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Mass Relationships in Chemical Reactions Chapter 3 HW 39-53 odd, 59, 66, 69,75, 78, 83, 86, 93-94 Micro-World atoms & molecules Macro-World grams Atomic mass is the mass of an atom in atomic mass units (amu) By definition and International agreement: 1 atom 12C “weighs” 12 amu On this scale 1H = 1.008 amu - which is 8.400% as massive 16O = 16.00 amu - which is 133.33% as massive 2 The average atomic mass is the weighted average of all of the naturally occurring isotopes of the element. 12121212121212121213 12121212121212121212 12121212121212121212 12121212121212121212 12121212121212121212 12121212121212121212 12121212121212121212 12121212121212121212 12121212121212121212 12121212121212121212 Carbon average atomic mass = 0.9890 * 12 amu + 0.0110 * 13.00335 amu = 12.01 amu *98.90% = 0.9890 Average Atomic Mass Calculation A random sample of Magnesium is composed of • 79.0% Mg-24 , 10.0% Mg-25 , and 11.0% Mg-26 If 100 atoms of Mg were randomly selected, 79 atoms would have 12 neutrons (N0), 10 atoms would have 13 N0, and 11 atoms would have 14 N0 • Find the weighted average for the mass of Magnesium by writing percentage as a decimal (e.g. 75% = 0.75) (0.790 x 24) + (0.100 x 25) + (0.11 x 26) = 24.3 amu Average Atomic Mass Calculation A random sample of chlorine is • 75.8% Cl-35 (34.97 amu) and • 24.2% Cl-37 (36.97 amu) • Find the weighted average for the mass of chlorine by writing percentage as a decimal (e.g. 50% = 0.50) (0.758 x 34.97) + (0.242 x 36.97) = 35.45 amu Example 3.1 Copper, a metal known since ancient times, is used in electrical cables and pennies, among other things. The atomic masses of its two stable isotopes: (62.93 amu) 69.09% (64.9278 amu) 30.9% Calculate the average atomic mass of copper. The relative abundances are given in parentheses. Example 3.1 Solution First the percents are converted to fractions: 69.09 percent to 69.09/100 or 0.6909 30.91 percent to 30.91/100 or 0.3091. We find the contribution to the average atomic mass for each isotope, then add the contributions together to obtain the average atomic mass. (0.6909) (62.93 amu) + (0.3091) (64.9278 amu) = 63.55 amu *Always average atomic mass shown of relative abundant isotopes The mass of a single carbon atom in grams: 0.0000000000000000000000199 The Mole (mol): A grouping unit to count numbers of particles Dozen = 12 Pair = 2 The mole (mol) is the amount of a substance that contains as many elementary units as there 12 are atoms in exactly 12.00 grams of C Some BIG Numbers… World population: ~7,000,000,000 Stars in our galaxy: ~300,000,000,000 Cells in our body: ~50,000,000,000,000 Avogadro’s Constant (number) 602,214,150,000,000,000,000,000 ~0.6 Septillion 6.02 x 1023 1 mole = NA = 6.02 x Jean Baptist Perrin 23 10 23 6.02*10 grains of sand could cover Texas in almost 3 feet 268,820 of sand miles2 23 6.02*10 © Starbursts would cover the surface of the entire US stacked and be 24.2 miles tall The Moon 7.3 × 1022 kg 6.02*1023 moles (mammal) Molar eggs shoes mass is the mass of 1 mole of marbles in grams atoms 1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g 1 12C atom = 12.00 amu 12.00 g 12C = 1 mole 12C atoms 65.39 g of Zn = 1 mole Zinc atoms 18.015 g of water = 1 mole water molecules 58.443 g of NaCl = 1 mole NaCl formula units *Read the masses on the periodic table as grams/mole 1 mol = NA = 6.02 x 23 10 1 12C atom 12.00 amu 12.00 g x 6.022 x 1023 12C atoms 1.66 x 10-24 g = 1 amu 1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu M = molar mass in g/mol NA = Avogadro’s number One Mole of: S C Mercury Copper Iron How to “Read” Chemical Equations 2 Mg + O2 2 MgO 2 atoms Mg + 1 molecule O2 makes 2 units of MgO or... (Scale everything up by NA) 2 moles Mg + 1 mole O2 makes 2 moles MgO so (using the Periodic Table we have) 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO MM: Molar