(\frac{{1}_{N-15}}{15amu}\right)=0.6022\times {10}^{24}=6.022\times {10}_{N-15}^{23}$$
The number of atoms in 12 g N-15 is less than Avogadro's number—in C-12,
however, it gives you Avogadro's number. For 15 g of N-15, the number of
N-15 is the same as Avogadro's number. Hence, when the atomic mass in *amu*
is the same as the gram quantity, you have 1 *mol*.

*amu*→ 12 g → Avogadro's number → 1

*mol*

*amu*→ 15 g → Avogadro's number → 1

*mol*

**Average Mass**

The average mass comes from averaging over all isotopic masses.

- (Isotope = has the same # of protons, but differing in # of neutrons)

*m*> is the average mass, and

*c*

_{1}is the fractional abundance of isotope 1 with the mass,

*m*

_{1}, etc.

If you add all the *c*_{1} + *c*_{2} +
*c*_{3} + *c*_{4} + … = 1

Here an easy example of average mass calculation.

** Example:** Copper has two naturally occuring isotopes. Cu-63
with its mass = 62.94 amu and its abundance = 69.17%, and Cu-65 with its
mass = 64.93 amu and its abundance = 30.83%. What is the average mass in
amu?

We have,

*< m > = c*

_{63}m_{63}+ c_{65}m_{65}to which we know all terms on the right-hand side of the equation. Only thing you need to worry about is the abundance expressed in percentage. The fractional abundance is obtained by diving the % by 100. Then,

Above is rather too easy. What about the following?

** Example:** Copper has two naturally occuring isotopes. Cu-63
with its mass = 62.94 amu and Cu-65 with its mass = 64.93 amu. If the
average mass of copper is 63.55 amu, what are the natural abundances
in percentage of each isotope?

This time, there are two unknowns, both the *c _{63}* and

*c*. One equation with two unkowns can not be solved. Nonetheless, we have again

_{65}*< m > = c*

_{63}m_{63}+ c_{65}m_{65}but we also have

*c*= 1

_{63}+ c_{65}

Using this, we now have two equations with two unknowns. This is solvable.
We can rearrange the second equation above to sove for *c _{63}*,
then substitute the

*c*equation into the average mass equation (1

_{63}^{st}eqn) to obtain

$${c}_{65}=\frac{\langle m\rangle -{m}_{63}}{{m}_{65}-{m}_{63}}$$

Once you get the *c _{65}* isolated, you can put all the numbers in.
Then, you get,
$${c}_{65}=\frac{\langle m\rangle -{m}_{63}}{{m}_{65}-{m}_{63}}=\frac{63.55amu-62.94amu}{64.93amu-62.94amu}=0.3065$$

Therefore, *c _{65}* = 30.65%. Then,

*c*= 69.35%. Comparing with the example above, the answer is reasonable.

_{63}

## Molar Mass

So the mystery of where the fraction in the average mass came from, is
solved. Let's go back to mole quantity. The average mass listed on the
Periodic Table, in the case of Xe, was 131.30 amu. It is much more covenient
for us to express this in macroscopic quantity. The mole and amu are defined
through C-12, 12 g is 1 mol, and we saw above that 15 g of 15 amu N-15
gives you 1 mol. Therefore, atomic mass in amu gives the equivalent
number in g, and always gives the Avogadro's number amount. The average
mass works exactly the same way. The average mass of Xe is 131.30 amu,
therefore in 131.30 g gives 1 mole. This is called *molar mass* of
element. The unit is *g/mol*.

** Example:** What are the molar masses of He, Ca, and Cu?

Just look at the Periodic Table and read off the average masses. For He, it is 4.003, therefore 4.003 g/mol. For Ca, it is 40.078, therefore 40.078 g/mol. Cu = 63.546 g/mol.

**Molar Mass of Molecules**

The average masses of atoms on the periodic table give the grams of element in one mole of that element. We call it, molar mass of element.

