2 - 1 Chapter 3. Stoichimetry Chapter 3. Stoichimetry Calculations with chemical formula and reactions are called Stoichiometry

2 - 2 KEY CONCEPTS Atomic, molecular and formula mass Mole, Avogadro's number and Molar mass Mass percent of elements Mass % from analytical data Mass % from formula Empirical formula from Mass % Chemical formula from composition Chemical formula from mass % Chemical Equations Stoichiometric coefficients Balancing chemical equations Stoichimetry Stoichiometric coefficients and Limiting reactant Yields of chemical reactions Actual yield and Thoretical yield Solutions concentration and dilution of solutions Solution stoichiometry

2 - 3 Question You were given a chance to pick 100 kg of gold(I)periodate - AuIO 4 or gold(I)nitrate- AuNO 3 Which one you would pick? Rationalize your choice.

2 - 4 Gold 1 oz = $ oz = g 1 kg = oz 1 kg Au = $ 14,109

Molecular mass vs. formula mass Formula mass Add the masses of all the atoms in formula - for molecular and ionic compounds. Molecular mass Calculated the same as formula mass - only valid for molecules. Both have units of either u or grams/mole.

2 - 6 Examples of M.W. H 3 PO 4 : H = 1.01 g/mol, P = g/mol, O = g/mol 1 amu is equal to 1 g/mol m.w. H 3 PO 4 = 3 x x x = g/mol

2 - 7 Examples of F.W. K 2 CO 3 : K = g/mol; C = g/mol; O = g/mol f.w. K 2 CO 3 = 2 x x x = g/mol

2 - 8 Mole Concept Mole Concept We use masses weighed in grams in chemical reaction. Need a conversion factor to convert grams to atoms and molecules. atomic weight or molecular weight taken in grams contains x atoms, molecules or particles. Avagadro’s Number The number x is called Mole or Avagadro’s Number

2 - 9 Can you guess this ? How long is 1 mole of seconds? x years 2 x billion years How deep is the layer of marbles, if 1mole of marbles are spread over the surface of earth? ~50 miles high

The Mole and Avogadro's number Number of atoms in grams of 12 C 1 mol = x atoms 1 g = x u 1 u = 1 g/mol Mole is the converssion factor between M.W and grams 1 mol = grams / molecular weight mole = g/m.w. Or g/f.w.

Conversion factors in Stoichiometry? x atoms = gram atomic weight x molecules = gram molecular weight x atoms C = grams of carbon (C) x molecules H 2 O = g of H 2 O x = 1 mol 1 g = x amu (u) 1 amu = 1 g/mol

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Atomic Masses Atomic mass in grams ? Mass of a copper atom in grams g Cu 1 mole x = x g Atomic mass in the periodic table divided by Avagadro’s Number

An atom weighs 7.47 x g. What is the name of the element this atom belongs to?

An atom weighs 7.47 x g. What is the name of the element this atom belongs to? Conversion factor: 1g = x u 7.47 x g x6.022 x u 1g = u or g/mole The element is Sc.

Example - (NH 4 ) 2 SO 4 How many atoms are in 20.0 grams of ammonium sulfate? Formula weight = grams/mol Atoms in formula= 15 atoms / unit moles = 20.0 g x = mol 1 mol g atoms = mol x 15 x 6.02 x10 23 atoms unit units mol atoms = 1.36 x10 24

How do you convert grams to mole and vice versa? Grams ---> mole grams /molecular (formula) weight Moles ----> grams moles x molecular (formula) weight

Calculations Calculate how many moles are in 100 g of sulfur-S 8 Calculate the grams of sugar in 3 moles of -glucose- C 6 H 12 O 6

% Element Composition in Compounds n x Atomic weight % mass = x 100 molecular weight n = subscript of the element in the formula

Calculation Al 2 (SO 4 ) 3 calculate % Al, % S and % O

2 - 21Examples a) Al 2 (SO 4 ) 3 :f.w. = g/mol, Atomic weight of oxygen = g/mol, n = x % O mass = x 100 = 56.12% oxygen c) CuSO 4 : f.w. = g/mol Atomic weight of oxygen = g/mol n = 4 4 x % O mass = = 40.01% oxygen

Calculate % composition from analysis A 12.5 g sample of a compound that contains only phosphorus and sulfur was analyzed and found to contain 7.04 g of phosphorus and 5.46 g of sulfur. Calculate the percent composition of elements.

