Report

Topic 3 Mass Relations and Stoichiometry Mass and Moles of a Substance Chemistry requires a method for determining the numbers of molecules in a given mass of a substance which has lead to the development of the mole (quantity of substance to be discussed later). – This allows the chemist to carry out “recipes” for compounds based on the relative numbers of atoms involved. – stoichiometry - the calculation involving the quantities of reactants and products in a chemical equation. 2 Molecular Weight and Formula Weight The molecular weight (for molecular substances) of a substance is the sum of the atomic weights of all the atoms in a molecule of the substance. – For, example, a molecule of H2O contains 2 hydrogen atoms (at 1.01 amu each) and 1 oxygen atom (16.00 amu), giving a molecular weight of 18.02 amu. H: 2 x 1.01 amu = 2.02 amu O: 1 x 16.00 amu = 16.00 amu 18.02 amu (mass of 1 molecule of H2O) 3 Molecular Weight and Formula Weight The formula weight of a substance is the sum of the atomic weights of all the atoms in one formula unit of the compound, whether molecular or not. one formula unit (FU) of NaCl: Na: 1 x 22.99 amu = 22.99 amu Cl: 1 x 35.45 amu = 35.45 amu 58.44 amu (mass of 1 FU NaCl) iron (III) sulfate, Fe2(SO4)3 Fe: 2 x 55.85 amu = 111.70 amu S: 3 x 32.07 amu = 96.21 amu O: 12 x 16.00 amu = 192.00 amu 399.91 amu (mass of 1 FU Fe2(SO4)3) Glucose, C6 H12 O6 Note: molecular wt and formula wt are calculated the same way. C: 6 x 12.01 amu = 72.06 amu H: 12 x 1.01 amu = 12.12 amu O: 6 x 16.00 amu = 96.00 amu 180.18 amu 4 (mass of 1 molecule) Mass and Moles of a Substance The Mole Concept – A mole is defined as the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 grams of carbon–12. – The number of atoms in a 12-gram sample of carbon–12 is called Avogadro’s number (Na). The value of Avogadro’s number is 6.022 x 1023. 1 mole = 6.022 x 1023 X X = ions, particles, atoms, molecules, items, etc. 1 mole Na2CO3 6.022 x 1023 FU Na2CO3 1 mole CO2 6.022 x 1023 molecules CO2 5 Mass and Moles of a Substance The molar mass (Mm) of a substance is the mass of one mole (6.022 x 1023 FU or molecules) of a substance. This is the term we will use the most in the course and is done the same way as formula wt. except using gram instead of amu. – For all substances, molar mass, in grams per mole, is numerically equal to the formula weight in atomic mass units. – That is, one mole of any element weighs its atomic mass in grams. 1 molecule of H2O - MW = 18.02 amu 1 mole of H2O - Mm = 18.02 g H2O 6.022 x 1023 molecules Note: we can use Mm as a conversion factor between grams and mols. Mm of 1 mole of MgSO4 . 7H2O Mg: 1 x 24.31 S: 1 x 32.07 O: 4 x 16.00 H:14 x 1.01 O: 7 x 16.00 g= g= g= g= g= 24.31 g 32.07 g Note: must account for 64.00 g mass of water in hydrates 14.14 g 112.00 g 6 . 246.52 g/mol MgSO4 7H2O How is it possible that Mm and FW/MW are the same value but different units?Avogadro’s number and amu are both based on carbon-12 and inverse values of each other. Let’s convert 1 Na atom g/mol A.)What is the mass in grams of one Cl atom? B.)in one HCl molecule? Notes: Converting mass to mols or vice versa will require using molar mass. Converting mass to atoms, molecules, ions, etc. or vice versa will require going through mols by using Avogadro’s number. 7 Mass and Moles of a Substance Mole calculations – Converting the number of moles of a given substance into its mass, and vice versa, is fundamental to understanding the quantitative nature of chemical equations. mass of " A" moles of " A" molar mass of " A" Converting mass to mols or vice versa will require using molar mass. Mm NaCl = 58.44 g/mol 58.44 g NaCl 1 mol NaCl 1 mol NaCl 58.44 g NaCl Note: we can use Mm as a conversion factor between grams and mols with g in numerator or denominator. 