1. The Mole through Percent Yield
STOICHIOMETRY
The Mole through Percent Yield

2. ATOMIC MASS, FORMULA MASS, and MOLECULAR MASS
These are all basically the same thing, calculated the same way.
They are for atoms (atomic mass), ionic compounds (formula mass), and covalent compounds (molecular mass).
Different atoms have different masses; THEREFORE different compounds have masses dependent upon the masses of the atoms that make them up.
You have to able to write formulas and count atoms to calculate these.

3. ATOMIC MASS, FORMULA MASS, and MOLECULAR MASS
This is the mass of an atom in atomic mass units (amu). It is the same as the average mass of the atoms of an element found on the periodic table.
Example:
C - single atom – amu
Na – single atom – amu
Br – single atom – amu

4. ATOMIC MASS, FORMULA MASS, and MOLECULAR MASS
Example:
H2O – molecular compound
H (2 x 1.01amu) = 2.02
O (1x16.00amu) = 16.00
18.02 amu

5. ATOMIC MASS, FORMULA MASS, and MOLECULAR MASS
Example: NaCl – ionic compound
Na (1 x 22.99amu) = 22.99
Cl (1 x 35.45amu) = 35.45
58.44 amu

6. Percent composition
What is the mass percent of hydrogen in sulfuric acid? %H in H2SO4?

7. THE MOLE
 The MOLE (mol) is the unit of measurement used to describe the amount of matter in a particular substance.
1 mole = x 1023 particles
= x 1023 particles
(particles are atoms, formula units, or molecules)
This is a rounded number.
It is the number of particles in exactly 12 grams of pure Carbon-12. Also known as Avogadro’s number.

8. Moles to moles How many moles of hydrogen are in 3.0 moles of water?
The formula is H2O H and 1 O
3.0 mol H2O mol H = mol H
1 mol H2O

9. MOLES AND MASS
We will use dimensional analysis (factor label) to make mole/mass conversions.
Use the conversion factor:
atomic
molecular MASS (g) = 1 MOLE
formula

10. PRACTICE
Always start with the number you are given and then find the conversion factor that gets you to the desired outcome.
A. Calculate the mass in grams of 2.0 moles of Na.
??g = 2.0 mol Na g Na = g Na
1 mol Na

11. MOLES AND PARTICLES
We use factor label to make mole/particles conversions.
Use the conversion factor:
1 mole = 6.02 x 1023 particles
(particles are atoms, formula units, or molecules)

12. practice
E. Determine the number of molecules in 2.23 moles of nitrogen, N2. ?? molecules = 2.23 mol N2 6.02x1023 molecules N2 1 mol N2 = molecules N2

13. MASS AND MOLE AND PARTICLE
1 mole mole
PARTICLES MOLE MASS
(atoms or x molar molecules mass (from
or ions, or periodic table)
formula units

14. EMPIRICAL FORMULA
The simplest whole number ratio of elements in a compound is called the EMPIRICAL FORMULA.
In contrast, the molecular formula of a compound is the actual number of atoms of each element in the compound.
It can be calculated from the mole ratio or from percentage composition data.

15. MOLE RATIO EXAMPLE
What is the empirical formula for a compound if a 2.50 g sample contains g of calcium and 1.60 g of chlorine?
Step One:
Determine the number of moles of Ca and Cl. (You must first determine the atomic mass of each.)
Ca g Ca mole Ca = mol
40.08g Ca
Cl 1.60g Cl mole Cl = mol
35.45g Cl

16. MOLE RATIO EXAMPLE Step Two:
To obtain the smallest ratio, divide both numbers of moles by the smaller number of moles ( mol).
Ca mol = Cl mol = 2.00
mol mol
These should be close to whole numbers MOST of the time.

17. The empirical formula is CaCl2
MOLE RATIO EXAMPLE
Step Three:
Look at the ratio of the two numbers. Round off to whole numbers (there are exceptions to this).
Ca becomes Ca 1
Cl becomes Cl 2
The empirical formula is CaCl2

18. ANOTHER WAY TO DO IT
Sometimes you are giving percentages, not masses and asked to determine the empirical formula.
If this is the case, you can change the percentages directly to grams because you assume a 100 g sample (100%)

19. RULES TO FOLLOW If the ratio is: X.1 or X.9, round up or down
If the ratio is X.2 to X.8, determine what to multiply it by to get a whole number.
Examples:
X.3 x 3 = X.9 which is rounded
X.5 x 2 = X.0 which is a whole number
X.4 x 5 = X.0 which is a whole number

20. MOLECULAR FORMULA
If you know the empirical formula of a substance and its molecular mass, you can determine its molecular formula.
STEPS:
Calculate empirical mass
Determine ratio of molecular mass to empirical mass. Apply to empirical formula.

