4.2. Formulas, Equations, and Stoichiometry notes.pdf

Transcript: Honors Chemistry Chemical Reactions: Formulas, Equations, and Stoichiometry Scene 1 The study of chemistry includes the properties, structure and reactions of matter. The mathematics of chemistry is used to determine chemical quantities involved in compounds and in chemical reactions. Experiments you conduct in the laboratory illustrate chemical reactions for learning purposes, so you practice the mathematics of chemical equations on a small scale basis. However, industries that manufacture chemical products are large scale businesses. The accuracy of chemical calculations on an industrial scale directly affects a company’s profit margin. Understanding the chemistry of compounds is essential to such ventures as medicine, agriculture, and environmental science. Throughout this program, we will use the ideas and terms presented in the programs “Introduction to Matter” and “Naming Chemical Compounds.” Therefore, it might be helpful to go back and review those programs before starting this one. Scene 2 Accurate chemical calculations depend on properly balanced chemical equations. In this program, you’ll learn how to interpret and balance chemical equations and use mathematics to describe them. The mathematics of chemistry is referred to as “stoichiometry.” Stoichiometry will be explored in this program after you’ve learned to write and balance chemical equations. Scene 3 Before you learn to balance equations, it will help to review some properties of matter. Compounds are substances that contain two or more elements, chemically combined in a fixed proportion. Compounds are described using chemical formulas. Chemical formulas consist of element symbols that indicate the atoms in a compound and numerical subscripts that indicate the number of each atom in that compound. For example, the chemical formula for aluminum oxide indicates there are two aluminum atoms for every three oxygen atoms. Look at the chemical formula for carbon dioxide. You’ll notice there is no subscript indicating the number of carbon atoms. When no subscript is provided, it should be understood there is only one atom of that element in the compound. Scene 4 A chemical formula indicates how many atoms of each element are in a compound. A more specific term – molecular formula – describes the composition of one molecule of a molecular compound, such as water, written as H2O. Ionic compounds, composed of positive and negative ions, are expressed in terms of formula units, not molecules. Comparing examples of molecular and ionic substances is a good way to learn the difference between molecules and formula units. Can you see how molecules of water form individual interacting units? In a sample of an ionic compound, such as sodium chloride, you cannot distinguish between individual units of interacting atoms, even though you know the compound consists of one sodium cation for every chloride anion. Scene 5 The actual weight of an atom in grams is an incredibly small number. For example, an atom of carbon 12, which contains six protons, six neutrons, and six electrons, weighs only 2.009 x 10-23 grams. Since these tiny numbers are difficult to work with, chemists have defined a new unit, the atomic mass unit or amu, for comparing the atomic weights of the approximately 112 known atoms. In 1961, the International Union of Pure and Applied Chemistry set a particular carbon atom, carbon 12, as the standard by which the weights of all the atomic elements are compared. By definition, one atomic mass unit or amu, is defined as 1/12 of the mass of a carbon 12 atom. Since most of the mass of the carbon 12 atom is in the six protons and six neutrons of its nucleus, one atomic mass unit is approximately the mass of one proton or neutron.

