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Chemistry 201 Lecture 1 Conversions Stoichiometry Balancing Equations NC State University The mass of a proton The mass of a proton is: 1.6726231 × 10−27 kg or 1.6726231 × 10−24 grams We know this value accurately because of mass spectrometry. The number cited here has eight significant figures. We do not usually need this precision, so we often write the value as 1.67 × 10−24 grams to three significant figures. Significant figures The number of significant figures is equal to the number of digits in a measured or calculated value that contribute to its precision. Precision refers to the ability to reproducibly measure or calculate a value. If we use three significant figures, it suggests that we can reproducibly measure the value to within about 1 part in 100 or with an accuracy of 1%. This is the most common number of significant figures and this will be the default in this course (unless otherwise specified). Example The mass of the electron is reported to be: 9.1093819 × 10-31 kg Write this number to three significant figures. Example The mass of the electron is reported to be: 9.1093819 × 10-31 kg Write this number to three significant figures. Solution: The specified value should be as close as possible to the true value so we should round-off the last digit. In this case we round up to obtain 9.11 × 10-31 kg Example The mass of a neutron is approximately equal to sum of the masses of an electron and a proton. Give the neutron mass to three significant figures. Example The mass of a neutron is approximately equal to sum of the masses of an electron and a proton. Give the neutron mass to three significant figures. Solution: Start with the accurate values: 1.6726231 × 10−27 kg 9.1093819 × 10-31 kg Example The mass of a neutron is approximately equal to sum of the masses of an electron and a proton. Give the neutron mass to three significant figures. Solution: Start with the accurate values: 1.6726231 × 10−27 kg 9.1093819 × 10-31 kg Add them together 1.6735340 × 10−27 kg Example The mass of a neutron is approximately equal to sum of the masses of an electron and a proton. Give the neutron mass to three significant figures. Solution: Start with the accurate values: 1.6726231 × 10−27 kg 9.1093819 × 10-31 kg Add them together 1.6735340 × 10−27 kg Round of to give 1.67 × 10−27 kg Example The actual value of the neutron mass is: 1.6749286 × 10−27 kg Calculate the difference between the value you obtained by summing the proton and electron masses. How many significant figures are possible in your answer? Example The actual value of the neutron mass is: 1.6749286 × 10−27 kg Calculate the difference between the value you obtained by summing the proton and electron masses. How many significant figures are possible in your answer? Solution: Neutron 1.6749286 × 10−27 kg Proton + electron 1.6735340 × 10−27 kg Difference 1.3946 × 10−30 kg There are 5 significant figures in the answer. Conversion factors for atomic mass The sum of the mass of a proton and an electron is the mass of a hydrogen atom. Question: how many hydrogen atoms are there in a gram of H atoms (to 3 significant figures)? Conversion factors for atomic mass The sum of the mass of a proton and an electron is the mass of a hydrogen atom. Question: how many hydrogen atoms are there in a gram of H atoms (to 3 significant figures)? Answer: The calculated value actually has units 1.6735340 × 10−27 kg/atom Or 1.6735340 × 10−24 grams/atom Therefore, we can invert it to find, 5.97(5) × 1023 atoms/gram Atomic mass unit When we consider all of the atoms in the periodic table, the average mass of a nucleon is considered to be: 1.6605388 × 10−24 grams/nucleon We call this the atomic mass unit. We can use this value to convert atomic masses to grams or vice versa. We write the conversion as, 1.6605388 × 10−24 grams/amu To employ this value we calculate the atomic mass of an atom or molecule and then we can calculate the weight in grams using this formula. Avagadro’s number If we invert average atomic mass ___________1_____________ 1.6605388 × 10−24 grams/amu We obtained the number of particles with a given amu per gram. This number is called Avagradro’s number and is given the symbol NA. NA = 6.022141 × 1023 amu/gram This number gives the number of particles for which an atomic mass has the the same value in grams. 