Unit 5 Chemical Formulas and Reactions
Rules for Writing Chemical Formulas
-Use subscripts to indicate the number of atomic or molecular units. -Use superscripts to indicate the charge on the monoatomic or polyatomic ion. The total charge of an ionic compound is zero.
Chemical Equation
A representation of a chemical reaction; the formulas of the reactants (on the left) are connected by an arrow with the formulas of the products (on the right).
Monatomic Ions
Consist of a single atom with a positive or negative charge resulting from the loss or gain of one or more valence electrons.
Balanced Equation
Each side of the equation has the same number of atoms of each element and mass is conserved.
How do you calculate the percent composition of a compound?
-The percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound, multiplied by 100%. % by mass of element = mass of element/mass of compound x 100% % by mass of element = mass of element in 1 mol compound/molar mass of compound x 100% Ex.) When a 13.60-g sample of a compound containing only magnesium and oxygen is decomposed, 5.40 g of oxygen is obtained. What is the percent composition of this compound? Knowns: mass of compound = 13.60 g mass of oxygen = 5.40 g O mass of magnesium = 13.60 g - 5.40 g O = 8.20 g Mg Unknowns: percent by mass of Mg = ?% Mg percent by mass of O = ?% O Determine the percent by mass of Mg in the compound. % Mg = 8.20 g/13.60 g (both g cancel) x 100% = 60.3% Mg Determine the percent by mass of O in the compound. % O = 5.40 g/13.60 g (both g cancel) x 100% = 39.7% O Ex.) Propane (C3H8), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum. Calculate the percent composition of propane. (Note: the g/mol is the element's atomic mass) Knowns: mass of C in 1 mol C3H8 = 3 mol x 12.0 g/mol = 36.0 g mass of H in 1 mol C3H8 = 8 mol x 1.0 g/mol = 8.0 g molar mass of C3H8 = 36.0 g/mol + 8.0 g/mol = 44.0 g/mol Unknowns: percent by mass of C =?% C percent by mass of H =?% H Determine the percent by mass of C in C3H8. %C = 36.0 g/44.0 g (both g cancel) x 100% = 81.8% C Determine the percent be mass of H in C3H8. %H = 8.0 g/44.0 g (both g cancel) x 100% = 18% H
What general guidelines can help you write the name and formula of a chemical compound?
1. Follow the rules for naming acids when H is the first element in the formula. 2. If the compound is binary, generally the name ends with the suffix -ide. If the compound is a molecular binary compound, use prefixes to indicate the number of atoms. 3. When a polyatomic ion that includes oxygen is in the formula, the compound name generally ends in -ite or -ate. 4. If the compound contains a metallic cation that can have different ionic charges, use a Roman numeral to indicate the numerical value of the ionic charge in the compound. Writing Chemical Formulas 1. An -ide ending generally indicates a binary compound. 2. An -ite or -ate ending means a polyatomic ion that includes oxygen is in the formula. 3. Prefixes in a name generally indicate that the compound is molecular. 4. A Roman numeral after the name of a cation shows the ionic charge of the cation.
How do you determine the name and formula of an acid?
1. When the name of the anion ends in -ide, the acid name begins with the prefix hydro-. The stem of the anion has the suffix -ic and is followed by the word acid. Therefore, HCI (X=chloride) is named hydrochloric acid. 2. When the anion name ends in -ite, the acid name is the stem of the anion with the suffix -ous, followed by the word acid. Thus, H2SO3 (X=sulfite) is named sulfurous acid. 3. When the anion name ends in -ate, the acid name is the stem of the anion with the suffix -ic, followed by the word acid. Thus, HNO3 (X=nitrate) is named nitric acid. To write the formula for an acid, use the rule for writing the name of the acid in reverse. Then balance the ionic charges just as you would for any ionic compounds. For example, consider hydrobromic acid. Following Rule 1, hydrobromic acid (hydro- prefix and -ic suffix) must be a combination of hydrgoen ion (H+) and bromide ion (Br-). So the formula of hydrobromic acid is HBr. Using Rule 2, hydrogen ion and phosphite ion (PO3-/3) must be the components of phosphorus acid. You need three hydrogen ions to balance the 3- charge of the phosphite anion. Thus, the formula for phosphorus acid is H3PO3. According to Rule 3, sulfuric acid (-ic ending) must be a combination of hydrogen ion and sulfate ion (SO 2-/4). The formula for sulfuric acid is H2SO4 because two hydrogen ions are needed to balance the 2- charge of the sulfate anion.