For molecules, molar mass, denoted as μ (mu of Greek character), is given as the total mass of the molecule in one mole of that molecule. Therefore, for water molecule,

*μ*

_{H2O}= 2

*m*

_{H}+

*m*

_{O}

** Example:** Molar mass of water:

Not much more complicated example...

** Example:** Molar mass of Prussian Blue (indigo dye):

**Chemical Calculations**

Now you're in the position to perform chemical calculations. For this
purpose, we follow * dimensional analysis!* What's important here
is that when comparing two chemical species, you must use stoichiometry, and
hence you need to have the balanced chemical equation.

Let's start simple.

** Example:** Given the following chemcial equation,
calculate various quatities.

_{4}+ 6Cl

_{2}→ 4PCl

_{3}

Let's start with:

How many moles of Cl_{2} do you need to obtain 2 moles of
PCl_{3}?

$$?mo{l}_{C{l}_{2}}=2.0mo{l}_{PC{l}_{3}}\left(\frac{6mo{l}_{C{l}_{2}}}{4mo{l}_{PC{l}_{3}}}\right)=3.0mo{l}_{C{l}_{2}}$$

How many moles of PCl_{3} can it be obtained if you have
2.0 mol of P_{4}?

$$?mo{l}_{PC{l}_{3}}=2.0mo{l}_{{P}_{4}}\left(\frac{4mo{l}_{PC{l}_{3}}}{1mo{l}_{{P}_{4}}}\right)=8.0mo{l}_{PC{l}_{3}}$$

If you have 2.00 g of P_{4}, how many mole of PCl_{3}
can it be obtained?

$$?mo{l}_{PC{l}_{3}}=2.00{g}_{{P}_{4}}\left(\frac{1mo{l}_{{P}_{4}}}{123.895{g}_{{P}_{4}}}\right)\left(\frac{4mo{l}_{PC{l}_{3}}}{1mo{l}_{{P}_{4}}}\right)=0.06457080mo{l}_{PC{l}_{3}}=0.0646mo{l}_{PC{l}_{3}}$$

If you have 2.00 g of P_{4}, how many grams of PCl_{3}
can it be obtained?
$$?{g}_{PC{l}_{3}}=2.00{g}_{{P}_{4}}\left(\frac{1mo{l}_{{P}_{4}}}{123.895{g}_{{P}_{4}}}\right)\left(\frac{4mo{l}_{PC{l}_{3}}}{1mo{l}_{{P}_{4}}}\right)\left(\frac{101.880{g}_{PC{l}_{3}}}{1mo{l}_{PC{l}_{3}}}\right)=6.578473{g}_{PC{l}_{3}}=6.58{g}_{PC{l}_{3}}$$

So, above examples shows the complexity of converting grams of cetrain reactant to grams of certain product is easily calculated.

**stoichiometry**
* g* to

*and*

**mol***to*

**mol***→ always done with the*

**g***molar mass*

## Percent Composition

**Percent composition of elements in a molecule**: We can use % by mass to
obtain the percentage of elements in compounds.

In water, 11.19% is H, and 88.81 is O. It then means that for a 100 g of
H_{2}O, there are 11.19 g H and 88.81 g O.

** Example:** Mass Percent

What is the mass % composition of H and O in H_{2}O?

Since there are 2 mol H and 1 mol O for 1 mol of H_{2} with

** Example:** Empirical formula from % composition

Using the above % for H_{2}O, obtain simplest formula (empirical
formula).

So, in 100 g H_{2}O, there are 88.81 g O and 11.19 g H. Since
the chemical formula canb be thought as the numerical mole ratio of
elements. Let's convert the mass of each element to # of moles.

$$\begin{array}{l}?mo{l}_{O}=88.81{g}_{O}\left(\frac{1mo{l}_{O}}{15.9994{g}_{O}}\right)=5.55083mo{l}_{O}\\ ?mo{l}_{H}=11.19{g}_{H}\left(\frac{1mo{l}_{H}}{1.0079{g}_{H}}\right)=11.10185mo{l}_{H}\end{array}$$

Molecular formula is simple numerical ratio between elements in
molecules. So, 11 mol H to 5.6 mol O reduces to 2 mol H and 1 mol O. This is
why water has formula H_{2}O.
$${H}_{11.10185}{O}_{5.55083}\to {H}_{\frac{11.10185}{5.55083}}{O}_{\frac{5.55083}{5.55083}}={H}_{2.00003}{O}_{1}={H}_{2}O$$

Basically, by writing down the number of moles calculated from % composition for each element, then divide the numbers by the smallest value of moles. Here the important thing it to calculate the number of moles from %.