Page 1 of 9

Transcript: Honors Chemistry Chemical Reactions: Formulas, Equations, and Stoichiometry Scene 6 Look at the value for the atomic mass of carbon as listed on the Periodic Table. You probably expect to see an atomic mass of 12 since you just learned the standard atomic mass is based on a carbon 12 atom. Actually, the atomic mass reported for carbon is 12.01. The reason for this discrepancy is because many elements, including carbon, occur naturally as several different isotopes. Isotopes are atoms of the same element that have different atomic masses. Isotopes possess the same number of protons, but differ in the number of neutrons in the nucleus. In the heavier isotope of carbon, carbon 13, one additional neutron increases the mass of the atom to approximately 13 amu. The atomic mass numbers as reported on the Periodic Table reflect the weighted average of the masses of all the naturally occurring isotopes of that element. Although 98.9% of all naturally occurring carbon atoms contain six neutrons in their nuclei, 1.1% contain seven neutrons in their nuclei. Much like one very tall basketball player raises the average height of a team, the contribution of these heavy carbon atoms increases the average atomic mass of the element carbon. Scene 7 By simply looking at the atomic weights as provided on the Periodic Table, you can determine that one hydrogen atom is about twelve times lighter than a carbon atom, and that a magnesium atom is approximately two times heavier than a carbon atom. With the exception of the hydrogen atom, an atom’s mass is roughly double the value of its atomic number. Although these terms sound similar, you must keep their meanings clear in your mind. Atomic numbers reflect only the number of protons in an atom’s nucleus, and therefore they are always whole numbers. Remember, the atomic mass values as reported on the Periodic Table reflect an average atomic mass of all naturally occurring isotopes of that element. Scene 8 The weight or mass of a compound is determined by adding up the atomic weights of the individual atoms within the compound. The term “molecular mass” describes the mass of individual molecules, such as a water molecule. The term “formula mass” describes the mass of a formula unit of an ionic compound, like sodium chloride. An all-encompassing term known as the “molar mass” describes the weight in grams of exactly one mole of any pure substance. You will learn more about the mole shortly. Scene 9 The relative mass of each element in a compound compared to the entire mass of a compound is known as its “percent composition.” The percent composition of a compound can be determined from its chemical formula. Let’s look at the chemical formula for sodium chloride, NaCl. To determine the percent composition for each element, first determine the molar mass of the compound. To calculate the molar mass of a compound, simply multiply the atomic mass of each atom by the number of atoms in the formula and report your answer in grams. Now, divide the molar mass of each element by the molar mass of the compound and multiply by one hundred to get a percent. Finally, check your work to make sure you have accounted for the entire mass of the substance.

Page 2 of 9

Transcript: Honors Chemistry Chemical Reactions: Formulas, Equations, and Stoichiometry Scene 10 Chemical formulas describe the type and number of atoms in a compound. Compounds may also be described using empirical formulas. You can think of an empirical formula as an abbreviated chemical formula. Empirical formulas describe the type of atoms and their ratios, but usually do not describe exactly how many atoms of that element are in the compound. Empirical formulas are easier to comprehend if you consider how they can be used to describe everyday situations around you. Consider the number of male and female students in a classroom. Imagine there are five males, abbreviated “M”, and 15 females, abbreviated “F.” The complete chemical formula for this class would be M5F15, whereas the empirical formula for this class would be MF3, which describes a ratio of three female students to every one male student. The empirical formula MF3 reflects a constant ratio of three female to every male student, but does not tell us exactly how many male and female students there are in the class. Scene 11 Let’s see how empirical formulas can be used to determine the chemical formula of some unknown substance. Acetylene, C2H2, is a gas used in the welding industry, where as benzene, C6H6, is a liquid solvent. Both compounds share the same empirical formula, CH, indicating a ratio of one carbon atom to every hydrogen atom in both of these compounds. Say you know the formula mass of one of these compounds is 78.11. By knowing the empirical formula CH, you can determine an empirical formula mass of 13.02. To determine the chemical formula for the unknown, simply divide the given formula mass of 78.11 by the empirical formula mass of 13.02, and you get a ratio of six. Because the empirical formula CH indicates a one to one ratio of carbon to hydrogen, there must be six carbon atoms and six hydrogen atoms in the molecular formula of the unknown compound, which in this case is benzene. Scene 12 Suppose you want to make molecules of water, H2O, from a supply of hydrogen and oxygen molecules in such a way that no hydrogen and oxygen molecules are leftover. If you took ten molecules of oxygen and 20 molecules of hydrogen, you could make exactly 20 molecules of water. This may seem easy until you actually try to do it. Have you ever seen an atom? Atoms are too tiny to count individually; therefore, a unit called the “mole” was established to count out atoms by weighing them. In chemistry, a mole is not a small furry rodent; instead, the mole concept serves as a bridge between the invisible universe of atoms and the visible world, where you measure quantities of atoms with a balance, or scale. Scene 13 A dozen eggs is a collection of 12 eggs, and a decade is a collection of ten years. A collection of particles known as a “mole” can consist of either atoms, molecules, or formula units. The number of particles in a mole has been determined as 6.02 x1023 particles, commonly referred to as “Avogadro’s Number.” Keep in mind our comparison – there are ten years in a decade and 12 eggs in a dozen. Similarly, there are 6.02 x 1023 particles in a mole. Scene 14 The mole concept is credited to an Italian chemist and physicist, Amadeus Avogadro. Interestingly, Avogadro did not actually determine the number associated with his name. Avogadro’s famous number, 6.02 x 1023, was actually determined several years after his death, but named in his honor in recognition of his development of the mole concept. Among his many contributions to chemistry, Avogadro determined that any gas occupies a volume of 22.4 liters at standard temperature and pressure; that is, zero degrees Celsius and one atmosphere of pressure.

