CHAPTER 3

STOICHIOMETRY Determination of quantities of materials consumed and produced in a chemical reaction.

CHEMICAL REACTION A + B Product Reactants

% 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

Composition of Compounds How many grams of silicon are there in 217.00 grams of SiO2? [Hint: Determine % composition first.]

Composition of Compounds Bonus: How many grams of sulfur are there in 69.26 grams of SO3?

What does this mean? If a compound has a % Composition of : 55.3% K

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 element atomic 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 MUST 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.

Problem A compound containing only sulfur and nitrogen is 69.9 % S by mass. The molar mass is 184 g/mol. Determine the empirical and molecular formulae of the compound.

Chemical Equations Terms: (s) = solid (l) = liquid (g) = gas D = 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.

Step 3: Balance other elements Balancing Equations Step 3: Balance other elements Step 4: Balance H’s and O’s last. Step 4: Double-check.

Sample Problem Cu + AgNO3 Ag + Cu(NO3)2 Balance the reaction: 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. Get the molar masses of each cpd in the equation. 2. If grams are given, convert grams to moles using the equation: mole = gram/molar mass

Solution (cont.) 3. Balance equation. 4. Get molar ratios from balanced equation. 5. Find actual moles using given masses.

Solution (cont.) 5. Re-adjust moles. 6. Convert moles to grams if required and to check that work is correct. Mole of reactants = moles of products

Steps in Stoichiometry 1. Get the molar masses of each cpd in the equation. 2. If grams are given, convert grams to moles using the equation: mole = gram/molar mass 3. Balance the equation. 3.

4. If only 1 mass is given (Case I), there is no limiting reagent 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. (Fractions x given mole) 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)

6. Convert moles to grams, if needed. Gram = mole x molar mass 7. Calculate % Yield and % Error, if needed. % Yield = Given mass or mole x 100 Theoretical Mass or Mole

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

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

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

Case II Stoichiometry 1. Find molar masses of each compound. 2. Convert both grams to moles. mole = g/molar mass Balance the reaction. To determine the limiting reagent, divide each calculated mole by the coefficient. The smaller value is the LR. This is for ID purposes only. Using the LR mole, re-adjust all molar ratios. Convert to moles to grams. Mass of Reactants = Mass of Products