8 Mass and Moles of a Substance Mole calculations Suppose we have 5.75 moles of magnesium. What is its mass? Note: we use Mm as a conversion factor between grams and mols with mol in this case being placed in the denominator to cancel with the 5.75 mols we are converting to g. 9 Mass and Moles of a Substance Mole calculations Suppose we have 100.0 grams of H2O. How many moles does this represent? 10 Mass and Moles and Number of Molecules or Atoms The number of molecules or atoms in a sample is related to the moles of the substance by Avogadro’s #: 1 mole HCl 6.02 10 HCl molecules 23 1 mole Fe 6.02 10 23 Fe atoms How many molecules are there in 56 mg HCN? HW 20-22 code: moles 11 Determining Chemical Formulas The percent composition of a compound is the mass percentage of each element in the compound. – We define the mass percentage of “A” as the parts of “A” per hundred parts of the total, by mass. That is, mass of " A" in whole mass % " A" 100% mass of the whole 20 g A 20 % A 100g sample If given the %A by mass, it is useful to put the percentage in ratio form for dimensional analysis calculations. 12 Mass Percentages from Formulas Let’s calculate the percent composition (%C, %H) of one molecule of butane, C4H10. First, we need the molecular wt of C4H10. 4 carbons @ 12.01 amu/atom 48.04 amu 10 hydrogens @ 1.01 amu/atom 10.10 amu 1 molecule of C 4 H 10 58.14 amu Now, we can calculate the percents (basically, part/whole). %C %H 48.04 amu C 58.14 amu total 10.10 amu H 58.14 amu total 100 % 82 .63 % C 100 % 17 .37 % H What if we wanted the %C & %H for 1 mol of butane? Same calculation and results except we would use molar mass, g/mol, instead of amu/atom. Note: % is a unit like g, etc. Don’t use the % button on calculator. 13 How many grams of carbon are there in 83.5 g of formaldehyde, CH2O, (40.0% C, 6.73% H, 53.3% O)? rewrite % into ratio = 33.4 g C 14 An unknown acid contains only C, H, O. A 4.24 mg sample of acid is completely burned. It gives 6.21 mg of CO2 and 2.54 mg H2O. What is mass% of each element in the unknown acid? goal of problem O2 %C, H, O in sample unknown acid CO2 + H2O C: has C, H, O 4.24 mg 6.21 mg 2.54 mg mol to mol ratio needed to convert from one species to another Note: We did the same calculation for H as we did for C except we did it in terms of mg and mmol to demonstrate how taking advantage of the prefix can simplify your work by adding m to the mols and g of the molar mass conversion factor (mmol/mg) and mol to mol ratio (mmol/mmol). This eliminates the useless conversion from mg to g and back again. H: O: Total mass = 4.24 mg = mass C + mass H + mass O = 1.69 mg + 0.285 mg + mass O mass O = 4.24 mg – 1.69 mg – 0.285 mg = 2.26 mg O %C: %H: %O: 100% - 39.9% - 6.72% = 53.4% O Alternatively, we could have just subtracted the 15 percent C and H from 100% to determine the %O. Determining Chemical Formulas We can determine the formula of a compound from the percent composition. – The percent composition of a compound leads directly to its empirical formula. – An empirical formula (or simplest formula) for a compound is the formula of the substance written with the smallest integer (whole number) subscripts. – From the empirical formula, we can find the molecular formula of a substance which will be a multiple of the empirical formula based on its 16 molar mass. Determining Chemical Formulas Benzoic acid is a white, crystalline powder used as a food preservative. The compound contains 68.8% C, 5.0% H, and 26.2% O by mass. What is its empirical formula? In other words, give the smallest whole-number ratio of the subscripts in the formula based on mols (C:H:O) Cx HyOz mols 17 Determining Chemical Formulas For the purposes of calculations of this type, we will assume we have 100.0 grams of