Page 3 of 9

Transcript: Honors Chemistry Chemical Reactions: Formulas, Equations, and Stoichiometry Scene 15 Comprehending the number of atoms represented by Avogadro’s Number can be mindboggling. One way to imagine a number as large as 6.02 x 1023 is to associate each particle represented by Avogadro’s Number with some small object you’re familiar with. Then multiply that object’s mass by Avogadro’s Number. For example, imagine an atom being the size of one grain of sand. In order to obtain Avogadro’s Number, or a mole, of sand grains, it would be necessary to dig the entire surface of the Sahara Desert, which is slightly smaller than the United States, to a depth of two meters, about six feet. A sand pile six feet deep and the size of the United States would represent one mole of sand. If Avogadro’s Number of sheets of paper were divided into a million equal piles, each pile would be so tall that it would stretch from the earth to the sun and beyond. Counting at a rate of one particle per second for 48 hours per week, it would take the entire population of the world ten million years in order to reach Avogadro’s Number. Compare these examples to a mole of carbon atoms, or a mole of water molecules. Can you see how the mole provides chemists with a convenient quantity to work with in the laboratory? Scene 16 You may recall atomic masses range from about one amu for the lightest atom, hydrogen, to 238 amu for the heaviest of natural atoms, uranium. Compare a light metallic atom, such as aluminum, with the egg of a small songbird, and a uranium atom to an ostrich egg. A dozen ostrich eggs would obviously weigh more and occupy more space than a dozen smaller eggs, but both collections contain the same number of units, 12 eggs. Again, though molar masses of different elements vary, the number of particles in a mole is always 6.02 x 1023. Logically, a mole of a heavy element such as uranium would weigh more and occupy more space than a mole of aluminum, but the number of atoms in a mole of uranium and a mole of aluminum are the same. How many atoms are in a mole of uranium or aluminum? By now you’ve probably guessed Avogadro’s Number, 6.02 x 1023, and you would be correct. Scene 17 A compound is just a specific arrangement of atoms to which chemists have assigned an appropriate name. For example, a molecule that consists of three carbon and eight hydrogen atoms is called “propane.” Propane is a common gas used to heat homes. To calculate the molecular weight of a propane molecule, simply multiply the number of each atom by its atomic weight, then add up the combined weights of all the atoms in the molecule. Do you see why propane has a molecular weight of 44.1? The term “molar mass” indicates the mass of one mole of a molecular compound; in this case, the weight of 6.02 x 1023 propane molecules. Since one mole of propane molecules consists of 3 moles of carbon and eight moles of hydrogen, we can simply report the molecular weight of propane in grams to arrive at molar mass of 44.1 grams. Scene 18 When a sample of any element is measured so that its mass in grams is equal to the element’s atomic mass, the same will always contain Avogadro’s number of atoms. Therefore, 12.01 grams of carbon has the same number of atoms as 196.97 grams of gold. This relationship also extends to compounds. For example, the molecular weight of water is 18.02 grams, so 18.02 grams of water will have the same number of molecules as 44.1 grams of propane, because the molecular weight of propane is 44.1 grams. How many molecules are in 18.02 grams of water, or 44.1 grams of propane gas? Again, the answer is Avogadro’s number, 6.02 x 1023.

