Unit 3: Stoichiometry The Mole, Molar Mass, % Composition, Balancing Chemical Equations, Limiting Reagent, Percent Yield

Unit 3 Topics 3.1 Counting by Weighing 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass 3.5 Learning to Solve Problems (Reading) 3.6 Percent Composition of Compounds 3.7 Determining the Formula of a Compound 3.8 Chemical Equations 3.9 Balancing Chemical Equations 3.10 Stoichiometric Calculations: Amounts of Reactants and Products 3.11 The Concept of Limiting Reagent

Chemical Stoichiometry Stoichiometry – The study of quantities of materials consumed and produced in chemical reactions. Section 3.1 – Counting by Weighing

Elements occur in nature as mixtures of isotopes. Carbon = 98 Elements occur in nature as mixtures of isotopes. Carbon = 98.89% 12C 1.11% 13C < 0.01% 14C Section 3.2 – Atomic Mass

Schematic Diagram of a Mass Spectrometer A mass spectrometer is used to compare the masses of atoms. Atoms are passed into a beam of high speed electrons, changing them into ions. An applied electric field accelerates these ions into a magnetic field. The ions are deflected based on their mass. The more massive ions are deflected the smallest amount.

Carbon 12 is defined as having a mass of exactly 12 atomic mass units (amu) All other atoms were compared to this standard Example Mass 13C = 1.0836129 Mass 12C Therefore, Mass 13C = (1.0836129) (12 amu) = 13.003355 amu

Average Atomic Mass (for Carbon) Average atomic mass: The weighted average of the masses of the isotopes for that element considering their relative abundance. = 98.89% of 12 amu + 1.11% of 13.0034 amu exact number = (0.9889)(12 amu) + (0.0111)(13.0034 amu) =12.01 amu

Average Atomic Mass for Carbon Even though natural carbon does not contain a single atom with mass 12.01, for stoichiometric purposes, we can consider carbon to be composed of only one type of atom with a mass of 12.01. This enables us to count atoms of natural carbon by weighing a sample of carbon.

Example An element consists of 62.60% of an isotope with mass 186.956 amu and 37.40% of an isotope with mass 184.953 amu. Calculate the average atomic mass and identify the element. (186.956amu)(.6260) + (184.953amu)(.3740) = 186.2 amu, Rhenium

Example: Gallium has two naturally occurring isotopes, 69Ga and 71Ga, with masses of 68.9257amu and 70.9249amu, respectively. Calculate the percent abundances of these isotopes of gallium. 69.72amu = (68.9257amu)(1-x) + (70.9249amu)(x) x = 0.3973 Ga-71 = 39.73% Ga-69 = 60.27%

The Mole Section 3.3 – The Mole The number equal to the number of carbon atoms in exactly 12 grams of pure 12C. 1 mole of anything = 6.022 x 1023 units of that thing (Avogadro’s number). 1 mole C = 6.022 x 1023 C atoms = 12.01 g C Section 3.3 – The Mole

Example Calculate the number of iron atoms in a 4.48 mole sample of iron. 4.48 mole x 6.02 x 1023 atoms = 1 mole 2.70×1024 Fe atoms

Example A silicon chip has a mass of 5.68 mg. How many silicon atoms are present in the chip? 1 g . 1000mg 1 mole . 28.09 gSi 6.02x1023 atoms 1 mole 5.68 mg x x x = 1.22x1020 atoms Si

Molar Mass Section 3.4 – Molar Mass Mass in grams of one mole of the substance: Examples: Molar Mass of N = 14.01 g/mol Molar Mass of H2O = 18.02 g/mol (2 × 1.01 g) + 16.00 g Molar Mass of Ba(NO3)2 = 261.35 g/mol 137.33 g + (2 × 14.01 g) + (6 × 16.00 g) Section 3.4 – Molar Mass

Percent Composition (%) Mass percent of an element: mass % = mass of element in compound mass of compound x100 Section 3.6 – Percent Composition of Compounds

Example Calculate the mass percent for iron in iron(III) oxide, (Fe2O3): 2 (55.85g) 2 (55.85g) + 3 (16.00g) Mass % Fe = = 69.94% Fe

Section 3.7 – Determining Formula of a Compound Formulas Empirical formula = CH Simplest whole-number ratio Molecular formula = (empirical formula)n , [n = integer] Molecular formula = C6H6 = (CH)6 Actual formula of the compound Section 3.7 – Determining Formula of a Compound

Determination of Empirical and Molecular Formulas If given grams of each element, go to step 2. If given percent of each element, assume 100.0g of the compound (change % into grams). Convert g moles for each element using molar mass. Divide each mole answer by the smallest mole answer. If you do not have a whole number, multiply by the smallest whole # to get a whole # ratio. These whole #s are the subscripts for each element. To find the molecular formula, divide the molecular molar mass by the empirical molar mass. Then multiply each of the subscripts by that answer.

