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Mass Relationships in Chemical Reactions Chapter 3 HW: Ch.3a 2, 5, 13-16, 20, 23, 26, 27, 39-53 odd (no 47) Ch.3b 59, 66, 69,75, 78, 83, 86, 93-94 Micro-World (atoms & molecules) Macro-World (grams) Atomic mass: mass of an atom in atomic mass units (amu) where 1 amu = 1.66 x 10-24 g 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 Example: Average Atomic Mass Calculation #1 A random sample of Magnesium is composed of • 79.0% Mg-24 , 10.0% Mg-25 , and 11.0% Mg-26 Find the weighted average for the mass of Magnesium 79.0% of 24 + 10.0% of 25 + 11.0% of 26 = Average atomic mass *Do Not Divide by the total Use the percentage in the decimal form (e.g. 75% = 0.75) (0.790 x 24) + (0.100 x 25) + (0.11 x 26) = 24.3 amu Example: Average Atomic Mass Calculation #2 A random sample of chlorine is 75.8% 35Cl (34.97 amu) & 24.2% 37Cl (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 Average atomic mass for relative abundant isotopes 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 6.022 x 1023 1 mole = NA = 6.022 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) 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 1 12C atom 12.00 amu 12.00 g x 6.022 x 1023 12C atoms 1 amu = 1.66 x 10-24 g 1.66 x 10-24 g = 1 amu or 1 g = 6.022 x 1023 amu M = molar mass in g/mol NA = Avogadro’s number Molar eggs mass is the mass of 1 mole of shoes 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 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 Ted-Ed: How big is a mole? www.youtube.com/watch?v=TEl4jeETVmg One Mole of: S C Mercury Copper Iron 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 grams/mole Converts between mass (g) & moles 23 10 NA = Avogadro’s number = 6.02 x Converts between moles & particles (atoms / molecules / formula units) Grams to Moles 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 6.46 g He 1 mol He x 4.00 g He = 1.61 moles He More Practice: 45.5 grams of Gold (Au) to moles Moles to Grams 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 = 65.4 g Zn 0.356 mol Zn 65.4 gram Zn x 1 mol Zn Zinc = 23.3 gram Zn More Practice: 0.56 mol Potassium to grams Moles to Atoms example How many atoms of Carbon are in 2.54 moles of C? The conversion factor is Avogadro’s Number 1 mole Carbon = 6.02 x 1023 atoms of Carbon Carbon 2.54 mol C 6.02 x 1023 atoms of C x 1 mol C = 1.53 x 1024 atoms of C More Practice: 0.074 mol U to atoms of U Grams to Atoms 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 16.3 g S 1 mol S x 32.1 g S 1 mol S = 6.02 x 1023 S atoms 6.02 x 1023 atoms of S x 1 mol S = 3.06 x 1023 atoms of S More Practice: 0.01 g of Carbon to atoms Atoms to Grams Example What would be the mass in grams 25 of 2.4 x 10 atoms of Copper? # atoms of Cu → moles of Cu 6.02 x 1023 Cu atoms = 1 mol Cu 2.4 x 1025 atoms Cu → grams Cu MM: 1 mol Cu = 63.55 g Cu 1 mol Cu x x 6.02 x 1023 atoms 63.55 g Cu 1 mol Cu = 2.5 x 103 g = 2.5 kg Cu More Practice: 9.8 x 1021 atoms of Uranium to grams More Practice Conversions Convert 24 grams Potassium to moles 24 grams K 1 mol K x 39.1 grams K = 0.61 moles K Convert 1.25 moles Gold to grams 1.25 mol Au 197.0 g Au x 1 mol Au = 246 grams Au Convert 16 grams Argon to atoms 16 g Ar 1 mol Ar x 39.9 g Ar 6.02 x 1023 Ar atoms = x 1 mol Ar 2.4 x 1023 Ar atoms Convert 3.4 x 1025 atoms of Boron to grams of Boron Bozeman Science Mole conversion tutorial: www.youtube.com/watch?v=xPdqEX_WMjo 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. x 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 Determining Molar Mass Examples 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 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” Compound Mass to 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 6.07 g CH4 1 mol CH4 x = 0.38 moles CH4 16.0 g CH4 More Practice: 32 grams of Isopropanol (C3H8O) to moles Atoms in a Compound 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 H atoms per Urea molecule So there are 4 moles of H per 1 mole urea grams of urea → moles of urea → moles of H → atoms of H 25.6 g of urea x 1 mol of urea 60.06 g urea x 4 mol of H atoms 1 mol of urea x 6.02 x 1023 H atoms 1 mole of H = 1.03 × 1024 H atoms More Practice: How many Fluorine atoms in 4.5 grams of AlF3? Bell Ringer Practice Conversions How many atoms of Phosphorous (P) are found in 6.30 g of Mg3(PO4)2? -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 Percent Composition #2 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 Molar Mass of H3PO4 H: 3 x 1.008 P : 1 x 30.97 O: 4 x 16.00 = 97.99 g More Practice: What is the % composition of Aluminum in Galaxite (MnAl2O4)? How many grams Al in 35.6 g of Galaxite? Example Using Percent Compositions #1 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 #1 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) 3 = 1.28 × 10 kg Using Percent Compositions Example #2 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 More Practice: What is the % composition of Aluminum in Galaxite (MnAl2O4)? How many grams Al in 35.6 g of Galaxite? 