1 H atom = 1 amu NA H atoms = 1 gram 1 C atom = 12 amu NA H atoms = 12 grams Example How much does a molecule of pyridine weigh? (Ummm… all right, what is its mass?) Example How much does a molecule of pyridine weigh? (Ummm… all right, what is its mass?) Solution: First, we find the chemical formula for pyridine. C5H5N H H H H H N Example How much does a molecule of pyridine weigh? (Ummm… all right, what is its mass?) Solution: Second, we look up the atomic masses in the periodic table. atomic mass = 5(12) + 5 + 14 = 79 amu Third, we use the conversion factor to calculate the mass in grams (79 amu)x (1.66 × 10−24 grams/amu) = 1.31 × 10−23 grams Avogadro’s number It is a bit difficult to weigh out 10−23 grams. H H H H H N Avogadro’s number Instead, let’s ask how many molecules it takes to convert the atomic mass to its value in grams. For example, using the grams/amu conversion, let’s calculate how many hydrogen atoms have the mass of 1 gram. Avogadro’s number Instead, let’s ask how many molecules it takes to convert the atomic mass to its value in grams. For example, using the grams/amu conversion, let’s calculate how many hydrogen atoms have the mass of 1 gram. Answer: since hydrogen weighs 1 amu, its mass is 1.6605388 × 10−24 grams/atom We can invert this value to find the number of atoms per gram. 6.0221417 × 1023 atoms/gram The mole Since this number converts from atoms to gram for hydrogen, we can see that it can be used to give the number of atoms for any formula weight (i.e. molecular weight of a compound given in grams). For example, pyridine has a formula weight of 91 grams. Therefore, There are 6.0221417 × 1023 molecules/91 grams of pyridine. Because of the importance of this number of atoms or molecules we give the name, mole. 1 mole = 6.0221417 × 1023 molecules or to 3 significant figures. 1 mole = 6.02 × 1023 molecules Avogadro’s number as a conversion factor Given the definition, 1 mole = 6.02 × 1023 molecules We can see that Avogadro’s number converts from molecules to moles. 6.02 × 1023 molecules/mole Atomic and molecular weight The atomic weight is the numerical value tabulated for the mass of each atom in the periodic table in atomic units. The use of the word “weight” is not precise here since weight in physics represents a force (w = mg). However, the name atomic weight is so ingrained that we will not attempt to change it. We use the periodic table to calculate the molar mass as follows: for H2SO4 we find the atomic weights, H = 1, S = 32 and O = 16. Molar mass = 2(1 g/mol) + 32 g/mol + 4(16 g/mol) = 98 g/mol Molecular and empirical formulae We can use the mass (weight) of a molecule as one constraint on the formula of a compound (molecule). Some molecules have a simpler empirical formula, which does not correspond to the molecular formula. For example, all carbohydrates have the empirical formula (CH2O). However, the various sugars and other Carbohydrates can differing total numbers of atoms so That the molecular formula is (CH2O)n, where n = 6 for glucose and many simple sugars. In general, Absolute Temperature We will use the absolute temperature scale (Kelvin) for all chemical calculations. Why? One important reason is that the absolute temperature is proportional to the kinetic energy of a substance. K stands for kinetic energy, R is the gas constant and n is the number of moles. N is the number of molecules. Temperature scales The absolute scale in Kelvins is offset from the Celsius scale by 273.16 degrees, meaning that 0 oC ~ 273 K to three significant figures. This value is accurate enough for our purposes. Therefore, we can use the formula The Celsius scale is used by every country in the world as the temperature scale (except the United States). We use the Fahrenheit scale. Conversion from Fahrenheit to Celsius The zero of the Celsius scale occurs at 32 oF and the The boiling point of water (100 oC) occurs as 212 oF. This means that one degree Celsius is exacly 9/5 times one degree Fahrenheit. For common values it is useful to have recall that 50 oF = 10 oC, 68 oF = 20 oC and 86 oF = 30 