CHEMICAL REACTIONS AND STOICHIOMETRY.pdf

CHEMICAL REACTIONS AND STOICHIOMETRY Ron Angelo R. Gatinga Special Science Teacher I TEST YOURSELF. TEST YOURSELF. CHEMICAL REACTIONS A chemical reaction is a process in which one set of substances, called reactants, is converted to a new set of substances, called products. In other wor...

CHEMICAL REACTIONS AND STOICHIOMETRY Ron Angelo R. Gatinga Special Science Teacher I TEST YOURSELF. TEST YOURSELF. CHEMICAL REACTIONS A chemical reaction is a process in which one set of substances, called reactants, is converted to a new set of substances, called products. In other words, a chemical reaction is the process by which a chemical change occurs. We need evidence before we can say that a reaction has occurred. Some of the types of physical evidence to look for are shown here: a color change formation of a solid (precipitate) within a clear solution evolution of a gas evolution or absorption of heat CHEMICAL REACTIONS A chemical reaction is a process in which one set of substances, called reactants, is converted to a new set of substances, called products. In other words, a chemical reaction is the process by which a chemical change occurs. We need evidence before we can say that a reaction has occurred. Some of the types of physical evidence to look for are shown here: a color change formation of a solid (precipitate) within a clear solution evolution of a gas evolution or absorption of heat STOICHIOMETRY Area of study that examines the quantities of substances consumed and produced in chemical reactions Based on the Law of Conservation of Mass (Antoine Lavoisier, 1789) “We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.” —Antoine Lavoisier CHEMICAL EQUATIONS Chemical equations are how chemists represent chemical reactions on paper. Arrows separate the starting materials (on the left), called reactants, from the ending materials (on the right), called products. “+” separates multiple starting or ending materials. BALANCING CHEMICAL EQUATIONS Follow the Law of Conservation of Mass, but how? Start with an element that is only in one reactant and product (C below). Balance it by changing coefficients, NOT subscripts. (Like in math, a “1” is not written, but it is assumed.) Move on to other elements, without changing coefficients that are set, until complete, checking all elements at the end (H, then O here; totals below). BALANCING CHEMICAL EQUATIONS Follow the Law of Conservation of Mass, but how? Start with an element that is only in one reactant and product (C below). Balance it by changing coefficients, NOT subscripts. (Like in math, a “1” is not written, but it is assumed.) Move on to other elements, without changing coefficients that are set, until complete, checking all elements at the end (H, then O here; totals below). WHY DO WE CHANGE COEFFICIENT INSTEAD OF SUBSCRIPTS TO BALANCE? Hydrogen and oxygen can make water OR hydrogen peroxide 2 H2(g) + O2(g) → 2 H2O(l) H2(g) + O2(g) → H2O2(l) We don’t change the formula because we don’t drink hydrogen peroxide. OTHER SYMBOLS IN CHEMICAL EQUATIONS CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) Δ The states of matter for the reactants and products are often written in parentheses to the right of each formula or symbol. (g) = gas; (l) = liquid; (s) = solid; (aq) = dissolve in aqueous (water) solution OTHER SYMBOLS IN CHEMICAL EQUATIONS CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) Δ Other symbols can be used to represent conditions during the chemical reaction. One example is the use of Δ over the reaction arrow, which means heat is needed for the reaction to take place. EXAMPLES Balance the following chemical equations: 1.Na(s) + H2O(l) → NaOH(aq) + H2(g) 2.Fe(s) + O2(g) → Fe2O3 3.Al(s) + HCl(aq) → AlCl3 (aq) + H2(g) YOUR TURN Balance the following chemical equations: 1.NH3 + O2 → N2 + H2O 2.H3PO4 + CaO → Ca3(PO4)2 + H2O 3.C3H8 + O2 → CO2 + H2O 4.NH3 + O2 → NO2 + H2O 5.NO2 + NH3 → N2 + H2O TYPES OF CHEMICAL REACTIONS 1.Combination/Synthesis Reaction 2.Analysis/Decomposition Reaction 3.Single Displacement Reaction 4.Double Displacement Reaction 5.Combustion Reaction SYNTHESIS REACTIONS Synthesis/Combination reaction is the direct union or combination of two substances to form a single compound. General Formula: A + B ⟶ AB EXAMPLES OF SYNTHESIS REACTIONS Nickel reacts with oxygen to form nickel oxide 2Ni + O2 ⟶ 2NiO Silver reacts with chlorine to form silver chloride. 