Example Calculation CH4 Determine the empirical formula of methane given that 6.00g of methane can be decomposed into 4.49g of carbon and 1.51g of hydrogen. 1 mol C 12.01gC .374 4.49gC 1.51gH x = = 1 .374 1 mol H 1.01 gH 1.50 x = = 4.01 ≈ 4 .374 CH4

Example P2O5 Empirical P4O10 Molecular 1 mol P 30.97gP 1.409 43.64gP A white powder is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass. The compound has a molar mass of 283.88g/mol. Determine the compound’s empirical and molecular formula. 1 mol P 30.97gP 1.409 43.64gP 56.36gO x = = 1 (2) = 2 1.409 P2O5 Empirical 1 mol O 16.00gO 3.523 x = = 2.500 (2) = 5 1.409 2P + 5O = 2(30.97) + 5(16.00) = 141.94g 283.88g 141.94g P4O10 Molecular = 2

Section 3.8 – Chemical Equations A representation of a chemical reaction: C2H5OH + 3O2  2CO2 + 3H2O reactants products Reactants are only placed on the left side of the arrow, products are only placed on the right side of the arrow. Section 3.8 – Chemical Equations

Section 3.9 – Balancing Chemical Equations C2H5OH + 3O2  2CO2 + 3H2O The equation is balanced. All atoms in the reactants are accounted for in the products. 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water. Section 3.9 – Balancing Chemical Equations

Animation Link: Balancing an Equation The balanced equation represents an overall ratio of reactants and products, not what actually “happens” during a reaction. Use the coefficients in the balanced equation to decide the amount of each reactant that is used, and the amount of each product that is formed.

Notice The number of atoms of each type of element must be the same on both sides of a balanced equation. Subscripts must not be changed to balance an equation. A balanced equation tells us the ratio of the number of molecules which react and are produced in a chemical reaction. Coefficients can be fractions, although they are usually given as lowest integer multiples.

Concept Check Which of the following are true concerning balanced chemical equations? There may be more than one true statement. I. The number of molecules is conserved. II. The coefficients tell you how much of each substance you have. III. Atoms are neither created nor destroyed. IV. The coefficients indicate the mass ratios of the substances used. V. The sum of the coefficients on the reactant side equals the sum of the coefficients on the product side.

Stoichiometric Calculations Chemical equations can be used to relate the masses of reacting chemicals. Section 3.10 – Stoichiometric Calculations

Calculating Masses of Reactants and Products in Reactions 1. Balance the equation for the reaction. 2. Convert the known mass of the reactant or product to moles of that substance. 3. Use the balanced equation to set up the appropriate mole ratios. 4. Use the appropriate mole ratios to calculate the number of moles of desired reactant or product. 5. Convert from moles back to grams if required by the problem.

Example Consider the following reaction: P4 (s) + 5 O2 (g)  2 P2O5 (s) If 6.25 g of phosphorus is burned, what mass of oxygen does it combine with? 6.25 gP4 1 1 mol P4 123.88gP4 5 mol O2 1 mol P4 32.00 gO2 1 mol O2 x x x = 8.07 gO2

Section 3.11 – Limiting Reagents Limiting Reactants Limiting reactant – the reactant that is consumed first and therefore limits the amounts of products that can be formed. Excess reactant – the reactant that is left over at the end of the reaction. Determine which reactant is limiting to calculate correctly the amounts of products that will be formed. Section 3.11 – Limiting Reagents

Example Consider the reaction of 80.0g of copper with 20.0 g of oxygen according to the following equation. What is the limiting reactant? 4 Cu + O2 → 2 Cu2O LR 80.0g Cu x 1 mole Cu 63.55g Cu = 1.26 mole Cu = 0.315 4 mol Cu 20.0g O2 x 1 mole O2 32.00g O2 = 0.625 mole O2 = 0.625 1 mol O2

Example If a sample containing 18.1g of NH3 is reacted with 90.4g of CuO, which is the limiting reactant and what mass of copper can be produced? The unbalanced equation is: NH3 + CuO → N2 + Cu + H2O 2 3 3 3 18.1g NH3 x 1 mole NH3 17.04g NH3 = 1.06 mole NH3 = 0.531 2 mol NH3 90.4g CuO x 1 mole CuO 79.55g CuO = 1.14 mole CuO = 0.379 LR 3 mol CuO Now use the LR to find the mass of Cu: 90.4g CuO x 1 mole CuO 79.55g CuO x 3 mole Cu 3 mole CuO x 63.55g Cu 1 mole Cu = 72.4 g Cu

To Calculate how much ER is left over: How much excess reactant remains after the reaction is complete? 2 NH3 + 3 CuO → N2 + 3 Cu + 3 H2O Use the LR to determine how much of the ER reacts with it. Subtract this from the original amount of the ER. 90.4g CuO x 1 mole CuO 79.55g CuO x 2 mole NH3 3 mole CuO x 17.04g NH3 1 mole NH3 = 12.9 g NH3 reacts 18.1g -12.9g = 5.2 g unreacted

Percent Yield Actual yield x 100 % yield = Theoretical yield An important indicator of the efficiency of a particular laboratory or industrial reaction. Theoretical yield is the amount of product formed when the LR is completely consumed according to stoichiometric calculations. Actual yield is the amount that is actually produced in a laboratory setting. % yield = Actual yield Theoretical yield x 100

Example Consider the following reaction: P4(s) + 6F2(g)  4PF3(g) What mass of P4 is needed to produce 85.0 g of PF3 if the reaction has a 64.9% yield? 85.0 gPF3 1 1 mol PF3 87.97gPF3 1 mol P4 4 mol PF3 123.88 gP4 1 mol P4 x x x = 29.9 gP4 64.9% = (85.0 g PF3 / x)(100); x = 130.97 g PF3 (130.97 g PF3)(1 mol PF3 / 87.97 g PF3)(1 mol P4 / 4 mol PF3)(123.88 g P4 / 1 mol P4) = 46.1 g of P4 needed x = 46.1 gP4 29.9 gP4 = 0.649 x