CHAPTER 3
CHAPTER 3
STOICHIOMETRY
• Determination of quantities of materials consumed and produced in a chemical reaction.
CHEMICAL REACTION
• A + B Product
–Reactants
Periodic Table
• Atomic Mass –number below the element
–not whole numbers because the masses are averages of the masses of the different isotopes of the elements
STOICHIOMETRY
–For example, the mass of C = 12.01 a.m.u is the average of the masses of 12C, 13C and 14C.
Determination of Aver. Mass
• Ave. Mass = [(% Abund./100) (atomic
mass)] + [(% Abund./100) (atomic mass)]
Take Note:• If there are more than 2
isotopes, then formula has to be re-adjusted
Sample Problem 1
• Assume that element Uus is synthesized and that it has the following stable isotopes:–284Uus (283.4 a.m.u.) 34.6 %–285Uus (284.7 a.m.u.) 21.2 %–288Uus (287.8 a.m.u.) 44.20 %
Solution
• Ave. Mass of Uus =• [284Uus] (283.4 a.m.u.)(0.346)• [285Uus] +(284.7 a.m.u.)(0.212)• [288Uus] +(287.8 a.m.u.)(0.4420)
• = 97.92 + 60.36 + 127.21 • = 285.49 a.m.u (FINAL ANS.)
For Your Benefit
• Do Problem 24 on page 123 of Zumdahl text.
The MOLE
• Amount of substance that contains as many entities as there are in exactly 12 grams of carbon-12.
The MOLE
• The mass of 1 mole = the atomic mass of the element in grams
Formula for Mole
Mole = mass of element
atomic mass of element
Sample Mole Calculations
1 mole of C = 12.011 grams
» 12.011 gm/mol
• 0.5 mole of C = 6.055 grams
» 12.011 gm/mol
Avogadro’s Number
• Way of counting atoms
• Avogadro’s number = 6.02 x 1023
Point to Remember
One mole of anything is 6.02 x 1023 units of that substance.
Avogadro’s Number and the Mole
• If one mole of anything is 6.02 x 1023 units of that substance, then:
• 1 mole of oranges = 6.02 x 1023
oranges
And……..
• 1 mole of C has the same number of atoms as one mole of any element
Also…..
• 1 mole of sand = 6.02 x 1023 particles
An Even Better Analogy…..
• 1 dozen = 12 entities
• a dozen apples has the same number of entities as a dozen oranges
Summary
• Avogadro’s Number gives the number of particles or atoms in a given number of moles
• 1 mole of anything = 6.02 x 10 23 atoms or particles
Sample Problem 2
• Compute the number of atoms and moles of atoms in a 10.0 gram sample of aluminum.
Solution
• PART I:
• Formula for Mole:
–Mole = mass of element
atomic mass of element
Solution (cont.)
• Part II: To determine # of atoms
• # atoms = moles x Avogadro’s number
Problem # 2
• A diamond contains 5.0 x 1021 atoms of carbon. How many moles of carbon and how many grams of carbon are in this diamond?
Molar Mass
• Often referred to as molecular mass
• Definition: –mass in grams of 1 mole of
the compound
Example Problem
• Determine the Molar Mass of C6H12O6
Solution
• Mass of 6 mole C = 6 x 12.01 = 72.06 g• Mass of 12 mole H = 12 x 1.008 = 12.096 g• Mass of 6 mole O = 6 x 16 = 96 g
• Mass of 1 mole C6H12O6
= 180.156 g
Problem # 2
• A diamond contains 5.0 x 1021 atoms of carbon. How many moles of carbon and how many grams of carbon are in this diamond?
Problem #3
• What is the molar mass of (NH4)3(PO4)?
% Mass Determination
Step I.
Total % Masses of atoms = 100 %
% Mass Determination
Step II. If formula is given, break the compound down and get total atomic masses of each element.
% Mass Determination
Step III.
Divide total atomic masses of each element by total molar mass to determine element contribution
% Mass Determination
Step IV.
Multiply by 100 to get percent
Sample Problem
• Find the % Mass of:
- FeO (% Fe = ? and % O = ?)
–Fe2O3
(% Fe = ? and % O = ?)
Composition of Compounds
How many grams of silicon are there in 217.00 grams of SiO2? [Hint: Determine % composition first.]
Empirical Formula
• Only gives the types of elements in the compound and the simplest ratio of the elements in the formula
Empirical Formula
• Does not tell exactly how many of the elements are in the compound
Molecular Formula• Gives the exact number of elements in
the compound as it exists.
• Gives you the exact elemental composition of the compound
• Formula of the compound as it would actually exist.