2Ag +Cl2 ⟶ 2AgCl Oxygen gas reacts to magnesium form magnesium oxide O2 + 2Mg ⟶ 2MgO DECOMPOSITION REACTIONS Decomposition reaction is a single substance broken into two or more different substances General Formula: AB ⟶ A + B EXAMPLE OF DECOMPOSITION REACTIONS Hydrogen peroxide decomposes to form water and oxygen gas 2H2O2 ⟶ 2 H2O + O2 SINGLE DISPLACEMENT REACTIONS In this type of chemical reaction, an element and a compound react and form a different element and a different compound. General Formula: Cation exchange: A + BC ⟶ B + AC Anion exchange: A + BC ⟶ C + BA ACTIVITY SERIES If an element is more reactive than its counterpart, then it is more likely to “steal” another cation/anion Can either be cation exchange or anion exchange. CATION EXCHANGE Fe + CuSO4 ⟶ Cu + FeSO4 Iron is more reactive than copper, that is why it was able to “steal” sulfate. CATION EXCHANGE Fe + CuSO4 ⟶ Cu + FeSO4 Iron is more reactive than copper, that is why it was able to “steal” sulfate. ANION EXCHANGE Cl2 + 2NaBr ⟶ 2NaCl + Br2 Chlorine is more reactive than bromine, that is why it was able to “steal” sodium. ANION EXCHANGE Cl2 + 2NaBr ⟶ 2NaCl + Br2 Chlorine is more reactive than bromine, that is why it was able to “steal” sodium. LESS REACTIVE ELEMENTS CANNOT STEAL IONS I2 + 2 KBr ⟶ NO REACTION CaO + Cu ⟶ NO REACTION DOUBLE DISPLACEMENT REACTIONS This happens when the positive and negative ions in the two compounds switch places forming two new compounds. General Formula: AB + CD ⟶ AD + CB EXAMPLES OF DOUBLE DISPLACEMENT REACTIONS HCl+ NaOH ⟶ NaCl+ H2O AgNO3 + KCl ⟶ AgCl + KNO3 COMBUSTION REACTIONS A chemical reaction in which a substance combines with oxygen and releases energy. Always produce carbon dioxide and water General Formula: CxHy + O2 ⟶ CO2 + H2O or CxHyOz + O2 ⟶ CO2 + H2O EXAMPLES OF COMBUSTION REACTIONS Cellular respiration/combustion of glucose Combustion of methane TYPES OF CHEMICAL REACTIONS 1. Combination/Synthesis Reaction Form: A+B→AB Two or more substances combine to form a single product. 2. Analysis/Decomposition Reaction Form: AB→A+B A single compound breaks down into two or more products. 3. Single Displacement Reaction Form: A+BC→AC+B One element replaces another in a compound. TYPES OF CHEMICAL REACTIONS 4. Double Displacement Reaction Form: AB+CD→AD+CB The ions of two compounds exchange places in an aqueous solution 5. Combustion Reaction Form: CxHy + O2 ⟶ CO2 + H2O or CxHyOz + O2 ⟶ CO2 + H2O A hydrocarbon reacts with oxygen to produce carbon dioxide and water, releasing energy. YOUR TURN Identify the types of chemical reactions. FORMULA WEIGHTS AND ELEMENTAL COMPOSITIONS OF SUBSTANCES FORMULA WEIGHT The formula weight (FW) of a substance is the sum of the atomic weights (AW) of the atoms in the chemical formula of the substance. Example: What is the formula weight of H2SO4? Solution: FW of H2SO4 = 2 (AW of H) + (AW of S) + 4(AW of O) = 2(1.008 amu) + (32.066 amu)+ 4(15.999 amu) = 98.078 amu -> 98.08 amu MORE EXAMPLES Find the formula weight of the following compounds: 1.Calcium chloride 2.Phosphoric acid 3.Potassium carbonate SOLUTION Calcium Chloride Phosphoric acid Potassium carbonate Formula: CaCl2 Formula: H3PO4 Formula: K2CO3 Formula Weight: Formula Weight: Formula Weight: Ca = 1 x 40.078 + H = 3 x 1.008 + K = 2 x 39.098 + Cl = 2 x 35.453 P = 1 x 30.974 + C = 1 x 12.011+ = 110.98 amu O = 4 x 15.999 O = 3 x 15.999 = 97.99 amu = 138.20 amu YOUR TURN Find the formula weight of the following compounds: 1.Sucrose (C12H22O11) 2.Calcium nitrate 3.Methanol (CH3OH) ELEMENTAL COMPOSITION One way to determine the identity of a substance is to measure its elemental composition and compare it to the calculated elemental compositions of possible candidate substances. We do these calculations based on the masses of each element in the compound: % mass composition of element (𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒂𝒕𝒐𝒎𝒔 𝒑𝒆𝒓 𝒆𝒍𝒆𝒎𝒆𝒏𝒕)(𝑨𝑾 𝒐𝒇 𝒆𝒍𝒆𝒎𝒆𝒏𝒕) = x 100% 𝑭𝑾 𝒐𝒇 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅 EXAMPLE Calculate the percentage of hydrogen, sulfur, and oxygen (by mass) in H2SO4 Solution: % mass composition of element (𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒂𝒕𝒐𝒎𝒔 𝒑𝒆𝒓 𝒆𝒍𝒆𝒎𝒆𝒏𝒕)(𝑨𝑾 𝒐𝒇 𝒆𝒍𝒆𝒎𝒆𝒏𝒕) = x 100% 𝑭𝑾 𝒐𝒇 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅 2 (1.008 𝑎𝑚𝑢) %H in H2SO4 = x 100 = 2.056% 98.078 𝑎𝑚𝑢 1 (32.066 𝑎𝑚𝑢) %S in H2SO4 = x 100 = 32.69% 98.078 𝑎𝑚𝑢 4 (15.999 𝑎𝑚𝑢) %O in H2SO4 = x 100 = 65.25% 98.078 𝑎𝑚𝑢 YOUR TURN Calculate the percentage