2 Atoms and the Atomic Theory

• Avogadro's constant is used to relate the Describing Chemical Compounds mass and molar mass to their elementary entities.

• Composition data can be used to determine the empirical and molecular formula of a chemical compound.

• The oxidation state of each element in a chemical compound can be determined using the rules for assigning oxidation states.

• The general rules can be used to name simple inorganic compounds.

• The general rules to name simple organic compounds include basic branched alkanes and alkanes with functional groups such as alcohols and carboxylic acids.

• The electron microscope has a scanning electron microscope.
• The names of chemical compounds and their formulas are discussed in this chapter.

• Water, ammonia, carbon monoxide, and carbon dioxide are simple chemical compounds.
• There are also acetylsalicylic acid and ascorbic acid.
• They are also chemical compounds.
• In this chapter, we will consider a number of ideas about compounds, as the study of chemistry is mostly about chemical compounds.

• All compounds are composed of two or more elements.
• The full range of compounds can be divided into a few broad categories by applying ideas from the periodic table.

• The symbols of their elements are derived from the chemical formulas that represent them.

• There is an overview of the relationship between names and formulas at the end of the chapter.

• Two fundamental kinds of chemical bonds hold together in a compound.
• The basic features of ionic and covalent bonding are not as clear-cut as we need them to be for the early chapters of the text.
• These statements imply.
• We will discuss chemical bonding in Chapters 10 and 11.

• The elements are marked by their sym bols in the formula for water.
• The number 1 is understood if no subscript is written.

• Another example of a chemical formula is CCl4.
• We can refer to water and carbon tetrachloride as compounds.

• The simplest whole-number ratio is what the subscripts in an empirical formula are reduced to.
• The formula P2O5 is the empirical formula for a compound.
• The empirical formula doesn't tell us a lot about a compound.
• The empirical formula CH2O is used to calculate acetic acid (C2H4O2), formaldehyde (CH2O), and blood sugar.

• In some cases, the empirical and molecular formulas are the same.
• The empirical formula is a multiple of the molecular formula.
• A molecule of acetic acid consists of eight atoms--two C atoms, four H atoms, and two O atoms--so the molecule's formula is C2H4O2.
• This is more than the number of atoms in the formula unit.
• The combining ratio of the atoms in the compound is shown in the formulas, but they don't show how the atoms are attached to each other.
• This information is conveyed by other types of formulas.

• The black spheres are carbon, the red are oxygen, and the white are hydrogen.
• The formula of acetic acid is written to show that one H atom in the molecule is fundamentally different from the other three.
• The formulas CH3COOH and CH3CO2H are used to show that the H atom is bonding to an O atom.
• Different versions of chemical formulas can be found in different sources.

• Two of the C atoms are bonding to one of the O atoms.
• There are differences between single and double bonds in the text.
• Think of a double bond as being a stronger bond than a single bond.

• The acetic acid molecule is represented by either CH3COOH or CH3CO2H.
• The different ways in which the H atoms are attached are still visible with this type of formula.

• Condensed structural formulas can be used to show how a group of atoms are attached.
• Consider C4H10 in Figure 3-2(b).
• The central carbon atom has a group of CH3 atoms attached to it.
• The formula shows this by enclosing the CH3 in parentheses to the right of the atom to which it is attached.
• If the central C atom isbonded to each of the other three C atoms, we can write a Condensed Structural formula.

• One way to simplify organic compounds is to write structures without showing the C and H atoms.
• The number of H atoms needed to complete each carbon atom's four bonds is assumed to be present, since a carbon atom exists wherever a line ends or meets another line.
• The symbols of other atoms and bond lines joining them to C atoms are written.

• Molecules occupy space and have a three-dimensional shape, but empirical and molecular formulas do not convey information about the spatial arrangements of atoms.
• Structural formulas can sometimes show this, but usually the only satisfactory way to represent the three-dimensional structure is with models.
• The models help us to see distances between the atoms and the shapes of the molecule.

• The visualization of butane, methylpropane, and testosterone can be somewhat misleading.
• Chemical bonds draw atoms into each other.
• A ball-and-stick model implies that the atoms are held apart.

• N molecule is magnified.

• The acetic acid molecule is made up of three different types of atoms and models of the molecule reflect this fact.

• The sizes of the colored spheres correspond to the size differences between the various atoms in the periodic table.

• One of the most important aspects of chemistry is the color scheme for shapes.

• Through empirical, HOOCCH2CH2COOH can be represented in the periodic table as line-angle formulas.