## Experimental Determination of Empirical Formula

Using the procedure to obtain empirical formula from % composition, we can
work out the so-called combustion analysis. **Combustion** of
hydrocarbon, species that contains only C and H, is a process where the
hyrdocarbon molecules react with O_{2} to produce carbon dioxide
and water.

_{x}H

_{y}+ O

_{2}→

*x*CO

_{2}+ (

*y*/2)H

_{2}O

So, if you know the amount of CO_{2} and H_{2}O produced in
the combustion process, we can calculate the values of *x* and
*y*, and therefore we can obtain empirical formula of the reactant.

** Example:** Combustion Analysis

Combustion of hydrocarbon produced 33.01 g CO_{2} and 13.51 g
H_{2}O. What is the empirical formula of the hydrocarbon?

From equation above, the number of moles of C in CO_{2} came from
C_{x}H_{y}, so as the number of moles of H in
H_{2}O. If we covert the mass of CO_{2} to mol of C,
and the mass of H_{2}O to mol of H, we can obtain the empirical
formula for the hydrocarbon.

$$\begin{array}{l}?mo{l}_{C}=33.01{g}_{C{O}_{2}}\left(\frac{1mo{l}_{C{O}_{2}}}{44.0095{g}_{C{O}_{2}}}\right)\left(\frac{1mo{l}_{C}}{1mo{l}_{C{O}_{2}}}\right)=0.750mo{l}_{C}\\ ?mo{l}_{H}=13.51{g}_{{H}_{2}O}\left(\frac{1mo{l}_{{H}_{2}O}}{18.01528{g}_{{H}_{2}O}}\right)\left(\frac{2mo{l}_{H}}{1mo{l}_{{H}_{2}O}}\right)=1.50mo{l}_{H}\end{array}$$

Above calculations show C_{0.750}H_{1.50}, and yields
CH_{2} by dividing both numbers by 0.750.

** Example:** Combustion Analysis #2

A 12.01 g tartaric acid contains C, H, and O. Upon combustion, 14.08 g
CO_{2} and 4.32 g H_{2}O are produced. Deduce the empirical
formula.

Now we have added complication of having O in the compound. Oxygen can come
from O_{2} and these oxygen becomes distributed into CO_{2}
and H_{2}O, so we must determine the mass of O in the sample somehow.

The way to calculate the mass of O in the sample is to calculate the masses of C and H and subtracted from the total sample mass (12.01 g). Then, you have the mass of oxygen from the original sample.

$$\begin{array}{l}?{g}_{C}=14.08{g}_{C{O}_{2}}\left(\frac{1mo{l}_{C{O}_{2}}}{44.0095{g}_{C{O}_{2}}}\right)\left(\frac{1mo{l}_{C}}{1mo{l}_{C{O}_{2}}}\right)\left(\frac{12.0107{g}_{C}}{1mo{l}_{C}}\right)=3.84259{g}_{C}\\ ?{g}_{H}=4.32{g}_{{H}_{2}O}\left(\frac{1mo{l}_{{H}_{2}O}}{18.01528{g}_{{H}_{2}O}}\right)\left(\frac{2mo{l}_{H}}{1mo{l}_{{H}_{2}O}}\right)\left(\frac{1.00794{g}_{H}}{1mo{l}_{H}}\right)=0.48340{g}_{H}\end{array}$$ $$12.01g-3.84259{g}_{C}-0.48340{g}_{H}=7.684{g}_{O}$$