of carbon, hydrogen, and oxygen (by mass) in sucrose (C12H22O11). SOLUTION Let’s first calculate We can now calculate the percent composition of each the formula weight element in sucrose: of sucrose (C12H22O11): Solution: % mass composition of element (𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒂𝒕𝒐𝒎𝒔 𝒑𝒆𝒓 𝒆𝒍𝒆𝒎𝒆𝒏𝒕)(𝑨𝑾 𝒐𝒇 𝒆𝒍𝒆𝒎𝒆𝒏𝒕) C = 12 x 12.011 + = x 100% 𝑭𝑾 𝒐𝒇 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅 H = 22 x 1.008 + %C in C12H22O11 = 12 (12.011 𝑎𝑚𝑢) x 100% = 42.11% 342.297 𝑎𝑚𝑢 O = 11 x 15.999 22 (1.008 𝑎𝑚𝑢) %H in C12H22O11 = x 100 = 6.479% 342.297 𝑎𝑚𝑢 = 342.297 amu 11 (15.999 𝑎𝑚𝑢) %O in C12H22O11 = x 100 = 51.41% 342.297 𝑎𝑚𝑢 THE MOLE AND AVOGADRO’S NUMBER In the lab, we cannot work with individual molecules. They are way smaller. In chemistry, the counting unit for numbers of atoms, ions, or molecules in a laboratory-size sample is the mole (abbreviated mol) One mole is the amount of matter that contains as many objects as the number of atoms in exactly 12 g of isotopically pure 12C. From experiments, scientists have determined this number to be 6.0221415 x 1023, which we usually round to 6.02 x 1023. Scientists call this value Avogadro’s number, NA, in honor of the Italian scientist Amedeo Avogadro (1776-1856), and it is often cited with units of reciprocal moles, 6.02 x 1023 mol-1. 1 mol of 12C = 6.02 x 1023 12C atoms 1 mol of H2O= 6.02 x 1023 H2O molecules 1 mol of NO3- = 6.02 x 1023 NO3- ions MOLAR MASS The molar mass is the mass of 1 mol of the substance (g/mol) The molar mass in grams per mole of any substance is numerically equal to its formula weight in atomic mass units. CONVERTING BETWEEN MOLES, MASS, AND ATOMS/MOLECULES/IONS EXAMPLE # 1 For example, let’s calculate how many copper atoms are in an old copper penny. Such a penny has a mass of about 3 grams, and let’s assume for simplicity that the penny is pure copper: 1 𝑚𝑜𝑙 𝐶𝑢 6.02 𝑥 1023 𝐶𝑢 𝑎𝑡𝑜𝑚𝑠 # of Cu atoms = 3 g Cu x x 1 𝑚𝑜𝑙 𝐶𝑢 63.546 𝑔 𝐶𝑢 = 3 x 10 22 Cu atoms EXAMPLE # 2 Calculate the number of moles of glucose (C6H12O6) in a 5.380 g sample. First, let’s calculate the Now let’s solve for the moles of molar mass of glucose: glucose: C = 6 x 12.011 1 𝑚𝑜𝑙 𝐶6𝐻12𝑂6 5.380 g C6H12O6 x H = 12 x 1.008 180.156 𝑔 𝐶6𝐻12𝑂6 O = 6 x 15.999 =0.02986 mol C6H12O6 = 180.156 g/mol EXAMPLE # 3 The volatile liquid ethyl mercaptan, C2H6S, is one of the most odoriferous substances known. It is sometimes added to natural gas to make gas leaks detectable. How many C2H6S molecules are contained in a 1 µL sample? The density of liquid ethyl mercaptan is 0.84 g/mL. SOLUTION (PART 1) First, let’s convert 1 µL to mL Let’s get the molar mass of since the density is in terms of liquid ethyl mercaptan (C2H6S) g/mL: C = 2 x 12.011 10¯6 𝐿 1000 𝑚𝐿 1 µL x x H = 6 x 1.008 1µ𝐿 1𝐿 = 0.001 mL S = 1 x 32.066 Next, let’s get the mass of ethyl = 62.136 g/mol mercaptan using its density: 0.84 𝑔 0.001 mL x 1 𝑚𝐿 = 8.4 x10-4 g SOLUTION (PART 2) We can now solve for the number of molecules of liquid ethyl mercaptan: 1 𝑚𝑜𝑙 𝐶2𝐻6𝑆 6.02 𝑥 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶2𝐻6𝑆 8.4 x10-4 g C2H6S x x 1 𝑚𝑜𝑙 𝐶2𝐻6𝑆 62.136 𝑔 𝐶2𝐻6𝑆 = 8 x 10 18 molecules of C2H6S YOUR TURN 1.What is the molar mass of cholesterol if 0.00105 mol has a mass of 0.406 g? 2.Gold has a density of 19.32 g/cm3. A piece of gold foil is 2.50 cm on each side and 0.100 mm thick. How many atoms of gold are in this piece of gold foil? EXAMPLE 1. Lead (II) chromate, PbCrO4, is a yellow paint pigment prepared by the precipitation reaction of lead (II) nitrate and potassium chromate solution. In a preparation, 45.6 g of lead (II) chromate is obtained as a precipitate. - Calculate the number of moles of PbCrO4 is produced in the reaction. FINDING EMPIRICAL FORMULA One can calculate the empirical formula from the percent composition. EXAMPLE # 1 The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.  We determine the simplest whole-number ratio of SOLUTION moles by dividing each number of moles by the smallest number of moles First, we assume percentages as actual grams of =7.002796979 → 7 5.104487553 components (totaling to 100 g of material): C= 0.7289212537 C = 61.31% → 61.31 g 5.099206349 H = 5.14% → 5.14 g H= 0.7289212537 =6.995551746 → 7 0.7289212537 N = 10.21% → 10.21 g N= 0.7289212537 =1 O = 23.33% → 23.33 g 1.458216139 Next, we calculate the number of moles of each O= 0.7289212537 =2.000512581 → 2 element: 1 𝑚𝑜𝑙 𝐶 C = 61.31 g x 12.011 𝑔 𝐶 =5.104487553 mol C 1 𝑚𝑜𝑙 𝐻 H = 5.14 g x 1.008 𝑔 𝐻 =5.099206349 mol H The empirical formula of N =10.21 g x 1 𝑚𝑜𝑙 𝑁 14.007 𝑔 𝑁 =0.7289212537 mol N para-aminobenzoic acid O =23.33 g x 1 𝑚𝑜𝑙 𝑂 =1.458216139 mol O or PABA is C7H7NO2. 