mass (grams/mol) Avogadro’s constant NA = 6.02 x 1023 Particles Atoms Molecules Formula units Remember Train Track Unit Conversions How many inches are in 3 miles? Need conversion factors: 1 mile = 5,280 ft; 1 foot = 12 in 3 miles 1 5280 feet 1 mile 12 inches 1 foot Multiply all numbers on the top Divide all numbers on the bottom 3 x 5280 x 12 1 x 1x 1 = 190,080 inches Chemistry Conversion Factors MM = molar mass in g/mol Converts between mass (g) & moles 23 10 NA = Avogadro’s number = 6.02 x Converts between moles & particles (atoms / molecules / formula units) Example How many moles of Helium atoms are in 6.46 g of He? Molar mass is the conversion factor and is found on the periodic table to be 4.003 g/mol. This can be written as 1 mol He = 4.003 g He Example How many grams of zinc are in 0.356 mole of Zn? The conversion factor is the molar mass of Zn. From the periodic table we find that 1 mol Zn = 65.39 g Zn Zinc Example How many atoms are in 16.3 g of Sulfur? Grams of S → moles of S → # atoms S MM: 1 mol S = 32.07 g 1 mol S = 6.02 x 1023 S atoms More Practice Conversions Convert 58 grams Calcium to moles 58 grams Ca 1 mol Ca x = 40.1 grams Ca 1.45 moles Ca Convert 0.2 moles Nickel to grams 0.2 moles Ni 58.7 grams Ni x = 1 moles Ni 11.7 grams Ni Convert 54 grams C to moles atoms 54 g C 6.02 x 1023 C atoms 1 mol C = x x 12 grams C 1 mol C 2.7 x 1024 C atoms Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. 1S SO2 2O SO2 32.07 amu + 2 x 16.00 amu 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2 Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. NaCl 1Na 22.99 amu 1Cl NaCl + 35.45 amu 58.44 amu For any ionic compound formula mass (amu) = molar mass (grams) 1 formula unit NaCl = 58.44 amu 1 mole NaCl = 58.44 g NaCl Example 3.5 Determine Molar Mass Calculate the molecular or formula masses (in g/mol) of the following compounds: (a) Water (H2O) (b) Isopropanol (C3H8O1) (rubbing alcohol) (c) (NH4)2S (d) Mg(NO3)2 (e) caffeine (C8H10N4O2) Example 3.5 Molar Mass Solutions a) H2O: 2*(1.01 g) + 1*(16.00 g) = 18.02 grams b) C3H8O1: 3*(12.01 g) + 8*(1.01 g) + 1*(16.00 g) = 60.05 grams c) (NH4)2S : 2*(14.01 g) + 8*(1.01 g) + 1*(32.07 g) = 68.17 grams d) Mg(NO3)2: 1*(24.31 g) + 2*(14.01 g) + 6*(16.00 g) = 148.33 grams e) C8H10N4O2: 8*(12.01 g) + 10*(1.01 g) + 4*(14.01 g) + 2*(16.00 g) = 194.20 grams Each Molar mass can be read with units “grams/mole” Example Methane (CH4) is the principal component of natural gas. How many moles of CH4 are present in 6.07 g of CH4? Molar mass of CH4 = 12.01 g + 4*(1.008 g) = 16.04 g 3.7 Example urea How many hydrogen atoms are present in 25.6 g of urea [(NH2)2CO] The molar mass of urea is 60.06 g. *Note there are 4 hydrogen atoms in each Urea molecule grams of urea → moles of urea → molecules of urea → atoms of H 25.6 g of urea x 1 mol of urea 60.06 g urea x 6.02 x 1023 molecules 1 mol of urea x 4 Hydrogen atoms 1 molecule of urea = 1.03 × 1024 H atoms Practice Conversions How many atoms of P are found in 6.30 mg of Mg3(PO4)2? What is the mass of 3.4 x Boron? 25 10 atoms of What is the % composition of Aluminum in Galaxite (MnAl2O4)? How many grams in 35.6 g galaxite? -Invented by Francis Aston in the 1920’s -Separates particles by e/m (charge/mass) -First demonstration of isotopes (Ne 20 & 22) -Led to precise determination of atomic masses and relative isotope abundancies. 