$$\begin{array}{l}?mo{l}_{C}=14.08{g}_{C{O}_{2}}\left(\frac{1mo{l}_{C{O}_{2}}}{44.0095{g}_{C{O}_{2}}}\right)\left(\frac{1mo{l}_{C}}{1mo{l}_{C{O}_{2}}}\right)=0.31993mo{l}_{C}\\ ?mo{l}_{H}=4.32{g}_{{H}_{2}O}\left(\frac{1mo{l}_{{H}_{2}O}}{18.01528{g}_{{H}_{2}O}}\right)\left(\frac{2mo{l}_{H}}{1mo{l}_{{H}_{2}O}}\right)=0.47959mo{l}_{H}\\ ?mo{l}_{O}=7.684{g}_{O}\left(\frac{1mo{l}_{{H}_{2}O}}{15.9994{g}_{O}}\right)=0.48027mo{l}_{O}\end{array}$$ $${C}_{0.31993}{H}_{0.47959}{O}_{0.48027}={C}_{1}{H}_{1.49905}{O}_{1.50117}={C}_{2}{H}_{3}{O}_{3}$$

If one calculates the molar mass for the empirical formula, you get 74.046 g/mol. The actual molar mass of tartaric acid is 148.092 g/mol. Therefore, if the empirical formula is multiplied by two, one gets the correct formula.

## Balancing Chemical Equation

**Chemical Equation**

Chemical equation shows the transformation of molecules starting with the
*reactants*, the left-hand side of the equation, and ending
in *products* on the right-hand side. The arrow separates the reactants
from the products. Below is an example of a reaction for water molecules
transformed into hydrogen molecules and oxygen molecules.

From the point of view of Conservation of Mass, it stands to reason that all atoms present in the reactants have to be present in the products.

*m*=

_{reactants}*m*

_{products}

There are 4 hydrogen atoms and 2 oxygen atoms on the left-hand side.

There are 4 hydrogen atoms and 2 oxygen atoms on the right-hand side.

When the chemical equation satisfys mass conservation, it is said to be
*balanced*.

**Balancing**

Some tips upon balancing chemical eqns.

** Example:** Balance the following:

_{2}S

_{3}

Since on the right-hand side, there are two Al and three S atoms, the left-hand side must have respectively the same Al and S atoms. So,

_{2}S

_{3}

Too easy? What about next one?

** Example:** Balance the following:

_{2}H

_{8}N

_{2}+ N

_{2}O

_{4}→ N

_{2}+ CO

_{2}+ H

_{2}O

Step by step.

First to match the number of C

_{2}H

_{8}N

_{2}+ N

_{2}O

_{4}→ N

_{2}+ 2CO

_{2}+ H

_{2}O

We need at least 8 H, so

_{2}H

_{8}N

_{2}+ N

_{2}O

_{4}→ N

_{2}+ 2CO

_{2}+ 4H

_{2}O

By doing above, you realize you have 8 oxygen on the right-hand side, but only 4 on the left, so

_{2}H

_{8}N

_{2}+ 2N

_{2}O

_{4}→ N

_{2}+ 2CO

_{2}+ 4H

_{2}O

Finally, adjust N by,

_{2}H

_{8}N

_{2}+ 2N

_{2}O

_{4}→ 3N

_{2}+ 2CO

_{2}+ 4H

_{2}O

However I explain it, you have to practice to get a hung of it!

## Stoichiometry

Consider the following:

_{4}+ 2H

_{2}O → Xe + O

_{2}+ 4HF

We can pictorially represent above chemical eqn as,

This is a microscopic view with each of the molecules in the reaction are shown.

How should we represent this microscopic knowledge onto the macroscopic view?
What I mean by is, say, you are asked to calculate the gram quantity of
XeF_{4} to make 10.0 g of HF. Now, it is no longer a single-molecule
picture. Remember that a mere 18.0 g of water contains 6.022 x 10^{23}
molecules!!??, i.e. 602 200 000 000 000 000 000 000 of them!!!!