• The atoms of metallic elements tend to lose electrons when they are combined with nonmetal atoms.
• The charge can be deduced from the group of the periodic table to which the element belongs.
• We can use the periodic table to write the formulas of ionic compounds.

• Each sodium atom gives up one electron to become a sodium ion, Na+, and each chlorine atom gives up one electron to become a chloride ion.
• The location of the elements in the periodic table and the charges on their simple ion are related.
• There must be one ion for each Cl Na+ ion.

• The chemical formula has the same ratio of atoms as the formula unit.
• A formula unit of an ionic compound does not exist as a distinct entity because it is buried in a crystal.
• It is not appropriate to call a formula unit of solid sodium chloride a molecule.

• There is a similar situation with magnesium chloride.
• In trace quantities of magnesium chloride, magnesium atoms lose two electrons to become magnesium ion.
• The formula unit must have at least two Cl- ion with a charge of -1, for it to work.
• There is a formula for magnesium chloride.

• The single ion NO 3 is joined by N atom bonds.

• A crystal is a formula unit of this compound.

• The formula based on this formula combination of one Na+ and unit is marked by enclosing NO3 in parentheses and the smallest 2 in the subscript.
• Section 3-6 talks about polyatomic ion.

• It's a formula unit.

• The metallic bond gives metals and compounds their characteristic properties.

• Adding up atomic described for atomic mass and atomic weight in the mass can be used to get Formula and molecular mass.

• As a result, we can apply the concept of a mole to and are related, they are not any quantity that we can represent by a symbol or formula.
• Mula units are referred to as molecules.

• We get a numerical value but different mass of 18.014 g H2O compared to 12 g for carbon-12 if we compare samples of water with the same molecule and carbon atoms.

• If we know the formula of a com pound, we can equate the following terms.

• Several different types of conversion factors can be applied in a variety of problem-solving situations.
• The strategy that works best for a particular problem depends on how the conversions are visualized.

• The main focus of a problem is the conversion of a mass in grams to moles or vice versa.
• Other conversions involving volumes, densities, percentages, and so on must be preceded or followed by this conversion.
• A conversion pathway is a helpful tool in problem solving.
• Table 3.1 summarizes the roles that density, molar mass, and Avogadro constant play in a conversion pathway.

• The central focus is the conversion of a measured quantity to an amount in moles.

• We can convert from moles to formula units with the Avogadro constant as a conversion factor.

• A conversion pathway that starts with the information and proceeds through a series of conversion factors is often helpful.

• We'll use a single line calculation to avoid having to write down intermediate results.

• The final answer is rounded to one significant figure because the sample's mass is given with one significant figure.
• There is one more significant figure than the measured quantity in the calculation above, which is why the molar mass is rounded to two significant figures.

• One of the most odoriferous substances is the volatile liquid ethyl mercaptan, C2H6S.
• It can be added to natural gas to detect gas leaks.
• The liquid ethyl mercaptan has a density of 0.84 g/mL.

• The conversion of a measured quantity to an amount in moles is the central focus again.
• It will be helpful to convert the volume to liters because of the density.
• The density can be used to get the mass in grams, and the molar mass can be used to get the mass in moles.
• The Avogadro constant can be used to convert the amount in moles to the number of molecules.
• The conversion pathway is described as mL : L : g : mol.

• The required conversions can be combined into a single line calculation.
• The calculation should be broken into three steps: (1) a conversion from volume to mass, (2) a conversion from mass to moles, and (3) a conversion from amount in moles to molecule.
• The roles played by density, molar mass, and Avogadro constant in the conversion pathway are emphasized in the three steps.

• 10-4 g C2H6S 1mol C converts from mass to moles.

• At the end.
• The volume and density are given with two significant figures.
• If the required conversions are combined into a single line calculation, rounding errors are avoided.

• The density of gold is 19.32 g> cm3.
• A piece of gold foil is 2.50 cm on each side.

• There is a limit of detectability.

• This is the only definition that can be used for elements like iron, magnesium, sodium, and copper, in which enormous numbers of individual spherical atoms are clustered together.
• Some elements are joined together to form a molecule.
• There are collections of molecule in the bulk samples of these elements.

• There are eight sulfur atoms in a sample of solid sulfur.
• There are four phosphorus atoms per molecule.

• For example, hydrogen has an atomic mass of 1.008u and a molecular mass of 2.016u; its molar mass can be expressed as 1.008 g H>mol H or 2.016 g H2>mol H2.

• If you don't do detailed calculations, you can determine which of the following quantities has the greatest mass and which has the smallest mass.

• A chemical formula gives a lot of information.

• There are other types of calculations that are based on the chemical formula.

• In this case, the factor is 2mol C.

• There is a conversion pathway for this problem.
• The volume of the sample needs to be converted to mass.
• The inverse of the molar mass as a conversion factor is required to convert the mass of halothane to moles.
• The formula of halothane is used to calculate the final conversion factor.

• The final conversion factor would have been 2 mol C/1 mol C2HBrClF3 if we had been asked for the number of moles.

• A sample of a new compound is sent to an analytical laboratory to determine its percent composition.
• The percent composition calculated from the formula of the expected compound is compared with the experimentally determined percent composition.
• The compound obtained can be compared to the one expected.

• The mass percent of an element in a com pound is calculated by the equation.

• Determine the mass of the compound.

• Determine the contribution of the element to the mass.

• This is the ratio of the numerator and the denominator.

• To get the mass percent of the element, you have to combine this ratio by 100%.

• The mass composition of a compound is a collection of mass percentages.

• The four-step method can be applied.
• Determine the molar mass first.
• Then convert the mass ratios to mass percents.
• The mass percents will be accurate if they are rounded to two decimal places.

• The mass of C2HBrF3 is more than 200 g>mol.

• The percent composition of halothane is as follows.

• The mass of carbon in a 100 g sample is equal to the mass percent of carbon.
• If you compare the calculation of g C with that for % C given earlier, you will see that both calculations involve the same factors but in a slightly different order.

• The main energy-storage molecule in cells is edinosine triphosphate.
• Its formula is C10H16N5P3O13.

• The percentages of elements in a compound can be used in two different ways.

• Ensure that the percentages total 1000% by checking the accuracy of the computations.
• The percentages of the elements should be determined.

• Sometimes a chemist has no idea what a chemical compound is.
• A report on the percent composition of the compound can be used to determine the formula.

• Consider the following five-step approach to determining a formula from the experimentally determined percent composition of the compound.
• The mass percent composition is 44.7% C, 7.52% H, and 47.7% O.

• The mass of the elements are equal to their percentages, which is 44.77 g C, 7.52 g H, and 47.71 g O.

• A tentative formula can be written based on the numbers of moles determined.

• The tentative formula has subscripts that can be converted to small whole numbers.
• The subscripts need to be divided by the smallest one.

• If all subscripts differ slightly from the whole numbers, they are rounded off to whole numbers, concluding the calculation at this point.

• If one or more subscripts are still not a whole number, use a small whole number to make them all integral.

• To find the factor, we must use the empirical formula with the true mass of the compound.
• By using methods introduced in Chapters 6 and 14 we can establish the mass from a separate experiment.
• The mass of 2-deoxyribose is 134 U.
• The empirical formula, C5H10O4, has a formula mass of 134.1 u.
• The measured mass is the same as the empirical mass.
• The formula is called C5H10O4.

• The five-step approach described in the flow diagram below will be applied to the next example, where we will find that the empirical formula and the molecular formula are not the same.

• Household ants and roaches are treated with dibutyl succinate.
• It has a composition of 62.58% C, 9.63% H, and 27.7% O.
• The mass is determined by experimentation.

• You can use the five-step approach.

• Determine the mass of each element in a 100.00 g sample.

• Round off any subscripts that are slightly different from the whole numbers.

• If you want to make the subscripts integral, use the factor 2 and write 2 * 5.49 as the empirical formula.

• Determine the empirical formula mass to establish the formula.

• The empirical formula mass is twice as large as the experimentally determined formula mass.

• Use numbers that are rounded off slightly to check the result.

• C12H22O4 (12 * 12 U/230 U) * 100% of the time, with a % H L of 22 and a % O L of 4.
• We can be confident that our answer is correct because the mass percents agree with those given in the problem.

• Sorbitol, used as a sweetener in some sugar-free foods, has a molecular mass of 182 U and a mass composition of 39.5% C, 7.74% H, and 52.70% O.

• The narcotic drug has a mass of 21.21% C, 2.22% H, and 17.6% O.
• Its mass is 726 U.

• The amount of rounding off depends on how the analysis is done.
• There isn't an ironclad rule on the matter.
• You can round off a subscript that is within a few hundredths of a whole number if you carry all the significant figures allowable in a calculation.
• If the deviation is more than this, you will need to adjust subscripts to integral values.
• If the appropriate constant is larger than a simple number, such as 2, 3, 4, or 5, you may find it easier to make the adjustment by multiplying twice, for example, by 2 and then by 8.