15.999 𝑔 𝑂 EXAMPLE # 2 Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula of ascorbic acid?  =1.00012157 → 1 3.406877029 C= SOLUTION H= 3.406462904 4.543650794 =1.33383246 → X 3.406462904 First, we assume percentages as actual grams of 3.406462904 components (totaling to 100 g of material): O= =1 3.406462904 C = 40.92% → 40.92 g The ratio for H is too far from 1 to attribute the H = 4.58% → 4.58 g difference to experimental error; in fact, it is 1 O = 54.50% → 54.50 g quite close to 1. This suggests we should 3 Next, we calculate the number of moles of each multiply the ratios by 3 to obtain whole element: numbers: C = 1(3) → 3 C = 40.92 g x 1 𝑚𝑜𝑙 𝐶 =3.406877029 mol C 12.011 𝑔 𝐶 H = 1.33383246(3) → 4 H = 4.58 g x 1 𝑚𝑜𝑙 𝐻 =4.543650794 mol H 1.008 𝑔 𝐻 O = 1(3) → 3 O = 54.50 g x 1 𝑚𝑜𝑙 𝑂 =3.406462904 mol O 15.999 𝑔 𝑂 The empirical formula of We determine the simplest whole-number ratio of moles by dividing each number of moles by the ascorbic acid (vitamin C) smallest number of moles. is C3H4O3. EXAMPLE # 3 A 2.144-g sample of phosgene, a compound used as a chemical warfare agent during World War I, contains 0.260 g of carbon, 0.347 g of oxygen, and 1.537 g of chlorine. What is the empirical formula of this substance? SOLUTION We do not have to assume the mass in terms of grams since they are already in grams. We can directly convert them to moles: =0.0216468237 mol C 1 𝑚𝑜𝑙 𝐶 C = 0.260 g x 12.011 𝑔 𝐶 O = 0.347 g x 1 𝑚𝑜𝑙 𝑂 =0.0216888555 mol O 15.999 𝑔 𝑂 The empirical formula of Cl = 1.537 g x 1 𝑚𝑜𝑙 𝐶𝑙 35.453 𝑔 𝐶𝑙 =0.0433531718 mol Cl phosgene is COCl2. We determine the simplest whole-number ratio of moles by dividing each number of moles by the smallest number of moles: C= 0.0216468237 0.0216468237 =1 O= 0.0216888555 0.0216468237 =1.001941708 → 1 Cl = 0.0433531718 0.0216468237 =2.002749798 → 2 YOUR TURN Dibutyl succinate is an insect repellent used against household ants and roaches. Its composition is 62.58% C, 9.63% H, and 27.79% O. What is the empirical formula of dibutyl succinate? DETERMINING THE MOLECULAR FORMULA Remember, the number of atoms in a molecular formula is a multiple of the number of atoms in an empirical formula. If we find the empirical formula and know a molar mass (molecular weight) for the compound, we can find the molecular formula. The subscripts in the molecular formula of a substance are always whole- number multiples of the subscripts in its empirical formula. EXAMPLE # 1 The empirical formula of a compound was found to be CH. It has a mass of 78 g/mol. What is its molecular formula? Solution: Whole-number multiple = 78/13 = 6 The molecular formula is C6H6. EXAMPLE # 2 Cyclohexane, a commonly used organic solvent, is 85.6% C and 14.4% H by mass with a molar mass of 84.2 g/mol. What is its molecular formula?  Get the whole number multiple: SOLUTION 84.2 First, let’s get the empirical formula: ≈6 14.027 C = 85.6 g x 1 𝑚𝑜𝑙 𝐶 12.011 𝑔 𝐶 =7.12680043 mol C This means that molecular formula H = 14.4 g x 1 𝑚𝑜𝑙 𝐻 =14.2857143 mol H of hexane is C(1)(6)H(2)(6) or C6H12. 1.008 𝑔 𝐻 C= 7.12680043 7.12680043 =1 =2.00450601 → 2 The molecular formula of 14.2857143 H= 7.12680043 Empirical Formula:CH2 cyclohexane is C6H12. Get the molar mass of the empirical formula: =14.027 g/mol. EXAMPLE # 3 The chlorine-containing narcotic drug pentaerythritol chloral has 21.51% C, 2.22% H, 58.63% Cl and 17.64% O, by mass, as its other elements. Its molecular weight is 726 g/mol. What are the empirical and molecular formulas of pentaerythritol chloral?  Multiply every ratio by 8: SOLUTION C = 1.624259821(8) → 13 First, let’s get the empirical formula: H = 2(8) → 16 1 𝑚𝑜𝑙 𝐶 Cl = 1.499896047(8) →12 C = 21.51 g x 12.011 𝑔 𝐶 =1.79085838 mol C 1 𝑚𝑜𝑙 𝐻 O = 1(8) → 8 H = 2.22 g x 1.008 𝑔 𝐻 =2.20238095 mol H Empirical Formula:C13H16Cl12O8 1 𝑚𝑜𝑙 𝐶𝑙 Cl = 58.63 g x 35.453 𝑔 𝐶𝑙 =1.65373875 mol Cl Get the molar mass of the empirical formula: 1 𝑚𝑜𝑙 𝑂 O = 17.64 g x 15.999 𝑔 𝑂 =1.10256891 mol O =725.699 g/mol. Get the whole number multiple: C= 1.79085838 1.10256891 =1.624259821 726 725.699 ≈1 This means that the empirical formula is also the =1.997499594 → 2 2.20238095 H= 1.10256891 molecular formula. 