Page 4 of 9

Transcript: Honors Chemistry Chemical Reactions: Formulas, Equations, and Stoichiometry Scene 19 Now you will see how to determine the molar mass of a more complex compound, such as magnesium nitrate. The chemical formula for magnesium nitrate is Mg(NO3)2. The subscript two is used to multiply each atom in parenthesis by two. Can you see how the chemical formula for magnesium nitrate describes a compound composed of one magnesium atom, two nitrogen atoms, and six oxygen atoms? Now we’re ready to determine the molecular weight of the compound. First, multiply the number of each atom in the compound by its atomic mass. Then combine the weights of all the atoms in the molecule. The result is a molecular weight of 148.314. To obtain the molar mass of this compound, simply report your findings in grams. The molar mass of magnesium nitrate would therefore be 148.314 grams. Scene 20 Chemists must be very careful when expressing chemical formulas. For example, just saying “a mole of oxygen” could refer to either a mole of oxygen atom or a mole of oxygen gas, and there is a significant difference. Gases such as nitrogen, oxygen, and hydrogen rarely exist in nature as single atoms. These atoms are found in the molecular state as “diatomic molecules” and are written as N2, O2, and H2 respectively. One mole of atomic oxygen consists of Avogadro’s number of oxygen atoms and has an atomic mass of approximately 16 grams, whereas one mole of oxygen gas consists of Avogadro’s number of oxygen gas molecules, each composed of two oxygen atoms. Therefore, one mole of oxygen gas would have a molecular mass of approximately 32 grams. Scene 21 Converting grams to moles and moles back to grams is a mathematical process that chemists perform on a daily basis. Grams are the practical value that is measured on a laboratory scale. Scales and balances read in grams, not in moles, so when designing chemical experiments, mole values first have to be converted to grams. For example, suppose you are told to prepare a 0.25 mole sample of sodium chloride for an experiment. The formula weight of sodium chloride can typically be found on the stock bottle from which the sample was obtained. Carefully observe how the calculation is set up so the mole units cancel. Multiplying the desired mole quantity, 0.25, by the formula weight of sodium chloride reflects the number of grams of sodium chloride to weigh out on a scale to prepare your sample. Scene 22 Suppose a chemistry experiment you want to conduct requires .1 moles of nickel as one of the reactants, but the stock room is down to its last 5.1 grams of nickel. Does 5.1 grams of nickel metal contain 0.1 moles of nickel atoms? To find out, set up the calculation so the gram units cancel out by dividing the amount of metal you have by the molar mass of nickel. Then multiply the numerator, which is one mole of nickel, to obtain your final answer in moles. Unfortunately, it looks like you do not have enough nickel to carry out your experiment. Scene 23 Suppose you know the weight in grams of a pure substance by measuring it on a laboratory balance, but want to know the number of atoms or molecules the sample contains. Calculating the number of particles is a simple two-step process. First, you must convert grams to moles. Then convert moles to molecules using Avogadro’s number. Let’s determine the number of molecules in ten grams of the sugar glucose. First, convert grams of glucose into moles of glucose using the molecular weight of the compound. Then convert moles of glucose to molecules of glucose using Avogadro’s number. The final answer will be reported in terms of molecules because glucose is a molecular compound. An ionic compound such as sodium chloride would be stated in terms of formula units, and an element such as gold would be given in atoms.

Page 5 of 9

Transcript: Honors Chemistry Chemical Reactions: Formulas, Equations, and Stoichiometry Scene 24 You may recall that chemical equations describe chemical reactions. Although reactions can be described in sentences, writing them out in this manner is cumbersome and time-consuming. In a chemical equation, an arrow separates the chemical formulas of the reactants on the left from the resulting chemical formulas of the products on the right. Notice that the letters in parenthesis are used to indicate the physical state of each substance. The letters (g) and (l) are abbreviations for the terms gas and liquid. The letter (s) is used to indicate the solid state, and the abbreviation (aq) for aqueous indicates a substance is dissolved in water. Scene 25 Chemical equations must be balanced to accurately portray chemical reactions. In other words, balanced equations reflect the law of conservation of mass, which states that the products formed in a reaction have the same total mass as that of the reactants present before the reaction took place. This is due to the fact that in a balanced equation, each side of the equation contains the same number of atoms of each element. Take a look at an example of a balanced equation. This equation is in balance because there is one nitrogen, one chlorine, and four hydrogen atoms are on either side of the equation. In other words, the types and number of atoms of each element on the reactant side of the equation equals the type and number of atoms on the product side of the equation. Scene 26 Sometimes one unit of each reactant forms one unit of each product. Most of the time, however, you must adjust the number of units to make the reaction balance. The most efficient way to balance chemical equations is to perform the following steps. First, determine and write out the correct unbalanced formula for the reactants and products. Second, count the number of atoms of each element on the left and right sides of the equation. Then begin balancing the equation by starting with the elements that appear only once on either side of the equation by using coefficients. A coefficient is a whole number that appears in front of the chemical formula in an equation. Continue this process until each substance balances and all coefficients are in their lowest possible ratio. Scene 27 Let’s work our way through a typical balancing problem. First, write out the correct chemical reaction for the formation of magnesium oxide. Second, count and list the number and type of each atom on each side of the equation. You will notice that oxygen is out of balance. Now begin balancing the equation using coefficients. First, place a two in front of magnesium oxide on the product side to bring oxygen into balance. Now to balance the magnesium. There are two magnesium atoms on the right and one on the left, so a coefficient of two will balance this element. Finally, check your work by accounting for atoms. As you see, there are two magnesium and two oxygen atoms on both sides of the equation. The equation is balanced. Scene 28 As you learned earlier, stoichiometry refers to the mathematical relationships that describe chemical formulas and chemical equations. “Stoichiometry” is taken from the Greek term “stoikheion,” meaning “element,” and “metron,” meaning “to measure.” Therefore, stoichiometry describes the process of measuring elements and elemental quantities. The major categories of stoichiometric problems are mass to mass conversions, mass to volume conversions, and volume to volume conversions.