1922 Nobel Prize in Chemistry Mass Spectrum of Ne Heavy Light Heavy Sample gets ionized Light Mass Spectrometer Percent composition of an element in a compound = n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound %C = %H = %O = Ethanol: C2H5OH Molecular weight: 46.07 g 2 x (12.01 g) 46.07 g 6 x (1.008 g) 46.07 g 1 x (16.00 g) 46.07 g x 100% = 52.14% x 100% = 13.13% x 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.0% Example 3.8 Percent Composition Phosphoric acid (H3PO4) is a colorless, syrupy liquid used in detergents, fertilizers, toothpastes, and in carbonated beverages for a “tangy” flavor. Calculate the percent composition by mass of H, P, and O in this compound. Example 3.8 Solution Solution The molar mass of H3PO4 is 97.99 g. The percent by mass of each of the elements in H3PO4 is calculated as follows: *Always make sure your numbers add back up to 100% Example 3.10 Compound Mass Percentages Chalcopyrite (CuFeS2) is a principal mineral of copper. Calculate the number of kilograms of Cu in 3.71×103 kg of chalcopyrite using percent compositions. Chalcopyrite. How do we calculate mass percent of an element? mass % = (Element Mass)*(Quantity) x 100% Compound Mass Example 3.10 Solution The molar masses of Cu and CuFeS2 are 63.55 g and 183.5 g, respectively. The percent composition of Cu is therefore: To calculate the mass of Cu in a 3.71 × 103 kg sample of CuFeS2, we need to convert the percentage to a fraction (that is, convert 34.63 percent to 34.63/100, or 0.3463) and write: mass of Cu in CuFeS2 = 0.3463 × (3.71 × 103 kg) = 1.28 × 103 kg Example Compound Mass Percentages Talc is a mineral composed of hydrated magnesium silicate with the chemical formula H2Mg3(SiO3)4 Determine the number of grams of pure Magnesium we could extract from 45.0 grams of the talc mineral using percent composition. (Mg Mass)*(3) % Mg = Total Mineral Mass x 100% 72.93 g Mg % Mg = = 19.23% Mg x 100% 379.29 g Mineral g Mg = % Mg 45.0 gram compound = 0.1923 x 45.0 g = 8.65 g Mg Percent Composition and Empirical Formulas Step 1&2 Step 3 Step 4 Early analysis of many molecules originally only yielded % composition Elemental analysis of compounds reveals the mass % of each element This allows us to derive the empirical formula (atom to atom ratio) Determining Empirical Formulas When the element % composition is known: 1. Assume 100g sample and change element percents to grams 2. Convert each to moles by dividing by molar mass of each atom 3. Divide each number by the smallest number If all Numbers are integers (or close) you are done; Put in formula 4. If not, multiply all numbers by the SAME smallest whole number so that all numbers are whole. We don’t expect numbers to be perfect integers based on experimental error, Round only if within one tenth. Empirical Formula determination An analysis of an unknown gas has determined that it contains 72.55% Oxygen and 27.45% Carbon by mass. What is the empirical formula of the compound? C#O# 72.55 g O 1 mol O 16 g O 27.45 g C 1 mol C 12 g C 4.53/2.29 = 1.98 = 4.53 mol O ~2O 2.29/2.29 = 1 C = 2.29 mol C CO2 Example 3.9 Determine the empirical formula of Ascorbic acid (vitamin C: cures/prevents scurvy). It is composed of 40.92% C, 4.58% H, and 54.50% O by mass. To determine the empirical formula, we will first assume a 100 g sample of Ascorbic Acid Example 3.9 Solution If we have 100 g of ascorbic acid, then each percentage can be converted directly to grams and from grams to moles of each element. Example 3.9 However, chemical formulas are written with whole numbers. Try to convert to whole numbers by dividing all the subscripts by the smallest subscript (3.406): where the sign means “approximately equal to.” This gives CH1.33O as the formula for ascorbic acid. Next, we need to convert 1.33, the subscript for H, into an integer. Multiply by the smallest whole number to obtain all integers. Example 3.9 This can be done by a trial-and-error procedure: 1.33 × 1 = 1.33 1.33 × 2 = 2.66 1.33 × 3 = 3.99 ~ 4 Because 1.33 × 3 gives us an integer (4), we multiply all the subscripts by 3 and obtain C3H4O3 as the empirical formula for ascorbic acid. Combust 11.5 g ethanol Collect 22.0 g CO2 and 13.5 g H2O “desiccant” SiO2 “scrubber” LiOH g C = % C x 22.0 g CO2 27.3% x 22.0 g = 6.0 g C 6.0 g C = 0.5 mol C g H = % H x 13.5 g H2O 11.1% x 13.5 g = 1.5 g H 1.5 g H = 1.5 mol H g O = (11.5 g sample) – (6 g C + 1.5 g H) = Empirical