1.65373875 Cl = 1.10256891 =1.499896047 The empirical formula and molecular formula of 1.10256891 pentaerythritol chloral is C13H16Cl12O8 O= 1.10256891 =1 MOLE RATIOS We multiply the ratios by a factor to get whole numbers when the calculated mole ratios of the elements in the compound are not whole numbers themselves. This often happens when the mole ratios are fractions. Here’s when and why you should multiply by a factor: 1. Fractional Ratios: If the mole ratios are fractions (e.g., 1.5, 2.5, etc.), it's necessary to multiply all ratios by a factor to convert them into whole numbers. This factor is typically chosen to eliminate the fraction and make all the numbers whole. 2. Common Factors: Common factors used are 2, 3, or 4, depending on the fraction. For example: o If a ratio is 1.5, you multiply by 2 to get whole numbers (1.5 × 2 = 3). o If a ratio is 0.5, you multiply by 2 (0.5 × 2 = 1). o If a ratio is 1.33, you might multiply by 3 (1.33 × 3 = 4). 3. Adjustments: After multiplying, double-check to ensure all ratios are close to whole numbers. Adjust if necessary to ensure the empirical formula reflects the simplest whole-number ratio of the elements. When to multiply mole ratio with a common factor? You know that you need to multiply the mole ratios by a common factor when the calculated mole ratios are not close to whole numbers. If any of the mole ratios are fractional or very close to a common fraction (like 1.5, 2.5, 1.33, etc.), you should multiply all the ratios by the smallest number that will convert those fractions into whole numbers. EXAMPLE # 3 The chlorine-containing narcotic drug pentaerythritol chloral has 21.51% C, 2.22% H, 58.63% Cl and 17.64% O, by mass, as its other elements. Its molecular weight is 726 g/mol. What are the empirical and molecular formulas of pentaerythritol chloral?  Multiply every ratio by 8: SOLUTION C = 1.624259821(8) → 13 First, let’s get the empirical formula: H = 2(8) → 16 1 𝑚𝑜𝑙 𝐶 Cl = 1.499896047(8) →12 C = 21.51 g x 12.011 𝑔 𝐶 =1.79085838 mol C 1 𝑚𝑜𝑙 𝐻 O = 1(8) → 8 H = 2.22 g x 1.008 𝑔 𝐻 =2.20238095 mol H Empirical Formula:C13H16Cl12O8 1 𝑚𝑜𝑙 𝐶𝑙 Cl = 58.63 g x 35.453 𝑔 𝐶𝑙 =1.65373875 mol Cl Get the molar mass of the empirical formula: 1 𝑚𝑜𝑙 𝑂 O = 17.64 g x 15.999 𝑔 𝑂 =1.10256891 mol O =725.699 g/mol. Get the whole number multiple: C= 1.79085838 1.10256891 =1.624259821 726 725.699 ≈1 This means that the empirical formula is also the =1.997499594 → 2 2.20238095 H= 1.10256891 molecular formula. 1.65373875 Cl = 1.10256891 =1.499896047 The empirical formula and molecular formula of 1.10256891 pentaerythritol chloral is C13H16Cl12O8 O= 1.10256891 =1 YOUR TURN Styrene, a compound used to make Styrofoam® cups and insulation, contains 92.3% C and 7.7% H by mass and has a molar mass of 104 g/mol. What is the empirical and molecular formula of styrene. COMBUSTION ANALYSIS Compounds containing C, H, and O are routinely analyzed through combustion in a chamber like the one shown below C is determined from the mass of CO2 produced. H is determined from the mass of H2O produced. O is determined by difference after the C and H have been determined. EXAMPLE # 1 Isobutyl propionate is the substance that flavors rum extract. Combustion of a 1.152 g sample of this carbon hydrogen oxygen compound yields 2.726 g CO2 and 1.116 g H2O. What is the empirical formula of isobutyl propionate? SOLUTION (PART 1) To expedite things, let’s calculate the molar mass of carbon dioxide and water: CO2 = 44.009 g/mol H2O = 18.015 g/mol Because all of the carbon in the sample is converted to CO2, we can use dimensional analysis and the following steps to calculate the mass C in the sample.: 1 𝑚𝑜𝑙 𝐶𝑂2 1 𝑚𝑜𝑙 𝐶 12.011 𝑔 𝐶 2.726 g CO2 x x x = 0.7439838669 g C 44.009 𝑔 𝐶𝑂2 1 𝑚𝑜𝑙 𝐶𝑂2 1 𝑚𝑜𝑙 𝐶 We do the same thing with H from H2O: 1 𝑚𝑜𝑙 𝐻2𝑂 2 𝑚𝑜𝑙 𝐻 1.008 𝑔 𝐻 1.116 g H2O x x 1 𝑚𝑜𝑙 𝐻 𝑂 x 1 𝑚𝑜𝑙 𝐻 = 0.1248879267 g H 18.015 𝑔 𝐻2𝑂 2 We now get the mass of O from masses of C and H: mass of O = mass of the sample – (mass of C + mass of H) mass of O = 1.152 g – (0.7439838669 g + 0.1248879267 g) = 0.2831282064 g O  We multiply by 2 to SOLUTION (PART 2) get the proper Now, let’s get the empirical formula: coefficients: 1 𝑚𝑜𝑙 𝐶 C = 0.7439838669 g x 12.011 𝑔 𝐶 =0.06194187552 mol C C=3.500209601 x 2 = 7 1 𝑚𝑜𝑙 𝐻 H = 0.1248879267 g x 1.008 𝑔 𝐻 =0.1238967527 mol H H=7 x 2 = 14 1 𝑚𝑜𝑙 𝑂 O = 0.2831282064 g x 15.999 𝑔 𝑂 =0.01769661894 mol O O=1 x 2 = 2 C= 0.06194187552 0.01769661894 =3.500209601 → x =7.0011539 → 7 0.1238967527 H= 0.01769661894 0.01769661894 O= 0.01769661894 =1 The empirical formula of isobutyl propionate