Examples Glucose has a molecular formula of C 6 H 12 O 6 (M.W g/mol). a) How many grams of C, H and O are available in 1 mole of glucose? b) Calculate mass percents of elements C, H and O in glucose.

mol C 6 H 12 O 6 = 6 mol C = 6 x =72.06g C 1 mol C 6 H 12 O 6 =12 mol H=12 x 1.01 = 12.12g H 1 mol C 6 H 12 O 6 =6 mol O =6 x = 96.00g O How many grams of C, H and O are available in 1 mole of glucose?

x 12 %C= x 100 = 40.00% C x 1.01 % H= x 100 = 6.73% H x %O= x 100 = 53.29% O % Mass percent of element. C 6 H 12 O 6 = g/mol

What is Empirical Formula Simple whole number ratio of each atom expressed in the subscript of the formula. Molecular Formula = C 6 H 12 O 6 of glucose Empirical Formula = CH 2 O Emiprical formula is calculated from % composition

How do you get Empirical Formula from % composition and vice versa?

% composition to Empirical formula Divide percent composition by atomic weight. Divide answer by lowest number to get simple ratio of moles or atoms. Multiply by a factor to get whole number ratio. Write a formula with whole number ratio as subscripts to get empirical formula.

2 - 29Examples A 12.5 g sample of a compound that contains only phosphorus and sulfur was analyzed and found to contain 7.04 g of phosphorus and 5.46 g of sulfur. a) Calculate the percent composition of elements. b) The empirical formula of the compound

2 - 30Calculation mass of element % Element = x 100 mass of sample 7.04 % P= x 100 = 56.3% P % S= x 100= 43.7% S

2 - 31Calculation % of P and S: Moles of P and S: 56.3g 43.7g = 1.82 mol P = 1.36 mol S Atom ratio: 1.82 atoms P 1.36 atoms S 1.33 atoms P 1.00 atoms S

2 - 32Calculation..continued 1.82 atoms P 1.36 atoms S Divide by lowest number: 1.33 atoms P 1.00 atoms S simple atom ratio: 1.33 atom P 1 atom S to get a simple whole number ratio. Multiply both numbers by 2, 3, and 4 etc. until 1.33 becomes close to a whole number.

2 - 33Calculation..continued To get simple whole number ratio: PS 1.33 atom 1 atom 2 x 1.33 = x x 1.33= x Therefore, Formula should contain 4 P atoms and 3 S atoms. Therefore, the empirical Formula of the compound is: P 4 S 3

More Example Glucose contains the elements C, H, and O in 39.99% C, 6.71% H and 53.28% O, respectively, by mass. a) Calculate the empirical formula of glucose. b) Molecular weight determination in solution for glucose has shown a molecular weight close to 180 g/mol. What is the molecular formula of glucose?

2 - 35CalculationEmpirical formula from % elemental composition. % composition: 39.99% C, 6.71% H, 53.28% O ii) mole ratio: C H O = 3.33; = 6.64; = mol C:6.64 mol H:3.33 mol O

2 - 36Calculation..continued Atom ratio 3.33 mol C:6.64 mol H:3.33 mol O Simple atom: Simple whole number number ratio of atom empirical formula The empirical formula of the compound is CH 2 O

2 - 37Calculation..continued Molecular Formula = n x empirical Formula Molecular weight 180 n = = = 6 Empirical Formula Weight 30 Molecular Formula = (CH 2 O) n = (CH 2 O) 6 Molecular Formula = C 6 H 12 O 6 of glucose

Question A molecular compound contains 92.3% carbon and 7.7% hydrogen by weight. What is its empirical formula?

Question A molecular compound has empirical formula CH. If mol of the compound weighs 3.90 g, what is its molecular formula?

Chemical Equation P 4 O 10 (s) + 6H 2 O (l) = 4 H 3 PO 4 (l) reactants reactants enter into a reaction. products products are formed by the reaction. Parantheses represent physical state stoichiometric coefficients coefficients are numbers in front of chemical formula gives the amounts (moles) of each substance used and each substance produced. Equation Must be balanced!

Chemical Reaction Could be described in words Chemical equation Chemical equation: Reactants? Products? reaction conditions? =, --->, or ? stoichiometric coefficients? Number in front of substances representing moles, atoms, molecules

Steps in Stoichiometric Calculations Check whether chemical equation is balanced get the moles from grams of materials find the limiting reactant calculate moles of products from the limiting reactant convert moles of the products to grams find the actual yield of the reaction calculate % yield of the reaction

Question How much hydrogen gas is produced when 1 kg of sodium reacts with water?

