General Chemistry PDF 2024/2025

‫الوعهد التكنىلىجي العالي‬ ‫جوهىريت هصر العربيت‬ ‫للعلىم الصحيت التطبيقيت ببني سىيف‬ ‫وزارة التعلين العالي‬ ‫‪General Chemistry‬‬ ‫‪Col...

‫الوعهد التكنىلىجي العالي‬ ‫جوهىريت هصر العربيت‬ ‫للعلىم الصحيت التطبيقيت ببني سىيف‬ ‫وزارة التعلين العالي‬ ‫‪General Chemistry‬‬ ‫‪Collected by‬‬ ‫‪Prof. Dr/ Hanafy M Abd El-Salam‬‬ ‫‪2024/2025‬‬  Contents Chapter 1: Some Basic Definitions: Mater, element, compound, physical and chemical properties. Chapter 2: Bonding between atoms and molecules Ionic bond, covalent bond, coordinate bond, metallic bond, H- bond and physical bonding Chapter 3: IS-Units their conversions and significant figures Chapter 4: Mole concepts. Mole conversions, chemical compositions, empirical and molecular formula, stoichiometry and chemical reaction. Chapter 5: Solutions Types of solutions, concentration expressions and colligative properties of solution. Practical part Referrences  Chapter 1: Some Basic Definitions: The definition of chemistry the study of the interactions of matter with other matter and with energy. Matters Matter is anything that has mass and takes up space. Physical properties They are characteristics that describe matter as it exists. Some physical characteristics of matter are shape, color, size, and temperature. An important physical property is the phase (or state) of matter. The three fundamental phases of matter are solid, liquid, and gas. Chemical properties They are characteristics of matter that describe how matter changes form in the presence of other matter. Does a sample of matter burn? Burning is a chemical property. Does it behave violently when put in water? This reaction is a chemical property. Physical change It occurs when a sample of matter changes one or more of its physical properties. For example, a solid may melt, or alcohol in a thermometer may change volume as the temperature changes. A physical change does not affect the chemical composition of matter. Chemical change It is the process of demonstrating a chemical property, such as the burning, its chemical composition changes, and new forms of matter with new physical properties are created. Note that chemical changes are frequently accompanied by physical changes, as the new matter will likely have different physical properties from the original matter. Substance It a sample of matter that has the same physical and chemical properties. Element It is the simplest type of chemical substance; it cannot be broken down into simpler chemical substances by ordinary chemical means. There are 118 elements known to science, of which 80 are stable. (The other elements are radioactive. Compound Compound is a combination of more than one element. The physical and chemical properties of a compound are different from the physical and chemical properties of its constituent elements; that is, it behaves as a completely different substance. There are over 50 million compounds known, and more are being discovered daily. Examples of compoundsinclude water, penicillin, and sodium,… The double helix structure maintained by hydrogen bonds between basepairs in helical chain is very important for life. The Kinetic Molecular Theory of Matter The postulates of the KMT: 1)Matter is composed of tiny particles (atoms, molecules or ions). 2)These particles are in constant, random motion and possess kinetic energy. 3) The particles also have potential energy due to intermolecularattractions. 4) The average kinetic energy increases as the temperature increases. 5)Energy is transferred from one particle to another during collisions, but at constant temperature the total energy of the system is constant. Kinetic energy is energy associated with motion; KE = ½mv2. If two particles of different masses are moving at the same speed, the heavier particle will have higher kinetic energy. Similarly, for particles of the same mass moving at the different speeds, the faster one will possess thegreater kinetic energy. Potential energy results from attractions of particles for each other. That is, particles under the influence of gravitational or electrical forces possess potential energy. Atoms in molecules possess potential energy because they are attracted to each other. Potential energy leads to cohesive forces in matters, which bring particles together, forming liquids and solids. Kinetic energy leads to disruptive forces in matters, which causes molecules to scatter and form gases. The state of a substance depends on the relative strength of the cohesive forces (potential energy) that hold particles together and the disruptive forces (kinetic energy) that tends to scatter them. Potential energy depends on molecular size and structures and is inherent properties of the molecules; they are independent of temperature. Whereas kinetic energy is temperature dependent (molecules move faster at higher temperature). Thus, temperature plays such an important role in determining the state ofmatter.  Chapter 2 Bonding between atoms and molecules Chemical Bonding Chemical bonding is Joining atoms together to form more complex structures. Bonds can be between atoms of the same element, or between atoms of different elements. There are several types of chemical bonding which have different properties and give rise to different structures. In many cases, atoms try to react to form valence shells containing eight electrons. The octet rule describes this, but it also has many exceptions Ionic bonding Ionic bonding occurs between positive ions (cations) and negative ions (anions). In an ionic solid, the ions arrange themselves into a rigid crystal lattice. NaCl (common salt) is an example of an ionic substance. In ionic bonding there is an attractive force established between large numbers of positive cations and negative anions, such that a neutral lattice is formed. This attraction between oppositely-charged ions is collective in nature and called ionic bonding. Covalent bonding Covalent bonding occurs when the orbitals of two non-metal atoms physically overlap and share electrons with each other. There are two types of structures to which this can give rise: molecules and covalent network solids. Methane (CH4) and water (H2O) are examples of covalently bonded molecules, and glass is a covalent network solid. Metallic bonding Metallic bonding occur between atoms that have few electrons compared to the number of accessible orbitals. This is true for the vast majority of chemical elements. In a metallically bonded substance, the atoms' outer electrons are able to freely move around - they are delocalised to form an 'electron pool'. Iron is a metallically bonded substance. Ionic Bonding Ions are atoms or molecules which are electrically charged. Cations are positively charged and anions carry a negative charge. Ions form when atoms gain or lose electrons. Since electrons are negatively charged, an atom that loses one or more electrons will become positively charged; an atom that gains one or more electrons becomes negatively charged. Ionic bonding occurs between positive ions (cations) and negative ions (anions). In an ionic solid, the ions arrange themselves into a rigid crystal lattice. NaCl (common salt) is an example of an ionic substance. In ionic bonding there is an attractive force established between large numbers of positive cations and negative anions, such that a neutral lattice is formed. This attraction between oppositely-charged ions is collective in nature and called ionic bonding. The smallest unit of an ionic compound is the formula unit, but this unit merely reflects that ratio of ions that leads to neutrality of the whole crystal, e.g. NaCl or MgCl2. One cannot distinguish individual NaCl or MgCl2 molecules in the structure. Covalent bonding. Covalent