is C7H14O2. EXAMPLE # 2 2-propanol, sold as rubbing alcohol, is composed of C, H, and O. Combustion of 0.255 g of 2-propanol produces 0.561 g of CO2 and 0.306 g of H2O. Determine the empirical formula of 2-propanol. SOLUTION (PART 1) To expedite things, let’s calculate the molar mass of carbon dioxide and water: CO2 = 44.009 g/mol H2O = 18.015 g/mol Because all of the carbon in the sample is converted to CO2, we can use dimensional analysis and the following steps to calculate the mass C in the sample.: 1 𝑚𝑜𝑙 𝐶𝑂2 1 𝑚𝑜𝑙 𝐶 12.011 𝑔 𝐶 0.561 g CO2 x x x = 0.1531089323 g C 44.009 𝑔 𝐶𝑂2 1 𝑚𝑜𝑙 𝐶𝑂2 1 𝑚𝑜𝑙 𝐶 We do the same thing with H from H2O: 1 𝑚𝑜𝑙 𝐻2𝑂 2 𝑚𝑜𝑙 𝐻 1.008 𝑔 𝐻 0.306 g H2O x x 1 𝑚𝑜𝑙 𝐻 𝑂 x 1 𝑚𝑜𝑙 𝐻 = 0.03424346378 g H 18.015 𝑔 𝐻2𝑂 2 We now get the mass of O from masses of C and H: mass of O = mass of the sample – (mass of C + mass of H) mass of O = 0.255 g – (0.1531089323 g + 0.03424346378 g) = 0.06764760392 g O SOLUTION (PART 2) Now, let’s get the empirical formula: 1 𝑚𝑜𝑙 𝐶 C = 0.1531089323 g x 12.011 𝑔 𝐶 =0.01274739258 mol C 1 𝑚𝑜𝑙 𝐻 H = 0.03424346378 g x 1.008 𝑔 𝐻 =0.03397169026 mol H 1 𝑚𝑜𝑙 𝑂 O = 0.06764760392 g x 15.999 𝑔 𝑂 =4.22823951 x10-3 mol O C= 0.01274739258 4.22823951 𝑥 10−3 =3.014822729 → 3 =8.03447633 → 8 0.03397169026 H= 4.22823951 𝑥 10−3 4.22823951 𝑥 10−3 O= 4.22823951 𝑥 10−3 =1 The empirical formula of 2-propanol is C3H8O. EXAMPLE # 3 The compound dioxane, which is used as a solvent in various industrial processes, is composed of C, H, and O atoms. Combustion of a 2.203-g sample of this compound produces 4.401 g CO2 and 1.802 g H2O. A separate experiment shows that it has a molar mass of 88.1 g/mol. What is the molecular formula for dioxane? SOLUTION (PART 1) To expedite things, let’s calculate the molar mass of carbon dioxide and water: CO2 = 44.009 g/mol H2O = 18.015 g/mol Because all of the carbon in the sample is converted to CO2, we can use dimensional analysis and the following steps to calculate the mass C in the sample.: 1 𝑚𝑜𝑙 𝐶𝑂2 1 𝑚𝑜𝑙 𝐶 12.011 𝑔 𝐶 4.401 g CO2 x x x = 1.201127992 g C 44.009 𝑔 𝐶𝑂2 1 𝑚𝑜𝑙 𝐶𝑂2 1 𝑚𝑜𝑙 𝐶 We do the same thing with H from H2O: 1 𝑚𝑜𝑙 𝐻2𝑂 2 𝑚𝑜𝑙 𝐻 1.008 𝑔 𝐻 1.802 g H2O x x 1 𝑚𝑜𝑙 𝐻 𝑂 x 1 𝑚𝑜𝑙 𝐻 = 0.2016559534 g H 18.015 𝑔 𝐻2𝑂 2 We now get the mass of O from masses of C and H: mass of O = mass of the sample – (mass of C + mass of H) mass of O = 2.203 g – (1.201127992 g + 0.2016559534 g) = 0.8002160546 g O  Get the whole SOLUTION (PART 2) number multiple: Now, let’s get the empirical formula: 88.1 1 𝑚𝑜𝑙 𝐶 ≈2 C = 1.201127992 g x 12.011 𝑔 𝐶 =0.1000023305 mol C 44.053 1 𝑚𝑜𝑙 𝐻 H = 0.2016559534 g x 1.008 𝑔 𝐻 =0.2000555093 mol H C=2 x 2 = 4 1 𝑚𝑜𝑙 𝑂 O = 0.8002160546 g x 15.999 𝑔 𝑂 =0.05001662945 mol O H=4 x 2 = 8 C= 0.1000023305 0.05001662945 =1.999381638 → 2 O=1 x 2 = 2 =3.999779903 → 4 0.2000555093 H= 0.05001662945 0.05001662945 O= 0.05001662945 =1 The molecular formula of The empirical formula of dioxane is C2H4O. dioxane is C4H8O2. Empirical Formula Weight: 44.053 g/mol YOUR TURN Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005-g sample of menthol is combusted, producing 0.2829 g of CO2 and 0.1159 g of H2O. a) What is the empirical formula for menthol? b) If menthol has a molar mass of 156 g/mol, what is its molecular formula? REACTION STOICHIOMETRY QUANTITATIVE INFORMATION FROM A BALANCED CHEMICAL EQUATION The coefficients in the balanced equation show – Relative numbers of molecules of reactants and products – Relative numbers of moles of reactants and products, which can be converted to mass STOICHIOMETRIC CALCULATIONS We have already seen in this unit how to convert from grams to moles or moles to grams. The new calculation is how to compare two different materials, using the mole ratio from the balanced equation. The mole ratio comes from the coefficients in the balanced equation. EXAMPLE # 1 How many grams of water can be produced from oxidation of 1.00 g of glucose? C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) SOLUTION C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) 1. First, we convert grams of C6H12O6 to moles using the molar mass of C6H12O6. 