1. Unit 4 Lecture 1- Balancing Eqns and the Mole
Stoichiometry
Balancing Equations
The Mole

2. Chemical Equations Mg(OH)2(s) + 2HCl(aq)  MgCl2(aq) + 2H2O(l)
special information goes here: Δ (heat), catalyst, spark, etc.
Mg(OH)2(s) + 2HCl(aq)  MgCl2(aq) + 2H2O(l)
reactants
products
the arrow in the equation means: “produce”, “yield”, “react to form”
Letters in parentheses show the physical state: (s) solid, (l) liquid, (g) gas, (aq) aqueous (in water)
Correctly written formulas for all reactants and products are a “must.”
Coefficients give the relative amounts of reactants used and products formed. Do NOT balance equations by changing the subscripts!!!

3. Balancing Chemical Equations
Gaseous nitrogen and hydrogen react to form ammonia:
Write the correct formulas for the reactants and products.
N2(g) H2(g)  NH3(g)
Coefficients (numbers in front of the formulas) are used to balance the numbers of atoms of each element.
2 atoms N atoms H atom N, 3 atoms H
N2(g) H2(g)  NH3(g)
2 atoms N atoms H atoms N, 6 atoms H
N2(g) H2(g)  NH3(g)
2 atoms N atoms H atoms N, 6 atoms H
The equation is now balanced.
The Law of Conservation of Mass has been satisfied.

4. Balancing Chemical Equations
Tip: Balance first the elements which appear least often.
NaOH(aq) H3PO4(aq)  Na3PO4(aq) H2O(l)
sodium hydroxide phosphoric acid sodium phosphate
Balance Na or P first.
3NaOH(aq) H3PO4(aq)  Na3PO4(aq) H2O(l)
3NaOH(aq) (1)H3PO4(aq)  (1)Na3PO4(aq) + H2O(l)
6 atoms of H, 7 atoms of O
2 atoms of H, 5 atoms of O
3NaOH(aq) (1)H3PO4(aq)  (1)Na3PO4(aq) + 3H2O(l)
Balanced equation:
3NaOH(aq) + H3PO4(aq)  Na3PO4(aq) + 3H2O(l)

5. Balancing Chemical Equations
CaC2 (s) H2O(l)  Ca(OH)2 (aq) C2H2 (g)
calcium carbide calcium hydroxide acetylene
CaC2 (s) H2O(l)  Ca(OH)2 (aq) C2H2 (g)
PCl3(g) H2O(l)  H3PO3 (aq) HCl (aq)
phosphorus trichloride phosphorous acid hydrochloric acid
PCl3 (g) H2O(l)  H3PO3 (aq) HCl (aq)
H2S(g) Fe(OH)3 (s)  Fe2S3 (s) H2O(l)
hydrogen iron(III) iron(III)
sulfide hydroxide sulfide
3 H2S(g) Fe(OH)3(s)  Fe2S3(s) H2O(l)

6. Mg(OH)2(s) + 2HCl(aq)  MgCl2(aq) + 2H2O(l)
Stoichiometry
The application of the laws of definite proportions and the conservation of mass to chemical reactions is called stoichiometry.
Mg(OH)2(s) + 2HCl(aq)  MgCl2(aq) + 2H2O(l)
Applications:
Computing the formula mass or molar mass of a compound from its chemical formula
Computing the % composition of a compound from its chemical formula
Computing the amounts of reactants required and products formed (the yield) for a specific chemical reaction

7. Mg(OH)2(s) + 2HCl(aq)  MgCl2(aq) + 2H2O(l)
Stoichiometry and the Mole
Mg(OH)2(s) + 2HCl(aq)  MgCl2(aq) + 2H2O(l)
2 molecules
2 molecules
1 formula unit
1 formula unit
In the laboratory, it is virtually impossible to work with single molecules or formula units. Instead, we generally work with amounts of chemicals large enough to see and manipulate.
A MOLE is a number (Avogadro’s number) large enough to give appreciable amounts of material.
A MOLE is x 1023 of anything…anything!!!
The mass of a mole of any element is its atomic mass in grams.
1 mole of magnesium has a mass of g (see periodic table).
1 atom of magnesium has a mass of amu.

