Chem

Consider theobromine, the active ingredient in chocolate: C7H8N4O2. If you eat one molecule of theobromine, then you've consumed? What if you eat one mole of theobromine? Did you still consume only 7 atoms of C?

7 atoms of C. No! You consumed one mole of molecules, and there are 7 atoms of C per molecule. So, you consumed 7 moles of C atoms! We can derive mole ratios from chemical formulas.

The Simple Gas Laws: Boyle's Law, Charles's Law, and Avogadro's Law

A sample of gas has four basic physical properties: pressure (P), volume (V), temperature (T), and amount in moles (n). These properties are interrelated—when one changes, it affects one or more of the others. The simple gas laws describe the relationships between pairs of these properties. For example, one simple gas law describes how volume varies with pressure at constant temperature and amount of gas; another law describes how vol-ume varies with temperature at constant pressure and amount of gas. These laws were deduced from observations in which two of the four basic properties were held constant in order to elucidate the relationship between the other two. Boyle's Law: Volume and Pressure In the early 1660s, the pioneering English scientist Robert Boyle (1627-1691) and his assistant Robert Hooke (1635-1703) used a J-tube (Figure 5.5▲) to measure the volume of a sample of gas at different pressures. After trapping a sample of air in the J-tube, they added mercury to increase the pressure on the gas. They found an inverse relationship between volume and pressure—an increase in one causes a decrease in the othe —as shown in Figure 5.6▶. This relationship is now known as Boyle's law. Boyle's law follows from the idea that pressure results from the collisions of the gas particles with the walls of their container. When the volume of a gas sample is decreased, the same number of gas particles is crowded into a smaller volume, resulting in more collisions with the walls and therefore an increase in pressure Scuba divers learn about Boyle's law during certification because it explains why a diver should not ascend toward the surface without continuous breathing. For every 10 m of depth that a diver descends in water, she experiences an additional 1 atm of pressure due to the weight of the water above her (Figure 5.8▼). The pressure regulator used in scuba diving delivers air at a pressure that matches the external pressure; otherwise the diver cannot inhale the air because the muscles that surround the chest cavity are not strong enough to expand the volume against the greatly increased external pressure. When a diver is at a depth of 20 m below the surface, the regulator delivers air at a pres- sure of 3 atm to match the 3 atm of pressure around the diver (1 atm due to normal atmo- spheric pressure and 2 additional atmospheres due to the weight of the water at 20 m). Suppose that a diver inhaled a lungful of air at a pressure of 3 atm and swam quickly to the surface (where the pressure drops to 1 atm) while holding her breath. What would happen to the volume of air in her lungs? Since the pressure decreases by a factor of 3, the volume of the air in her lungs would increase by a factor of 3, severely damaging her lungs and possibly killing her. We can use Boyle's law to calculate the volume of a gas following a pressure change or the pressure of a gas following a volume change as long as the temperature and the amount of gas remain constant. For these types of calculations, we write Boyle's law in a slightly different way. This relationship shows that if the pressure increases, the volume decreases, but the prod- uct P * V is always equal to the same constant. For two different sets of conditions, we can say that P1V1 = constant = P2V2

Molarity (M)

A unit of concentration in moles of solute per liter of solution. Molarity (M) = mole of solute/liter of solution

Certain groups and periods have specific names. Name of Group 1A:

Alkali Metals

Certain groups and periods have specific names. Name of Group 2A:

Alkaline earth Metals

Determining Empirical Formulas

An empirical formula can be determined through chemical stoichiometry by determining which elements are present in the molecule and in what ratio. The ratio of elements is determined by comparing the number of moles of each element present.

Parts of a Chemical Equation

Chemical equations Designed to represent the transformation of one or more chemical species into new substances. Reactants The starting components of a reaction Products The ending components. Chemical equations are always written: Reactants Products

When covalent compounds dissolve, the molecules themselves stay intact.

Covalent compounds (nonelectrolytes) C12H22O11(s) -->C12H22O11(aq)

Predicting Ionic Charges: Charge of Group 2A:

2+

Predicting Ionic Charges: Charge of Group 6A:

2-

Predicting Ionic Charges: Charge of Group 5A:

3-

Strong electrolytes

Dissociate completely (soluble ionic compounds, strong acids, and strong bases)

Weak electrolytes

Dissociate only a little bit (weak acids and weak bases)

Nonelectrolytes

Do not dissociate (covalent molecules)

Scientific knowledge is

Empirical Based on observation

The Structure of the Atom: Plum-pudding Model

John Dalton and the Atomic Theory

Converting Units

King Henry Did (L) Drink Chocolate Milk K= kilogram H = hectogram D = Deca D= Deci C = Centi M = milli

Compounds containing the following ions are generally soluble

Li⁺, Na⁺, K⁺, and NH₄⁺ = no exceptions NO₃⁻ AND C₂H₃O₂⁻ = no exceptions Cl⁻, Br⁻, and I⁻ = When these ions pair with Ag⁺, Hg₂⁺, or Pb²⁺, the resulting compounds are insoluble SO₄²⁻ = When SO₄²⁻ pairs with Sr²⁺, Ba²⁺, Pb²⁺, Ag⁺, or Ca²⁺, the resulting compound is insoluble

Precision

Precision refers to how close the measured value is to the actual value

Atoms are the basic unit of a chemical element and are composed of

Protons, neutrons, and electrons. Atoms are neutral, so they always have the same number of protons and electrons.

Aqueous solutions

Result when something is dissolved in water. Components of a solution:

Example: Use the solubility table to predict which ionic compounds are soluble in water are room temperature. KClO3 CaCO3 BaSO4 NaMnO4

Soluble because of K+ Insoluble because of CO₃²⁻ Insoluble—Ba2+ is an exception to the rule that sulfates usually are soluble Soluble because of Na+

Periods

The horizontal rows of the periodic table

Theoretical yield

The ideal amount of product that a reaction can make mathematically.

Empirical formula

The lowest whole number ratio of atoms in a compound

Concentration

The measure of the amount of solute dissolved in the solvent The larger the value the more solute is dissolved per unit amount of solvent.

Dissolution

The process of a substance dissolving in solvent. The molecules of the solute are surrounded by molecules of solvent in a process called solvation.

Dilution

The process of lowering the concentration of a solution by adding more solvent. The number of moles of solute does not change during a dilution. M₁xV₁ =M₂xV₂

Percent yield

The ratio of actual yield to theoretical yield. Percentage yield = actual yield/theoretical yield x 100%

In most reactions, one reactant will run out before the others.

The reactant that runs out first is called the limiting reactant (limiting reagent) The other reactant is called the excess reactant (excess reagent).

It is impossible to count a mole of substance.

Use molar masses (g/mol) to convert from mass to moles or vice versa.