What guidelines are used to write the name and formula of a binary molecular compound?
1. Write the names of the elements in hte order listed in the formula. 2. Use prefixes appropriately to indicate the number of each kind of atom. If just one atom of the first element is in the formula, omit the prefix mono- for that element. Also, the vowel at the end of a prefix is sometimes dropped when the name of the element begins with a vowel. 3. End the name of the second element with the suffix -ide. To write the formula of a binary molecular compound, first use the prefixes in the name to tell you the subscript of each element in the formula. Then write the correct symbols for the two elements with the appropriate subscripts.
Base
A base is generally an ionic compound that produces hydroxide ions when dissolved in water.
Decomposition Reaction
A chemical change in which a single compound breaks down into two or more simpler products. Decomposition reactions involve only one reactant and two or more products. Most decomposition reactions require energy in the form of heat, light, or electricity. Ex.) 2HgO(s) --> 2Hg(l) + O2(g)
Combustion Reaction
A chemical change in which an element or a compound reacts with oxygen, often producing energy in the form of heat and light. A combustion reaction always involves oxygen as a reactant. Often the other reactant is a hydrocarbon, which is a compound composed of hydrogen and carbon. The complete combustion of a hydrocarbon produces carbon dioxide and water. Elemental carbon (soot) and toxic carbon monoxide gas may be additional products. The complete combustion of hydrocarbon releases a large amount of energy as heat.
Single-Replacement Reactions
A chemical change in which one element replaces a second element in a compound. You can identify a single-replacement reaction by noting that both the reactants and the products consist of an element and a compound. Whether one metal will displace another metal from a compound depends upon the relative reactivities of the two metals. Ex.) Zn(s) + Cu(NO3)2(aq) --> Cu(s) + Zn(NO3)2(aq)
Combination Reaction
A chemical change in which two or more substances react to form a single new substance. When a Group A metal and a nonmetal react, the product is a binary ionic compound. 2K(s) + Cl2(g) --> 2KCl(s) When two nonmetals react in a combination reaction, more than one product is often possible. S(s) + O2(g) --> SO2(g) sulfur dioxide 2S(s) + 3O2(g) --> 2SO3(g) sulfur trioxide More than one product may also result from the combination reaction of a transition metals (II) and (III). Some nonmetal oxides react with water to produce an acid. Some metallic oxides react with water to give a base, or a compound containing hydroxide ions. Ex.) 2Mg (s) + O2(g) --> 2MgO(s) In this reaction, the product is a single substance (MgO), which is a compound.
Double-Replacement Reactions
A chemical change involving an exchange of positive ions between two compounds. They generally take place in aqueous solution and often produce a precipitate, a gas, or a molecular compound such as water. For a double-replacement reaction to occur, one of the following is usually true: 1. One of the products is only slightly soluble and precipitates from solution. For example, the reaction of aqueous solutions of sodium sulfide and cadmium nitrate produces a yellow precipitate of camium sulfide. Na2S(aq) + Cd(NO3)2(aq) --> CdS(s) + 2NaNO3(aq) 2. One of the products is a gas. Poisonous hydrogen cyanide gas is produced when aqueous sodium cyanide, also a poison, is mixed with sulfuric acid. 2NaCN(aq) + H2SO4(aq) --> 2HCN(g) + Na2SO4(aq) 3. One product is a molecular compound such as water. Combining solutions of calcium hydroxide and hydrochloric acid produces water. Ca(OH)2(aq) + 2HCI(aq) --> CaCl2(aq) + 2H2O(l)
Skeleton Equation
A chemical equation that does not indicate the relative amounts of the reactants and products. The first step in writing a complete chemical equation is to write the skeleton equation.
Binary Compound
A compound composed of two elements. Binary compounds can be ionic compounds or molecular compounds. If you know the name of a binary ionic compound, you can write the formula.
Acid
A compound that contains one or more hydrogen atoms and produces hydrogen ions when dissolved in water. When naming an acid, you can consider the acid to consist of an anioin combined with as many hydrogen ions as needed to make the molecule electrically neutral. Therefore, the chemical formulas of acids are in the general form HnX, where X is a monatomic or polyatomic anion, and n is a subscript indicating the number of hydrogen ions that are combined with the anion.
Avogadro's Hypothesis
Avogadro's hypothesis states that equal volumes of gases at the same temperature and pressure contain equal numbers of particles. The particles that make up different gases are not the same size. However, the particles in all gases are so far apart that a collection of relatively large particles does not require much more space than the same number of relatively small particles.
How do you determine the name and formula of a base?
Bases are named in the same way as ionic compounds--the name of the cation is followed by the name of the anion. To write the formula for a base, first write the symbol for the metal cation followed by the formula for the hydroxide ion. Then, balance the ionic charges just as you would for any ionic compound. For example, aluminum hydroxide consists of the aluminum cation (Al3+) and the hydroxide anion (OH-). You need three hydroxide ions to balance the 3+ charge of the aluminum cation. Thus, the formula for aluminum hydroxide is Al(OH)3.
How is the law of definite proportions consistent with Dalton's atomic theory?
Dalton postulated that atoms combine in simple whole-number ratios. If the ratio of atoms of each element in a compound is fixed, then it follows that the ratio of their masses is also fixed.
Writing a Skeleton Equation
Hydrochloric acid reacts with solid sodium hydrogen carbonate. The products formed are aqueous sodium chloride, water, and carbon dioxide gas. Write a skeleton equation for this chemical reaction. Reactants: sodium hydrogen carbonate (solid), hydrochloric acid (aqueous) NaHCO3(s), HCI(aq) Products: sodium chloride (aqueous), water (liquid), carbon dioxide (gas) NACI(aq), H2O(l), CO2(g) Separate the reactants from the products with an arrow. NaHCO3(s) + HCI(aq) --> NaCl(aq) + H2O(l) + CO2(g)
Catalyst
In many chemical reactions, a catalyst is added to the reaction mixture.A substance that speeds up the reaction, but is not used up in the reaction. A catalyst is neither a reactant nor a product, so its formula is written above the arrow in a chemical equation.
Law of Multiple Proportions
In the early 1800s, Dalton and others studied pairs of compounds that contain the same elements but have different physical and chemical properties. Using the results from these studies, Dalton stated the law of multiple proportions: Whenever the same two elements form more than one compound, the different masses of one element that combine with the same mass of the other element are in the ratio of small whole numbers. For example, CuCl and CuCl2.
Crisscross Method
In this method, the numerical value of the charge of each ion is crossed over and becomes the subscript for the other ion. Notice that the signs of the charges are dropped. Fe2O3 2(3+)+3(2-)=0 The formula is correct because the overall charge of the formula is zero, and the subscripts are expressed in the lowest whole-number ratio.
Calculating the Molar Mass of a Gas at STP
Known: Denisty = 1.964 g/L an 1 mol of gas at STP = 22.4 L Unknown: molar mass = ? g/mol First, identify the conversion factor needed to convert density to molar mass. 22.4 L/1 mol Multiply the given density by the conversion factor. 1.964 g/1 L(L cancels) x 22.4 L(L cancels)/1 mol =44.0 g mol
Converting Mass to Moles
Known: mass = 92.2 g Fe2O3 Unknown: number of moles = ? mol Fe2O3 First, determine the mass of 1 mol of Fe2O3. 2 mol Fe x 55.8 g Fe/1 mol Fe = 111.6 g Fe = 111.6 Fe 3 mol O x 16.0 g O/1 mol O = 48.0 g O 1 mol Fe2O3 = 111.6 g Fe + 48.0 g O = 159.6 g Fe2O3 Identify the conversion factor relating grams of Fe2O3 to moles of Fe2O3. 1 mol Fe2O3/159.6 Fe2O3 Multiply the given mass by the conversion factor. 92.2 g Fe2O3 x 1 mol Fe2O3/159.6 g Fe2O3 =0.578 mol Fe2O3
Converting Moles to Mass