formula C0.5H1.5O0.25 Divide by smallest subscript (0.25) Empirical formula C2H6O 4.0 g O = 0.25 mol O The molecular formula can be determined with both % Composition and Molar Mass A sample of a compound contains 30.46% Nitrogen and 69.54% Oxygen by mass In a separate experiment, the molar mass of the compound is estimated to be between 90 g and 95 g. Determine the molecular formula and the accurate molar mass of the compound. Example We first assume 100 g and convert % to grams. We then convert each element from grams to moles. To convert to whole numbers we divide each mole by the smaller subscript (2.174). After rounding off, we obtain NO2 as the empirical formula. The Molecular formula must have a Molar Mass that is a multiple of the Empirical formula’s Molar Mass (NO2, N2O4, N3O6, N4O8…) Example Empirical molar mass = 14.01 g + 2(16.00 g) = 46.01 g Next, we determine the ratio between the molar mass and the empirical molar mass The molar mass is twice the empirical molar mass. This means that there are two NO2 units in each molecule of the compound, and the molecular formula is N2O4. Chemical Reaction: A process in which one or more substances is changed into one or more new substances. A chemical equation uses chemical symbols to show what happens during a chemical reaction: reactants products 3 ways of representing the reaction of H2 with O2 to form H2O Equations Symbols Symbol Meaning + “and” “yields” or “produces” → ↔ Reversible or (g) Gas Vapor, H2O(g) (l) Liquid Water, H2O(l) (s) Solid Ice, H2O(s) (aq) Aqueous (dissolved in H2O) NaCl(s) + H2O(l) Na+(aq) + Cl-(aq) + H2O(l) Phase State : represent what state of matter a substance is in. How to “Read” Chemical Equations 2 Mg + O2 2 MgO 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO Or 2 moles Mg + 1 mole O2 makes 2 moles MgO Or 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO NOT 2 grams Mg + 1 gram O2 makes 2 g MgO We can not compare mass to mass, it must be in particles or Moles In 1 mole of HBr there are 80 grams of Br to 1 gram H, Yet still 1 atom Br to 1 atom H Law of Conservation of Matter/Mass • Chemical reactions only rearrange bonded atoms with no atoms created or destroyed • Every atom from the reactants must be equal in the products Fire converts/Not destroys 2C2H6 + 7O2 → 4CO2 + 6H2O Reactants 4 Carbons 12 Hydrogens 14 Oxygens = Products 4 Carbons 12 Hydrogens 14 Oxygens Balancing Chemical Equations 1. Write the correct formula(s) for the reactants on the left side and the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C2H6 + O2 CO2 + H2O We only change the coefficients (number in front) to make the number of atoms of each element the same on both sides. Never change the subscript (that would change the chemicals) 2C2H6 NOT C4H12 Balancing Chemical Equations 2. Count the number of atoms on each side. Check to see if it’s already balanced. LiBr + KNO3 → LiNO3 + KBr • On the reactant side, there is 1 Li, 1 Br, 1 K,1 N, and 3 O atoms. • On the product side, there is 1 Li, 1 Br, 1 K,1 N, and 3 O atoms. • This equation is balanced. Balancing Chemical Equations 3. Leaving Hydrogen and Oxygen till last,place a coefficient to balance one element. Continue balancing the others elements with coefficients __H2SO4 + __NaOH → __Na2SO4 + __H 2 2 2O Reactants Reactants Sulfur 11 Sulfur Sodium 21 Sodium Hydrogen 43 Hydrogen Oxygen 65 Oxygen == Products 1 Sulfur 2 Sodium 24 Hydrogen 56 Oxygen Combustion of Glucose 6 2 → __CO 6 6 __C6H12O6 + __O + __H 2 2O The balanced equation shows it take 6 molecules of O2 to react with 1 molecule glucose to produce 6 molecules of OR CO2 and H2O each. 6 mole of O2 reacts with 1 mole glucose to produce 6 moles of CO2 and H2O each. Reactants 6 Carbon 12 Hydrogen 188 Oxygen = Products 61 Carbon 122 Hydrogen 13 183 Oxygen Balance these Chemical equations: • H3PO4 + NaOH → Na3PO4 + H2O • Mg + O2 → MgO • NH4NO2 → N2 + H2O • P2O5 + H2O → H3PO4 • NaHCO3 → Na2CO3 + CO2 + H2O • Li + H2O → LiOH + H2 Balancing Difficult Chemical Equations Equations can be difficult if the reactants don’t react in a simple 1:2, 1:3, 1:4 etc. ratio. C2H6 + O2 CO2 + H2O C2H6 + O2 2CO2 + H2O C2H6 + O2 2CO2 + 3H2O 2 oxygen atoms *multiply CO2 by 2 *multiply H2O by 3 7 oxygen atoms Because it’s O2, no whole number coefficient can give us 7 atoms on the reactants (only multiples of 2) Balancing Chemical Equations C2H6 + _O2 2CO2 + 3H2O 7 oxygen atoms Balance using the smallest fraction possible that would give the desired quantity *multiply O2 by 7 to balance 2 7 C 2H6 + O2 2 2CO2 + 3H2O Because each particle must be an integer, remove the fraction by multiplying the whole reaction by the denominator x2 2C2H6 + 7O2 4CO2 + 6H2O Example 3.12 When aluminum metal is exposed to air, a protective layer of aluminum oxide (Al2O3) forms on its surface. This layer prevents further reaction between Al and O2, and it is why Al does not corrode. Write a balanced equation for the formation of Al2O3. An atomic scale image of aluminum oxide. Use the simplest fraction to balance Oxygen Example 3.12 Multiplying both sides of the equation by 2 gives whole-number coefficients. Check The equation is balanced. Also, the coefficients are reduced to the simplest set of whole numbers. Stoichiometry Stoichiometry is the quantitative relationship found between reactants and products in a reaction. • 2 hydrogen molecules and 1 oxygen molecules produce 2 water molecules Multiply all by Avogadro's # • 2 moles hydrogen and 1 mole oxygen produce 2 moles water Stoichiometry Stoichiometry is the "recipe of a reaction". From the equation we see that 1 mole CH4 2 mole O2 = “Stoichiometric equivalents” 1x 2x 2x 1x Mole Ratio 1C2H6O + 3O2 → 2CO2 + 3H2O • The coefficients in a balanced reaction tell us the mole ratio. • It takes 1 mole of ethanol to react with 3 moles of oxygen. This produces 2 moles of carbon dioxide and 3 moles of water. • The mole ratio will act as our conversion factor for calculations. Combustion of Ethanol (C2H6O) 1C2H6O + 3O2 → 2CO2 + 3H2O • If we have 4 moles of ethanol, how many moles oxygen are needed for the reaction? 3 mol oxygen 4 mol ethanol x = 12 mol oxygen 1 mol ethanol Mole ratio conversion factor • How many moles CO2 are made from reacting 0.6 moles ethanol? 0.6 mol ethanol x 2 mol CO2 1 mol ethanol = 1.2 mol CO2 Mole ratio conversion factor Calculations can go forward or backward, always looking at the mole to mole ratio 2C2H6 + 7O2 → 4CO2 + 6H2O How many moles of oxygen does it take to produce 3.5 moles of water (with excess ethane)? 3.5 mol water x 7 mol oxygen 6 mol water = 4.1 mol oxygen How many moles of ethane (C2H6) does it take to produce 3.5 moles of water (with excess oxygen)? 3.5 mol water x 2 mol ethane 6 mol water = 1.2 mol ethane Amounts of Reactants and Products 1. 2. 3. 4. Write balanced chemical equation Convert quantities of known substances into moles Use coefficients in balanced equation to calculate the number of moles of the sought quantity Convert moles of sought quantity into desired units Example 3.14 Mass to Mole to Mole to Mass (Products) to (Reactants) All alkali metals react in water to produce hydrogen gas and the corresponding alkali metal hydroxide. How many grams of Li are needed to produce 9.89 g of H2? Lithium reacting with water to produce hydrogen gas. Example 3.14 Solution ? g of Li 9.89 g of H2 How many grams of water are produced from the combustion of 35 g of propane? (C3H8) 35 g C3H8 x 1 mol C3H8 x 4 mol H2O x 18 g H2O = 57.3 44 g C3H8 1 mol C3H8 1 mol H2O g H2O How many grams of CO2 are produced from the combustion of 35 g of propane? 35 g C3H8 x 1 mol C3H8 x 3 mol CO2 x 44 g CO2 = 105 g 44 g C3H8 1 mol C3H8 1 mol CO2 CO2 *Note: Different amounts of each product are made by the same amount of reactant Practice problems • How many grams are produced of each product in the following decomposition reactions? 