Chapter 4 Chemical Calculations
(and Chapters 12 and 13 of the book by Smith and Vukovich)

The Chapter Story: Calculating the number of Vit-B12 molecules

The amount of 125 m g of vit-B12 in a tablet is a very small amount compared to a 1-g vit-C tablet! Find out how many vit-B12 molecules is in that tiny 125 m g of vit-B12! The formula of vit-B12 is C63H88CoN14O14P.  (The molar mass of vit-B12 is 1355 g/mol.)
Thought process:  mass ---> # mole ---> # molecule

Let’s start with the simple molecule H2O!

  1. What is the meaning of the "formula" of water?
  2. What is the meaning of atomic mass in amu, and in g?
  3. What is the meaning of "formula mass"?
  4. What is the "formula mass" in unit of amu of a single water molecule?
  5. What is the "meaning" of the Avogadro’s number, 6.02 ´ 1023?
  6. What is the formula mass in unit of g of water?
  7. What is the definition of "mole" (in "mol" as a unit)?
Formula mass (review atomic mass in Chapter 2!!)

The average mass in the unit of amu or molar mass (g/mol) of a representative unit of any pure substance

NaCl has a latticed structure in which each Na+ is surrounded by 6 Cl anions and likewise each Cl is surrounded by 6 Na+ cations with the representative unit NaCl, i.e., one Na+ and one Cl. Its formula mass is 22.9898 (of Na) + 35.453 (of Cl) = 58.443 amu or g/mol.

(The lattice can be seen at

How about water and glucose (C6H12O6)?

• A glucose molecule is 10 times heavier than a water molecule;
• With a fixed amount of water and glucose (e.g. 180 g), there are more water molecules (10 times more) than glucose molecules;
• 18 g water and 180 g glucose contain the same number of molecules
What is a mole?
Definition: The number of carbon atoms contained in 12.011-g carbon is 1 mole (1 mol), which is 6.02 ´ 1023 atoms and is called the Avogagro's number. (C-12 is used for defining amu.)

Or a more "general definition" as in the textbook as "The number of atoms (or formula units or molecules) contained in the atomic mass in g (or formula mass in g) of an element (or a compound)".

Molar mass: mass in g of 1 mol of the substance

Examples (see more on Table 4.1)
a) water, 18 g/mol (2 mol H + 1 mol O = 2 g + 16 g), or mol/18 g
    2 mol H atoms react with 1 mol O atoms to give 1 mol water molecules
    (But, remember "all gases are diatomic, except noble gases." Thus, instead of having 2H + O ® H2O, you MUST write 2H2 + O2 ® 2H2O.)

b) CO2, 44 g/mol (1 mol C + 2 mol O = 12 g + 16 g ´ 2), or mol/44 g
    1 mol C reacts with 2 mol O (1 mol O2) to give 1 mol CO2;

c) A small size protein with ~100 amino acids, ~10,000 g/mol.

The "larger" the molecule, the more the mass is in a mole of the molecule.

Exercises: (Chapter 12, Smith & Vukovich)
a) What is the weight in grams of 0.25 mol sucrose (C12H22O11)?
The formula mass of sucrose is _______(Y) g/mol.

Thus, 0.25 mol ´ ______ (Y) g/mol = _______ g

(To think algebraically, X mol of a compound with a formula mass Y (g/mol) has an amount (g) of X (mol) ´ Y (g/mol).)

b) A tablet of an antacid contains 500.0 mg Al(OH)3 (What is its electron dot structure?). How many moles of Al(OH)3 are in one tablet? How many Al(OH)3 units are there? How many moles of Al, O, and H are there?

The formula mass of Al(OH)3 is _______ (Y) g/mol.
500.0 mg = 0.5000 g
0.5000 (g)/Y (g/mol) = _______ mol Al(OH)3
Each mol Al(OH)3 has __ mol Al, __ mol O and __ mol H.

(How many moles (or molecules) of sucrose, water, the 10,000 g/mol protein above, ethanol C2H6O, or CO2 are in 0.5 g of each?)

Empirical formula
The simplest whole-number ratio of the constituents of a compound (which can be determined with elemental analysis techniques).
Thus, acetic acid (CH3COOH) has an empirical formula of CH2O, which describes that the ratio of C:H:O = 2:4:2 = 1:2:1

Examples 4.8
A compound has 36.04% Ca and 63.96% Cl. Determine its empirical formula.

First, find the percent ratios of each constituent. Then, convert the percent ratios into "moles". Finally, find the simplest ratio of the mole ratio.

36.04% Ca is equivalent to 36.04 g in 100 g of the compound
63.96% Cl = 63.96 g Cl/100 g compound
(How many grams of Ca or Cl is in 40 g of the compound?)

Find the molar ratio of Ca:Cl
36.04 g Ca ´ 1 mol/40.08 g Ca = 0.8992 mol Ca
63.96 g Cl ´ 1 mol/35.45 g Cl = 1.804 mol Cl
Ca:Cl = 0.8992:1.804 = (0.8992/0.8992):(1.804/0.8992) =1.000:2.006 ~ 1:2
(Do you get the same answer if you start with 40 g of material?)
A few more examples can be found at: