CEMICAL FRMULAS AND EQUATINS Name Date Period Assign oxidation numbers to each element in the following compounds or polyatomic ions. To really keep on your toes, see if you can name them as well. MgBr 2 Fe 2 3 AlN S 3 P 4 3- Cr 2 7 2- Cl 2 CuS 4 Name the following compounds. Use roman numerals when appropriate 2. Ca() 2 3. KF 1. AlCl 3 4. Pb(N 2 ) 2 5. SrC 3 6. Ag 2 Naming Covalent Compounds Worksheet Write the formulas for the following covalent compounds: 1) antimony tribromide 2) hexaboron silicide 3) chlorine dioxide 4) hydrogen iodide 5) iodine pentafluoride 6) dinitrogen trioxide 7) ammonia 8) phosphorus triiodide
Write the names for the following covalent compounds: 9) P 4 S 5 10) 2 11) SeF 6 12) Si 2 Br 6 13) SCl 4 14) C 4 15) B 2 Si 16) NF 3
xidation Numbers The chemical formula for water is 2. Carbon Dioxide is C 2. Why does oxygen combine in different ratios, in different compounds? Do Chemistry students need to memorize the chemical formulas for each of the millions of known compounds? Is there a way to predict the ratio by which elements will combine in a given situation? Fortunately, that is what oxidation numbers are for. You probably recall learning about ions in Biology. An ion is a charged particle formed when a neutral atom or group of atoms gain or lose one or more electrons. When a single atom forms an ion, as in the case of Al +3, it is called a monatomic ion. When of group of atoms that are covalently bonded together form an ion, as in the case of N 4 +, it is called a polyatomic ion. Sometimes ions with opposite charges are attracted together and will form ionic compounds. Table Salt, NaCl is such a compound formed from Na + ions and Cl - ions. Neutral atoms can also form compounds when they join together, as in the case of water ( 2 ). owever, since these compounds are not composed of ions, they are called molecular compounds. You will learn more about these types of compounds in lesson 5-3. Regardless as to whether a compound is made up of ions or not, each atom in the compound has an apparent charge. This apparent charge, called the oxidation number, represents the charge that an atom would have if electrons were transferred completely to the atom with the greater attraction for them in a given situation. These oxidation numbers can be used to predict the ratio by which atoms will combine when they form compounds. The following rules help us assign the oxidation number of elements: Table 5-2a - Predicting xidation Numbers 1. In free elements (that is, in uncombined state), each atom has an oxidation number of zero. Ex. In 2, the oxidation number of each oxygen atom is zero. 2. For ions composed of only one atom, the oxidation number is equal to the charge on the ion. Ex. The oxidation number of Ca 2+ is +2. 3. All alkali metals (elements in column 1of the periodic table, with the exception of hydrogen) have an oxidation number of +1. Ex. The oxidation numbers of Li, K, and Na will always be +1. 4. All alkaline earth metals (elements in column 2 of the periodic table) have an oxidation number of +2. Ex. The oxidation number of Ba is +2. 5. The oxidation number of Aluminum (Al) is always +3. 6. The oxidation number of oxygen in most compounds (such as 2 and C 2 ) is -2. In hydrogen peroxide ( 2 2 ) and peroxide ( 2 2- ) oxygen shows a -1 oxidation number. 7. The oxidation number of hydrogen is +1, except when in is bonded to a metal as a negative ion, in which case it is -1. Ex. 2 shows hydrogen as +1. Na shows hydrogen as -1. 8. When halogens (elements in column 17 on the periodic table) form negative ions,
they will have an oxidation number of -1. Ex. NaCl and CaCl 2 both show chlorine with a -1 oxidation number. 9. In a neutral molecule, the sum of the oxidation numbers of all of the atoms must be zero. Ex. In 2, each hydrogen is +1 and the oxygen is -2. So, (2 x +1) + (-2) = 0. 10. In a polyatomic ion, the sum of oxidation numbers of all the elements in the ion must be equal to the net charge of the ion. Ex. In the polyatomic ion known as hydroxide ( - ), the oxygen is -2 and the hydrogen is +1. So, (-2) + (+1) = -1, the same as the charge on the hydroxide ion ( - ) Now, in time you will find it easy to predict many oxidation numbers, as you become more familiar with the periodic table and the rules above. Until that time, you should make use of reference tables that list the oxidation numbers of common ions. Depending on your teacher, he or she may allow you to make use of such tables for quizzes and exams. For your convenience, I will provide examples of these tables below. Keep in mind that the table that your teacher uses may differ from the ones provided below. Table 5-2b - xidation Numbers of Some Common Monatomic Ions CARGE IN +1 +2 +3 +4 NNE -1-2 Aluminum (Al) Argon (Ar) Barium (Ba) Bromide (Br) Cadmium (Cd) Calcium (Ca) Cesium (Cs) Chloride (Cl) Fluoride (F) ydride () ydrogen () Iodide (I) Lithium (Li) Magnesium (Mg) Neon (Ne) xide () Potassium (K) Sodium (Na) Silver (Ag) Strontium (Sr)