bonding occurs when pairs of electrons are shared by atoms. By sharing their outer most (valence) electrons, atoms can fill up their outer electron shell and gain stability. Nonmetals will readily form covalent bonds with other nonmetals between one to three covalent bonds with other nonmetals depending on how many valence electrons they posses. The Octet Rule requires all atoms in a molecule to have 8 valence electrons--either by sharing, losing or gaining electrons--to become stable. For Covalent bonds, atoms tend to share their electrons with each other to satisfy the Octet Rule. It requires 8 electrons because that is the amount of electrons needed to fill a s- and p- orbital (electron configuration); also known as a noble gas configuration. Each atom wants to become as stable as the noble gases that have their outer valence shell filled because noble gases have a charge of 0. Although it is important to remember the "magic number", 8, note that there are many Octet rule exceptions. Types of covalent bonds Single covalent Bonds A single bond is formed when two (one pair) of electrons are shared between two atoms. It is depicted by a single line between the two atoms. Although this form of bond is weaker and has a smaller density than a double bond and a triple bond, it is the most stable because it has a lower level of reactivity. Example 1: HCl (H-Cl) Hydrogen Chloride has 1 Hydrogen atom and 1 Chlorine atom. Hydrogen has only 1 valence electron whereas Chlorine has 7 valence electrons. To satisfy the Octet Rule, each atom gives out 1 electron to share with each other; thus making a single bond. Double covalent Bonds A Double bond is formed when two atoms share two pairs of electrons with each other. It is depicted by two horizontal lines between two atoms in a molecule. This type of bond is much stronger than a single bond, but less stable; this is due to its greater amount of reactivity compared to a single bond. Example 2: CO2 (O=C=O) Carbon dioxide has a total of 1 Carbon atom and 2 Oxygen atoms. Each Oxygen atom has 6 valence electrons whereas the Carbon atom only has 4 valence electrons. To satisfy the Octet Rule, Carbon needs 4 more valence electrons. Since each Oxygen atom has 3 lone pairs of electrons, they can each share 1 pair of electrons with Carbon; as a result, filling Carbon's outer valence shell (Satisfying the Octet Rule). Triple covalent Bonds A Triple bond is formed when three pairs of electrons are shared between two atoms in a molecule. It is the least stable out of the three general types of covalent bonds. Example 3: Acetylene H-C≡C-H Acetylene has a total of 2 Carbon atoms and 2 Hydrogen atoms. Each Hydrogen atom has 1 valence electron whereas each Carbon atom has 4 valence electrons. Each Carbon needs 4 more electrons and each Hydrogen needs 1 more electron. Hydrogen shares its only electron with Carbon to get a full valence shell. Now Carbon has 5 electrons. Because each Carbon atom has 5 electrons--1 single bond and 3 unpaired electrons--the two Carbons can share their unpaired electrons, forming a triple bond. Now all the atoms are happy with their full outer valence shell. Theories of Formation of Covalent bonding 1- Valence Bond Theory The theory assumes that electrons occupy atomic orbitals of individual atoms are attracted to the nucleus of another atom. This attraction increases as the atoms approach one another until the atoms reach a minimum distance where the electron density begins to cause repulsion. The covalent bond is formed by combination of the single electron occupied atomic orbitals in a chemical bonding In polyatomic covalent molecules, the central atom prepares a set of molecular orbitals equal to the number of required bonds to get the octet state via a process known as hyberdization Hybridization Hybridization is a simple model that deals with mixing orbitals to from new, hybridized, orbitals. Properties of hybrid orbitals The following ideas are important in understanding hybridization: 1. Hybrid orbitals do not exist in isolated atoms. They are formed only in covalently bonded atoms. 2. Hybrid orbitals have shapes and orientations that are very different from those of the atomic orbitals in isolated atoms. 3. A set of hybrid orbitals is generated by combining atomic orbitals. The number of hybrid orbitals in a set is equal to the number of atomic orbitals that were combined to produce the set. 4. All orbitals in a set of hybrid orbitals are equivalent in shape and energy. 5. The type of hybrid orbitals formed in a bonded atom depends on its electron-pair geometry as predicted by the VSEPR theory. 6. Hybrid orbitals overlap to form σ bonds. Unhybridized orbitals overlap to form π bonds. Types of Hybridization For polyatomic molecules we would like to be able to explain: The number of bonds formed Their geometries sp3 Hybridization An example of this is methane (CH4). This is the combination of one s orbital and three p orbitals. The hybrid orbital has 25%s and 75% p sp2 Hybridization An example of this is BF3. This is the combination of one s orbital and two p orbitals. The hybrid orbital has 33.4%s and 66.6% p Boron electron configuration: The three sp2 hybrid orbitals have a trigonal planar arrangement to minimize electron repulsion. sp Hybrid Orbitals sp Hybridization An example of this is BeF2. This combines one s orbital with one p orbital. This means that the s and p characteristics are equal and the hybrid orbital has 50%s and 50% p Molecular Orbitals Theory Chemical bonding occurs when the net attractive forces between an electron and two nuclei exceeds the electrostatic repulsion between the two nuclei. For this to happen, the electron must be in a region of space which we call the binding region. Conversely, if the electron is off to one side, in an anti-binding region, it actually adds to the repulsion between the two nuclei and helps push them away antibonding orbital empty. Since any orbital can hold a maximum of two electrons, the bonding orbital in H2+is only half-full nuclei by 270 kJ/mole, so we might expect two electrons to produce twicethis much stabilization, or 540 kJ/mole. Bond order is defined as the difference between the number of electron pairs occupying bonding and nonbonding orbitals in the molecule. A bond order of unity corresponds to a conventional "single bond".must go into the next higher slot— which turns out to be the sigma antibonding orbital. The presence of an electron in this orbital, as we have seen, gives rise to a repulsive component which acts against, and partially cancels out, the attractive effect of the filled bonding orbita. Taking our building-up process one step further, we can look at the possibilities of combining to helium atoms to form dihelium. You should now be able to predict that He2 cannot be a stable molecule; the reason, of course, is that we now have antibonding orbital, the attractive forces win out and we have a stable molecule. The bond energy of dilithium is 110 kJ/mole; notice that this value is less of its 1s orbital to a value well below that of the 1s hydrogen orbital. There are two occupied atomic orbitals on the lithium atom, and only antibonding orbitals— so we would expect the dicarbon molecule to be stable, and it is. (But being extremely reactive, it is known only in the gas phase.) You will recall that one pair of electrons shared between two atoms constitutes a ―single‖ smaller than the 945 kJ bond energy of N2— not surprising, considering that oxygen has two electrons in an antibonding orbital, compared to nitrogen’s one. The two unpaired electrons of the dioxygen molecule give this substanc Nonpolar Covalent Bond A Nonpolar Covalent Bond is created when atoms share their electrons equally. This usually occurs