oC. Body temperature is exactly 98.6 oF = 37.00 oC. Ideal gas law The number of moles, n, appears in the ideal gas law: This is an equation of state, which means it relates the variables of state, pressure, volume, and temperature. R is a constant known as the universal gas constant. R = 8.31 J/mol-K or R = 0.08206 L-atm/mol-K Note that the units of the ideal gas law are units of energy. The ideal gas law as an energy equation At first PV may not look like an energy. However, when a fuel is combusted (for example in the cylinder of a car engine), it builds up a pressure, which causes an expansion, an increase in volume. Pressure-volume work is the how engines propel cars. Temperature also represents an energy. The temperature of a gas is proportional to the kinetic energy of the gas molecules. nRT is actually an energy term as can be seen from the units of R. Solving for the number of moles We can use the ideal gas law to obtain the number of moles of a gas, provided we are given P, V and T. The formula is: For example, how moles are there in 22.4 liters of gas at 273 K and at sea level. Solution: At sea level, P = 1 atm so Molar gas volume The solution to the problem is n = 1.0. We conclude that one mole has a volume of 22.4 liters at 273 K. We call this the molar gas volume. Note that the molar gas volume changes with temperature. Problem: Calculate the molar gas volume at 373 K. Density The density of a material is defined as the mass per unit volume. Density is often given in grams per cubic centimeter (cc), which is the same thing as grams per milliliter (mL). Density is a material property, r. r(H2O) = 1 gm/mL. r(C2H5OH) = 0.789 gm/mL. r(Pb) = 11.34 gm/mL. r(Au) = 19.30 gm/mL. Archimedes’ dilemma ? Archimedes’ dilemma It is easy to find the mass of the crown, but how does one measure its density? Archimedes’ solution If he seems unnecessarily excited, remember that the penalty for failing to solve the problem was death. Archimedes’ principle The crown displaces a volume of water equal to its own volume. It is easy to figure out the volume of the displaced water. Next Archimedes realized that this explains how ships can float. The buoyancy of an object is equal the weight of the water it displaces. Concentration: molarity The molarity is defined as the number of moles per liter. We most often consider the molarity of the solute. For example, if we dissolve 40 grams of NaCl in one liter of H2O we have: In thinking about conversions it can also be useful to understand the molarity of the solvent. What is the molarity of H2O? Concentration: molality The molality is defined as the number of moles per kg of solvent. You might ask why there are two different units of concentration (molarity and molality). The answer is that these have different applications. The molality is used for the colligative properities while the molarity is used for many solutions in the laboratory where the volume is the common means of measuring and dispensing solutions. Concentration: mole fraction The mole fraction is the number of moles of a given substance divided by the total number of moles present in the system. The convert from one of these to the other remember that the ratio of each of these quantities are equal: Chemical equations A chemical equation tells us how one set of species converts into another. As bonds are broken and reformed the numbers of atoms in a compound will change. The number of molecules that combine must account for this so that there is conservation of mass. H2 + O2 H2O This equation is not balanced since there is an extra O atom on the left hand side. Balancing chemical equations Think of the equation as an algebraic equation a H2 + b O2 = x H2O In order for this equation to be satisfied, the following must be true: H: 2a = 2x O: 2b = x Actually, there are an infinite number of solutions. We just want the most practical one. Let x = 2, then b = 1 and a = 2. Balancing chemical equations The balanced equation is: 2 H2 + O2 = 2 H2O Of course, we could have chosen x = 1. Then, the equation would be: H2 + ½ O2 = H2O Often we choose the value that gives all integers for the coefficients, but this is not required. General Method: Example Let’s consider the combustion