Sulfide (S) Zinc (Zn) Now, some elements show different positive oxidation numbers, in different situations. The stock system, which you will learn more about in lessons 4-3 and 4-4, uses Roman numerals to show the oxidation number of the element. For example, Lead(II) is lead with an oxidation number of +2. Chromium(III) is Chromium with an oxidation number of +3. The oxidation number for these types of elements will always be positive. I provide a table below, but once you understand the stock system you will not need the table any longer. Chromium(III) Cobalt(II) Copper(I) Copper(II) Iron(II) Iron(III) Lead(II) Lead(IV) Manganese(II) Mercury(II) Nickel(II) Tin(II) Table 5-2c - Stock System xidation Numbers of Metals with Multiple xidation States CARGE IN +1 +2 +3 +4 Table 5-2d - xidation Numbers of Some Common Polyatomic Ions CARGE IN +1 +2-1 -2-3 Acetate, (C3C - ) Ammonium (N 4 + ) Carbonate (C 3 2- )
Chlorate (Cl - 3 ) Chromate (Cr 2-4 ) Cyanide (CN - ) Dichromate (Cr 2 2-7 ) ydroxide ( - ) ypoclorite (Cl - ) Iodate (I 3 -) Nitrate (N - 3 ) Nitrite (N - 2 ) xalate (C 2 2-4 ) Perchlorate (Cl - 4 ) Permanganate (Mn - 4 ) Peroxide ( 2-2 ) Phosphate (P 3-4 ) Silicate (Si 2-3 ) Sulfate (S 2-4 ) Sulfite (S 2-3 ) Tartrate (C 4 4 2-6 ) Tetraborate (B 4 2-7 ) Thiosulfate (S 2 2-3 )
Writing Chemical Formulas A chemical formula is a combination of elemental symbols and subscript numbers that is used to show the composition of a compound. Depending of the type of compound that the formula represents, the information that it provides will vary slightly. Before we go about learning how to write chemical formulas, it is important that you clearly understand the difference between molecular compounds and ionic compounds. Ionic compounds are composed of charged ions that are held together by electrostatic forces. A typical type of ionic compound, called a binary compound because it is made up of two elements, will be composed of metallic positive ions (cations) and nonmetal negative ions (anions). Another type of ionic compound, called a ternary compound as it contain three elements, is composed of monatomic ions and polyatomic ions. When dealing with ionic formulas it is very important to remember that the formula does not show how the compound actually exists in nature. It only shows the ratio by which the individual ions combine. For example, the ionic formula for calcium chloride is CaCl 2. Since calcium chloride is an ionic compound, this formula does not mean that there are actually two chlorine atoms floating around attached to one calcium atom. Ionic compounds are actually continuous, lacking the discrete units that make up a sample of a molecular substance. Rather, the formula shows that a sample of calcium chloride contains twice as many chlorine atoms as calcium atoms. Remember that ionic compounds are not molecules, so the formula CaCl 2 is said to represent one formula unit of calcium chloride. Molecular compounds are held together by covalent bonds, or shared pairs of electrons. Molecular formulas do show these molecules as they actually exist as discrete units in nature. When we say that the molecular formula of water is 2, we can see that the molecules of water are made up of three atoms, two hydrogen atoms are covalently bonded to each oxygen atom. A special type of chemical formula, called an empirical formula, shows the composition of a molecule not as it actually exists, but in a simple whole number ratio. The difference between empirical and molecular formulas will be explained in lesson 5-5. This lesson will concentrate on writing simple chemical formulas when given a formula name. In learning how to write chemical formulas, you will make use of the oxidation numbers that you learned about in lesson 5-2. For your convenience, print out the tables from lesson 5-2 before you
continue with this lesson, as they will be referred to from time to time. Writing Ionic Formulas I. Binary Compounds - Binary compounds are compounds that are composed of only two elements. When you write the formulas for binary compounds, they will consist of two elemental symbols, and they may also have one or two subscript numbers, if the elements don't combine in a one to one ratio. You are probably familiar with the formula NaCl for table salt. This formula shows no subscripts because one ion of Na will be present for each ion of Cl, in any sample of table salt. You will be given the name of a binary compound and you will be expected to be able to write the proper formula for the compound. There will be two sources of information for writing the correct formula. The compounds name will give you the elements that make up the compound. The oxidation numbers of the ions involved will show you the ratio by which they combine. Let's go through an example; Example 1. Write the correct formula for Barium Fluoride. Step one - Write the symbols for the elements in the compound. If you need to review the elemental symbols, see lesson 5-1. Note that the ending "ide" is used for fluoride to show that it is a negative ion of fluorine. Barium = Ba Fluoride = F Step two - Look up the oxidation numbers of the elements involved (in table 5-2b or some similar table), and write them as superscripts to the right of the elemental symbols. Note that when no number accompanies a charge symbol, as in the case of fluoride below, they charge value is understood to be "1". Barium = Ba 2+ Fluoride = F - Step three - Use the correct combination of ions to produce a compound with a net charge of zero. In this case, (2+) + 2(-1) = 0. So, two fluoride ions will cancel out one barium ion. Since it would take two fluoride ions (each with a charge of negative one) to cancel out one barium ion (with a charge of plus two) we use a subscript of two after the symbol for fluorine to show the ratio. BaF 2 If this seems confusing to you, it will get simpler over time. Example 2. Write the proper formula for the ionic compound lithium bromide. Step one - Write the symbols for the elements in the compound. Note that the ending "ide" is used for bromide to show that it is a negative ion of bromine. Lithium = Li Bromide = Br Step two - Look up the oxidation numbers of the elements involved (in table 5-2b or some similar table), and write them as superscripts to the right of the elemental symbols. Note that when no number accompanies a charge symbol, as in the case of fluoride below, they charge value is
understood to be "1". Lithium = Li + Bromide = Br - Step three - Use the correct combination of ions to produce a compound with a net charge of zero. In this case, (+1) + (-1) = 0. so, one lithium ion will cancel out the charge of one bromide ion. This means that the two elements will combine in a one to one ratio, and know subscripts will be needed. LiBr II. Ternary Compounds - Ternary compounds are composed of three different elements. The most common types of ternary compounds consist of a metallic cation (positive ion) and a polyatomic anion (negative ion). The only common polyatomic ion with a positive charge is the ammonium ion. At any rate, To write these formulas you will want to have reference tables with the information provided on tables 5-2b and 5-2d. Example 1. Write the proper chemical formula for potassium hydroxide. Step one - Write the symbols for the monatomic and polyatomic ions in the compound. You will find the symbol potassium on table 5-2b. ydroxide is a polyatomic ion, which will be found on table 5-2d. Eventually you will recognize the name of a polyatomic ion, but for now if you can't find an ion on one table, look on the other. Potassium = K ydroxide = Step two - Look up the oxidation numbers of the ions involved (in table 5-2b and 5-2d, or some similar tables), and write them as superscripts to the right of the elemental symbols. Potassium = K + ydroxide = - Step three - Use the correct combination of ions to produce a compound with a net charge of zero. Parenthesis must be used if you need more than one of a polyatomic ion. In this case, (+1) + (-1) = 0. So, only one of each ion is used. No subscripts are necessary. If you needed more than one hydroxide ion, it would be put in parenthesis with the subscript on the outside. K Note the importance of upper and lower case Example 2. Show the correct formula for Calcium Nitrate. Step one - Write the symbols for the monatomic and polyatomic ions in the compound. Calcium = Ca Nitrate = N 3 Step two - Look up the oxidation numbers of the ions involved (in table 5-2b and 5-2d, or some similar tables), and write them as superscripts to the right of the elemental symbols.
Calcium = Ca 2+ Nitrate = N 3 - Step three - Use the correct combination of ions to produce a compound with a net charge of zero. Parenthesis must be used if you need more than one of a polyatomic ion. In this case (+2) + 2(- 1) = 0. We need to show two nitrate ions in our formula. The subscript is put on the outside of the parenthesis to show that the entire polyatomic ion is doubled. Ca(N 3 ) 2 The correct use of parenthesis will seem hard at first, but you must master this skill with practice! III. The Stock System - Some elements, like iron and lead, have more than one oxidation number. If you were given a compound name like lead chloride, you would not know if you should used an oxidation number of +2 or +4 for the lead. The stock system is used to specify which form of an element, that shows multiple oxidation numbers, is used in a particular compound. A roman numeral is shown after the name of the positive ion (cation) to indicate the oxidation number of the positive ion. Example 1. Show the correct formula for lead(iv) nitrate. Step one - Write the symbols for the ions in the compound. Lead = Pb Nitrate = N 3 Step two - Look up the oxidation number of the negative ion involved (in table 5-2b and 5-2d, or some similar tables). The positive ion will have a positive oxidation number equal to the roman numeral. Write the numbers as superscripts to the right of the elemental symbols. Lead = Pb 4+ Nitrate = N 3 - Step three - Use the correct combination of ions to produce a compound with a net charge of zero. Parenthesis must be used if you need more than one of a polyatomic ion. Pb(N 3 ) 4 Example 2. Show the correct formula for Copper(II) Fluoride Step one - Write the symbols for the ions in the compound. Copper = Cu Fluoride = F Step two - Look up the oxidation number of the negative ion involved (in table 5-2b and 5-2d, or some similar tables). The positive ion will have a positive oxidation number equal to the roman numeral. Write the numbers as superscripts to the right of the elemental symbols. Copper = Cu 2+ Fluoride = F -
Step three - Use the correct combination of ions to produce a compound with a net charge of zero. Parenthesis must be used if you need more than one of a polyatomic ion. Writing Molecular Formulas CuF 2 I. Binary Molecular Compounds - The standard method for naming binary molecular compounds has changed over the years. Currently, the stock system is commonly used for naming molecular compounds. Names like "carbon dioxide", "carbon monoxide", and "dinitrogen pentoxide" are really remnants of an older system that used prefixes to identify the number of elements involved. When you are writing the formula for a molecular compound using the stock system, you will not really notice any difference from the methods described above, until you study bonding. You should be aware that you are not dealing with ions when you are working with molecular formulas, rather you are looking up what might be called the apparent charge on each atom. Example 1. Write the correct formula for nitrogen(iv) oxide. Step one - Write the symbols for the elements involved. Nitrogen = N xide = Step two - Use the roman numeral as the apparent charge of the first element. Find the apparent chart of the second element by looking on reference tables such as 5-2a. Nitrogen = N 4+ xide = 2- Step three - Determine the ratio by which the elements will bond to show a net charge of zero. Use subscripts to indicate the number of atoms of each element present. In this case, (+4) + 2(-2) = 0. N 2 Example 2. Write the correct formula for nitrogen(iii) oxide. Step one - Write the symbols for the elements involved. Nitrogen = N xide = Step two - Use the roman numeral as the apparent charge of the first element. Find the apparent charge of the second element by looking on reference tables such as 5-2a. Nitrogen = N 3+ xide = 2- Step three - Determine the ratio by which the elements will bond to show a net charge of zero. Use subscripts to indicate the number of atoms of each element present. In this case, 2(+3) + 3(-2) = 0.
N 2 3 II. ther Molecular Formulas - There are other types of molecular formulas, besides binary, which you will eventually be required to write. These lessons will be presented at other times. Naming Compounds As first mentioned in an early lesson, there are two main types of compounds, ionic and molecular. Some of the compounds that you will learn about this year will require special systems for naming, and we will learn about them at a later time. For example, we will learn how to correctly name the various types of acids and bases when we study the chapters on acids and bases. In this lesson you will learn enough to name most of the compounds that you will come in contact with in your laboratory activities this year. References will be made to the tables from lesson 5-2, so it would be wise to have them handy as you go over this material. I. Binary Compounds. As you know, binary compounds consist of only two elements. The formula for a binary compound may contain more than two letters, but it will contain only two capital letters. When naming a binary compound, regardless of whether it is ionic or molecular, follow the following steps: 1. Write the name of the element represented by the first symbol in the formula. 2. Write the name of the element represented by the second symbol in the formula, but change the ending of the element's name to "ide". 3. Check a reference table to determine the number of positive oxidation numbers that the first element forms. If it only forms one then you are done. 4. If the first element shows more than one oxidation number, than use the stock system. Determine the oxidation number that the first element is showing and write that roman numeral inbetween the two elemental names. Example 1. What is the correct name for the compound AlBr 3? Step 1. Write the name of the element represented by the first symbol in the formula. aluminum Step 2. Write the name of the element represented by the second symbol in the formula, but change the ending of the element's name to "ide". In this case, bromine becomes bromide. aluminum bromide Step 3. Check a reference table to determine the number of positive oxidation numbers that the first element forms. If it only forms one then you are done. Aluminum always has an oxidation number of +3, therefore there is no need for a roman numeral. ur answer is; aluminum bromide
Example 2. What is the correct name for the element NiS? Step 1. Write the name of the element represented by the first symbol in the formula. nickel Step 2. Write the name of the element represented by the second symbol in the formula, but change the ending of the element's name to "ide". In this case, sulfur becomes sulfide. nickel sulfide Step 3. Check a reference table to determine the number of positive oxidation numbers that the first element forms. If it only forms one then you are done. Nickel forms oxidation numbers of +2, +3 and +4, so we must go to the next step. Step 4. If the first element shows more than one oxidation number, than use the stock system. Determine the oxidation number that the first element is showing and write that roman numeral inbetween the two elemental names. We check the oxidation number of sulfide and find that it is -2. If one nickel is canceling out one sulfur than the apparent charge on the nickel must be +2. (+2) + (-2) = 0. nickel(ii) sulfide Example 3. What is the correct name for the compound Fe 2 3? Step 1. Write the name of the element represented by the first symbol in the formula. iron Step 2. Write the name of the element represented by the second symbol in the formula, but change the ending of the element's name to "ide". So, oxygen becomes oxide. iron oxide Step 3. Check a reference table to determine the number of positive oxidation numbers that the first element forms. If it only forms one then you are done. Iron can be +2 or +3, so we must go on to step 4. Step 4. If the first element shows more than one oxidation number, than use the stock system. Determine the oxidation number that the first element is showing and write that roman numeral inbetween the two elemental names. We know that oxygen is -2 in this case. Since we have 3 atoms of oxygen, each with a charge of -2, then the total negative charge is -6. We must have +6 to balance out the -6. Since there are two iron atoms to make up a total of +6, each must be +3. 2(+3) + 3(-2) = 0. iron(iii) oxide