2 - 44Examples Calculate the following using the chemical equation given below: 4 NH 3 (g) + 5 O 2 (g) ----> 4 NO(g) + 6 H 2 O(g) a) moles of NO(g) from 2 moles of NH 3 (g) and excess O 2 (g). b) moles of H 2 O(g) from 3 moles of O 2 (g) and excess NH 3 (g).

Examples How many moles of H 2 O will be produced by 0.80 mole of O 2 with excess H 2 according to the equation? 2H 2 (g) + O 2 (g) = 2 H 2 O(l)

Question 2Al(s) + 6HCl(aq)--> 2AlCl 3 (aq) +3H 2 (g) According to the equation above, how many grams of aluminum are needed to react with mol of hydrochloric acid?

g Al

What is the limiting reagent? Limiting reagent is the reactant, which is used up first. To find the limiting reactant you have to compare the amounts of reactants in moles.

Examples A g sample of phosphorus ( M.W g/mol) burns in g of oxygen (M.W g/mol) according to following equation: P 4 (s) + 5O 2 (g) = P 4 O 10 (s) What is the limiting reagent?

Theoretical yield Theoretical yield is the amount (grams) of products formed according to chemical equation. Use the limiting reagent to calculate the moles of the product and then convert moles to grams.

Actual Yield Actual yield is the grams of the product obtained by an experiment. Actual yield should be less than the theoretical yield if the experiment was carried out meticulously. If the products are contaminated with impurities or the formula of product was wrong the actual yield could be higher.

% Yield actual yield % yield = x 100 theoretical yield If the products are contaminated with impurities or the formula of product was wrong the % yield could be higher than 100%.

2 - 53Question Sulfur trioxide, SO 3, is made from the oxidation of SO 2 and the reaction is represented by the equation 2SO 2 + O > 2SO 3 A 16.0-g sample of SO 2 gives 18.0 g of SO 3. The percent yield of SO 3 is

2 - 54Examples A g sample of phosphorus ( M.W g/mol) burns in g of oxygen (M.W g/mol) according to following equation: P 4 (s) + 5O 2 (g) = P 4 O 10 (s) a) What is the limiting reagent? b) How many moles of P 4 O 10 are produced theoretically? c) If 612 g of P 4 O 10 is actually produced in this reaction, calculate the percent yield.

How do you calculate moles of substances in solutions Use concentration of solution to convert L or mL of solution in to moles What is concentration of a solution? The relative amounts of solute and solvent There are so many ways to show amount: g, mole, equivalents,volume

Concentration Units a) Molarity (M) b) Normality (N) c) Molality (m) f) Mole fraction (a)(a) g) Mass percent (% weight) h) Volume percent (% volume) i) "Proof" j) ppm and ppb

Molarity M = moles solutemol liters of solution L = [ ] - special symbol which means molar ( mol/L )

Molarity What’s the molarity of a solution that has g HCl in 2.0 liters? First, you need the FM of HCl. FM HCl = x 1 H x 1 Cl = g/mol Next, find the number of moles. moles HCl = g HCl / g/mol = 0.50 mol Finally, divide by the volume. M HCl = 0.50 mol / 2.0 L = 0.25 M

2 - 59Examples Calculate the molarity of a solution prepared by dissolving g of K 2 SO 4 in enough water to make mL solution. moles of solute Molarity(M) = Liters of solution

solute = K 2 SO 4 ; F.W. = g/mol; mass= 200g moles of K 2 SO 4 = ? 200 g / g K 2 SO 4 = mol K 2 SO mL = ? Liters of solution = 0.5 L Molarity? mol K 2 SO 4 Molarity of K 2 SO 4 sol. = Liters of solution = 2.30 mole/Liter = 2.30 M (M = moles/liters)

Normality (N) Equivalents of solute Normality(N) = Liters of solution moles Equivalents = n n is # of H, OH in or e- in acid or base N(normality) = Molarity x n (# of H, OH or e-)e-)

Examples How many grams of KNO 3 are contained in 500 mL of a M solution of potassium nitrate? How many mL of 2.00 M solution of HNO 3 are required with water to make a 250 mL of 1.50 M nitric acid solution?

