CH110B

Chapter 3 Notes

How can we represent the law of conservation of mass? Chemical equations are the SHORTHAND used.

Molecular mass

The sum of the atomic masses of the atoms in a compound

Ex- CO₂ is 12.011 + 2(15.9994) AMU

Ex- Mg(NO₃)₂ is 24.3050 + 2(14.0067 + 3(15.9994))

FORMULA MASS is not a topic we will discuss in detail yet

The mole

Abbreviated mol NOT m which is short for meter

Number of atoms in exactly 12g of cabon-12....same number as in 16g of oxygen-16, 1g of hydrogen, etc.

6.02x10²³ = Avogadro's number = HUGE number

Molar mass

The mass of one mole of atoms or molecules

Ex- CO₂ is 12.011 + 2(15.9994) GRAMS = 6.02x10²³ molecules of CO₂

Percent composition

What is the mass composition of C in CO₂?

Not just a calculation of one mole of C per 3 moles of atoms in one mole of CO₂...must calculate the percent by weight (mass) of the 12g carbon per mole in the 44g per mole CO₂

The 5-10-5 fertilizer discussion in the box on p.92 is a good practical example

Given the mass percents, you should be able to determine the empirical formula...given the molecular mass, you should be able to determine the chemical formula

Determined experimentally by a technique called ELEMENTAL ANALYSIS

Equations

REACTANTS on the left, PRODUCTS on the right

Stoichiometry- QUANTITATIVE relationship between (or measurement of) substances

In the reaction of CO with H₂ to produce CH₃OH, one MOLECULE of carbon monoxide is stoichiometrically equivalent to two MOLECULES of hydrogen and one MOLECULE of methanol...This molecule ratio of 1:2:1 applies to MOLES as well and allows us to generate conversion factors which are extremely useful (see Figure 3.9)

Yield

If one reactant runs out before another, it is the LIMITING REACTANT or REAGENT (See Figure 3.11 for a nice practical example)

Reactions do not always produce 100% of the product potential or THEORETICAL YIELD

Solutions

MOLARITY or MOLAR CONCENTRATION is the amount of SOLUTE in moles per LITER of SOLUTION

We will rarely deal with MOLALITY, moles of solute per KILOGRAM of solution

Dilution does not change the amount of solute, but DOES change concentration

M₁V₁ = M₂V₂

Given the concentration of a reactant in solution, there is simply one more conversion (M to mol) involved in stoichiometric calculations