when two atoms have similar or the same electron affinity. The closer the values of their electron affinity, the stronger the attraction. This occurs in gas molecules; also known as diatomic elements where all the atoms have the same electronegativity since they are the same kind of element Examples of gas molecules that have a nonpolar covalent bond: Hydrogen gas atom, Nitrogen gas atoms, etc. Polar Covalent Bond A Polar Covalent Bond is created when the shared electrons between atoms are not equally shared. This occurs when one atom has a higher electronegativity than the atom it is sharing with. The atom with the higher electronegativity will have a stronger pull for electrons. As a result, the shared electrons will be closer to the atom with the higher electronegativity, making it unequally shared. A polar covalent bond will result in the molecule having a slightly positive side (the side containing the atom with a lower electronegativity) and a slightly negative side (containing the atom with the higher electronegativity) As a result of polar covalent bonds, the covalent compound will be slightly polar and have an electrostatic potential allowing it to form weak bonds with other polar molecules. Example: Water, Sulfide, Ozone, etc. Bonding between molecules Intermolecular forces- forces of attraction and repulsion between molecules that hold molecules, ions, and atoms together. There are many types of bonds that can be formed between two or more molecules, ions or atoms. Intermolecular forces cause molecules to be attracted or repulsed by each other. Often, these define some of the physical characteristics (such as the melting point) of a substance. A large difference in electronegativity between two bonded atoms will cause a permanent charge separation, or dipole, in a molecule or ion. Two or more molecules or ions with permanent dipoles can interact within dipole-dipole interactions. The bonding electrons in a molecule or ion will, on average, be closer to the more electronegative atom more frequently than the less electronegative one, giving rise to partial charges on each atom, and causing electrostatic forces between molecules or ions. Hydrogen bond The large difference in electronegativities between hydrogen and highly electronegative atoms such as fluorine, nitrogen and oxygen, coupled with their lone pairs of electrons cause strong electrostatic forces between molecules. Hydrogen bonds are responsible for the high boiling points of water and ammonia with respect to their heavier analogues. Types of hydrogen bond Intermolecular hydrogen bond This occurs when the hydrogen bonding is between H-atom of one molecule and an atom of the electronegative element of another molecule. For example (i) Hydrogen bond between the molecules of hydrogen fluoride. (ii) Hydrogen bond in alcohol or water molecules. Intermolecular hydrogen bond results into association of molecules. Hence, it usually increases the melting point, boiling point, viscosity, surface tension, solubility, surface tension, solubility, etc. Intremolecular hydrogen bond Intramolecular hydrogen bond This bond is formed between the hydrogen atom and an atom of the electronegative element (F, O, N) , of the same molecule. Intramolecular hydrogen bond results in the cyclization of the molecules and prevents their association. Consequently, the effect of this bond on the physical properties is negligible. For example, intramolecular hydrogen bonds are present in molecules such as o-nitrophenol, o-nitrobenzoic acid, etc. Intramolecular hydrogen bond Van der Waals Dispersion forces: The origin of van der Waals dispersion forces (London Forces) Temporary fluctuating dipoles In a symmetrical molecule like hydrogen, however, there is no electrical distortion to produce positive or negative parts. At an instance, one molecules is attracted to the nucleus of another one. This attraction increases until they reach a minimum distance where the electron density begins to cause repulsion. Due to this repulsion, one end of the molecule attains high electron desity and become - and the other end will be temporarily short of electrons and so becomes +. An instant later the electrons may well have moved up to the other end, reversing the polarity of the molecule. This even happens in monatomic molecules - molecules of noble gases, like helium, which consist of a single atom. This sets up an induced dipole in the approaching molecule, which is orientated in such a way that the + end of one is attracted to the - end of the other. The strength of dispersion forces Dispersion forces between molecules are much weaker than the covalent bonds within molecules. It isn't possible to give any exact value, because the size of the attraction varies considerably with the size of the molecule and its shape. How molecular size affects the strength of the dispersion forces The boiling points of the noble gases are helium -269°C neon -246°C argon -186°C krypton -152°C xenon -108°C radon -62°C All of these elements have monatomic molecules. The reason that the boiling points increase as you go down the group is that the number of electrons increases, and so also does the radius of the atom. The more electrons you have, and the more distance over which they can move, the bigger the possible temporary dipoles and therefore the bigger the dispersion forces. Because of the greater temporary dipoles, xenon molecules are "stickier" than neon molecules. Neon molecules will break away from each other at much lower temperatures than xenon molecules - hence neon has the lower boiling point. This is the reason that (all other things being equal) bigger molecules have higher boiling points than small ones. Bigger molecules have more electrons and more distance over which temporary dipoles can develop - and so the bigger molecules are "stickier".   Larger and heavier atoms and molecules exhibit stronger dispersion forces than smaller and lighter ones.  In a larger atom or molecule, the valence electrons are, on average, farther from the nuclei than in a smaller atom or molecule. They are less tightly held and can more easily form temporary dipoles.  The ease with which the electron distribution around an atom or molecule can be distorted is called the polarizability. London dispersion forces tend to be:  stronger between molecules that are easily polarized.  weaker between molecules that are not easily polarized.  Chapter 3 Unit Conversion & Significant Figures The process of conversion depends on the specific situation and the intended purpose. This may be governed by regulation, contract, technical specifications or other published standards. Engineering judgment may include such factors as:  The precision and accuracy of measurement and the associated uncertainty of measurement.  The statistical confidence interval or tolerance interval of the initial measurement.  The number of significant figures of the measurement.  The intended use of the measurement including the engineering tolerances.  