• The water vapor and carbon dioxide gas are absorbed by substances.
• The mass of these absorbers increases with the amount of water and carbon dioxide.
• As shown below, we can think of the matter.

• The carbon atoms in the sample are found in the CO2.

• The H2O contains all the hydrogen atoms.
• The sample being analyzed was the only source of carbon and hydrogen atoms.
• Oxygen atoms in the CO2 and H2O could have come from either the sample or the oxygen gas consumed in the combustion.
• The quantity of oxygen in the sample has to be determined indirectly.

• Oxygen gas passes through the tube to the sample being analyzed.
• A portion of the apparatus is in a high temperature furnace.

• Magnesium perchlorate and carbon dioxide gas are absorbed as they leave the furnace.
• H2O and CO2 can be produced in the combustion reaction if there are differences in the mass of the absorbers.

• Oxygen is present at the end of the reaction because of the excess of oxygen.

• There is a need to conserve mass.

• The prevention of scurvy depends on the composition of chemical compounds.

• The carbon and hydrogen atoms in the vitamin C sample are both in CO2 and H2O.
• Oxygen atoms in CO2 and H2O come partly from the sample and partly from the oxygen gas consumed in the combustion.
• We focus on carbon and hydrogen first in the determination of the percent composition.
• The mole ratios and the amounts of C, H, and O are used to calculate the empirical formula.

• 44.009 g CO2 1 mol CO2 was sent to g C.

• We have two options at this point.
• The number of moles of C and H has already been determined in the 0.2000 g sample.

• The mass percent composition does not need to be determined as a preliminary calculation for the determination of the empirical formula.
• If the number of moles of the different atoms in the sample can be determined, the empirical formula can be based on a sample of any size.

• The substance that flavors rum is isobutyl propionate.

• The greatest mass of both CO2 and H2O is produced when there is an excess of oxygen.

• Some samples can't be easily burned, but we have seen how combustion reactions can be used to analyze Chemi cal substances.
• Other types of reactions can be used for chemical analyses.
• Some of the methods will be cited later in the text.

• A some ena is a basic concept in chemistry that deals with measurable properties.
• This is the case with the monatomic ion.

• There are polyatomic ion and mol other atoms in compounds.

• An Na atom, a metal, loses one electron to a nonmetal in this compound.

• Na+ is in a state of oxidation and Cl + 1 is in a state of -1.

• An Mg atom loses two electrons to become Mg2+, and each Cl atom gains one electron to become Cl-.
• The oxidation state of Cl is -1, while the oxidation state of Mg is +2.

• The oxidation state of the two Cl atoms should be the same.
• If their total is zero, each oxidation state must be 0.
• The oxidation state of an atom can vary depending on the compounds in which it occurs.
• We assign H the oxidation state of +2 in the molecule H2O.
• The total of the oxidation states of the atoms must be zero.

• The term oxidation number refers to a number.
• The two terms will be used interchangeably.

• We need some rules for assigning oxidation states from these examples.

• It's MnO 4.

• The group 1 metals have an O.S.

• H2O usually has an O.S.

• Oxygen has an O.S.
• in its compounds.

• Group 17 elements have an O.S.

• The formula represents a molecule.

• The total of the oxidation states of the atoms in this formula unit is zero.
• Three O atoms have a total of 6.
• Two Al atoms have a total of 6.

• This formula is used for permanganate ion.
• The oxidation states of the atoms in the ion are summed up by rule 2.
• The four O atoms have a total of -8.

• The formula unit of the compound is NaH.
• Rule 5 states that H should have an O.S.
• Both atoms would have an O.S.
• Na has an O.S.

• H2O2 is hydrogen peroxide.
• Rule 5 states that H has an O.S.
• The oxidation states of the two H atoms are +2 and the oxidation states of the two O atoms are -2.

• The oxidation states of four O atoms are -8.
• The sum of the oxidation states must be +8 for three Fe atoms.

• Without writing down any arithmetic expressions, you should be able to do the math associated with assigning oxidation states in your head.
• Make sure the oxidation states are equal to the charge on the atom, molecule, or ion in order to check your result.

• The oxidation states are equal to the charge on the MnO 7 + 4(-2).

• We only saw values for oxidation states before that.
• The assumption is that all the atoms of an element have the same oxidation state.
• Fe3O4 is probably better represented as FeO # Fe2O3 through a combination of two simpler formula units.

• The Fe atoms are +3.

• We may need to "fragment" a formula into its parts before assigning oxidation states.

• It is more useful to know the oxidation states of the individual N atoms than it is to know the average oxidation state of the two N atoms.

• The naming of chemical com pounds in the next section is the first use of oxidation states.

• A compound with a mass of 32 g>mol has N in a higher oxidation state than NH3.

• We have referred to compounds mostly by their formu, but we need to give them names.
• When we know that a com contains the same elements, we can look up its properties in a handbook, find a chemical in a storeroom shelf, or discuss an experiment with a colleague.
• There are cases in which different compounds have the same formula.
• We need to distinguish among compounds by name in these names and formulas.

• If all compounds were referred to by the color yellow.

• That has its own set of rules.

• In Section 3-7, we will introduce the naming of organic compounds.

• The approach is illustrated below.

• The charge of the ion in a formula unit must be zero.

• The names and symbols of simple ion are listed in Table 3.3.
• When writing names and formulas of metals and nonmetals, you will find this list useful.
• It is important to distinguish between the compounds of some metals.
• The oxidation state of the metal is indicated by the Roman numeral immediately following its name.

• Two different word endings are used to distinguish between two compounds with the same two elements but in different proportions, such as Cu2O and CuO.
• The oxidation state of copper in Cu2O is +2.

• Once you understand the pattern, there are some Simple Ions.

• The compound is a molecule if the two elements are both nonmetals.
• The method of naming these compounds is similar.

• In each case, identify the cations and their charges, based on periodic table group numbers or on oxidation states that are Roman numerals.
• O2, F-, and S2 are the anions.

• The cation in Cu2O is Cu+, copper(I), which is a different form of copper.

• It's difficult to know when to use Roman numerals and when not to.
• Roman numerals are not used in naming compounds because the metals of groups 1 and 2 have only one ionic form.

• We need to distinguish between pairs of nonmetals.

• Several substances have common or trivial names that are so well established that their systematic names are almost never used.

• We sometimes want to emphasize that their solutions are acids, even though we use names like hydrogen chloride.
• In the text, acids will be discussed.
• In water, HCl ionizes into hydrogen ion and chloride ion.

• It doesn't show a tendency to produce H+ under any conditions.

• In Chapter 5, we will see that bases yield OH- in solutions.

• Below is a list of the most important binary acids.

• It is common among the nonmetals.
• Table 3.5 contains a number of polyatomic ion and compound names.

• There are more polyatomic anions than there are polyatomic cations.

• The only ones that do arehydroxide ion andcyanide ion.

• Some nonmetals form a series of oxoanions with different numbers of oxygen atoms.

• The species produced in the solution is more complex than the simple ion H+.

• The H+ combines with an H2O molecule to produce an ion called the hydronium ion, H3O+.

• We will use H+ in place of H3O+ until we discuss this more fully in Chapter 5.

• It will become obvious.

• For example, hydrogencarbonate, hydrogenphosphate, and so forth are written as a single word.

• I carry a charge of -1.

• Various numbers of H atoms are found in some series of oxoanions.

• There are several oxoacids in Table 3.6.
• The names and formulas of compounds in which the hydrogen of the oxoacid has been replaced by a metal are listed.
• Acids and salts are both ionic compounds.

• formulas are often written to reflect this fact, for example, HOCl instead of HClO and HOClO instead of HClO2.

• CuCl2 and Ca(H2PO4)2 are ionic compounds.
• We need to identify and name the compounds.

• HIO4 and ClO2 are compounds.
• HIO4 is a compound of two nonmetals.

• The oxidation state of Cu is + 2.
• We must clearly distinguish between the two possible chlorides because Cu can also exist in the oxidation state of + 1.
• CuCl2 is a metal.

• Both are nonmetals.
• There is a compound called chlorine dioxide.

• The oxidation state of I is 7.
• The compound periodic acid should be referred to as "purr-eye-oh-dic" acid.

• The compound calcium dihydrogenphosphate has two of these ion present.

• It is associated with naming compounds correctly.
• It takes a lot of practice to master this subject.

• The copper compound that Joseph Proust used to establish the law of constant composition is referred to in different ways.
• You have simpler formulas.

• N4S4 is the formula.

• The subscript 2 should be placed around NH + 4.
• The formula is (NH4)2CrO.

• In Table 3.6, there is a comparison of bromide acid to HClO3.

• This leads to a formula.

• The placement of parentheses in writing formulas is important.

• You should be able to understand a formula even if it is complex.

• Hydrates are some of the complex substances you are likely to encounter.
• This doesn't mean that the compounds are wet.
• The solid structure of the compound has water in it.
• The six H2O mole cules per formula unit are shown.