2. Next, we convert moles of C6H12O6 to moles of H2O using the stoichiometric relationship that 1 mol C6H12O6 = 6 mol H2O. 3. Finally, we convert moles of H2O to grams using the molar mass of H2O. Solution: 1 𝑚𝑜𝑙 𝐶6𝐻12𝑂6 6 𝑚𝑜𝑙 𝐻2𝑂 18.015 𝑔 𝐻2𝑂 1.00 g C6H12O6 x x 1 𝑚𝑜𝑙 𝐶 𝐻 𝑂 x 180.156 𝑔 𝐶6𝐻12𝑂6 6 12 6 1 𝑚𝑜𝑙 𝐻2𝑂 = 0.5999800173 ≈ 0.600 g H2O = The amount of water produced from oxidation of 1.00 g of glucose is 0.600 g. EXAMPLE # 2 Propane, C3H8 , is a common gas fuel used for cooking and home heating. What mass of O2 is consumed in the combustion of 3.50 g of propane? Before solving this, elucidate the actual chemical equation first and balance it: C3H8 (g)+ O2 (g)→ CO2 (g) + H2O (g) Balancing the equation, we get: C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g) SOLUTION C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g) 1. First, we convert grams of C3H8 to moles using the molar mass of C3H8. 2. Next, we convert moles of C3H8 to moles of O2 using the stoichiometric relationship that 1 mol C3H8 = 5 mol O2. 3. Finally, we convert moles of O2 to grams using the molar mass of O2. Solution: 1 𝑚𝑜𝑙 𝐶3𝐻8 5 𝑚𝑜𝑙 𝑂2 31.998 𝑔 𝑂2 3.50 g C3H8 x x x = 44.097 𝑔 𝐶3𝐻8 1 𝑚𝑜𝑙 𝐶3𝐻8 1 𝑚𝑜𝑙 𝑂2 = 12.69848289 ≈ 12.7 g O2 The amount of O2 consumed from combustion of 3.50 g of propane is 12.7 g. EXAMPLE # 3 Solid lithium hydroxide is used in space vehicles to remove the carbon dioxide gas exhaled by astronauts. The hydroxide reacts with the carbon dioxide to form solid lithium carbonate and liquid water. How many grams of carbon dioxide can be absorbed by 1.00 g of lithium hydroxide? Before solving this, elucidate the actual chemical equation first and balance it: LiOH(s) + CO2(g) → Li2CO3 (s) + H2O(l) Balancing the equation, we get: 2 LiOH(s) + CO2(g) → Li2CO3 (s) + H2O(l) SOLUTION 2 LiOH(s) + CO2(g) → Li2CO3 (s) + H2O(l) 1. First, we convert grams of LiOH to moles using the molar mass of LiOH. 2. Next, we convert moles of LiOH to moles of CO2 using the stoichiometric relationship that 2 mol LiOH = 1 mol CO2. 3. Finally, we convert moles of CO2 to grams using the molar mass of CO2. Solution: 1 𝑚𝑜𝑙 𝐿𝑖𝑂𝐻 1 𝑚𝑜𝑙 𝐶𝑂2 44.009 𝑔 𝐶𝑂2 1.00 g LiOH x x x 23.948 𝑔 𝐿𝑖𝑂𝐻 2 𝑚𝑜𝑙 𝐿𝑖𝑂𝐻 1 𝑚𝑜𝑙 𝐶𝑂2 = 0.9188449975 ≈ 0.919 g CO2 = The amount of CO2 absorbed by 1.00 g of lithium hydroxide is 0.919 g. YOUR TURN Calcium hydride reacts with water to form calcium hydroxide and hydrogen gas. How many grams of calcium hydride are needed to form 4.50 g of hydrogen? CaH2 + H2O → Ca(OH)2 + H2 SOLUTION CaH2 +2 H2O → Ca(OH)2 + 2 H2 1. First, we convert grams of H2 to moles using the molar mass of H2. 2. Next, we convert moles of H2 to moles of CaH2 using the stoichiometric relationship that 2 mol H2 = 1 mol CaH2. 3. Finally, we convert moles of CaH2 to grams using the molar mass of CaH2. Solution: 1 𝑚𝑜𝑙 𝐻2 1 𝑚𝑜𝑙 𝐶𝑎𝐻2 42.094 𝑔 𝐶𝑎𝐻2 4.50 g H2 x x x 1 𝑚𝑜𝑙 𝐶𝑎𝐻 2.016 𝑔 𝐻2 2 𝑚𝑜𝑙 𝐻2 2 = 46.97991071 ≈ 47.0 g CaH2 = The amount of CaH2 needed to produce 4.50 g of hydrogen gas is 47.0 g. LIMITING AND EXCESS REACTANTS Often, the reactants used in a chemical reaction are not present in precise stoichiometric amounts. For example, a natural gas-fired power plant generates electricity by producing hot gases that drive turbines, predominantly through the following combustion reaction: Power plants typically operate with an excess of O2(g) to achieve the maximum energy from the hydrocarbon fuel and minimize production of harmful byproducts, like carbon monoxide, that result from incomplete combustion. Consequently, the amount of CH4 introduced determines how much CO2 and H2O water are produced, as well as the amount of energy released LIMITING AND EXCESS REACTANTS The limiting reactant is the reactant present in the smallest stoichiometric amount. The other reactants are sometimes called excess reactants. EXAMPLE # 1 The most important commercial process for converting N2 from the air into nitrogen-containing compounds is based on the reaction of N2 and H2 to form ammonia (NH3): N2(g) + 3H2(g) → 2NH3(g) How many moles of NH3 can be formed from 3.0 mol of N2 and 6.0 mol of H2? SOLUTION: N2(g) + 3H2(g) → 2NH3(g) If we assume one reactant is completely consumed, we can calculate how much of the second reactant is needed. By comparing the calculated quantity of the second reactant with the amount available, we can determine which reactant is limiting. We then proceed with the calculation, using the quantity of the limiting reactant. 3 𝑚𝑜𝑙 𝐻2 3.0 mol N2 x 1 𝑚𝑜𝑙 𝑁 = 9.0 mol H2 2 Because only 6.0 mol H2 is available, we will run out of H2 before the N2 is gone, which tells us that H2 is the limiting reactant. Therefore, we use the quantity of H2 to calculate the quantity of NH3 produced. 