Historical definitions of the units and their derivatives used in old measurements; e.g., international foot vs. US survey foot. Some conversions from one system of units to another need to be exact, without increasing or decreasing the precision of the first measurement, this is sometimes called soft conversion. It does not involve changing the physical configuration of the item being measured. By contrast, a hard conversion or an adaptive conversion may not be exactly equivalent. It changes the measurement to convenient and workable numbers and units in the new system. It sometimes involves a slightly different configuration, or size substitution, of the item. Nominal values are sometimes allowed and used. Conversion factors A conversion factor is used to change the units of a measured quantity without changing its value. The unity bracket method of unit conversion consists of a fraction in which the denominator is equal to the numerator, but they are in different units. Because of the identity property of multiplication, the value of a quantity will not change as long as it is multiplied by one. Also, if the numerator and denominator of a fraction are equal to each other, then the fraction is equal to one. So as long as the numerator and denominator of the fraction are equivalent, they will not affect the value of the measured quantity. The following example demonstrates how the unity bracket method is used to convert the rate 5 kilometers per second to meters per second. The symbols km, m, and s represent kilometer, meter, and second, respectively. (5 km / s )x (1000 m / 1 km) = (5000 m / s) 22 Thus, it is found that 5 kilometers per second is equal to 5000 meters per second. The Standard Units Scientists have agreed on a set of international standard units for comparing all our measurements called the SI units.  Système International = International System Quantity Unit Symbol length meter m mass kilogram kg time second s temperature kelvin K Length: 1) measure of the two-dimensional distance an object covers 2) often need to measure lengths that are very long (distances between stars) or very short (distances between atoms) 3) SI unit = meter 4) 1 meter = distance traveled by light in a specific period of time 5) commonly use centimeters (cm)  1 m = 100 cm  1 cm = 0.01 m = 10 mm  1 inch = 2.54 cm (exactly) Mass 1) measure of the amount of matter present in an object  Weight measures the gravitational pull on an object, which depends on its mass. 2) SI unit = kilogram (kg) 3) commonly measure mass in grams (g) or milligrams (mg)  1 kg = 2.2046 lb, 1 lb = 453.59 g  1 kg = 1000 g = 103 g  1 g = 1000 mg = 103 mg  1 g = 0.001 kg = 10−3 kg  1 mg = 0.001 g = 10−3 g Time 1) measure of the duration of an event 2) SI units = second (s) 3) 1 s is defined as the period of time it takes for a specific number of 23  radiation events of a specific transition from cesium-133. Temperature 1) measure of the average amount of kinetic energy  higher temperature = larger average kinetic energy 2) Heat flows from the matter that has high thermal energy into matter that has low thermal energy.  until they reach the same temperature 3) Heat is exchanged through molecular Temperature Scales Fahrenheit scale, °F  used in the U.S. Celsius scale, °C  used in all other countries Kelvin scale, K  absolute scale, no negative numbers  directly proportional to average amount of kinetic energy  0 K = absolute zero A Celsius degree is 1.8 times larger than a Fahrenheit degree. The standard used for 0 ° on the Fahrenheit scale is a lower temperature than the standard used for 0 ° on the Celsius scale. 24 The size of a ―degree‖ on the Kelvin scale is the same as on the Celsius scale. Though, technically, we don’t call the divisions on the Kelvin scale degrees; we call them kelvins! So 1 kelvin is 1.8 times larger than 1 °F. The 0 standard on the Kelvin scale is a much lower temperature than on the Celsius scale. Example 1. Convert 40.00 °C into K and °F Practice Convert 0.0 °F into Kelvin Related Units in the SI System All units in the SI system are related to the standard unit by a power of 10. 1) The power of 10 is indicated by a prefix multiplier. The prefix multipliers are always the same, regardless of the standard unit. Common Prefix Multipliers in the SI System Prefix Symbol Decimal Equivalent Power of 10 mega- M 1,000,000 Base x 106 kilo- k 1,000 Base x 103 deci- d 0.1 Base x 10−1 centi- c 0.01 Base x 10−2 milli- m 0.001 Base x 10−3 micro- m or mc 0.000 001 Base x 10−6 nano- n 0.000 000 001 Base x 10−9 0.000 000 000 pico p Base x 10−12 001 Volume measure of the amount of space occupied SI unit = cubic meter (m3) 25  commonly measure solid volume in cubic centimeters (cm3)  1 m3 = 106 cm3  1 cm3 = 10−6 m3 = 0.000 001 m3 commonly measure liquid or gas volume in milliliters (mL)  1 L is slightly larger than 1 quart  1 L = 1 dm3 = 1000 mL = 103 mL  1 mL = 0.001 L = 10−3 L  1 mL = 1 cm3 Common Units and Their Equivalents Length 1 kilometer (km) = 0.6214 mile (mi) 1 meter (m) = 39.37 inches (in) 1 meter (m) = 1.094 yards (yd) 1 foot (ft) = 30.48 centimeters (cm) 1 inch (in) = 2.54 centimeters (cm) exactly Mass 1 kilogram (km) = 2.205 pounds (lb) 1 pound (lb) = 453.59 grams (g) 1 ounce (oz) = 28.35 grams (g) Volume 1 liter (L) = 1000 milliliters (mL) 1 liter (L) = 1000 cubic centimeters (cm3) 26 1 liter (L) = 1.057 quarts (qt) 1 U.S. gallon (gal) = 3.785 liters (L) Significant Figures The non-place-holding digits in a reported measurement are called significant figures.  Some zeroes in a written number are there only to help you locate the decimal point. Significant figures tell us the range of values to expect for repeated measurements.  The more significant figures there are in a measurement, the smaller the range of values is. 12.30 cm 12.3 cm.has 4 sig. figs.has 3 sig. figs and its range is and its range is to 12.31 cm 12.29 to 12.4 cm 12.2 Rules of Counting Significant Figures 1) All nonzero digits are significant. So , 1.5 has 2 sig. figs. 2) Interior zeroes are significant. So, 1.05 has 3 sig. figs. 3) Leading zeroes are NOT significant. So, 0.001050 has 4 sig. figs.  1.050 x 10−3 4) Trailing zeroes may or may not be significant.  Trailing zeroes after a decimal point are significant.  1.050 has 4 sig. figs.  Trailing zeroes before a decimal point are significant if the decimal point is written.  150.0 has 4 sig. figs.  Zeroes at the end of a number without a written decimal point are ambiguous and should be avoided by using scientific notation. 27   if 150 has 2 sig. figs. then 1.5 x 102  but if 150 has 3 sig. figs. then 1.50 x 102 5) A number whose value is known with complete certainty is exact.  from counting individual objects  from definitions  1 cm is exactly equal to 0.01 m  from integer values in equations  in the equation for the radius of a circle, the 2 is exact Exact numbers have an unlimited number of significant figures. Example Determining the number of significant figures in a number How many significant figures are in each of the following? 0.04450 m 5.0003 km 10 dm = 1 m 1.000 × 105 s 0.00002 mm 10,000 m Multiplication and Division with Significant Figures When multiplying or dividing measurements with significant figures, the result has the same number of significant figures as the measurement with the fewest number of significant figures. 5.02 × 89.665 × 0.10 = 45.0118 = 45 3 sig. figs. 5 sig. figs. 2 sig. figs. 2 sig. figs. 5.892 ÷ 6.10 = 0.96590 = 0.966 28  4 sig. figs. 3 sig. figs. 3 sig. figs. Addition and Subtraction with Significant Figures When adding or subtracting measurements with significant figures, the result has the same number of decimal places as the measurement with the fewest number of decimal places. 5.74 + 0.823 + 2.651 = 9.214 = 9.21 2 dec. pl. 3 dec. pl. 3 dec. pl. 2 dec. pl. 4.8 − 3.965 = 0.835 = 0.8 1 dec. pl. 3 dec. pl. 1 dec. pl. Rounding When rounding to the correct number of significant figures, if the number after the place of the last significant figure is a) 0 to 4, round down Drop all digits after the last sig. fig. and leave the last sig. fig. alone. Add insignificant zeroes to keep the value if necessary. b) 5 to 9, round up Drop all digits after the last sig. fig. and increase the last sig. fig. by one.  