For macroscopic quantities, we define ** mole**, abbreviated as mol.

^{23}

^{12}C atoms in exactly 12.00 g of

^{12}C.

It means that there are 12.011 g of naturally occuring carbon, containing different isotopes. It means that the average masses of the elements on the Periodic Table are the masses of each element in 1 mol of that element.

Therefore, the number of molecules in the microscopic picture also corresponds
to the amount of moles of each molecules. The coefficients in front of
chemical formula are so-called * stoichiometric coefficients*.
They represent the number of moles!

## Yield of Product

**Acutal vs. Theoretical Yields**

The amount of product actually produced is called * yield* (or

*actual yield*). Since the reaction of interest may have other pathways so that the the product may not form as much as it should, or even your filteration system may have reduced your products. So, the actual yield may be lower than the theoretically possible amount of product. The

*is the maximum amount allowed by the stoichiometry.*

**theoretical yield**
**Percent Yield**

$${\%}_{yield}=\frac{{g}_{actual}}{{g}_{theoretical}}\times 100\%$$

** Example:** % yield of reaction:

Given the following reaction.

_{4}+ 6Cl

_{2}→ 4PCl

_{3}

_{3}produced from 2.00 g of P

_{4}, what is the % yield?

$$?{g}_{PC{l}_{3}}=2.00{g}_{{P}_{4}}\left(\frac{1mo{l}_{{P}_{4}}}{123.895{g}_{{P}_{4}}}\right)\left(\frac{4mo{l}_{PC{l}_{3}}}{1mo{l}_{{P}_{4}}}\right)\left(\frac{101.880{g}_{PC{l}_{3}}}{1mo{l}_{PC{l}_{3}}}\right)=6.578473{g}_{PC{l}_{3}}=6.58{g}_{PC{l}_{3}}$$ $${\%}_{yield}=\frac{6.44g}{6.58g}\times 100\%=97.8723\%=97.9\%$$

%_{yield} can be used as the conversion factor in the dimensional
analysis!

__It means that some # of g_{something actual} =
100 g_{something theoretical}__

## Limiting Reactant

Here is a recipe for pizza:

Item | Qty |

Dough | 1 |

Tomato sauce | 5 oz |

Cheese | 2 cups |

So, if you have above quantity of items, you can make a pizza. If you have two
doughs, but you still have a 5-oz tomato sauce, and 4 cups of cheese, how many
complete pizza can you make? Obviously only one still, because of the
5 oz tomato sauce. In chemical terms, the equivalent of tomato sauce here is
what we call * limiting reactant*, and this yield only one pizza.

** Example:** Limiting Reactant: What is the mass of PCl

_{3 }formed from 125 g of P

_{4}and 323 g of Cl

_{2}in the following reaction?

_{4}+ 6Cl

_{2}→ 4PCl

_{3}

$$\begin{array}{l}?{g}_{PC{l}_{3}}=125{g}_{{P}_{4}}\left(\frac{1mo{l}_{{P}_{4}}}{123.9{g}_{{P}_{4}}}\right)\left(\frac{4mo{l}_{PC{l}_{3}}}{1mo{l}_{{P}_{4}}}\right)\left(\frac{137.3{g}_{PC{l}_{3}}}{1mo{l}_{PC{l}_{3}}}\right)=554{g}_{PC{l}_{3}}\\ ?{g}_{PC{l}_{3}}=323{g}_{C{l}_{2}}\left(\frac{1mo{l}_{C{l}_{2}}}{70.91{g}_{{P}_{4}}}\right)\left(\frac{4mo{l}_{PC{l}_{3}}}{6mo{l}_{C{l}_{2}}}\right)\left(\frac{137.3{g}_{PC{l}_{3}}}{1mo{l}_{PC{l}_{3}}}\right)=417{g}_{PC{l}_{3}}\end{array}$$

Therefore, Cl_{2} yields smaller quantity of product. Hence,
Cl_{2} is the limiting reactant, yielding 417 g.