sample; therefore, benzoic acid at 68.8% C means which means the mass of each element equals the numerical value of the percentage. Since x, y, and z in our formula represent mole-mole ratios, we must first convert these masses to moles. Cx HyOz 18 Determining Chemical Formulas Our assumed 100.0 grams of benzoic acid (68.8%C, 5.0%H, 26.2%O) would contain: 1 mol C 68.8 g C 5.72(8) mol C 12.01 g 1 mol H 5 .0 g H 4.9(5) mol H 1.01 g 1 mol O 26.2 g O 1.63(7 ) mol O 16.00 g This isn’t quite a whole number ratio, but if we divide each number by the smallest of the three, a better ratio will emerge. We are dividing by the smallest number to get the species all on the same scale – “normalizing” the values. Note: 68.8g + 5.0 g + 26.2 g = assumed 100.0 g sample 19 Determining Chemical Formulas Next step is to get all species on the same scale, basically normalizing to moles of smallest species (in this case, 1.63 mols of O) for the purpose of getting values that can easily be converted to whole number ratios. 1 .63( 7 ) m ol O 1.63(7) 1.00 5 .728 m ol C 1.63(7) 3.50 4 .95 m ol H 1.63(7) 3.0 2 x C3.5 H3 O1 C7H6O2 now it’s not too difficult to see that the smallest whole number ratio is 7:6:2 (multiple everything by 2 to get whole number. The empirical formula is C7H6O2 . This is the empirical as well as molecular formula of benzoic acid because the molar mass of the empirical formula is the same as the molecular formula of 20 benzoic acid. Determining Chemical Formulas An empirical formula gives only the smallest whole-number ratio of atoms in a formula. The molecular formula should be a multiple of the empirical formula (since both have the same percent composition). C 2 H 3 O2 empirical formula (lowest whole # ratio) C 4 H 6 O4 molecular formula (multiple of 2 x emp) C8H12O8 molecular formula (multiple of 4 x emp) Which is not an empirical formula meaning not lowest whole #? CH4 CH4O C2H4O2 C2H6O C2H4O2 is divisible by 2; CH2O would be its empirical formula To determine the molecular formula, we must know the molecular weight (molar mass) of the compound. 21 Determining Chemical Formulas Determining the molecular formula from the empirical formula. molar mass = n x empirical formula mass where n is the multiple factor n = molar mass empirical mass 22 Acetic Acid contains 39.9% C, 6.7% H, 53.4% O. Determine empirical formula. The molecular mass of acetic acid is 60.0 g/mol. What is the molecular formula? We assume 100.0 g sample meaning we have 39.9 g C, 6.7 g H, 53.4 g O and calculate the mols of each. Next, we get them on same scale by dividing each by the smallest mols, 3.33 mols empirical formula = CH2O Mm= 30.03 g/mol 2x HW 23 code: formula CH2O C2H4O2 23 Stoichiometry: Quantitative Relations in Chemical Reactions Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction. – It is based on the balanced chemical equation and on the relationship between mass and moles (molar mass is an important concept here; g mol, mol g) and mol to mol ratios. – Such calculations are fundamental to most quantitative work in chemistry. What we are discussing deals with mol to mol ratios (very important concept) and is the basis of many calculations. 24 Molar Interpretation of a Chemical Equation The balanced chemical equation can be interpreted in numbers of molecules, but generally chemists interpret equations as “mole-to-mole” relationships. Lets look at the following reaction: N 2 (g) 3H 2 (g) 2 NH 3 ( g ) 1 molecule N2 + 3 molecules H2 1 mol N 2 3 mol H 2 28.02 g N2 + 3(2.02 g) H2 2 (17.04 g) NH3 28.02 g 6.06 g 34.08 g = 34.08 g + 34.08 g 2 molecules NH3 2 mol NH 3 If the number of each particular atom is the same on both sides, the Law of Conservation of Mass will be preserved and the mass of reactants will be equal to the mass of the products. 