Add insignificant zeroes to keep the value if necessary. To avoid accumulating extra error from rounding, round only at the end, keeping track of the last sig. fig. for intermediate calculations. rounding to two significant figures 2.34 rounds to 2.3 because the 3 is where the last sig. fig. will be and the number after it is 4 or less 2.37 rounds to 2.4 because the 3 is where the last sig. fig. will be and the number after it is 5 or greater 2.349865 rounds to 2.3 because the 3 is where the last sig. fig. will be and the number after it is 4 or less rounding to two significant figures 0.0234 rounds to 0.023 or 2.3 × 10−2 because the 3 is where the last sig. fig. will be and the number after it is 4 or less 0.0237 rounds to 0.024 or 2.4 × 10−2 because the 3 is where the last sig. fig. will be and the number after it is 5 or greater 0.02349865 rounds to 0.023 or 2.3 × 10−2 because the 3 is where the last sig. fig. will be and the number after it is 4 or less Both Multiplication/Division and Addition/Subtraction with Significant Figures: 29 When doing different kinds of operations with measurements with significant figures, do whatever is in parentheses first, evaluate the significant figures in the intermediate answer, then do the remaining steps. 3.489 × (5.67 – 2.3) = 3.489 × 3.37 = 12 2 dp 1 dp 4 sf 1 dp & 2 sf 2 sf 30  Chapter 4 MOLE CONCEPT The Mole A mole is a quantity that contains the same number of items (atoms, molecules, ions, etc.) equal to the number of carbon atoms in exactly 12 grams of 12C.  1 mole = 6.022 x 1023 items. 6.022 x 1023 is also called the Avogadro’s number; thus, a mole contains the Avogadro’s number of items. The mass of one 12C atom is exactly 12 u; while the mass of one mole of 12C is exactly 12 g (also called gram-atomic mass). That is, 12 g of carbon- 12 contains 6.022 x 1023 12C-atoms.  Therefore, 12 g = 6.022 x 1023 x 12 u;  1 g = 6.022 x 1023 u; or 1 u = 1.6605 x 10-24 g The gram-atomic mass of an element implies the mass of one mole of that element. Element Atomic mass Gram-atomic Mass Number of Atoms in (u) (g) one gram-atomic mass Hydrogen 1.008 1.008 6.022 x 1023 Carbon 12.01 12.01 6.022 x 1023 Oxygen 16.00 16.00 6.022 x 1023 Aluminum 26.98 26.98 6.022 x 1023 Gold 197.0 197.0 6.022 x 1023 The gram-atomic mass is also the molar mass of the element expressed in grams. Molar Mass The molar mass of a compound is the mass of 1 mole of the compound in grams. The molar masses of water, salt (NaCl), and table sugar (C12H22O11) are calculated as follows:  Molar mass of H2O = (2 x 1.008 g) + (16.00 g) = 18.02 g/mol; Molar mass of NaCl = (22.99 g + 35.45 g) = 58.44 g/mol; Molar mass of C12H22O11 =(12 x 12.01g) +(22 x 1.008g) +(11 x16.00 g)= 342.3 g/mol 31 Mole (mol OR n) This is a SI (Le Système International d’ Unités) unit (remember from Unit 1) that is used to represent the amount of a substance. It can be written to show the number of atoms or molecules in a working sample of some element, compound, or molecule, such as sucrose (table sugar) – C6H12O6 This concept is very important because scientists, teachers and students cannot work with individual atoms or molecules because they are very, very small and can’t be handled one at a time. So the mole was conceived to represent a working amount of a substance. How to calculate a mole: 1) Determine the total atomic or molecular mass of the substance you are working with, using the chemical formula and Periodic Table. 2) Then weigh out, using an electronic balance and weigh boat, that calculated amount. 3) Congratulations, you have just weighed out 1 mole of that substance! For example: Atomic Mass Aluminum has an atomic mass of 26.98 AMUs. So you would weigh out 26.98 grams of Aluminum to get a workable amount called a mole. Molecular Mass Salt (NaCl) has an atomic mass of: Sodium 23.00 AMUs Chlorine 35.5 AMUs Total AMUs = 23.00 + 35.5 = 58.5 AMUs Weigh out 58.5 grams of salt The unit for of measurement is g/mol. Basic Measurements or Unit Conversions involving the Mole concept: More than a mole: the basic concept is: amount you have/ amount of a mole = # of moles  You have 21.6 grams of Boron (B). How many moles do you have? 1 mole of Boron = 10.8 grams so… 21.6/10.8 = 2.0 moles.  You have 77.25 grams of Phosphorus (P). How many moles do you have?  1 mole of Phosphorus = 30.9 grams so… 77.25/30.9 = 2.5 moles.  You have 16.03 grams of Sulfur (S). How many moles do you have?  1 mole of Sulfur = 32.06 grams so… 16.03/32.06 = 0.5 moles  You have 2.43 grams of Magnesium (Mg). How many moles do you have?  1 mole of Magnesium = 24.30 grams so… 2.43/24.30 = 0.1 moles Conversions from one unit to another unit involving the mole concept: 32 The basic concept is: Unit given x 𝒖𝒏𝒊𝒕 𝒘𝒂𝒏𝒕𝒆𝒅 = Unit wanted 𝒖𝒏𝒊𝒕 𝒈𝒊𝒗𝒆𝒏 The given unit cancels out and leaves you with the unit wanted. Moles Atoms/Molecules a. You have 2.0 moles of Copper. How many atoms of Copper do you have? 1 mole = 6.022 x 1023 atoms 6.022 X 1023 = 12.044 x 1023 atoms So: 2 moles x 𝑎𝑡𝑜𝑚𝑠 1 𝑚𝑜𝑙𝑒 33 But using your rules for scientific notation, it becomes 1.2044 x 1024 atoms  You have 0.25 moles of Oxygen. How many atoms of Oxygen do you have? 1 mole = 6.022 x 1023 atoms 23 𝑎𝑡𝑜𝑚𝑠 So: 0.25 moles x 6.022 X 10 = 1.505 x 1023 atoms 1 𝑚𝑜𝑙𝑒 Atoms/Molecules moles You have 1.806 x 1024 atoms of Zinc (Zn). How many moles of Zinc do you have? Step 1: Convert 1.806 x 1024 to 18.06 x 1023 (It must have the Avogadro exponent of 23.) Step 2: 18.06 x 1023 Atoms x 1 𝑚𝑜𝑙𝑒 = 3 moles 6.022 X 1023 𝑎𝑡𝑜𝑚𝑠  You have 5.9 x 1022 atoms of Titanium (Ti) How many moles of Titanium do you have? Step 1: Convert 5.9 x 1022 to 0.59 x 1023 (It must have the Avogadro exponent of 23.) Step 2: 0.59 x 1023 Atoms x 1 𝑚𝑜𝑙𝑒 = 0.098 moles 6.022 X 10 𝑎𝑡𝑜𝑚𝑠 23 Grams Moles You have 54.0 grams of Carbon (C). How many moles of Carbon do you have? 1 𝑚𝑜𝑙𝑒 54.0 grams x = 4.5 moles 12.0 𝑔𝑟𝑎𝑚𝑠 You have 10.0 grams of Nickel (Ni). How many moles of Nickel do you have? 10.0 grams x 𝟏 𝒎𝒐𝒍𝒆 = 0.17 moles Moles Grams 𝟓𝟖.𝟔𝟗 𝐠𝐫𝐚𝐦𝐬 You have 8.5 moles of Fluorine (F) gas. How grams of Fluorine do you 34 have? 8.5 moles x 𝟏𝟗.𝟎𝟎 𝒈𝒓𝒂𝒎𝒔 = 161.5 grams 𝟏 𝐦𝐨𝐥𝐞 You have 0.45 moles of Scandium (Sc) gas. How grams of Scandium do you have? 0.45 moles x 𝟒𝟒.𝟗𝟔 𝒈𝒓𝒂𝒎𝒔 = 20.23 grams 𝟏 𝐦𝐨𝐥𝐞 What is the mass of a 256.0 ml sample of carbon dioxide gas at STP conditions? 35 How many carbon atoms are in the sample described in the above example? 1 mole 1 mole C 6.022 x 1023 atoms of C 0.2560 liters x x x 1 mole C 22.4 liters 1 mole 𝐶𝑂2 21 = 6.88 x10 How many atoms total are there in the gas sample of CO2? 1 mole 3 moles atoms 6.022 x 1023 atoms 0.2560 liters x x 22.4 liters x 1 mole atoms 1 mole 𝐶𝑂2 22 = 2.06 x 10 What is the mass, in grams, of a single molecule of carbon dioxide gas? 1 mole 𝐶𝑂2 44.0 grams 1 molecule of 𝐶𝑂2 x x 6.022 x 1023 molecules 1 mole 𝐶𝑂2 = 7.30 x 10−23𝑔𝑟𝑎𝑚𝑠 1. What is the percentage of metal in a 5.00-gram sample of potassium dichromate? K2Cr2O7 - note: the 5.00-gram sample does not matter; the percentage of metal in any size sample is the same. 2(39.1 g/mole) + 2(52.0 g/mole) % metal = x100 = 61.93 % 2(39.1 g/mole )+2(52.0 g/mole)+7(16.0 g/mole) Empirical & Molecular Formulas Determining the Formula of a Compound To determine the formula of a compound we first have to know its elemental composition and their mass percentages. This information can be obtained by performing elemental analyses on a sample of the compound. Knowledge of the elemental composition will yield the empirical formula, and if the molar mass (or molecular mass) is also known, the molecular formula can be deduced. The empirical formula is a formula that shows the simplest mole ratio of elements in the compound. The molecular formula of a compound is the true formula that indicates the actual number of atoms of each type in a molecule (or the number of moles of each type of element in 1 mole of the compound). But it does not tell how atoms are connected in the molecule. 