Page 6 of 9

Transcript: Honors Chemistry Chemical Reactions: Formulas, Equations, and Stoichiometry Scene 29 Most stoichiometric problems are solved in a similar manner. First, you must convert from grams or liters into moles, then convert moles of reactants into moles of products, and finally convert moles of the product back to grams or liters. This process becomes more clear when you’re actually given numbers and problems to solve, as the following situations will illustrate. Scene 30 In mass to mass problems, you’re provided with a known mass of one substance, usually a reactant, and are required to determine the mass of another substance, usually the product in a reaction. Watch how this common problem is solved. Your assignment is to determine the amount of sodium chloride produced by reacting exactly five grams of sodium metal with an excess amount of chlorine gas. The first step in the process is to make sure the equation is balanced. Then convert the five grams of sodium into moles of sodium. Next, convert the moles of sodium to moles of the product, in this case sodium chloride. And finally, convert moles of sodium chloride to grams of sodium chloride to obtain your answer in units of grams. The final answer shows that reacting five grams of sodium metal with excess chlorine gas yields 12.71 grams of sodium chloride salt. Scene 31 Avogadro was known primarily for his work on gases. Avogadro determined that a mole of any gas, regardless of its chemical composition, has a volume of 22.4 liters at 0 degrees Celsius and at atmospheric pressure. Therefore, one mole of gas, regardless of what kind of gas it is, will occupy 22.4 liters of space and contain 6.02 x 1023 molecules. Next you will learn how this fundamental principle simplifies the process of comparing volumes of different gases. Scene 32 Many chemical reactions involve reactants and products in the form of gases. Gases are measured in terms of volume. The amount of space, or volume, the number of gas molecules occupies is proportional to the number of gas molecules in that sample. A good example of a mass to volume reaction is the decomposition of water. Water can decompose when electrical current is passed through it; that is to say, water molecules break down, forming hydrogen and oxygen gases. Scene 33 Let’s take 50 grams of water and calculate the volume of oxygen and hydrogen gas that could theoretically be produced from its complete decomposition. First convert the given 50 grams of water to moles of water. Then use this number to determine moles of the products oxygen and hydrogen. Then, knowing the moles of hydrogen and oxygen allows you to calculate the volumes in liters of oxygen and hydrogen gas produced. Finally, state the answer. The decomposition of 50 grams water results in the formation of 31.1 liters of oxygen gas and 62.2 liters of hydrogen gas. Our answer makes sense because there is a two-to-one ratio of hydrogen to oxygen in every water molecule, and one mole of any gas – whether that gas is hydrogen or oxygen – occupies the same volume: 22.4 liters. Scene 34 Since a mole of any gas occupies a volume of 22.4 liters, volume to volume ratios are the same as mole to mole ratios. Therefore, balanced equations directly indicate volume ratios. Take for example the formation of ammonia from hydrogen and nitrogen gas. The balanced equation shows that one mole of nitrogen reacts with three moles of hydrogen to form two moles of ammonia. Since each of these species are gases, you can translate the balanced equation into volume ratios and restate the equation as 22.4 liters of nitrogen would completely react with 67.2 liters of hydrogen to form 44.8 liters of ammonia.

