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CHAPTER 3

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

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CHEMICAL REACTION A+BProduct –Reactants

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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

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STOICHIOMETRY –For example, the mass of C = 12.01 a.m.u is the average of the masses of 12 C, 13 C and 14 C.

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Determination of Aver. Mass Ave. Mass = [(% Abund./100) (atomic mass)] + [(% Abund./100) (atomic mass)]

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Take Note: If there are more than 2 isotopes, then formula has to be re-adjusted

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Sample Problem 1 Assume that element Uus is synthesized and that it has the following stable isotopes: – 284 Uus (283.4 a.m.u.)34.6 % – 285 Uus (284.7 a.m.u.)21.2 % – 288 Uus (287.8 a.m.u.) 44.20 %

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Solution Ave. Mass of Uus = [ 284 Uus](283.4 a.m.u.)(0.346) [ 285 Uus] +(284.7 a.m.u.)(0.212) [ 288 Uus] +(287.8 a.m.u.)(0.4420) = 97.92 + 60.36 + 127.21 = 285.49 a.m.u (FINAL ANS.)

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For Your Benefit Do Problem 24 on page 123 of Zumdahl text.

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The MOLE Amount of substance that contains as many entities as there are in exactly 12 grams of carbon-12.

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The MOLE The mass of 1 mole = the atomic mass of the element in grams

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Formula for Mole Mole = mass of element atomic mass of element

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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

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Avogadro’s Number Way of counting atoms Avogadro’s number = 6.02 x 10 23

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Point to Remember One mole of anything is 6.02 x 10 23 units of that substance.

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Avogadro’s Number and the Mole If one mole of anything is 6.02 x 10 23 units of that substance, then: 1 mole of oranges = 6.02 x 10 23 oranges

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And…….. 1 mole of C has the same number of atoms as one mole of any element

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Also….. 1 mole of sand = 6.02 x 1023 particles

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An Even Better Analogy….. 1 dozen = 12 entities a dozen apples has the same number of entities as a dozen oranges

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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

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Sample Problem 2 Compute the number of atoms and moles of atoms in a 10.0 gram sample of aluminum.

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Solution PART I: Formula for Mole: –Mole = mass of element atomic mass of element

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Solution (cont.) Part II:To determine # of atoms # atoms = moles x Avogadro’s number

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Problem # 2 A diamond contains 5.0 x 10 21 atoms of carbon. How many moles of carbon and how many grams of carbon are in this diamond?

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Molar Mass Often referred to as molecular mass Definition: –mass in grams of 1 mole of the compound

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Example Problem Determine the Molar Mass of C 6 H 12 O 6

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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 C 6 H 12 O 6 = 180.156 g

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Problem # 2 A diamond contains 5.0 x 10 21 atoms of carbon. How many moles of carbon and how many grams of carbon are in this diamond?

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Problem #3 What is the molar mass of (NH 4 ) 3 (PO 4 )?

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% Mass Determination Step I. Total % Masses of atoms = 100 %

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% Mass Determination Step II. If formula is given, break the compound down and get total atomic masses of each element.

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% Mass Determination Step III. Divide total atomic masses of each element by total molar mass to determine element contribution

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% Mass Determination Step IV. Multiply by 100 to get percent

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Sample Problem Find the % Mass of: - FeO (% Fe = ? and % O = ?) –Fe 2 O 3 (% Fe = ? and % O = ?)

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Composition of Compounds How many grams of silicon are there in 217.00 grams of SiO 2 ? [Hint: Determine % composition first.]

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Composition of Compounds Bonus: How many grams of sulfur are there in 69.26 grams of SO 3 ? [Hint: Determine % composition first.]

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Empirical Formula Only gives the types of elements in the compound and the simplest ratio of the elements in the formula

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Empirical Formula Does not tell exactly how many of the elements are in the compound

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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.

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EF vs. MF Sucrose or table sugar: Molecular Formula = C 6 H 12 O 6 Empirical Formula = CH 2 O

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Empirical Formula EF Determination when % Masses are given

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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.

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Step 2. Convert Mass % to grams. Step 3. Convert all grams to moles using the equation: mole = gram of element atomic mass of element

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Step 4. Divide all calculated moles by the smallest calculated mole to get a simplest ratio of 1.

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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.

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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.

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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.

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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.

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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.

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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.

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Chemical Equations Terms: (s) = solid (l) = liquid (g) = gas = heat (aq) = aqueous solution

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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 H 2 O is present on the other side, break up water into H and OH.

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

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Sample Problem Balance the reaction: Cu + AgNO 3 Ag + Cu(NO 3 ) 2 Ca(OH) 2 + H 3 PO 4 H 2 O + Ca 3 (PO 4 ) 2

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Stoichiometric Calculations Given the reaction: C 3 H 8 +5O 2 CO 2 + 4H 2 O Info:molar ratios

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Problem C 3 H 8 +O 2 CO 2 + 4H 2 O If 25 grams of C 3 H 8 is used, how much O 2 is needed?

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Solution 1. Balance equation. 2. Get molar ratios from balanced equation. 3. Find actual moles using given masses.

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Solution (cont.) 4. Re-adjust moles. 5. Convert moles to grams if required.

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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

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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

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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)!

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5. Convert moles to grams, if needed. Gram = mole x molar mass 6. Calculate % Yield and % Error, if needed.

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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!

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Yields Theoretical Yield –the amount of product formed when the limiting reagent is totally consumed

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

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