36 The structural formula shows how atoms are bonded in the actual molecule. - the empirical formula of a substance is the simplest whole-number mole ratio of the elements that make up the atom - the molecular formula is the actual number of elements that make up a compound – it will always be a whole number multiple of the empirical formula - for example, the simple sugar – glucose – has a molecular formula of C6H12O6. Its empirical formula is CH2O. - a special group of compounds called ―hydrates‖ have a definite number of water molecules attached to the ―anhydrous‖ salt. - Analysis of the compounds may in several different units – grams, percentages, moles, number of particles, etc. In all cases, the quantity needs to be converted into moles. - Once the number of moles of each substance is determined, divide all moles by the smallest mole quantity. If fractions are produced, then multiply by a factor that will give whole numbers. - If percentages are given, assume that you have 100 grams; then the percentage will be equivalent to grams. Example Problems: 1. A compound of boron and hydrogen contains 18.9% hydrogen and 81.1% boron. What is the empirical formula of this compound? no of moles of H: 18.9 g / 1.0 g/mole = 18.9 moles no of moles of B: 81.1 g / 10.8 g/mole = 7.51 moles (18.9 moles H) / (7.51 moles B) = 2.52 (7.51 moles B) / (7.51 moles B) = 1.00 mole ratio is 2.5 : 1 x 2 = 5.0 : 2 B2H5 2. A certain compound used for fertilizer analyzes 5.0% hydrogen, 35.0% nitrogen, and 60.0% oxygen. What is its empirical formula? for H: 5.0 g / 1.0 g/mole = 5.0 moles of hydrogen for N: 35.0 g / 14.0 g/mole = 2.50 moles of nitrogen for O: 60.0 g / 16.0 g/mole = 3.75 moles of oxygen divide all three mole quantities by 2.50 H : N : O = 2.0 : 1.0 : 1.50 multiply all three quantities by 2 H:N:O=4:2:3 Formula = H4N2O3 3. If the compound in #2 is known to have a molecular weigh of 160.0 g/mole, what is the true, or molecular, formula? 37 4. Molecular weight of the empirical formula = 80.0 g/mole 160.0 g/mole / 80.0 g/mole = 2; then true molecular formula = (H4N2O3) x 2 = H8N4O6 5. The complete combustion of a hydrocarbon (compound composed only of carbon & hydrogen) in oxygen produced 176 grams of CO2 and 90.0 grams of H2O. What is the empirical formula of this hydrocarbon? CxHy + O2 = CO2 + H2O Solution: 6. If the compound in #5 is know to be a gas at STP conditions, and a 2.50-gram sample is known to occupy 96.6 ml, what is the true, or molecular, formula of this compound? (2.50 g / 0.0966 liters) (22.4 liters / 1 mole) = 60.0 g/mole = molecular weight. molecular weight of C2H5 = 29.0 g/mole 60.0 g/mole / 29.0 g/mole ≈ 2; then (C2H5) x 2 = C4H10 = butane Determination of Empirical Formula and Molecular Formula Example: Elemental analysis showed that an organic compound contains the elements C, H, N, and O. When a 1.279-g sample was completely burned in excess of oxygen, 1.60 g of CO2 and 0.77 g H2O were obtained. In a separate analysis, a 1.625-g sample yields 0.216 g of nitrogen. Determine the empirical formula of the compound. Solution: The masses of carbon and hydrogen in the sample are calculated as follows: Mass of carbon = 1.60 g CO2 x (1 mol/44.0 g) x (1 mol C/1 mol CO2)x (12.0 g C/ 1 mol C) = 0.436 g Percent C in sample = (0.436 g C/1.279 g sample) x 100% = 34.1% Mass of hydrogen = 0.77 g H2O x (1 mol/ 18.0 g )x (2 mol H/1 mol H2O ) x (1.008 g H/ 1 mol) = 0.086 g Percent H in sample = (0.086 g H/1.279 g sample) x 100% = 6.7% Percent N in sample = (0.216 g N/1.625 g sample) x 100% = 13.3% Percent O in sample = 100% – 34.1% C – 6.7% H – 13.3% N = 45.9% If a 100-g sample of the compound, there will be 34.1 g C, 6.7 g H, 13.3 g N, and 45.9 g O. Converting the mass of each element into its moles, we will obtain Mole of C = (34.1 g C)/(12.0 g/mol) = 2.84 mol 38  Mole of H = (6.7 g)/(1.008 g/mol) = 6.6 mol Mole of N = (13.3 g N)/(14.0 g/mol) = 0.950 mol N Mole of O = (45.9 g O)/(16.0 g/mol) = 2.87 mol O Dividing throughout by the mole of N (the smallest mole in the compound), we obtain the following mole ratio: 2.99 C-to-6.9 H-to-1 N-to-3.02 O, which simplifies to 3C:7H:1N:3O; This mole ratio yields the empirical formula C3H7NO3. Deriving Empirical and Molecular Formula from percent Composition and Molar Mass Example: A compound with molar mass 180 g/mol consists of 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. Determine the empirical and molecular formula of the compound. Solution: Assuming a 100-g sample, the masses of carbon hydrogen and oxygen in the sample are: 40.0 g, 6.7 g, and 53.3 g, respectively. Moles of C = (40.0 g C)/(12.0 g/mol) = 3.33 mol C; Moles of H = (6.7 g H)/(1.008 g/mol) = 6.6 mol H; Moles of O = (53.3 g O)/(16.0 g/mol) = 3.33 mol O. Dividing by the smallest mole yields a mole ratio: 1C:2H:1O, which yields an empirical formula of CH2O. The molecular formula is (CH2O)n, where n = Molar mass = 180 g/mol = 6 Empirical formula mass 30 g/mol This yields a molecular formula of C6H12O6. Exercises 1) A 2.32-g sample of an organic compound containing carbon, hydrogen, and oxygen was burned completely in excess of oxygen, which yields 5.28 g CO2 and 2.16 g water. Determine the empirical formula of the compound. (Answer: C3H6O) 2.A compound consisting of 82.66% carbon and 17.34% hydrogen has a molecular mass of 58.1 u. Determine its molecular formula. (Answer: C4H10) Percent Composition of Compounds A pure substance always contains the same elements combined in the same proportion by mass. The composition of a compound is often expressed in terms of the percentages (by mass) of its elements. If the formula of the compound is known, the mass percents of their elements can be calculated. 39 Mass%of Element =(Total mass of that Element per mole of compound) x100% Molar mass of the Compound For example, the mass percent of hydrogen, nitrogen, and oxygen in ammonium nitrate, NH4NO3 are calculated as follows: Molar mass of NH4NO3 =(4 x 1.008 g) +(2 x 14.01 g) + (3 x 16.00 g) = 80.05 g/mol Total mass of H/mol NH4NO3 = 4 x 1.008 g = 4.03 g  Mass Percent of H = (4.03 g/80.05 g) x 100% = 5.04% Total mass of N per mol NH4NO3 = 2 x 14.01 g = 28.02 g  Mass Percent of N = (28.02 g/80.05 g) x 100% = 35.00% Total mass of O/mol NH4NO3 = 3 x 16.00 g = 48.00 g  Mass Percent of O = (48.00 g/80.05 g) x 100% = 59.9 Chemical Equations and What They Mean When substances react, old bonds are broken and new ones formed; atoms are reorganized in different ways. The equation must show that the total number of each type of atoms stays the same after the reaction. A chemical equation provides two types of information: the nature of the reactants and products and the relative numbers of each type of atoms. An equation must also indicate the physical states of the substances: (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous solution. Balancing Chemical Equations In chemical reactions atoms are not created or destroyed, and this is indicated in the balanced equation. Balancing equation is done by introducing the smallest integer coefficients to one or more of the reactants and/or products involved in the reaction without changing their formula. Thus, the formula of each substance in an equation must be written correctly first before any attempt to balance the equation is made. The following steps may be helpful in balancing an equation. 1. Begin with the compound that contains the most atoms or types of atoms. 2. Balance elements that appear only once on each side of the arrow. 