Télécharger la présentation
## CHAPTER 3

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**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 (% Fe = ? and % O = ?)**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 • 14.6% P • 30.1% O**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.**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.**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. 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. 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**Yield**• Actual Yield - often given as percent yield % Yield = actual yieldX 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