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.

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

Yield

• Actual Yield - often given as percent yield% Yield = actual yield X 100

• theoretical yield