Emily Scheerer
Section 3.1–Chemical Equations
Section 3.2–Patterns of Chemical Reactivity
Section 3.3–Atomic and Molecular Weights
Section 3.4–The Mole
Section 3.5–Empirical Formulas from Analyses
Section 3.6–Quantitative Information from Balanced
Equations
Section 3.7 – Limiting Reactants

~ Law of conservation of mass: total mass of all substances after a chemical reaction is the same as the total mass before the reaction.
~ Balancing Chemical Equations (a few tips)
Δ above the arrow means the addition of heat
Never change subscripts, only coefficients
Use the symbols (g), (l), (s), (aq)

•All alkali metals react with water to form hydrogen and hydroxide compounds. (M represents an alkali metal)
2M(s) + 2H
2
O(l)

2MOH(aq) + H
2
(g)
•Combustion reactions are rapid reactions that produce flame, generally involving O
2.
Example: C
3
H
8
+5O
2

3CO
2
+4H
2
O
•In combination reactions, two or more substances form one product.
Example: 2Mg(s) + O
2
(g)

2MgO(s)
In decomposition reactions one substance produces 2 or more other substances.
Example: CaCO
3
(s)

CaO(s) + CO
2

We can measure the masses of individual atoms with the atomic mass unit, amu . The amu is defined by assigning the mass of exactly 12 amu to the 12 C isotope of carbon.
1 amu = 1.66054 x 10 -24 g and 1g = 6.02214 x 10 23 amu
Most elements are a mixture of isotopes. The average atomic mass of each element (expressed in amu’s) is known as it’s atomic weight.
The atomic weights are in the periodic table.

The formula weight of a substance is the sum of the atomic weights of each atom in its chemical formula.
If the chemical formula is merely the chemical symbol of an element, such as Na, then the formula weight equals the atomic weight of the element. If the chemical formula is that of a molecule, then the formula weight is also called the molecular weight.
Percentage Composition from Formulas
Occasionally we must calculate the percentage composition of a compound
(that is, the percentage by mass contributed by each element in the substance). Here’s how:

• Mole The amount of matter that contains as many objects as as the number of atoms in 12g of
12 C.
• This number is 6.02214199 x 10 23 and is called
Avogadro’s number.
(use 6. 02 x 10 23 )
•
Molar Mass : The mass in grams of 1 mole is called the molar mass (in grams) and is numerically equal to it’s formula weight (in amu)

Example: What is the mass in grams of 1.000 mol of glucose, C
6
H
12
O
6
?
Solution
Analyze: We are given the chemical formula for glucose and asked to determine its molar mass.
Plan: The molar mass of a substance is found by adding the atomic weights of its component atoms.
Solve:
Because glucose has a formula weight of 180.0 amu, 1 mol of this substance has a mass of 180.0 g. In other words, C
6
H
12
O
6
? has a molar mass of 180.0

Calculate the number of H atoms in 0.350 mol of C
6
H
12
O
6
Solution
Analyze: We are given both the amount of the substance (0.350 mol) and its chemical formula (C
6
H
12
O
6
). The unknown is the number of H atoms.
Plan: Avogadro's number provides the conversion factor between the number of moles of and the number of molecules of C
6
H
12
O
6
. Once we know the number of molecules of we can use the chemical formula, which tells us that each molecule of contains 12 H atoms. Thus, we convert moles of to molecules of and then determine the number of atoms of H from the number of molecules of C
6
H
12
O
6
Solve:

Interconverting Masses, Moles, and Numbers of Particles (also called dimensional analysis)
Calculate the number of moles of glucose(C
6
H
12
O
6)
C
6
H
12
O
6.
Solution in 5.380 g of
Analyze: We are given the number of grams of C
6
H
12
O
6 calculate the number of moles.
and asked to
Plan: The molar mass of a substance provides the conversion factor for converting grams to moles. The molar mass of C
6
H
12
O
6 is 180.0
Solve: Using 1 mol C
6
H
12
O
6 we have to write the appropriate conversion factor,

The empirical formula tells the relative number of atoms of each element it contains. The general procedure for finding empirical formula is seen below:

• Mesitylene, a hydrocarbon that occurs in small amounts in crude oil, has an empirical formula of C
3
H
4
. The experimentally determined molecular weight of this substance is 121 amu. What is the molecular formula of mesitylene?
•
Solution
• Analyze: We are given the empirical formula and molecular weight of mesitylene and asked to determine its molecular formula.
• Plan: The subscripts in a molecular formula are wholenumber multiples of the subscripts in its empirical formula.
To find the appropriate multiple, we must compare the molecular weight with the formula weight of the empirical formula.
• Solve: First we calculate the formula weight of the empirical formula, C
3
H
4

Next we divide the molecular weight by the empirical formula weight to obtain the factor used to multiply the subscripts in C
3
H
4
Only whole-number ratios make physical sense because we must be dealing with whole atoms. We therefore multiply each subscript in the empirical formula by 3 to give the molecular formula: C
9
H
12

The coefficients in a balanced equation can be interpreted as the relative number of molecules and as the relative number of moles involved in the reaction.
consider the combustion of butane (C
4
H
10
), the fuel in disposable cigarette lighters: the conversion sequence is: combined in a single sequence of factors:

Outline of the procedure used to calculate the number of grams of a reactant consumed or of a product formed in a reaction, starting with the number of grams of one of the other reactants or products.

The reactant that is completely consumed in in a reaction is called the limiting reactant .
Real World Example:
You have 10 pieces of bread and 7 pieces of cheese. You want to make grilled cheese sandwiches with two pieces of bread and 1 piece of cheese, which will run out first?
The bread will run out first, because even though there is less cheese, you need more bread for each sandwich, making bread the limiting reactant.

The quantity of the product that is calculated to form when all of the limiting reactant has reacted is the theoretical yield . The amount of product actually obtained is the actual yield . The percent yield relates the actual yield to the theoretical.

• Adipic acid, H
2
C
6
H
8
O
4 is used to produce nylon.
It is made commercially by a controlled reaction between cyclohexane (C
6
H
12
) and O
2
.
•
(a) Assume that you carry out this reaction starting with 25.0 g of cyclohexane, and that cyclohexane is the limiting reactant. What is the theoretical yield of adipic acid?
• (b) If you obtain 33.5 g of adipic acid from your reaction, what is the percent yield of adipic acid?

Solution
Plan: (a) The theoretical yield is the calculated quantity of adipic acid formed in the reaction. We carry out the following conversions: g C
6
H
12 mol H
2
C
6
H
8
O
4
 g H
2
C
6
H
8
O
4.
(b)
 mol C
6
H
12

Once we have calculated the theoretical yield, we use Equation 3.13 to calculate the percent yield.
Solve: (a)
(b)

~ How to balance equations
~ common chemical reactions (like decomposition, combustion, combination, etc)
~ amu’s, molecular weight, percent composition
~ Avogadro’s number and how to use it
~ dimensional analysis
~ Empirical formulas: how to get them, how to get molecular formulas from them, how to get theoretical yield from them.
~ Limiting reactants and percent yield formula