21. BALANCING CHEMICAL EQUATIONS
Chemical reactions involve a rearrangement of atoms. Chemical equations are expressions in symbols and formulas that represent a chemical reaction.
EXAMPLE:
Magnesium + Oxygen yields Magnesium Oxide + energy
(word equation)
Mg O2  MgO + heat
(skeleton equation)
2 Mg O2  2 MgO + heat
(balanced equation)
A balanced chemical equation has the same number of atoms of each element on both sides of the equation.

22. RULES TO FOLLOW Change coefficients only.
Never change a symbol, formula, or subscript.
Place coefficients in front of the entire chemical formula.

23. STEPS TO BALANCING AN EQUATION
1. Write the chemical equation for the reaction.
Verify that the chemical symbols and formulas are correct.
2. Count the number of atoms of each element on both sides of the arrow.
If the atom numbers are equal, then the equation is balanced.
3. Balance the equation using coefficients. A coefficient is a whole number placed in front of a chemical formula to indicate the number of molecules present. Balance metals, then nonmetals, then hydrogen, then oxygen
Look for the least common multiple (LCM).
4. Check your work by recounting the number of atoms on each side of the arrow.

24. Word Equations
Write the correct formulas for each of the chemicals named. Determine which are the reactant(s) and which are the product(s).
All the other steps are the same.

25. Helpful Hints Symbols you should know: means “yields”
(s) means solid Ag (s)
(g) means gas H2 (g)
(l) means liquid H2O (l)
(aq) means aqueous NaCl (aq)
dissolved in water

26. Helpful Hints All metals are singular
Non-noble gases are diatomic – travel in pairs. The diatomic gases are N2, O2, F2, Cl2, Br2, I2, H2

27. HELPFUL HINTS
Any word problem that you are converting to a balanced equation needs symbols if:
it is an acid (aq)
It is an ionic substance that you can tell its solubility by the rules (aq) or (s)
It explicitly states what it is
It is one of the seven diatomic gases (g)
A singular metal (s) or in the case of Hg (l)

28. STOICHIOMETRY
Chemical reactions stop when one of the reactants is used up.
You need to know how many grams of one reactant will react with a known mass of another reactant OR if you have 10.0g X, how many grams of XY will be produced with an excess of Y.
Stoichiometry is the study of quantitative relationships between amounts of reactants used and the products formed by a chemical reactions.

29. STEPS TO FOLLOW
1. Write down what you know and what you need to find out.
2. Write the balanced chemical equation.
3. Determine the number of moles of the given substance. You may have to convert it from grams to moles.
4. Determine the mole ratio between what is known and what is unknown.
5. Convert the unknown moles to grams if needed.

30. PRACTICE
Sulfuric acid is formed when sulfur dioxide reacts with oxygen and water. How many moles of oxygen gas are needed when you have moles of SO2?

31. LIMITING REACTANTS
Rarely are reactants in a chemical reaction present in the EXACT ratios specified by the balanced equation.
Usually one or more are in excess while one is limited.
The substance that is completely consumed (used up) in a reaction is the limiting reactant.
The reaction cannot continue without more of it.
Excess reactants are the reactants that are left over, in excess.

32. HOW TO DETERMINE THE LIMITING REACTANT
Treat these problems as mass to mass problems and then compare the answers to determine the limiting reactant.
The limiting reactant will be the one that converted to the lesser amount.

33. PRACTICE
Nickel reacts with hydrochloric acid to produce nickel (II) chloride and hydrogen gas. If 5.00g nickel and 2.50g hydrochloric acid react, determine the limiting reactant and the mass of nickel (II) chloride produced.

34. PERCENT YIELD
We can calculate theoretical amounts of chemicals produced in a chemical reaction by doing limiting reactant problems.
In reality, the actual experimental amounts yielded in the laboratory are much less due to human error or other factors.
Chemists want to compare the theoretical yield to the actual yield to see how efficient they are.

35. PERCENT YIELD Percent Yield = actual amt of product x 100
theoretical amt
Theoretical amt – maximum amount of product that CAN be produced from a given amount of reactant.
Actual amt – amount of product actually produced when the chemical reaction is carried out (the experimental amount).