3. Next balance elements that appear more than once on either side. 4. Balance free elements last. 5. Finally, check that smallest whole number coefficients are used. For example, the combustion of ethanol (C2H5OH) to form carbon dioxide 40 and water vapor is represented by the following equation (not balanced): C2H5OH(l) + O2(g) = CO2(g) + H2O(g) Since carbon and hydrogen appear only once on each side of the equation, we balance these elements first by introducing coefficients 2 and 3 in front of CO2 and H2O, respectively. C2H5OH(l) + O2(g) = 2 CO2(g) + 3H2O(g) Next we balance the oxygen by introducing a coefficient 3 in front of O2: C2H5OH(l) + 3 O2(g) CO2(g) + 3H2O(g) A correct balanced equation should not contain fractions as coefficients. If a fraction appears after all atoms have been balanced, the entire equation is multiplied by a simplest factor that will eliminate the fraction. For example, when the following equation is balanced, a fraction of 2 appears in the coefficient of O2: C8H18(l) + 25/2 O2(g) 8CO2(g) + 9H2O(g);(8C, 18H, 25 O) (8C, 18H, 25 O) To remove this fraction, we multiply the entire equation by 2, which yield the proper equation: 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g); Exercise-: Balance the following equations using the simplest integer coefficient possible: 1. C6H6(l) + O2(g) = CO2(g) + H2O(g); 2. (NH4)2Cr2O7(s) = Cr2O3(s) + N2(g) + H2O(g): 3. C6H14(l) + O2(g) = CO2(g) + H2O(g) 4. Mg3N2(s) + H2O(l) = Mg(OH)2(s) + NH3(g) Stoichiometric Calculations: Amounts of Reactants and Products Balanced Equations in Reaction Stoichiometry Stoichiometry implies the quantitative relationships between substances in a chemical reaction. The coefficients in chemical equations represent numbers of molecules or moles of substances, not the mass of a substance. Yet, when actual 41 reactions, the amounts of reactants and products are measured in units for mass. The number of moles of each substance must be calculated from its mass. Mole-to-Mole Relationships in Stoichiometric Calculations The mole is the key to quantitative relationships between substances undergoing chemical reactions. Consider the equation: N2(g) + 3H2(g) 2NH3(g) At molecular quantity, this equation implies that 1 N2 molecule will react with 3 H2 molecules to form 2 NH3 molecules. In molar quantity, it implies 1 mole of N2 react with 3 moles of H2 to yield 2 moles of NH3. If the amounts of nitrogen and hydrogen are given, we can calculate how much ammonia will be formed at the end of the reaction. Likewise, knowing how much NH3 we want to produce, we can calculate the amount of each reactant that will be required. For example, if we want to know how many moles of H2 are needed to react with 5 moles of N2 and how many moles of NH3 will be formed, we can perform the calculation as follows: Mol of H2 = 5 mol N2 x 3 mol H2 = 15 mol H2 1 mol N2 Mol of NH3 = 5 mol N2 x 2 mol NH3 = 10 mol NH3 1 mol N2 The factor (3 mol H2/1 mol N2) and (2 mol NH3/1 mol N2) are called the stoichiometric ratios, or stoichiometric factors. The following stoichiometric calculations apply to the above reaction: 1. To calculate the mol of H2 needed and mol of NH3 expected from the given moles of N2: a) ? mol of H2 = Given mol of N2 x (3 mol H2/1 mol N2) b) ? mol of NH3 = Given mol of N2 x (2 mol NH3/1 mol N2) 2. To calculate the mol of N2 needed and mol of NH3 expected from the given moles of H2: a) ? mol N2 = Given mol of H2 x (1 mol N2/3 mol H2) b) ? mol NH3 = Given mol of H2 x (2 mol NH3/3 mole H2) 3. To calculate the mol of H2 and N2 needed, respectively, from the given moles of NH3:: a) ? mol N2 = Given mol NH3 x (1 mol N2/2 mol NH3) b) ? mol H2 = Given mol NH3 x (3 mol H2/2 mol NH3) Exercise: 1. According to the equation: 42  2 C8H18(l) + 25 O2(g) = 16 CO2(g) + 18 H2O(g), (a) how many moles of CO2 are produced when 5.5 moles of octane (C8H18) are completely burned in air? (b) How many moles of water are also formed? (c) How many moles of oxygen are reacted by 5.5 moles of octane? (Answer: (a) 44 moles of CO2; (b) 50. moles of H2O) Mass-to-Mole-to-Mole-to-Mass Relationships When actual reactions are carried out, the quantities of substances are measured in mass units. We have to convert their masses into moles so that they can be directly related to each other using the coefficients in the balanced equation. Again consider the reaction: N2(g) + 3H2(g) = 2NH3(g) Suppose that 454 g N2 is available and we want to know how many grams of H2 are needed to react completely with this amount of N2 and how many grams of NH3 will be produced. To calculate the mass of H2 needed: 454 g N2 x 1 mol N2 x 3 mol H2 x 2.016 g H2 = 98.0 g H2 28.02 g 1 mol N2 1 mol H2 To calculate the mass of NH3 formed: 454 g N2 x 1 mol N2 x 2 mol NH3 x 17.03 g NH3 = 552 g NH3 28.02 g 1 mol N2 1 mol NH3 Useful Steps in Calculating Masses of Reactants and Products in Chemical Reactions: 1. Balance the equation if it is not already balanced; 2. Convert the given mass of a reactant or product to moles of the substance; 3. Use the balance equation to set up the appropriate mole ratios; 4. Use appropriate mole ratios to calculate the moles of the desired reactant or product; 5. Convert from moles back to grams of the desired substance. In general, for a reaction: aA + bB cC, 1) Given grams of A, calculate grams of B: grams of A x 1 mol A x b mol B x Molar mass of B = grams of B Molar mass of A a mol A 1 mol B 2) Given grams of A, calculate grams of C: Grams of A x 1 mol A x c mol C x Molar mass of C = 43  grams of C Molar mass of A a mol A 1 mol C 3) Given grams of B, calculate grams of A: Grams of B x 1 mol Bx a mol A x Molar mass of A = grams of A Molar mass of B b mol B 1 mol A 4) Given grams of B, calculate grams of C: Grams of B x 1 mol Bx c mol C x Molar mass of C = grams of C Molar mass of B b mol B 1 mol C 5) Given grams of C, calculate grams of A: Grams of C x 1 mol C x a mol A x Molar mass of A Molar mass of C c mol C 1 mol A 6) Given grams of C, calculate grams of B: Grams of C x 1 mol C x b mol B x Molar mass of B Molar mass of C c mol C 1 mol B Exercise- 1. According to the equation: N2(g) + 3H2(g) 2NH3(g), (a) how many grams of N2 are required to react completely with 45.5 g of H2? (b) How many grams of ammonia are produced when 45.5 g of H2 are completely reacted with enough N2? (Answer: (a) 211 g N2; (b) 256 g NH3) Limiting Reactant Chemical reactions carried out in laboratories or manufacturing plants are not always carried out such that reactants are mixed in the exact stoichiometric quantities. One of the reactants maybe present in a limited amount and get used up first. It is referred to as the limiting reagent. When a reaction involves a limiting reagent, the amount of products formed is dependent on the quantity of the limiting reagent that is available. a) You will see the word "excess" used in this section and in the problems.It is used several different ways: "compound A reacts with an excess of compound B" - In this case, mentally set compound B aside for the moment. Since it is "in excess," this means there is more than enough of it. Some other compound (maybeA) will run out first. b) "20 grams of A and 20 grams of B react. Which is in excess?" What we will do below is find out which substance runs out first (called the limiting reagent). Obviously (I hope), the other compound is seen to be in excess. c) "after 20 gm. of A and 20 gm. of B react, how much of the excess 44 compound remains?" To answer this problem, we would subtract the limiting reagent amount from the excess amount. What is the Limiting Reagent? It is simply the substance in a chemical reaction that runs out first. It seems to simple, but it does cause people problems. Let's try a simple example. Reactant A is a test tube. I have 20 of them. Reactant B is a stopper. I have 30 of them. Product C is a stoppered test tube. The reaction is: A + B ---> C or: test tube plus stopper gives stoppered test tube. So now we let them "react." The first stopper goes in, the second goes in and so on. Step by step we use up stoppers and test tubes (the amounts go down) and make stoppered test tubes (the amount goes up). Suddenly, we