8. Mg(OH)2(s) + 2HCl(aq)  MgCl2(aq) + 2H2O(l)
Stoichiometry and the Mole
Mg(OH)2(s) + 2HCl(aq)  MgCl2(aq) + 2H2O(l)
1 mole
2 moles
1 mole
2 moles
One way to tell if we have a mole of a chemical is to weigh it.
The mass of a mole (aka the molar mass) of any element is its atomic mass in grams.
The mass of a mole (aka the molar mass) of any compound is its formula mass in grams.

9. Stoichiometry and the Mole
Mg(OH)2(s) + 2HCl(aq)  MgCl2(aq) + 2H2O(l)
1 mole
2 moles
1 mole
2 moles
1 formula unit of Mg(OH)2 has:
1 ion Mg amu
2 atoms O 2 x = amu
2 atoms H 2 x = amu
formula mass amu

10. Mg(OH)2(s) + 2HCl(aq)  MgCl2(aq) + 2H2O(l)
Stoichiometry and the Mole
Mg(OH)2(s) + 2HCl(aq)  MgCl2(aq) + 2H2O(l)
1 mole
2 moles
1 mole
2 moles
1 mole of Mg(OH)2 has:
1 mole of Mg2+ ions g
2 moles of O atoms 2 x = g
2 moles of H atoms 2 x = g
molar mass g

11. Mg(OH)2(s) + 2HCl(aq)  MgCl2(aq) + 2H2O(l)
Stoichiometry and the Mole
Mg(OH)2(s) + 2HCl(aq)  MgCl2(aq) + 2H2O(l)
1 mole
2 moles
1 mole
2 moles g
2 x g
= g g
2 x g
= g
Balanced chemical equations tell us the proportions in which chemicals react. These are generally in MOLES.
One way to measure moles in the lab is to weigh the chemical.
Molar masses:
Mg(OH) g
HCl g
MgCl g
H2O g

12. Converting from Mass to Moles
Once we know the molar mass of a compound, we can calculate the number of moles present in ANY MASS of that compound. Simply treat the molar mass as a conversion factor.
1 mol Mg(OH)2 = g
How many moles of Mg(OH)2 are in g of the compound?
29.16 g x 1 mol Mg(OH)2 = mol Mg(OH)2 g
How many moles of Mg(OH)2 are in 4.07 x 10-3 g of the compound?
4.07 x 10-3 g x 1 mol Mg(OH)2 = x mol Mg(OH)2

13. Converting from Moles to Mass
If we can convert from mass to moles, we can also convert from moles to mass.
What is the mass in grams of 1.5 moles of Mg(OH)2 ?
1.5 mol Mg(OH)2 x g = g
1 mol Mg(OH)2

14. Unit 4 Lecture 1- Balancing Eqns and the Mole
Examples
1. What is the mass in grams of mol of MgCl2?
2. How many moles of NH4Cl are in 76.5 g?
3. How many moles of nitrate ions are present in 2.5 mol of Al(NO3)3?
4. How many grams of nitrate ions are present in 10.0 g of Al(NO3)3?

15. Converting from Moles to Numbers of Formula Units, Molecules, or Atoms
Moles can be converted into numbers of formula units (for ionic compounds) OR numbers of molecules OR numbers of atoms.
How many formula units of Mg(OH)2 are in 1.5 moles of the compound?
1.5 mol Mg(OH)2 x x 1023 formula units = x formula units
1 mol Mg(OH)2
How many hydroxide ions (OH-) are in 1.5 moles of Mg(OH)2?
1.5 mol Mg(OH)2 x 2 mol OH- ions x x 1023 ions = x ions
1 mol Mg(OH) mol OH- ions

16. The Map Number of things Moles Mass conversion factor:
1 mol = molar mass
conversion factor:
1 mol = x 1023

17. Examples
1. How many molecules of HCOOH are present in mol of the compound?
2. Calculate the mass of x 1026 formula units of NH4Cl.