STOICHIOMETRY
• Determination of quantities of materials consumed and produced in a chemical reaction.
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)]
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.)
The MOLE
• Amount of substance that contains as many entities as there are in exactly 12 grams of carbon-12.
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 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
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.
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
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?
% 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
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
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.
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 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.
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.
Solution
• 1. Balance equation.
• 2. Get molar ratios from balanced equation.
• 3. Find actual moles using given masses.
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
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, 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)!
• 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
Top Related