II. Ternary Compounds - Ternary compounds contain three elements. The only type of ternary compounds that we will learn how to name in this chapter are those that consist of one polyatomic ion and one monatomic ion. The vast majority of these types of compounds consist of a positive monatomic ion and a negative polyatomic ion. For this type of compound you follow the steps below: Step 1. Write the name of the element represented by the first symbol in the compound. Step 2. Write the name of the polyatomic ion, without changing the ending. Step 3. Check a reference table to determine the number of positive oxidation numbers that the first element forms. If it only forms one then you are done. Step 4. If the first element shows more than one oxidation number, than use the stock system. Determine the oxidation number that the first element is showing and write that roman numeral inbetween the two names. Example 1. Name the compound Ca(CN) 2? Step 1. Write the name of the element represented by the first symbol in the compound. Calcium Step 2. Write the name of the polyatomic ion, without changing the ending. Calcium Cyanide Step 3. Check a reference table to determine the number of positive oxidation numbers that the first element forms. If it only forms one then you are done. Calcium is always +2, so the final answer is as below: Calcium Cyanide Example 2. What is the name of the compound Fe(N 3 ) 2? Step 1. Write the name of the element represented by the first symbol in the compound. iron Step 2. Write the name of the polyatomic ion, without changing the ending. iron nitrate Step 3. Check a reference table to determine the number of positive oxidation numbers that the first element forms. If it only forms one then you are done. Iron can be +2 or +3, so we must go on to
step 4. Step 4. If the first element shows more than one oxidation number, than use the stock system. Determine the oxidation number that the first element is showing and write that roman numeral inbetween the two names. Nitrate shows an oxidation number of -1. Since there are two nitrate ions in the compound, the total negative charge is -2. Therefore, the iron must be +2. (+2) + 2(-1) = 0. iron(ii) nitrate Special Exception: The Ammonium ion (N 4 + ) is a positive polyatomic ion. When it combines with a negative monatomic ion, you change the ending of the negative ion to "ide". When it combines with a negative polyatomic ion, you just name both ions. (N 4 ) 2 S is called ammonium sulfide N 4 is called ammonium hydroxide
Balancing Chemical Equations According to the law of conservation of mass, a chemical equation must be balanced. This means that the total number of atoms on the reactant side must be equal to the total number of atoms on the product side. This really involves three skills; interpreting a chemical formula, determining whether or not a chemical equation is balanced, and balancing the equation. I. Interpreting a Chemical Formula. If you can read a chemical formula correctly, you can check the balance on a chemical equation. ne of the biggest problems for new Chemistry students is correctly reading the number of atoms inside parenthesis. Let us practice this skill first, with a couple of examples. Example 1. ow many atoms of each element are there in one formula unit of ammonium sulfide? Ammonium Sulfide is (N 4 ) 2 S Remember that a subscript pertains only to the element that precedes it, unless it precedes parenthesis, in which case it is a multiplier for each element in the parenthesis. In the example above, the subscript 4 only pertains to hydrogen, while the subscript 2 acts as a multiplier for both nitrogen and hydrogen, giving us as a final tally; (N 4 ) 2 S of nitrogen; 8 atoms of hydrogen; and 1 atom of sulfur. Example 2. ow many atoms of each element are there in one formula unit of barium nitrate? Barium Nitrate = Ba(N 3 ) 2 Now, the subscript 3 pertains only to the oxygen, but the subscript 2 becomes a multiplier for each element in the parenthesis. Therefore; Ba(N 3 ) 2 1 atom of barium; of nitrogen; and 6 atoms of oxygen. II. Checking the Balance of a Chemical Equation. When we write chemical equations for a chemical reaction, we use special numbers called coefficients to represent multiple molecules or formula units. For example; 6 2 As before, the subscript 2 pertains only to the hydrogen. owever, the coefficient 6