• The piece of filter paper was removed by heating.
• Anhydrous com cobalt(II) chloride is allowed to dry.
• In the use of anhydrous magnesium dry air, the paper is blue perchlorate, which can be used as water absorbers.
• The color change in humid water indicates this.
• The paper is pink because of anhydrous air.
• This fact can be used to make something.

• There are many organic compounds in nature.
• The foods we eat are mostly made up of organic compounds, which include energy- producing fats and carbohydrates, as well as trace compounds that impart color, odor, and flavor to these foods.
• Most of the fuels used to power automobiles, trucks, trains, or airplanes are mixtures of organic compounds.
• Most of the drugs produced by pharmaceutical companies are complex organic compounds.

• The diversity of organic compounds is due to the ability of carbon atoms to combine with other carbon atoms and with other elements.
• A framework of chains or rings is formed when carbon atoms join together.
• Most of the organic compounds contain hydrogen atoms, and there are many common four covalent bonds.

• The possibilities allow for many different organic compounds.
• Some of the organic compounds are ionic.

• There are millions of organic compounds.
• Their names are not easy to understand.
• The rules for naming inorganic compounds are of little use here, as a systematic approach to naming these compounds is crucial.
• The name is a-D-glucopyranosyl-b-D-fructofuranoside.
• We only need to recognize organic compounds and use their common names together with an occasional systematic name at this point.
• In Chapter 26 we will look at the systematic nomenclature of organic compounds.

• The simplest hydrocarbon has one carbon atom and four hydrogen atoms.
• The number of hydrogen atoms increases in a systematic way as the number of carbon atoms increases.
• Carbon atoms can form chains and rings, and the nature of the chemical bonds between the carbon atoms can vary, making organic chemistry complex.
• The simplest alkane is methane, followed by ethane, and then propane.
• The Number of Carbon ber of the alkane series is formed by the addition of one C atom and two H atoms to the preceding member.

• Common names are reflected in the number of the first four word stems in Table 3.7.

• Heptane is 7.

• The possibilities for isomerism increase very rapidly as organic molecules become more complex.

• They have the same formula but different structures.
• We check to see if the formulas are the same.
• The formulas represent different compounds if they are not the same.

• The first compound is called C7H16, while the second is called C8H18.

• The molecule are notomers.

• The same formula, C6H14, is used for these molecules.
• They are related.
• In the first structure, a side chain is on the middle carbon atom of a five-carbon chain, and in the second structure it is on the second carbon atom.

• There are clearly different compounds shown in part a.
• These compounds are expected to have different properties.
• The compounds shown have the same formula but different structures.
• They have different properties.
• The compound shown on the left has a slightly higher boiling point than the compound shown on the right.

• The framework of organic compounds is provided by carbon chains, other atoms or groups of atoms replace one or more of the hydrogen atoms to form different compounds.
• There is a common alcohol molecule in beer, wine, and spirits.

• The common name of one com hydroxide ion is often spelled that way.

• Methanol has a common name.

• The grain alcohol in beer and wine is safe to consume in moderate quantities, whereas wood alcohol is a dangerous poison.

• Some alcohols have their atoms attached to carbon chains or rings of organic molecule and give them their characteristic properties.
• There are compounds that have the same functional group.
• Functional groups are discussed in more detail in Chapter 26.

• The possibility of isomers is increased by the presence of functional groups.

• There is only one propane molecule.
• If one of the H atoms is replaced by a hydroxyl group, there are two possibilities for the point of attachment: at one of the end C atoms or the middle C atom.
• There are two isomers.
• The first or end C atom is where the OH group is.

• The OH group is on the second C atom.

• COOH confers acidic properties on a molecule.
• The two O atoms are bound to the C atom in two different ways.
• One bond is a single bond to an oxygen atom that is also attached to a H atom, and the other is a double bond to a lone O atom.

• The burning sensation that accompanies a bite from an ant is caused by formic acid being injected.

• acetic acid is in water.
• There are examples of substituting for one or more H atoms with the introduction of an additional functional group.

• To determine which functional group is present, look at the formula.
• Some of the carbon-carbon bonds may be single bonds.

• The bonds of carbon-carbon are all single.
• The compound is called alkane.

• One H atom has been replaced by a Cl atom, and there are only single bonds in its molecule.
• A chloroalkane is a com pound.

• CO2H is a carboxylic acid.

• This compound is an alcohol.

• Carbon-hydrogen and carbon-carbon bonds are notreactive.