Known: number of moles = 0.45 mol Al2O3 Unknown: mass = ? g Al2O3 First, determine the mass of 1 mol of Al2O3. 2 mol AI (mol Al cancels) x 27.0 g Al (atomic mass)/1 mol Al (mol Al cancels) = 54.0 g Al 3 mol O x 16.0 g O/1 mol O = 48.0 g O 1 mol Al2O3 = 54.0 g Al + 48.0 g O = 102.0 g Al2O3 Identify the conversion factor relating moles of Al2O3 to grams of Al2O3. 102.0 g Al2O3/1 mol Al203 Multiply by the given number of moles by the conversion factor. 9.45 mole Al2O3 (mole Al2O3 cancels) x 102.0 g Al2O3/1 mol Al2O3 (mol Al2O3 cancels) =964 g Al2O3
Calculating Gas Quantities at STP
Known: number of moles = 0.60 mol SO2 and 1 mol SO2 = 22.4 L SO2 at STP Unknown: volume = ? L SO2 First, identify the conversion factor relating moles of SO2 to volume of SO2 at STP. 22.4 L SO2/1 mol SO2 Multiply the given number of moles by the conversion factor. 0.50 mol SO2(mol SO2 cancels) x 22.4 L SO2/1 mol SO2(1 mol SO2 cancels) =13 L SO2
Prefixes Used in Naming Binary Molecular Compounds
Mono - 1 Di - 2 Tri - 3 Tetra - 4 Penta - 5 Hexa - 6 Hepta - 7 Octa - 8 Nona - 9 Deca - 10
Single Replacement Reaction Example
One example of a single replacement reaction occurs when a piece of copper metal is dipped into an aqueous solution of silver nitrate. You may recall that metals, such as copper and silver, tend to lose electrons and form cations, or positively charged ions. During a single replacement reaction, positive ions always replace positive ions. Therefore, in this example, copper takes the place of silver in the compound, and a new compound is formed. The compound will probably be formed with copper (II), the most common ion. The ions of silver form metallic silver. This reaction can be described as follows: solid copper+aqueous silver nitrate yield solid silver+aqueous copper (II) nitrate The reaction can be more precisely described by writing it as a balanced chemical equation: Cu(s)+2AgNO3(aq)-->2Ag(s)+Cu(NO3)2(aq) Since two nitrate ions are needed on the product side for the copper (II) ion, the reactant side will need two silver nitrate ions. The product side then needs to have two solid silver atoms. The letter s, written behind copper on the reactant side and silver on the product side, stands for the word solid. The letters a q, written behind silver nitrate on the reactant side and copper (II) nitrate on the product side, stand for the word aqueous.
Coefficients
Small whole numbers that are placed in front of the formulas in an equation in order to balance it.
Polyatomic Example
Some ions, such as the sulfate ion, are called polyatomic ions. The sulfate anion consists of one sulfur atom and four oxygen atoms. These five atoms together comprise a single anion with an overall 2- charge. The formula is written SO2-/4. Note that the names of most polyatomic anions end in -ite or -ate.
STP
Standard temperature and pressure (STP) means a temperature of O degrees C and a pressure of 101.3 kPa, or 1 atmosphere (atm). At STP, or 6.02 x 10^23 representative particles, of any gas occupies a volume of 22.4. The quantity, 22.4 L, is called the molar volume of a gas.
Law of Definite Proportions
States that in samples of any chemical compound, the masses of the elements are always in the same proportions.
Balanced Binary Ionic Compound Example
The binary compound calcium bromide is composed of calcium cations (Ca^2+) and bromide anions (BR^-). The two ions do not have equal numerical charges. Thus, each calcium ion with its 2+ charge must combine with (or balanced by) two bromide ions, each with a 1- charge. That means that the ions must combine in a 1:2 ratio, so the formula for calcium bromide is CaBr2. The net ionic charge of the formula unit is zero.
-ite and -ate
The charge on each polyatomic ion in a given pair is the same. The -ite ending indicates one less oxygen atom than the -ate ending. However, the ending does not tell you the actual number of oxygen atoms in the ion. All anions ending with names ending in -ite or -ate contain oxygen.
Empirical Formula
The empirical formula of a compound gives the lowest whole-number ratio of the atoms or moles of the elements in a compound. An empirical formula may or may not be the same as a molecular formula. For example, the lowest ratio of hydrogen to oxygen in hydrogen peroxide is 1:1. Thus, the empirical formula of hydrogen peroxide is HO. The molecular formula of hydrogen peroxide, H2O2, has twice the number of atoms as the empirical formula, but the ratio is still the same, 1:1.