2.8 g 2KClO3 → 2KCl + 3O2 12.0 g CaCO3 → CaO + CO2 • How many grams of H2O are required for complete reaction with 37 g of the other reactant? K2O + SiCl4 + H2O → 4H2O → 2KOH H4SiO4 + 4HCl • How many grams of Aluminum are required to obtain 100 g Iron? 2Al + 3FeO → Al2O3 + 3Fe ? g of Al 100 g of Fe 3.13 Example The food we eat is broken down, in our bodies to provide energy. The overall equation for this complex process represents the degradation of glucose (C6H12O6). If 86 g of C6H12O6 is consumed by a person, what is the mass of CO2 produced? (180 g/mol) (1 mol C6H12O6 6 mol CO2) 86 g Glucose x (44 g/mol) 1 mol Glucose x 6 mol CO2 x 44 g CO2 = 126 g CO 2 180 g Glucose 1 mol Glucose 1 mol CO2 Stoichiometry: Chemistry for Massive Creatures - Crash Course Chemistry #6 www.youtube.com/watch?v=UL1jmJaUkaQ Reaction Percent Yield Not every reaction goes to 100% completion every time Or we can’t always collect (purify) all of the product. Theoretical Yield is the amount of product that would result if all the limiting reagents reacted. This is what you have already been calculating. Actual Yield is the amount of product actually obtained from a reaction. This is an observed value and you must be told this amount. Actual Yield % Yield = Theoretical Yield x 100% 2Mg + O2 → 2MgO If we react 10 grams of O2 with excess Magnesium, how much MgO would be theoretically expect to obtain? 10 g O2 x 1 mol O2 x 2 mol MgO x 40.3 g MgO = 25.2 g MgO 32 g O2 1 mol O2 1 mol MgO Theoretical yield = 25.2 g MgO If we actually only ended up with 20.3 grams MgO, what is our percent yield? % Yield = Actual Yield Theoretical Yield x100% = 20.3 g MgO 25.2 MgO x 100% = 80.6% We obtained a reaction yield of 80.6% MgO from this reaction. Limiting Reagents The reaction can only proceed as far as the limiting reagent will allow it. Once it is gone, the other reactant has nothing to react with so the reaction stops. To determine, we must calculate the product yield from both reactants and choose the smaller amount. Limited by the number of car bodies so production stops Limiting Reagents: Reactant used up first in the reaction to result in its completion. Unless all reactants are in exact stoichiometric equivalents, one will limit the reaction (i.e. 2 moles of N2 and 3 moles of H2) N2 + 3H2 → 2NH3 H2 is limiting N2 Excess 1P2O5 + 3H2O → 2H3PO4 If we react 44 grams of P2O5 with 25 grams of water, which is the limiting reagent and how many grams of H3PO4 will be produced? We first find out how many grams of product could be made in each case? 44 g P2O5 x 1 mol P2O5 x 2 mol H3PO4 x 98 g H3PO4 = 61.2 g 141 g P2O5 1 mol P2O5 1 mol H3PO4 25 g H2O x 1 mol H2O x 2 mol H3PO4 x 98 g H3PO4 = 90.7 g 18 g H2O 3 mol H2O 1 mol H3PO4 We then choose the smaller amount of product: 61.2 g of H3PO4 P2O5 is the limiting reagent and there is excess water. Example: Limiting Reagent and Excess Urea [(NH2)2CO] is prepared by reacting ammonia with CO2: 637.2 g of NH3 are reacted with 1142 g of CO2. (a) Which of the two reactants is the limiting reagent? (b) Calculate the mass of product (NH2)2CO formed. (c) How much excess reagent (in grams) is left at the end of the reaction? Example 3.15 Solutions Solution We carry out two separate calculations. First, starting with 637.2 g of NH3, we calculate the number of moles of (NH2)2CO that could be produced if all the NH3 reacted according to the following conversions: Combining these conversions in one step, we write Example 3.15 Solutions Second, for 1142 g of CO2, the conversions are The number of moles of (NH2)2CO that could be produced if all the CO2 reacted is It follows, therefore, that NH3 must be the limiting reagent because it produces a smaller amount of (NH2)2CO. Example 3.15 Solutions Solution The molar mass of (NH2)2CO is 60.06 g. We use this as a conversion