pertains to every element in the compound, whether or not they are found in parenthesis. The 6 tells us that there are six molecules of water, with a total of 1 of hydrogen and 6 atoms of oxygen. nce again, a coefficient pertains to every element in the compound, regardless of parenthesis. You will need to keep this in mind when you check the balance of an equation. Example 1. Determine if the following reaction is balanced or not. Ca() 2(cr) ---> Ca (cr) + 2 (g) Let us make an organized tally table and compare both sides of the equation; Ca() 2(cr) ---> Ca (cr) + 2 (g) Reactant Side Product Side Elements Ca 1 atom Elements Ca 1 atom As you can see, this reaction is balanced, so no coefficients are necessary. Example 2. Check the balance on the following chemical reaction; Ca() 2(aq) + Cl (aq) ---> CaCl 2(aq) + 2 (l) Ca() 2(aq) + Cl (aq) ---> CaCl 2(aq) + 2 (l) Reactant Side Product Side Ca Cl Ca Cl 1 atom 3 atoms 1 atom 1 atom 1 atom As you can see, this reaction is not balanced. You are not allowed to change any subscripts, but coefficients may be added in order to obtain balance. III. Balancing Chemical Equations Balancing chemical equations is a skill that only develops with practice, but for starters, look at the tally above. Notice that you need more Cl on the reactant side. What would the tally look like if we add a coefficient of 2 to the Cl on the reactant side? Ca() 2(aq) + 2Cl (aq) ---> CaCl 2(aq) + 2 (l) Ca() 2(aq) + 2Cl (aq) ---> CaCl 2(aq) + 2 (l) Reactant Side Product Side
Ca Cl Ca Cl 1 atom 4 atoms 1 atom 1 atom Now we need more oxygen and more hydrogen on the product side. Let's add a coefficient of 2 to the 2 on the product side and check the balance again. Ca() 2(aq) + 2Cl (aq) ---> CaCl 2(aq) + 2 2 (l) Ca Ca() 2(aq) + 2Cl (aq) ---> CaCl 2(aq) + 2 2 (l) Reactant Side Product Side Cl Ca Cl 1 atom 4 atoms 1 atom 2 atom 4 atoms Now the equation is balanced. Example 2. Write a balanced chemical equation for the reaction below; Propane reacts with oxygen gas to yield carbon dioxide and water. First, you need to be able to turn a word equation into a chemical equation. The one above would become; Now, let us tally the information in a table: C 3 8(g) + 2(g) ----> C 2(g) + 2 (g) C 3 8(g) + 2(g) ----> C 2(g) + 2 (g) Reactant Side Product Side C C 3 atoms 8 atoms 1 atom 3 atoms Well, a quick look shows us that we will need more hydrogen and more carbon on the right hand side. Let us start by multiplying the number of hydrogen on the product side by four, giving us a total of 8 atoms of hydrogen. Be aware that this will also change the number of oxygen atoms on the product side. Let us look at how a coefficient of 4 in front of water changes things.
C 3 8(g) + 2(g) ----> C 2(g) + 4 2 (g) C C 3 8(g) + 2(g) ----> C 2(g) + 4 2 (g) Reactant Side Product Side C 3 atoms 8 atoms 1 atom 8 atoms 6 atoms Now we have a match with the number of hydrogen atoms. Let us balance the carbon atoms next, because in order to change the carbon atoms on the product side, it will also affect the number of oxygen atoms. We need to multiply the number of carbon atoms on the product side by three, so we will place a coefficient of three in front of the carbon dioxide and check the tally again. C 3 8(g) + 2(g) ----> 3C 2(g) + 4 2 (g) C 3 8(g) + 2(g) ----> 3C 2(g) + 4 2 (g) Reactant Side Product Side C C 3 atoms 8 atoms 3 atom 8 atoms 10 atoms Now, we have matched the number of atoms for two of the elements. A subscript of 5 in front of the oxygen on the reactant side should finish the job.
C 3 8(g ) + 5 2(g) ----> 3C 2(g) + 4 2 (g) C C 3 8(g) + 5 2(g) ----> 3C 2(g) + 4 2 (g) Reactant Side Product Side C 3 atoms 8 atoms 10 atoms 3 atom 8 atoms 10 atoms We have achieved proper balance! In practice, the process is not nearly as long and tedious as this may have appeared. nce you gain some experience, you will find that you can balance these equations quickly and painlessly. Start practicing with the worksheets below, and be sure to browse the links for more information.
Writing Chemical Formulas for Binary Compounds Name Section This page is designed to help students practice written problems, and is meant to be printed out. it the print command and show all work in the spaces provided. Show the following steps; Step one - Write the symbols for the elements in the compound. Step two - Look up the oxidation numbers of the elements involved and write them as superscripts to the right of the elemental symbols. Step three - Use the correct combination of ions to produce a compound with a net charge of zero. Multiple ions are indicated with subscripts. Write the correct chemical formulas for the following: 1. lithium oxide 2. potassium chloride 3. calcium oxide 4. barium bromide
Naming Compounds Name Section This page is designed to help students practice written problems, and is meant to be printed out. it the print command and show all work in the spaces provided. Name the following compounds. Use roman numerals when appropriate 1. AlCl 3 2. Ca() 2 3. KF 4. Pb(N 2 ) 2 5. SrC 3 6. Ag 2 Answers 1. aluminum chloride 2. calcium hydroxide 3. potassium fluoride 4. lead(ii) nitrate 5. strontium carbonate 6. silver oxide
Balancing Chemical Equations Name Section 1. Balance the following chemical equations. Mg + N 2 ---> Mg 3 N 2 2. KN 3(s) ---> KN 2(s) + 2(g) 3. Fe + 2 S 4 ---> Fe 2 (S 4 ) 3 + 2 4. MgC 3(s) ---> Mg (s) + C 2(g)