run out of one of the "reactants." Which one? That's right. We run out of test tubes first. Seems obvious, doesn't it? We had 20 test tubes, but we had 30 stoppers. So when the test tubes are used up, we have 10 stoppers sitting there unused. And we also have 20 test tubes with stoppers firmly inserted. So, which "reactant" is limiting and which is in excess? Exercise Here's a nice limiting reagent problem we will use for discussion. Consider the reaction: 2 Al + 3 I2 ------ > 2 AlI3 Determine the limiting reagent and the theoretical yield of the product if one starts with: a) 1.20 mol Al and 2.40 mol iodine. b) 1.20 g Al and 2.40 g iodine c) How many grams of Al are left over in part b? Part A Solution: we already have moles as the unit, so we use those numbers directly. Here is how to find out the limiting reagent: take the moles of each substance and divide it by the coefficient of the balanced equation. For aluminum: 1.20 / 2 = 0.60 and For iodine: 2.40 / 3 = 0.80 The lowest number indicates the limiting reagent. Aluminum will run out first in part a. 45 The second part of the question "theoretical yield" depends on finding out the limiting reagent. Once we do that, it becomes a stoichiometric calculation. Al and AlI3 stand in a one-to-one molar relationship, so 1.20 mol of Al produces 1.20 mol of AlI3. Notice that the amount of I2 does not play a role, since it is in excess. Part B Solution: since we have grams, we must first convert to moles. The we solve just as we did in part a just above. For the mole calculation: aluminum is 1.20 g / 26.98 g mol¯1 = 0.04477 mol iodine is 2.4 g / 253.8 g mol¯1 = 0.009456 mol To determine the limiting reagent: aluminum is 0.04477 / 2 = 0.02238 and iodine is 0.009456 / 3 = 0.003152 The lower number is iodine, so we have identified the limiting reagent. Finally, we have to do a calculation and it will involve the iodine, NOT the aluminum. I2 and AlI3 stand in a three-to-two molar relationship, so 0.009456 mol of I2 produces 0.006304 mol of AlI3. Again, notice that the amount of Al does not play a role, since it is in excess. From here figure out the grams of AlI3 and you have your answer. Part C Solution: since we have mole, we calculate directly and then convert to grams. Al and I2 stand in a two-to-three molar relationship, so 0.009456 mol of I2 uses 0.006304 mol of Al. Convert this aluminum amount to grams and subtract it from 1.20 g and that's the answer. For example, consider the reaction: H2SO4(aq) + 2NH3(g) (NH4)2SO4(s) Suppose that 35 g of H2SO4 and 15 g of NH3 are allowed to react. To determine the limiting reactant we calculate the mole of each reactant as follows: mole of NH3 = 15 g NH3 x 1 mol NH3 = 0.88 mol NH3 17.03 g NH3 mole of H2SO4 = 35 g H2SO4 x 1 mol H2SO4 = 0.36 mol H2SO4 46  98.08 g H2SO4 mole ratio: 0.88 mol NH3 = 2.4 mol NH3 > 2/1 0.36 mol H2SO4 1 mol H2SO4 Since the mole ratio of NH3 to H2SO4 is larger than the stoichiometric ratio according to the balanced equation, it means that there are more NH3 than is needed, which makes H2SO4 as the limiting reagent. The amount of ammonium sulfate, (NH4)2SO4 is calculated as follows: 0.36 mol H2SO4 x 1 mol (NH4)2SO4 x 132.1 g (NH4)2SO4 = 47 g (NH4)2SO4 1 mol H2SO4 1 mol (NH4)2SO4 We can also calculate how much of the NH3 is reacted: 0.36 mol H2SO4 x 2 mol NH3 x 17.03 g NH3 = 12 g NH3 1 mol H2SO4 1 mol NH3 On the other hand, if a reaction mixture contains 45 g H2SO4 and 15 g NH3, their number of moles are: mole of NH3 = 15 g NH3 x 1 mol NH3 = 0.88 mol NH3 17.03 g NH3 mole of H2SO4 = 45 g H2SO4 x 1 mol H2SO4 = 0.46 mol H2SO4 98.08 g H2SO4 The mole ratio = 0.88 mol NH3 =1.9 /1< 2/1, which makes NH3 as the limiting reagent 0.46 mol H2SO4 grams of (NH4)2SO4 = 0.88 mol NH3 x 1 mol (NH4)2SO4 x 132.1 g (NH4)2SO4 = 58 g 2 mol NH3 1 mol (NH4)2SO4 Alternatively, we can obtain the amount of product by performing two sets of stoichiometric calculations based on each reactant as the limiting reagent. For example, suppose that a mixture consists of 25 g of NH3 and 70. g of H2SO4. We calculate the amount of (NH4)2SO4 based on each reactant as the limiting reactant as follows: grams of (NH4)2SO4= 25 gNH3x 1 mol NH3 x 1 mol (NH4)2SO4x132.1 g (NH4)2SO4,= 97 g 17.03 g NH3 2 mol NH3 1 mol (NH4)2SO4 grams (NH4)2SO4=70.gH2SO4x1molH2SO4x1mol(NH4)2SO4x132.1g (NH4)2SO4, = 94 g 98.08 g H2SO4 1 mol H2SO4 1 mol (NH4)2SO4 Since the calculation based on available H2SO4 yields less ammonium 47 sulfate, it means that H2SO4 is the limiting reagent and the maximum amount of (NH4)2SO4 expected is 94 g. Summary of Steps involved in Solving Stoichiometry Problems 1. Write the balanced equation for the reaction. 2. If the masses of more than one reactant are given, convert them to moles using the molar masses of each substances; 3. Find the mole ratios and determine which reactant is the limiting reagent. 4. Calculate the amount of products formed based on the limiting reagent as follows: a. Convert grams of limiting reactant to moles b. Calculate the mole of product by multiplying mole of limiting reagent with the stoichiometric ratio of the product to this reactant. c. Convert moles of product to mass by multiplying the mole with its molar mass. d. Round off answer to the correct number of significant figures, that is the one given for the limiting reactant. Exercise-: According to the reaction: N2(g) + 3 H2(g) = 2 NH3(g), if 81.3 g N2 and 18.7 g H2 gases are reacted, (a) determine which reactant is the limiting reagent? (b) How many grams of NH3 are formed based on this limiting reagent ? (c) How many grams of the excess reactant remains after the reaction? (Answer: (a) limiting reactant is N2; (b) 98.8 g NH3 formed; (c) 1.2 g of H2 remains) 2. Methanol is produced by the following reaction: CO(g) + 2 H2(g) = CH3OH(l) In an experiment, 158 g of carbon monoxide is reacted with 25.0 g of hydrogen gas. (a) How many grams of methanol will be produced? (b) Which reactant is in excess? (Answer: (a) 181 g of methanol; (b) H2 is in excess (excess by 2.3 g Actual, Theoretical, and Percent Yield The theoretical yield is the amount of product calculated based on the balanced equation and the limiting reagent available. In actual experiments, the amount of products obtained are often less than those calculated from the stoichiometric relationships. This is because of several factors that affect reaction yields, such as: (i) reactions never go to completion; they reach a state of equilibrium; (ii) impure reactants were used; (iii) some of the products is lost during separation and purification, and (iv) side-reactions occur that result in a lower product yields. The amounts of products actually obtained from a reaction is called the actual yield. The Percent Yield for a reaction calculated as follows: 48 Percent Yield = 𝐴𝑐𝑡𝑢𝑎𝑙 𝑌𝑖𝑒𝑙𝑑 x 100 Theoretical 49 1. 75.0 g aluminum oxide is reacted with excess sulfuric acid to produce aluminum phosphate, Al2(SO4)3, according to the following equation (not balanced), Al2O3(s) + H2SO4(aq) Al2(SO4)3 (s) + H2O(l) (a) What is the theoretical yield of Al2(SO4)3 in this reaction? (b) If the experiment yields only 233 g of Al 2(SO4)3, what is the percent yield of Al2(SO4)3? (Answer: (a) theoretical yield = 252 g Al2(SO4)3; (b) percent yield = 92.6%) References 2- N. S. Akhmetov, General and Inorganic Chemistry, (1983), Translated from Russian by A. Rosinkin, MIR Publishers – Moscow. 3- K. F. Purcell and J. C. Kotz, An introduction to Inorganic Chemistry, (1980), Saunders collage Publishing, Philadelphia 4- P. Matthews, Advanced Chemistry, (1996), Cambridge UniversityPress, UK 5- G. C. Hill and J. S. Holman, Chemistry in Context (1987) 2nd ed. ELBS 6- F. A. Cotton, G. Wilkinson and P. L. Gaus, Basic Inorganic Chemistry (1994), 3rd ed., John Wiley and sons 50