25 Molar Interpretation of a Chemical Equation Suppose we wished to determine the number of moles of NH3 we could obtain from 4.8 mol H2. We will assume N2 is in excess which means H2 is limited and will dictate the amount of product that will be formed; once the H2 is gone, the reaction will stop. N 2 (g) 3H 2 (g) 2 NH 3 ( g ) We should realize that we have conversion factors that exist from the mol to mol ratios available from the balanced equation to go from one species to another. 26 Mass Relationships in Chemical Equations How many grams of HCl are required to react with 5.00 grams MnO2 according to this equation? 4 HCl(aq) MnO 2 (s ) 2 H 2O(l) MnCl 2 (aq) Cl 2 ( g ) ?g 5.00 g Note: To get from one substance to another requires the use of mol to mol ratios. 27 Mass Relationships in Chemical Equations How many grams of CO2 gas can be produced from 1.00 kg Fe2O3? Fe2O3 (s) + 3 CO (g) 2 Fe (s) + 3 CO2 (g) 1.00 kg assume excess ?g HW 24 code: stoich 28 Limiting Reagent • The limiting reactant (or limiting reagent) is the reactant that is entirely consumed when the reaction goes to completion. • The limiting reagent ultimately determines how much product stops when one component , can be obtained. R + R P Reaction limiting reagent, is exhausted – For example, bicycles require one frame and two wheels. If you have 20 wheels but only 5 frames, how many bicycles can be made? oo oo oo oo oo oo oo oo oo oo 20 wheels 10 wheels in excess; therefore, frames are limiting factor 1 frame + 2 wheels 1 bike Smaller value is correct answer or you can determine which is the limiting reagent and only calculate that value. Basically, we must find out which species will run out first because it will dictate how much 29 product will form. Limiting Reagent If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, how many grams of H2 are produced? Zn(s) 2 HCl(aq) ZnCl 2 (aq) H 2 ( g ) 0.30 mol 0.52 mol ?g First, we calculate how many mols of H2 is possible for each reactant. limiting reagent HCl We use the mols of H2 possible from the limiting reagent, HCl, to determine the mass of H2. The maximum product possible is based on the limiting reagent; smaller mols of product is maximum possible based on the limiting reagent which is HCl in this case. Note: The limiting reagent is not necessarily the species with the smaller mass or 30 #mols; you must account for the mol-to-mol ratio as demonstrated in this problem. If 7.36 g Zn was heated with 6.45 g sulfur, what amount of ZnS was produced? 8 Zn(s) S 8 8 ZnS 7.36 g mols produced by each reactant: 6.45 g ?g smaller mols; limiting reagent This is known as the theoretical yield. 31 Theoretical and Percent Yield The theoretical yield of product is the maximum amount of product that can be obtained from given amounts of reactants. The percentage yield is the actual yield (experimentally determined) expressed as a percentage of the theoretical yield (calculated like in previous example). actual yield (exp) %Yield 100% theoretical yield (calc) Once again, % is a unit. 32 Theoretical and Percent Yield To illustrate the calculation of percentage yield, recall that the theoretical yield of ZnS in the previous example was 11.0 g ZnS. If the actual yield experimentally for the reaction is 9.32 g ZnS, what is the %yield? experimental value 9.32 g ZnS % Yield 100% 84.7% 11.0 g ZnS % is unit, not button theoretical value; calculated value on calculator 33 If 11.0 g CH3OH are mixed with 10.0 g CO, what is the theoretical yield of HC2H3O2 in the following reaction? If the actual yield was 19.1 g, what is the %yield of HC2H3O2? CH3OH (l) + CO (g) HC2H3O2 (l) 11.0 g 10.0 g mols produced by each reactant: ?g smaller mols; limiting reagent theoretical yield experimental value theoretical value; calculated value HW 25 code: limiting 34