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 due to some experimental error, Generally round only if within one tenth. Example: Empirical Formula Determination #1 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 C(1 mol)O(2 mol) = 4.53 mol O 4.53/2.29 = 1.98 mol ~ 2 mol O 2.29/2.29 = 1 mol C = 2.29 mol C C(1 atom)O(2 atom) CO2 Example: Empirical Formula #2 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 2nd: convert each mass to mole 40.92 g C 4.58 g H 54.50 g O Example: Empirical Formula #2 Solution 3rd:We then take the smallest value of moles and divide each number by that same smallest value • This gives CH1.33O as the formula for ascorbic acid. • If any # is more than 0.1 away from whole, it is too far to round. • Multiply all values by integers until all values are whole numbers Because 1.33 × 3 gives us an integer (4), we 1.33 × 1 = 1.33 multiply all the subscripts by 3 and obtain 1.33 × 2 = 2.66 C3H4O3 as the empirical formula for 1.33 × 3 = 3.99 ~ 4 ascorbic acid. More Practice: CxHyNz : 42.1% C, 8.8% H & 49.1% N 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 Example: Molecular Formula 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: Molecular Formula Solution 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: Molecular Formula Solution 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. More practice: Find the Molecular formula PxOy : 43.7% P & 56.3% O with a molar mass between 280 – 290 g/mol. Bell Ringer: Molecular Formula Aconitic acid has a mass composition of: 41.4% Carbon 3.4% Hydrogen 55.2% Oxygen It also has a molar mass of 174 grams/mole What is the molecular formula (CxHyOz)of Aconitic acid? 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 1 Sulfur 21 Sodium 43 Hydrogen 65 Oxygen = Products 1 Sulfur 2 Sodium 2 Hydrogen 4 5 Oxygen 6 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. More Practice: _C3H6O2 + _O2 → _CO2 + _H2O Stoichiometry is the quantitative relationship found between reactants and/or 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 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 → 2CO 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 More Practice: 1) moles of CO2 from 1.5 moles C2H6 2) moles of H2O from 0.34 moles of O2 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 Stoichiometry Example #1 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. Stoichiometry Example #1 Solution ? g of Li 9.89 g of H2 More Practice: Grams of LiOH produced from 2.9 grams Li Stoichiometry Example #2 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 Stoichiometry Example #3 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 44 g C3H8 1 mol C3H8 1 mol CO2 *Note: Different amounts of each product are made by the same amount of reactant = 105 g CO2 More Practice How many grams of Barium phosphate are produced when adding 1.3 grams of Barium hydroxide to a solution of phosphoric acid? __Ba(OH)2 + __ H3PO4 → __ Ba3(PO4)2 + __H2O 1.3 grams Excess ? grams 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 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% Example: Percent Yield 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 Bozeman Science Stoichiometry Tutorial: www.youtube.com/watch?v=LQq203gyftA Example #1: Limiting Reagent and 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 #2: Limiting Reagent and Excess Urea [(NH2)2CO] is prepared by reacting ammonia with CO2: 63.7 g of NH3 are reacted with 114.2 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: Limiting Reagent and Excess 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 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 3a) Find the Molecular formula PxOy : 43.7% P & 56.3% O with a molar mass between 280 – 290 g/mol. 3b) 5.8 g of PxOy react with 15 g water to form H3PO4. Balance Rxn and find the theoretical yield. 3c) What is the % yield if only 5.4 grams of product is recovered?