• The characteristic properties of organic compounds are determined by functional groups.
• We will look at organic compounds in more detail in Chapters 26 and 27.

• These compounds are named.

• Determine the type of compound.
• To form the name, count the number of carbon atoms and select the stem from Table 3.7.
• The carbon number is placed before the part of the name that relates to the functional group.

• The structure of the compound is that of an alkane molecule with a five-carbon chain.

• 2-fluorobutane is a compound.

• The end C atom of the carbon chain is in a carboxyl group.
• propanoic acid is a compound.

• The compound is called pentan.

• In naming the compound, we stated that the F atom was attached to the second C atom in a fourcarbon chain.
• We can number the C atoms in two different ways, so there is some ambiguity in that statement.

• The correct name is 2-fluorobutane.

• Give plausible names for the molecules that correspond to the ball-and-stick models.

• First, identify the number of carbon atoms in the chain, and then determine the type and position of the functional group.

• The formula is CH31 CH222 CH3 because no functional groups are indicated.

• The formula is called CH31 CH222CO2H.

• The substitution of a chlorine atom for a H atom means that it is on the first C atom of the carbon chain.
• The formula for the structure is CH31 CH223 CH2Cl.

• The carbon chain is six C atoms long.
• The formula for the structure is CH31 CH224 CH2 OH.

• To get a structural formula from the name, we split it into component pieces: stem, prefix, and suffix.
• Information about the structure of the molecule is provided by all three components.

• The two main classes of chemical compounds are ball-and-stick and space-filling.

• Overall, that is neutral.
• The concept of the Avogadro can be written in many different ways.

• The symbols C and H are mostly omitted from the CHEM shown except for those between C and H atoms.
• The percent composition of the compound was determined by the relative sizes.
• The names and symbols of some simple ion are used for organic compounds.
• There are empirical formulas listed in the table.
• There are some ion simplest formulas that can be written.
• There is a determination of the determined molecular mass.

• One of the uses of the expres concept is to aid in naming sion 3.3, and the names and formulas of oxoanions are given.

• Adding names to the formulas of water molecule associated with their formula units.

• A 2.250 g sample of a dicarboxylic acid was burned in excess of oxygen and yielded 4.548 g CO2 and 1.629 g H2O.
• The mass of the acid was found in a separate experiment.

• The percent composition of the compound will need to be determined using the combustion data.

• Store intermediate results in your calculator without rounding off, if you use the molar mass with one more significant figure.

• The mass percent C is followed by chemical compounds in the sample of dicarboxylic acid.

• The empirical formula mass is twice as large as the experimentally determined one.

• The two C atoms, two H atoms, and four O atoms are accounted for.
• The structure is based on C6H12

• There are many other possibilities after we found a plausible structural formula.
• We can't identify the H atom on the chain.

• A sample of a liquid was heated to remove the water of hydration.
• The solid was analyzed and found to be 27.74% Mg, 23.57% P, and 48.69% O by mass.

• An unknown solid hydrate was analyzed and found to be 17.15% Cu, 19.14% Cl, and 60.45% O, by mass.

• The ball ball-and-stick models of the molecules are given here.
• The and-stick models are given here.

• The amount of liquid bromine in the blood is about 3.10 g.

• 0.50 mol of gaseous is supplied in a cylindrical form.
• There is 75 mL of liquid thio.
• C4H4S 1d is the amount of white phosphorus in a cylinder.

• It has four times the number of H atoms as O atoms.

• The proportions of C and O are the same.

• H is the highest percentage by mass.

• It has a mass ratio of 3 and 4.

• All of the minerals are semiprecious.

• Determine the percent H2O in the element.

• Be in beryl and quinine.

• Determine the percent, by mass, of the indicated ing in order of increasing %.

• There are two oxides of sulfur that have the same molecule.
• What is its mass?

• One has 28.8% O by mass and the other has a formula unit.
• The formula of this compound is 37.8% O.

• The indigo dye has a composi hydrogen-oxygen compound with a percent of it's value by mass.

• The mass of adenine is 4.8% H, 8.3% N, 37.8% O.
• The empir percent composition is 44.45% C, 3.73% H, and 51.82% N.

• Its mass is 135.14 U.

• The compound is formed by the element X, which contains 3.81% H, 20.73% N, 11.84% O, and 23.72% S.

• The chlorophyll has a mass of 2.72%.

• The molec of the two compounds was found to be 38.3% C, 1.49% H, 52.28% Cl, and 7.86% O by mass.

• The mass is 94.34% C and 5.66% H.