What are the five general types of reactions?
The five general types of reactions include combination, decomposition, single-replacement, double-replacement, and combustion.
How does the molecular formula of a compound compare with the empirical formula?
The molecular formula of a compound is either the same as its experimentally determined empirical formula, or it is a simiple whole-number multiple of its empirical formula. Ex.) Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N. Divide the molar mass by the empirical formula mass to obtain a whole number. Multiply the empirical formula subscripts by this value to get the molecular formula. Knowns: empirical formula = CH4N molar mass = 60.0 g/mol Unknown: molecular formula = C?H?N? First, calculate the empirical formula mass. efm of CH4N = 12.0 g/mol + 4(1.0 g/mol) + 14.0 g/mol = 30.0 g/mol Divide the molar mass by the empirical formula mass. molar mass/efm = 60.0 g/mol/30.0 g/mol (g/mol cancel) = 2 Multiply the formula subscripts by this value. (CH4N) x 2 = C2H8N2.
Naming Anions
The name of an anion is not the same as the element's name. Anion names start with the stem of the element name and end in -ide.
Naming Cations
The names of the cations of the Group 1A, Group 2A, and Group 3A are the same as the name of the metal, followed by the word ion or cation.
How can you calculate the empirical formula of a compound?
The percent composition of a compound can be used to calculate the empirical formula of that compound. The percent composition tells the ratio of masses of the elements in a compound. The ratio of masses can be changed to a ratio of moles by using conversion factors based on the molar mass of each element. The mole ratio is then reduced to the lowest whole-number ratio to obtain the empirical formula of the compound. Ex.) A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula of the compound? Knowns: percent by mass of N = 25.9% N percent by mass of O = 74.1% O Unknown: empirical formula = N?O? (Note: g N and g O are atomic mass) Convert the percent by mass of each element to moles. 25.9 g N x 1 mol N/14.0 g N (g N cancel) = 1.85 mol N 74.1 g O x 1 mol O/16.0 g O = 4.63 mol O The mole ratio of N to O is N1.85O4.63. Divide each molar quantity by the smaller number of moles to get 1 mol for the element with the smaller number of moles. 1.85 mol N/1.85 = 1 mol N 4.63 mol O/1.85 = 2.50 mol O The mole ratio of N to O is now N1O2.5. Multiply each part of the ratio by the smallest whole number that will convert both subscripts to whole numbers. 1 mol N x 2 = 2 mol N 2.5 mol O x 2 = 5 mol O The empirical formula is N2O5.
Percent Composition
The relative amounts of the elements in a compound are expressed as the percent composition, or the percent by mass of each element in the compound. These percents must total 100%. The percent composition of a compound is always the same. If you know the relative masses of each element in a compound, you can calculate the percent composition of the compound. Ex.) Potassium Chromate K2CrO4, K=40.3%, Cr=26.8%, and O=32.9%.
Single Replacement Reaction and Aqueous Solutions
The type of reaction that occurs when an ion or element reacts with a compound. This most often occurs in aqueous solutions. An aqueous solution is formed when an ionic compound is dissolved in water. Although ionic solutions are hard and have a high melting point, they dissolve readily in water due to the positive charge of water's hydrogen atoms and negative charge of water's oxygen atoms. During dissolution, the cation and anions separate.
What are the steps for writing and balancing a chemical equation?
To write a balanced chemical equation, first write the skeleton equation. Then use coefficients to balance the equation so that it obeys the law of conservation of mass.
How do you write a skeleton equation?
To write a skeleton equation, write the chemical formulas for the reactants to the left of the yields sign (arrow) and the formulas for the products to the right. To add more information to the equation, you can indicate the physical states of substances by putting a symbol after each formula. Use (s) for solid, (l) for liquid, (g) for gas, and (aq) for a substance in aqueous solution (a substance dissolved in water).
How do you determine the formula and name of a compound with a polyatomic ion?
To write the formula for a compound with a polyatomic ion first write the symbol (or formula) for the cation followed by the symbol (or formula) for the anion. Then, add subscripts as needed to balance the charges. Whenever more than one polyatomic ion is needed to balance the charges in an ionic compound, use parentheses to set off the polyatomic ion in the formula.