Mole fraction (  a ) moles of solute (substance)  a = moles of solute + solvent

Calculate the mole fraction of benzene in a benzene(C 6 H 6 )- chloroform(CHCl 3 ) solution which contains 60 g of benzene and 30 g of chloroform. M.W. = (C 6 H 6 ) M.W. = (CHCl 3 )

Weight/Weight % Weight/Weight %= Mass Solute Total Mass x 100 If a ham contained 5 grams of fat in 200 g of ham, what is the % wt/wt? 5 g / 200g * 100=2.5 wt/wt% On the label, it would say 97.5 % fat free. Use the same units for both

Volume/Volume % Volume/Volume %= Volume Solute Total Volume x 100 If 10 ml of alcohol is dissolved in water to make 200 ml of solution, what is the concentration? 10 ml / 200 ml * 100=5 V/V% Alcohol in wine is measured as a V/V%. Use the same units for both

Weight/Volume % Weight/Volume %= Mass solute Total Volume x 100 If 5 grams of NaCl is dissolved in water to make 200 ml of solution, what is the concentration? 5 g / 200 ml * 100=2.5 wt/v% Saline is a 0.9 wt/v% solution of NaCl in water. use g and ml

Very low concentrations Pollutants in air and water are typically found at very low concentrations. Two common units are used to express these trace amounts. Parts per million - ppm Parts per billion - ppb Both are modifications of the % system which could be viewed as parts per hundred - pph. Both mass and volume % systems are used.

Low concentrations in air Trace amounts in are are expressed as volume/volume ratios. ppm= x 10 6 ppb= x 10 9 Example. Example. One cm 3 of SO 2 in one m 3 of air would be expressed as 1 ppm or 1000 ppb. volume solute volume solution volume solute volume solution

Low concentrations in water Mass percentages are used for water pollutants. ppm =x 10 6 ppb =x 10 9 Example. Example. One ppm of a toxin in water is the same as 1 mg / liter since one liter of water has a mass of approximately 10 6 mg. mass solute mass solution mass solute mass solution

Solution preparation Solutions are typically prepared by: Dissolving the proper amount of solute and diluting to volume. Dilution of a concentrated solution. Lets look at an example of the calculations required to prepare known molar solutions using both approaches.

Dilution Problems Why we dilute solutions? Preparing solutions by adding water to concentrated solutions moles before = moles after M i V i = MfVfMfVf M i = initial molarity V i = initial volume M f = final molarity V f = final molarity

2 - 73Examples How many mL of 2.00 M solution of HNO 3 are required with water to make a 250 mL of 1.50 M nitric acid solution? M i V i = MfVfMfVf Mi Mi = 2.00 Vi Vi = ? Mf Mf = 1.50 Vf Vf = 250 mL MfVfMfVf 1.50 x 250 V i = = = mL M i 2.00

Stoichiometric calculations of solutions reactions Check whether chemical equation is balanced from volume of solutionsget the moles from volume of solutions find the limiting reactant calculate moles of products from the limiting reactant convert moles of the products to grams find the actual yield of the reaction calculate % yield of the reaction

Solution stoichiometry example Determine the volume of M HCl that must be added to completely react with 250 ml of 2.50 M NaOH Balanced chemical equation HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O (l) The first step is to determine how many moles of NaOH we have.

Solution stoichiometry example We have 250 ml of a 2.50 M solution. mol NaOH = L x 2.50 mol/L = mol NaOH From the balanced chemical equation, we know that we need one mole of HCl for each mole of NaOH. That means we need mol HCl.

Solution stoichiometry example Now we can determine what volume of our M HCl solution is required. L= mol HCl / M HCl = mol = 6.26 L 1 L mol ( )

2 - 78Examples How many mLs of M BaCl 2 are required to react completely with 25 mL of M Fe 2 (SO 4 ) 3 ? 3 BaCl 2 (aq) + Fe 2 (SO 4 ) 3 (aq)---> 3 BaSO 4 (s) + 2 Fe Cl 3 (aq) 3 BaCl 2 = 1 Fe 2 (SO 4 ) 3

Calculate the moles of Fe 2 (SO 4 ) 3 : moles = Molarity x Liters of solution M x L = mole Fe 2 (SO 4 ) 3 Then convert Fe 2 (SO 4 ) 3 to BaCl 2 mole, BaCl 2 moles to liters and liters to mL mole Fe 2 (SO 4 ) > BaCl 2 moles mol Fe 2 (SO 4 ) 3 x 3 = BaCl 2 moles = Molarity x Liters of solution = x Liters BaCl 2 = 0.15L = 150 mL of BaCl 2

2 - 80Examples How many mLs of M NaOH are required to react with 500 mL of M H 2 SO 4 completely? Acid Base Reactions. N a xV a = N b x V b. N a = Normality of acid, N b = Normality of base V= volume N(normality) = Molarity x n (# of H or OH) N a (0.170 x 2) xV a (500) = N b (0.300x 1)x V b (?). V b = x 2 x 500/ = mL NaOH