Page 7 of 9

Transcript: Honors Chemistry Chemical Reactions: Formulas, Equations, and Stoichiometry Scene 35 What if you have ten liters of nitrogen gas and needed to know the volume of hydrogen gas needed to completely react with the nitrogen? Because the molar ratio of hydrogen to nitrogen is three to one, three times the volume of hydrogen would be required to completely react with the nitrogen. Therefore, the answer should be three times the volume of nitrogen, which it is. Now determine how much ammonia, NH3, would be produced as a result of such a reaction. Can you see why the answer is 20 liters? Knowing how molar equivalents relate to volume quantities can save you a lot of time when calculating volume to volume relationships. Scene 36 Often it is best to introduce new concepts by comparing them to everyday situations that you’re familiar with. Let’s say you work in a bike shop, and one of your jobs is to assemble bicycles. You would like to assemble as many bikes as possible for display in the store, but the shipment of tires you’re expecting has not arrived. You have plenty of time to assemble bikes, and plenty of bicycle frames, but you are limited in the number of bikes you can put together because of the limited supply of tires. The same concept can be applied to chemical equations. The amount of the products formed by a reaction is dependent on the amount of available reactants. Scene 37 When the quantities of reactants are combined in the proportions described by a balanced equation, all the reactants are consumed in the reaction and there are no reactants remaining after the product is formed. Most of the time, however, one of the reactants will be depleted and others will be left over. Leftover reactants are described as being in “excess.” The first reactant to be completely converted to product is called the “limiting reactant.” In other words, the reactant you run out of first limits the amount of product the reaction is capable of forming. Remember, when the supply of tires ran out, bicycle production came to a halt. Scene 38 The first reactant completely converted to product is the limiting reactant. Therefore, the quantity of product formed in a reaction, known as the “yield,” is dependent on the quantity of the limiting reactant. Determining limiting reactants is a four step process. Again, let’s examine the familiar sodium chloride reaction. First, make sure the equation is balanced. Then convert the quantity of each reactant into moles of each reactant. The third step is to convert moles of each reactant into moles of product. Moles of product can then be converted back into grams. Finally, compare the amount of product formed by each reactant. The limiting reactant – in this case, chlorine gas – is identified as the one capable of being converted to the least amount of product. The amount of product formed by the limiting reactant is known as the “yield.” Scene 39 Up to this point you have seen chemical calculations performed as though things never go wrong in chemical reactions. This assumption is as faulty as assuming you will score 100% on every test you take. One hundred percent efficiency is a rare achievement for any human endeavor, including chemistry. Chemists account for inefficiency in chemical reactions by comparing theoretical calculated yields to actual yield. When the chemical equation is used to calculate the amount of product that will be formed in a reaction, a value for the theoretical yield is obtained. This is the maximum amount of product that can be expected to form from the reactants.

Page 8 of 9

Transcript: Honors Chemistry Chemical Reactions: Formulas, Equations, and Stoichiometry Scene 40 In contrast to theoretical yield, actual yields reflect the amount of product that is obtained from laboratory experiments. Different chemical reactions will vary in the actual yield that can be obtained. Simply put, some experiments can be conducted more efficiently than others. Two of the many factors that contribute to decreased efficiency are impure reactants and sloppy laboratory procedures. Scene 41 The percent yield compares the amount of product expected to the amount of product actually obtained. For example, in our sodium chloride reaction, the theoretical yield – 16.5 grams – would be produced by reacting ten grams of sodium metal with ten grams of chlorine gas. However, after conducting the experiment, the actual yield of sodium chloride turned out to be only 14.5 grams. Watch how the percent yield is determined for this reaction. First, divide the actual yield by the theoretical yield, and multiply by 100 to convert the decimal to a percent. The percent yield for this reaction is 87.9%, meaning that approximately 88% of the theoretical yield was actually obtained. Scene 42 In this program you observed how elements can be rearranged to form compounds as described by chemical equations. While reviewing atomic structure, we considered the fact that atoms are too tiny to actually count. The mole concept was then discussed in order to bridge the gap between individual atoms and practical laboratory quantities, as weighed in grams. Stoichiometry involves calculations that relate quantities of substances. The mole concept allows for a quantity of a substance, as measured in grams, to be expressed in units of atoms, molecules, or moles. Scene 43 A quantity of a given substance as reported in grams is easily converted to units of moles and vice versa. However, before performing stoichiometric calculations, you must start with a balanced equation. Balanced equations properly reflect mole ratios between the reactants and products in a chemical reaction. Since balanced equations reflect mole ratios, you are now capable of calculating the amounts of product formed from a known quantity of reactants. After all, chemistry is all about the manufacturing of compounds, researching new and useful compounds, and furthering knowledge of the structure and properties of matter.

Page 9 of 9