How do you determine the formula and name of a binary ionic compound?
To write the formula of a binary ionic compound, first write the symbol of the cation and then the anion. Then add subscripts as needed to balance the charges. The positive charge of the cation must balance the negative charge of the anion so that the net ionic charge of the formula is zero. If you know the formula for a binary ionic compound, you can write its name. First you must verify that the compound is composed of a monatomic metallic cation and a monatomic nonmetallic anion. To name any binary ionic compound, place the cation name first, followed by the anion name. If the metallic element in a binary ionic compound has more than one common ionic charge, a Roman numeral must be included in the cation name.
What describes proportions?
Two laws--the law of definite proportions and the law of multiple proprtions--describe the proportions in which elements combine to form compounds.
Roman Numerals and Suffix
Two methods are used to name ions with more than one common ionic charge. The preferred method is called the Stock system. In the Stock system, you place a Roman numeral in parentheses after the name of the element to indicate the numerical value of the charge. For example, the cation Fe^2+ is named iron (II) ion. An older, less useful method for naming these cations uses a root word with different suffixes at the end of the word. The older, or classical, name of the element is used to form the root name for the element.
How do polyatomic ions differ from monatomic ions? How are they similar?
Unlike a monatomic ion, a polyatomic ion is composed of more than one atom. But like a monatomic ion, a polyatomic ion behaves as a unit and carries a charge.
Flowchart
Use Q(x)R(y). 1. Q+H? yes=acid 2. no= >2 Elements? yes= Q=Group A? yes=Name the ions and no=Name the ions; use a Roman numeral with the cation. 3. no= Q=Metal? no=Binary molecular, use prefixes in the name. 4. yes= Q=Group A? yes=Name the ions. no=Name the ions; use a roman numeral with the cation.
How do you convert the mass of a substance to the number of moles of the substance?
Use the molar mass of an element or compound to convert between the mass of a substance and the moles of the substance. The conversion factors for these calculations are based on the relationship: molar mass = 1 mol. molar mass/1 mol and 1 mol/molar mass The molar volume is used to convert between the number of moles of gas and the volume of the gas at STP. The conversion factors for these calculations are based on the relationship 22.4 L = 1 mol at STP. 22.4 L/1 mol and 1 mol/22.4 L
How can you determine the charges of monoatomic ions?
When the metals in Groups 1A, 2A, and 3A lose electrons, they form cations with positive charges equal to their group number. The charge of a Group A nonmetal is determined by subtracting 8 from the group number. The charges of the cations of many transition metal ions must be determined from the number of electrons lost.
Percent Composition as a Conversion Factor Example
You can use percent composition to calculate the number of grams of any element in a specific mass of a compound. To do this, multiply the mass of the compound by a conversion factr based on the percent composition of the element in the compound. In the sample problem above, you found that propane is 81.8 percent carbon and 18 percent hydrogen. That means that in a 100-g sample of propane, you would have 81.8 g of carbon and 18 g of hydrogen. That means that in a 100-g sample of propane, you would have 81.8 g of carbon and 18 g of hydrogen. You can use the following conversion factors to solve for the mass of carbon or hydrogen contained in a specific amount of propane. 81.8 g C/100 g C3H8 and 18 g H/100 g C3H8 Known: mass of C3H8 = 82.0 g Unknowns: mass of carbon = ? g C mass of hydrogen = ? g H To calculate the mass of C, first write the conversion factor to convert from mass of C3H8 to mass of C. 81.8 g C/100 g C3H8 Multiply the mass of C3H8 by the conversion factor. 82.0 g C3H8 x 81.8 g C/100 g C3H8 (g C3H8 cancel) = 67.1 g C To calculate the mass of H, first write the conversion factor to convert from mass of C3H8 to mass of H. 18 g H/100 g C3H8 Multiply the mass of C3H8 by the conversion factor. 82.0 g C3H8 x 18 g H/100 g C3H8 (g C3H8 cancel) = 15 g H
Transition Metals with One Ionic Charge
the names of these cations do not have a Roman numeral. These exceptions include silver, with cations that have a 1+ charge, as well as cadmium and zinc, with cations that has a 2+ charge.