2 𝑚𝑜𝑙 𝑁𝐻3 6.0 mol H2 x 3 𝑚𝑜𝑙 𝐻 = 4.0 mol NH3 2 4.0 mol of NH3 will be produced from 3.0 mol of N2 and 6.0 mol of H2 EXAMPLE # 2 The reaction 2 H2(g) + O2(g) → 2 H2O(g) is used to produce electricity in a hydrogen fuel cell. Suppose a fuel cell contains 150 g of H2(g) and 1500 g of O2(g). How many grams of water can form? SOLUTION 2 H2(g) + O2(g) → 2 H2O(g) First, let’s get the moles for each reactant: 1 𝑚𝑜𝑙 𝐻2 1 𝑚𝑜𝑙 𝑂2 150 g H2 x 1500 g O2 x 2.016 𝑔 𝐻2 31.998 𝑔 𝑂2 =74.4047619 mol H2 =46.87792987 mol O2 Now we get the mole/coefficient (MC) ratio. The smallest ratio is the limiting reactant: 74.4047619 𝑚𝑜𝑙 𝐻2 46.87792987 𝑚𝑜𝑙 𝑂2 2 𝑚𝑜𝑙 𝐻2 1 𝑚𝑜𝑙 𝑂2 =37.20238095 =46.87792987 Since we know that H2 is the limiting reactant, we use it for the stoichiometric calculation: 1 𝑚𝑜𝑙 𝐻2 2 𝑚𝑜𝑙 𝐻2𝑂 18.015 𝑔 𝐻 𝑂 The amount of water that 150 g H2 x x x 2 can be formed from 150 g 2.016 𝑔 𝐻2 2 𝑚𝑜𝑙 𝐻2 1 𝑚𝑜𝑙 𝐻2𝑂 of H2 and 1500 of O2 is 1.3 1340.401786 ≈ 1.3 x 103 g H O 2 = x 103 g YOUR TURN Molten gallium reacts with arsenic to form the semiconductor, gallium arsenide, GaAs, used in light–emitting diodes and solar cells: Ga(l) + As(s) → GaAs(s) If 4.00 g of gallium is reacted with 5.50 g of arsenic, how many grams of the excess reactant are left at the end of the reaction? SOLUTION 1 Ga(l) + As(s) → GaAs(s) First, let’s get the moles for each reactant: 1 𝑚𝑜𝑙 𝐺𝑎 1 𝑚𝑜𝑙 𝐴𝑠 4.00 g Ga x 5.50 g As x 69.723 𝑔 𝐺𝑎 74.922 𝑔 𝐴𝑠 =0.05736987795 mol Ga = 0.0734096794 mol As Now we get the mole/coefficient (MC) ratio. The smallest ratio is the limiting reactant: 0.05736987795 𝑚𝑜𝑙 𝐺𝑎 0.0734096794 𝑚𝑜𝑙 𝐴𝑠 1 𝑚𝑜𝑙 𝐺𝑎 1 𝑚𝑜𝑙 𝐴𝑠 = 0.05736987795 = 0.0734096794 Since Ga is the limiting reactant, we need to compute the amount of As used up in the reaction: 1 𝑚𝑜𝑙 𝐺𝑎 1 𝑚𝑜𝑙 𝐴𝑠 74.922 𝑔 𝐴𝑠 4.00 g Ga x x 1 𝑚𝑜𝑙 𝐺𝑎 x =4.298265996 g As = 69.723 𝑔 𝐺𝑎 1 𝑚𝑜𝑙 𝐴𝑠 The amount of excess Subtract the amount of As used up in the reaction from the initial amount: arsenic is 1.20 g. 5.50 g As - 4.298265996 g As =1.201734004 ≈ 1.20 g As SOLUTION 2 Ga(l) + As(s) → GaAs(s) First, let’s get the moles for each reactant: 1 𝑚𝑜𝑙 𝐺𝑎 1 𝑚𝑜𝑙 𝐴𝑠 4.00 g Ga x 5.50 g As x 69.723 𝑔 𝐺𝑎 74.922 𝑔 𝐴𝑠 =0.05736987795 mol Ga = 0.0734096794 mol As Now we get the mole/coefficient (MC) ratio. The smallest ratio is the limiting reactant: 0.05736987795 𝑚𝑜𝑙 𝐺𝑎 0.0734096794 𝑚𝑜𝑙 𝐴𝑠 1 𝑚𝑜𝑙 𝐺𝑎 1 𝑚𝑜𝑙 𝐴𝑠 = 0.05736987795 = 0.0734096794 Since Ga is the limiting reactant, we need to subtract the MC ratio of Ga from excess As: 0.0734096794 - 0.05736987795 = 0.01603980145 This will serve as a multiplier for the computation of mass of excess As: The amount of excess arsenic is 1.20 g. 0.01603980145 x 1 mol As x 74.922 g As/mol As 1.201734004 ≈ 1.20 g As THEORETICAL AND ACTUAL YIELD The quantity of product calculated to form when all the limiting reactant is consumed is called the theoretical yield. The amount of product actually obtained, called the actual yield , is almost always less than (and can never be greater than) the theoretical yield. There are many reasons for this difference. Some of the reactants may not react, for example, or they may react in a way different from that desired (side reactions). In addition, it is not always possible to recover all of the product from the reaction mixture. The percent yield of a reaction is the actual yield divided by the theoretical yield, multiplied by 100 to convert to percent: EXAMPLE # 1 Adipic acid, H2C6H8O4, used to produce nylon, is made commercially by a reaction between cyclohexane (C6H12) and O2: 2 C6H12(l) + 5 O2(g) → 2 H2C6H8O4(l) + 2 H2O (g) a) Assume that you carry out this reaction with 25.0 g of cyclohexane and that cyclohexane is the limiting reactant. What is the theoretical yield of adipic acid? b) If you obtain 33.5 g of adipic acid, what is the percent yield for the reaction? SOLUTION FOR (a) 2 C6H12(l) + 5 O2(g) → 2 H2C6H8O4(l) + 2 H2O (g) The theoretical yield is just a simple stoichiometric calculation: 1 𝑚𝑜𝑙 𝐶6𝐻12 2 𝑚𝑜𝑙 𝐻2𝐶6𝐻8𝑂4 146.142 𝑔 𝐻2𝐶6𝐻8𝑂4 25.0 g C6H12 x x x 84.162 𝑔 𝐶6𝐻12 2 𝑚𝑜𝑙 𝐶6𝐻12 1 𝑚𝑜𝑙 𝐻2𝐶6𝐻8𝑂4 = = 43.41092179 ≈ 43.4 g H2C6H8O4 The theoretical yield of adipic acid upon carrying out the reaction with 25.0 g cyclohexane is 43.4 g. SOLUTION FOR (b) The percent yield is calculated by using the equation below to compare the given actual yield (33.5 g) with the theoretical yield. 33.5 𝑔 𝐻2𝐶6𝐻8𝑂4 x 100 % = 43.41092179 𝑔 𝐻2𝐶6𝐻8𝑂4 = 77.16952006 ≈ 77.2 % The percent yield of adipic acid is 77.2% YOUR TURN If 3.00 g of titanium metal is reacted with 6.00 g of chlorine gas, Cl2, to form 7.7 g of titanium(IV) chloride in a combination reaction, what is the percent yield of the product?