EF vs. MF
Sucrose or table sugar:
Molecular Formula = C6H12O6
Empirical Formula = CH2O
Empirical Formula
• EF Determination when % Masses are given
Steps in Determining EF• Step 1. Sum up all given percentages.
• If sum of percentages = 100 % or very close to it, proceed to Step 2.
• If sum is < 100 %, the missing percentage is often due to oxygen or the missing element present in the elemental analysis.
• Step 2. Convert Mass % to grams.
• Step 3. Convert all grams to moles using the equation:
mole = gram of elementatomic mass of
element
• Step 4. Divide all calculated moles by the smallest calculated mole to get a simplest ratio of 1.
• Step 5. If the ratios are whole numbers, you now have the Empirical Formula. The ratios are the subscripts of the elements in the empirical formula.
• If the ratios are not whole numbers, follow the rule of rounding.
Rule of Rounding Molar Ratios
• Mole ratios can only be rounded to the nearest whole number if they are < 0.2 away from the nearest whole number. For ex: 1.95 = 2; 3.18 = 3 and 4. 13 = 4.
• If the mole ratio is > 0.2 away from the nearest whole number, multiply the mole ratio by a certain integer to get it close to the nearest whole number. For ex: 3.5 x “2” = 7; 6.33 x “3” = 18.99 = 19; 4.25 x “4” = 11.
Please Remember• If you have to multiply a mole ratio by an
integer to get close to a whole number, you MUST multiply all the other mole ratios by the same integer.
• “In short, what you do to one mole ratio, you also do to the rest.”
• The ratios give you the subscripts in the EF.
Steps To Determine the Molecular Formula
• Step 1. Now that you have the empirical formula, get the ratio of the “given” molar mass to the empirical formula mass. Ratio = Given Molar Mass
Empirical Formula Mass
* Round ratio to the nearest whole number.
• Please note that the Empirical formula Mass is the sum of the atomic masses of all the elements in the Empirical Formula.
• Step 2. Once the ratio has been determined, multiply all the subscripts in the empirical formula by the ratio. This gives you the Molecular Formula.
Chemical Equations
Terms:(s) = solid(l) = liquid(g) = gas
= heat(aq) = aqueous solution
Balancing Equations• * Use coefficients to balance equations!
• Step 1: Balance metals first.
• Step 2: If possible, consider poly-atomic ions as a group. If “OH” is present on one side and H2O is present on the other side, break up water into H and OH.
Balancing Equations
• Step 3: Balance other elements
Step 4: Balance H’s and O’s last.
• Step 4: Double-check.
Sample Problem
• Balance the reaction:
• Cu + AgNO3 Ag + Cu(NO3)2
• Ca(OH)2 + H3PO4 H2O + Ca3(PO4)2
Stoichiometric Calculations
• Given the reaction:
• C3H8 + 5O2 CO2 + 4H2O
• Info: molar ratios
Problem
• C3H8 + O2 CO2 + 4H2O
• If 25 grams of C3H8 is used, how much O2 is needed?
Solution
• 1. Balance equation.
• 2. Get molar ratios from balanced equation.
• 3. Find actual moles using given masses.
Solution (cont.)
• 4. Re-adjust moles.
• 5. Convert moles to grams if required.
Steps in Stoichiometry• 1. Get the molar masses of each cpd in
the equation.
• 2. Balance the equation.
• 3. If grams are given, convert grams to moles using the equation: mole = gram/molar mass
• 4. If only 1 mass is given (Case I), there is no limiting reagent. Re-adjust each mole using the molar ratios from the balanced equation.
• 5. If more than 1 mass is given, there is a LIMITING REAGENT! Base all actual moles of needed reactant and desired product on the Limiting Reagent (not on the Excess)! (Case II)
Limiting and Excess Reagents
• Limiting reagent = limits the amt. of product that can form
• Excess Reagent = reagent that is over and above what is needed
Case II Stoichiometry
• Has a limiting and excess reagent
• Case II applies when there are 2 or more given masses or moles
• 5. Convert moles to grams, if needed.Gram = mole x molar
mass
• 6. Calculate % Yield and % Error, if needed.
Determining the Limiting Reagent
• To determine the limiting reagent, divide all calculated moles by the coefficients in the balanced reaction. The smallest value is the Limiting Reagent.
• Please note: Do not use these values for the rest of your calculations. This is only for the IDENTIFICATION of the Limiting Reagent!
Yields
• Theoretical Yield
–the amount of product formed when the limiting reagent is totally consumed
Yield
• Actual Yield - often given as percent yield% Yield = actual yield X 100
• theoretical yield