• A 0.1888 g sample of a hydrocarbon produces a total of 0.6260 g in combustion analysis, a 1.3020 g sample of thio CO2 and 0.1602 g H2O in combustion analysis.
• The phene produces CO2 and H2O with a mass of 106u.

• A sample of 1.00 mol of the compound is burned in excess of this carbon-hydrogen-oxygen compound, which yields oxygen: CH4, C2H5OH, C10H8 and C6H5OH.

• Its mass is 108.1 U.

• A 1.562 g sample of alcohol is used in rocket fuels.
• When burned in excess of oxygen.

• The density of sample is converted to 0.226 g N2.
• How much is the empir?

• The organic solvent thiophene is a sulfur compound that can be burned in excess of oxygen and produce CO2, H2O, and SO2 in the process.

• The states of FeO 2 dation are +3, +4, and +6.

• Two states are +2, +3, and +4, respectively.

• Oxygen oxidation following cases remind us that there is a state of -2.
• There are exceptions.
• There are exceptions.

• Oxygen and hydrogen have oxidation states of +2 and -2 in their compounds.

• The charge is -2.

• A hydrate has the composition H2O by mass.

• An 8.129 g sample of MgSO4 x H2O is heated until it can be removed from the liquid.
• The water of hydration is taken off.

• To deposit one mole of Ag.
• One faraday is the percentage of solution containing Ag+ that requires a quantity of elec atoms.

• The inside back cover of a particular type of brass contains Cu, Sn, Pb, and stants.
• A sample of 1.1713 g is treated in this way.

• The compound contained PbSO4 and Zn to 0.246 g.

What is the name of a drug?

• The sulfate M21SO423 is formed by the metal M.

• A 0.732 g mixture of methane, CH4, and ethane is found to have a mass of 61.10%.

• C2H6 yields 2.064 g CO2.

• The mixture of H2SO4 and water has a density of 0.0640 g per liter.
• 2.686 g of a second oxide of lead is contained in the mixture.
• What will be determined by converting H2SO4 to 1NH422SO4.

• The IUPAC pound MI is removed through precipitation.
• The recommended atomic mass is 1.186 g I.
• What is it?

• A range of values is reported for an oxoacid with the formula HxEyOz.

• The lower bound and oxygen and chlorine are used to calculate the dieldrin.

• Half as many chlorine atoms as carbon can be found in S 381 U.
• The average person can smell it.

• A sample of Na2SO4 has a limit of detectability of ethyl mercaptan in parts per exposed to the atmosphere.

• The Na2SO4 10 H2O mixture has a density of 1.2 g>L at room temperature.

• The atomic mass of Bi is determined by con ties.
• The compound Bi1C6H523 is verting to Bi2O3.

• Dry air is treated with fluorine gas.
• The mole per treatment converts all the gold in the foil to a small amount of krypton.
• It was given a gold fluoride.
• The gold density is 19.3 g/ cm3.

• A public water supply had a small amount of a compound.

• How much CHCl3 would be present in amol of water.

• A sample of the compound was 0.733 g.

• The black substance is oxide nitrogen, phosphorus, and potassium.
• The percent composition and tilizer have numbers on them.

• There are some substances that are slightly smilly smilly smilly smilly smilly smilly smilly smilly smilly smilly smilly smilly smil The molecule thick is the P. The K as KCl is a practical use of this phenomenon.
• The expressions to cover ponds were created in the 19th century.
• A monolayer is formed on water.
• The mol century, before the nature of chemical compounds, was fully understood.
• To convert from % P2O5 to another, rather like pencils tightly packed and standing % P and from % K2O to % K, the factors 2 mol P>mol upright in a coffee mug.
• The model below shows the stearic acid molecule in the monolayer.

What should the proportions of the 1.00 mL solution be?

• A layer of copper sulfate goes on the water.
• The changes suggested by the 85 cm2 are part of the area covered by the monolayer.
• Assume that the pictures are of oleic acid.

• A 2.574 g sample of CuSO # 4 x H2O was cooled and reweighed.
• The solid was cooled and reweighed after being reheated.

• This solid was heated and cooled before being reweighed.

• CuCO # 3 Cu(OH)2 is one of the formulas that might be used.
• What is the formula?

• Ibuprofen is used in drugs.
• The 2.174 g sample is burned in an excess of oxygen and will die.
• Red blood cells yield 6.029 g CO2 and 1.709 g H2O as the sole source of blood for an adult human, and the blood contains ucts.
• What is the percent composition, by mass, of a total of iron?

• The mass of XF3 is 65% F.