factor to convert from moles of (NH2)2CO to grams of (NH2)2CO: Example 3.15 Solutions (c) Strategy Working backward, we can determine the amount of CO2 that reacted to produce 18.71 moles of (NH2)2CO. The amount of CO2 left over is the difference between the initial amount and the amount reacted. Solution Starting with 18.71 moles of (NH2)2CO, we can determine the mass of CO2 that reacted using the mole ratio from the balanced equation and the molar mass of CO2. The conversion steps are Example 3.15 Solutions Combining these conversions in one step, we write The amount of CO2 remaining (in excess) is the difference between the initial amount (1142 g) and the amount reacted (823.4 g): mass of CO2 remaining = 1142 g − 823.4 g = 319 g 40 g of NO are mixed with 8 g O2 to form NO2: 2NO + O2 → 2NO2 Which reactant is limiting the reaction? How much NO2 is produced? How much of the other reactant is left? If only 18 g is recovered from the reaction, what’s the percent yield? Fe + CuSO4 → FeSO4 + Cu How much Iron was added to Copper (II) Sulfate to actually obtain 32.1 g Copper (a 73.1% reaction yield)? % Yield = Actual Yield Theoretical Yield x 100% = 32.1 g Cu ? g Cu x 100% = 73.1% The theoretical yield must have been 43.9 g Cu 43.9 g Cu x 1 mol Cu x 1 mol Fe x 55.8 g Fe = 38.6 g Fe 63.5 g Cu 1 mol Cu 1 mol Fe 38.6 grams of Iron was added to CuSO4 Example 3.17 Titanium is a strong, lightweight, corrosion-resistant metal that is used in rockets, aircraft, jet engines, and bicycle frames. It is prepared by the reaction of titanium (IV) chloride with molten magnesium between 950°C and 1150°C: In a certain industrial operation 3.54 × 107 g of TiCl4 are reacted with 1.13 × 107 g of Mg. • Calculate the theoretical yield of Ti in grams. • Calculate the percent yield if 7.91 × 106 g of Ti are actually obtained. Example 3.17 Solution Carry out two separate calculations to see which of the two reactants is the limiting reagent. First, starting with 3.54 × 107 g of TiCl4, calculate the number of moles of Ti that could be produced if all the TiCl4 reacted. The conversions are so that Example 3.17 Next, we calculate the number of moles of Ti formed from1.13 × 107 g of Mg. The conversion steps are And we write Therefore, TiCl4 is the limiting reagent because it produces a smaller amount of Ti. Example 3.17 The mass of Ti formed is (b) Strategy The mass of Ti determined in part (a) is the theoretical yield. The amount given in part (b) is the actual yield of the reaction. Example 3.17 Solution The percent yield is given by Check Should the percent yield be less than 100 percent? Fun Practice Problems How many oxygen atoms are found in 20 g of CaCO3? A compound analysis yields the following data: 44.4% C, 6.21% H, 39.5% Sulfur, 9.86% Oxygen What is the molecular formula if the molar mass ~120? Fun? Practice Problems CO + O2 → CO2 Balance the above reaction and then calculate the mass of oxygen needed to produce 27 g CO2 with excess CO If 10 g methane CH4 is combusted (uses O2) to form CO2 and H2O. How many grams of each product are formed by the reaction and how many grams of oxygen were used? 40 g of NO are mixed with 8 g O2 to form NO2, which reactant is limiting the reaction, how much NO2 is produced, and how much of the other reactant is left? (Write equation first) If only18 g is recovered from the reaction, what’s the % yield? __H3PO4 + __NaOH → __Na3PO4 + __H2O What is the theoretical yield of water if 12.5 grams of H3PO4 reacts with 18.4 grams of NaOH? The actual yield was 5.2 grams water, what is the percent yield? Mass Relationships Exam Review 1) Average Atomic Mass: 80.1% 11B and 19.9% 10B 2) How many Phosphorous atoms are in 2.4 g Mg3(PO4)2 2) Find the Molecular formula PxOy : 43.7% P & 56.3% O with a molar mass between 280 – 290 g/mol. 3) 5.8 g of PxOy react with 15 g water to form H3PO4. Balance Rxn and find the theoretical yield.