of ethane: C2H6 + O2 = H2O + CO2 Step 1. apply coefficients a C2H6 + b O2 = x H2O + y CO2 Step 2. Write down an equation for each element C: 2a = y H: 6a = 2x O: 2b = x + 2y Step 3. Choose a value that will give a an integer answer (if possible on the first try). General Method: Example We are balancing a C2H6 + b O2 = x H2O + y CO2 We are on step 3. Make a first trial guess… C: 2a = y H: 6a = 2x x = 3 gives a = 1, and y = 2 O: 2b = x + 2y General Method: Example We are balancing a C2H6 + b O2 = x H2O + y CO2 We are on step 3. Make a first trial guess… C: 2a = y H: 6a = 2x x = 3 gives a = 1, and y = 2 O: 2b = x + 2y therefore 2b = 3 + 2(2) = 7 C2H6 + 7/2 O2 = 3 H2O + 2 CO2 Or 2 C2H6 + 7 O2 = 6 H2O + 4 CO2 General Method: Example Let’s consider the combustion of pyridine: C5H5N + O2 = H2O + CO2 + NO Step 1. apply coefficients a C5H5N + b O2 = x H2O + y CO2 + z NO Step 2. Write down an equation for each element C: 5a = y H: 5a = 2x O: 2b = x + 2y + z N: a = z Step 3. Choose a value that will give a an integer answer (if possible on the first try). General Method: Example We are balancing a C6H5N + b O2 = x H2O + y CO2 + z NO We are on step 3. Make a first trial guess for the coefficient. C: 5a = y H: 5a = 2x O: 2b = x + 2y + z N: a = z General Method: Example We are balancing a C6H5N + b O2 = x H2O + y CO2 + z NO We are on step 3. Make a first trial guess… C: 5a = y H: 5a = 2x x = 5 gives a = 2, and z = 2, y = 10 O: 2b = x + 2y + z N: a = z General Method: Example We are balancing a C6H5N + b O2 = x H2O + y CO2 + z NO We are on step 3. Make a first try. C: 5a = y H: 5a = 2x x = 5 gives a = 2, and z = 2, y = 10 O: 2b = x + 2y + z 2b = 5 + 20 + 2 = 29, b = 27/2 N: a = z We have acceptable values if we can live with noninteger coefficients. 2 C6H5N + 27/2 O2 = 5 H2O + 10 CO2 + 2 NO General Method: Example We can multiply through by 2 to eliminate the noninteger coefficient. 4 C6H5N + 27 O2 = 10 H2O + 20 CO2 + 4 NO While there is some arbitrariness in the choice, this is inherent since the solutions are not unique. Only the relative values of the coefficients are determined by the stoichiometry, but not their absolute values. DMSO Example The combustion of dimethyl sulfoxide is given by C2H6SO + O2 = H2O + CO2 + SO3 DMSO Example The combustion of dimethyl sulfoxide is given by C2H6SO + O2 = H2O + CO2 + SO3 Step 1. Apply coefficients a C2H6SO + b O2 = x H2O + y CO2 + z SO3 Step 2. Write an equation for each element C: 2a = y H: 6a = 2x S: a = z O: a + 2b = x + 2y + 3z DMSO Example We are balancing a C2H6SO + b O2 = x H2O + y CO2 + z SO3 Step 3. Make a guess to try and get integer values C: 2a = y H: 6a = 2x S: a = z O: a + 2b = x + 2y + 3z DMSO Example We are balancing a C2H6SO + b O2 = x H2O + y CO2 + z SO3 Step 3. Make a guess to try and get integer values C: 2a = y H: 6a = 2x x = 3, a = 1 S: a = z O: a + 2b = x + 2y + 3z DMSO Example We are balancing a C2H6SO + b O2 = x H2O + y CO2 + z SO3 Step 3. Make a guess to try and get integer values C: 2a = y H: 6a = 2x x = 3, a = 1, then y = 2, z = 1 S: a = z O: a + 2b = x + 2y + 3z DMSO Example We are balancing a C2H6SO + b O2 = x H2O + y CO2 + z SO3 Step 3. Make a guess to try and get integer values C: 2a = y H: 6a = 2x x = 3, a = 1, then y = 2, z = 1 S: a = z O: a + 2b = x + 2y + 3z 2b = x + 2y + 3z – a = 3 + 2(2) + 3(1) – 1 = 9, so b = 9/2 DMSO Example We are balancing a C2H6SO + b O2 = x H2O + y CO2 + z SO3 Step 3. Make a guess to try and get integer values C: 2a = y H: 6a = 2x x = 6, a = 2, then y = 4, z = 2 S: a = z O: a + 2b = x + 2y + 3z 2b = x + 2y + 3z – a = 6 + 2(4) + 3(2) – 2 = 18, so b = 9 and 2 C2H6SO + 9 O2 = 6 H2O + 4 CO2 + 2 SO3 Stoichiometric ratios Once we have balanced a chemical equation, we have determined the fixed ratios in which the various compounds must react in order for the reaction to proceed. These known as stoichiometric ratios. For example, if we return to the reaction, 2 C2H6SO + 9 O2 = 6 H2O + 4 CO2 + 2 SO3 We see that 9 O2 must react for every 2 C2H6SO. The ratio is: Stoichiometric ratios Similarly, we can state that 6 moles of H2O are produced for every two moles of C2H6SO consumed. Thus, the stoichiometric ratio is 3 for H2O relative to C2H6SO. 2 C2H6SO + 9 O2 = 6 H2O + 4 CO2 + 2 SO3