stoichiometry flow chart

Wow! That's confusing! But look at the red density conversion on the flowchart above. We are converting from volume to mass, so we use density as the conversion factor. Example 2. How many mL is 1.50 kg of gasoline? If you look up the density of gasoline, it is 0.688 g/mL (pure isoocane, no ethanol). First convert 1.50 kg to grams, and obviously that is 1.50x103g. Part II. Converting from mass to moles and from moles to mass: Molar mass is our conversion factor. Molar mass is often called molecular weight or formula weight, depending on the type of compound (molecular or ionic). Example 3. Convert 100. g of water to moles of water. This is a simple enough calculation and I could have left out "H2O", but I find it helpful when you're converting from the mass of one compound to the mass of another, so I often include the chemical formula. Likewise you can convert moles to grams. Try to do this one yourself: Example 4. Convert 10.5 mole of methane, CH4, to grams of methane. The answer is 168 g CH4. A good tool to calculate molar masses is: http://www.tutor-homework.com/Chemistry_Help/molar_mass_calculator.html. Part III. Converting from moles of one compound to moles of another: We use the stoichiometric coefficients. NOTE: This is likely the most important step in stoichiometric calculations, but one that students often forget or miscalculate. Ane we ALWAYS NEED A BALANCED REACTION!!! Example 5. How many moles of aluminum oxide can be made from 18.7 moles of Aluminum metal, assuming we have more than enough oxygen gas to react? To first answer this, we need a balanced reaction: 4 Al(s) + 3 O2(g) --> 2 Al2O3 And how many moles of oxygen gas are needed ("consumed")? Part IV. Converting between moles and number of molecules: Use Avogadro's number, 6.022x1023. This is very much like converting between "dozen eggs" and just eggs. If you have 10 doz. eggs, then you have 120 eggs, if you have 3.5 doz. eggs, you have 42 eggs. Example 6. How many molecules are in 10.3 moles of hydrogen gas? Part V. Converting between molecules and atoms. Use the molecular formula. Example 7. We'll pick up from where we left off above. How many atoms of hydrogen are there from the step above? Part VI. Converting volume of a solution to the moles present: We use molarity (written M or mol/L) as the conversion factor. Example 8. If you have 50.3 mL of 0.206 M solution of sodium chloride, how many moles of sodium chloride are there? Part VII. Combining it all! Example 9. How many atoms of gold are in 103.2 lbs of the compound [Au(CH2)2P(C6H5)2]2Cl2? To simplify things, the condensed formula is Au2C28H28P2Cl2 and its molar mass is 891.33 g/mol. Notice that we first converted pounds to grams, then grams to moles, moles to molecules, and finally molecules to atoms. There are no shortcuts to go from the first step to the last, that's why I made that flowchart above to show how to go from one quantity to another. Example 10. If we have 31.8 grams of hydrogen gas, and more than enough oxygen gas, how many grams of water can be made? Grams of compound "A" to grams of compound "B". This is one of the most common types of problems! First we need the balanced reaction! 2 H2 + O2 --> 2 H2O Example 11. If you have 100.0 g of Al and 100.0 g of O2, how much (in grams) aluminum oxide, Al2O3, can be made? This is a limiting reactants a.k.a. limiting reagents (pronounced "re-agents") problem and THESE ARE DIFFICULT. Balanced reaction!: 4 Al(s) + 3 O2(g) --> 2 Al2O3(s) These are the steps to follow: (a) Convert each reactant to moles: 100.0 g Al (1 mol / 26.98 g) = 3.706 mol Al 100.0 g O2 (1 mol / 32.00 g) = 3.125 mol O2 (b) Divide the moles by their stoichiometric coefficients in the balanced reaction. (This is how I tell which reactant is the limiting reactant). Al: 3.706 / 4 = 0.9265 O2: 3.125 / 3 = 1.042 You can see that the aluminum has the smallest mole/stoichiometric coefficient ratio; therefore aluminum is the limiting reactant! (c) The amount of products made and reactants remaining all depends on the limiting reactant. Example 12. If it takes 38.56 mL of 0.05004 M NaOH to titrate (neutralize) 50.00 mL of H2SO4, what is the concentration of the acid? This is a type of Solution Stoichiometry also called a Titration. Titration calculations are fairly straightforward when the reactants are in a one-to-one mole ratio in the balanced chemical reaction, but sometimes they are not, like in this reaction. Balanced reaction: H2SO4(aq) + 2 NaOH(aq) --> 2 H2O(l) + Na2SO4(aq) Titration is just another way to say "neutralization", but done in such a way that we can analyze the concentration of a solution, sulfuric acid in this case. The easiest way to answer this is with a formula, the "titration formula": C: Concentration (mol/L) V: Volume (L or mL) subscript a: acid subscript b: base a: the acid stoichiometric coefficient b: the base stoichiometric coefficient.

A mole is

a means of counting the large number of particles in samples. One mole is the number of atoms in exactly 12 grams of 12C (carbon-12). 1 mole contains Avogadro's number (6.022 x 1023 particles/mole) of particles.

In covalent bonds

bonding atoms share electrons in pairs. The shared electrons have lower potential energy than they would in the isolated atoms because they interact with the nuclei of both atoms. Covalently bonded atoms are called molecules.

Compounds are composed of atoms held together by

chemical bonds. The properties of the compound are totally different from the constituent elements.

The discovery of radioactivity made possible

experiments that disproved the "plum pudding" model.

The average atomic mass

for an element can be found on the periodic table along with the atomic number. Also called atomic mass, atomic weight, or average atomic weight

The mixture of aqueous solutions of two soluble ionic compounds results in a solution containing

four ions. Two new compounds can form when the cation from one ionic compound pairs with the anion from the other and vice versa.

Stoichiometry

is a term used to describe quantitative relationships in chemistry. Any chemistry question that asks "How much?" of a particular substance consumed or formed in a chemical reaction is a stoichiometry problem. All stoichiometry problems must start with a balanced chemical equation. Stoichiometric coefficients from balanced chemical equations provide mole ratios that can be used a conversion factors from one substance to another in the equation.

A precipitation reaction

is one in which the mixing of two solutions results in the formation of a solid, or precipitate. Precipitation reactions do not always occur when two aqueous solutions are mixed. Only insoluble compounds form precipitates. If both products are soluble, then NO REACTION occurs.

Mass percent

is one way to express how much of an element is in a given compound. Mass percent of element x = (mass of element X in 1 mole of compound/mass of 1 mole of the compound) x 100

Aqueous chemical reactions can be written as a

molecular equation. -The complete formula for each compound is shown. Complete ionic equation -Emphasize dissociation of reactants and products by writing all strong electrolytes in their dissociated forms REMEMBER: Only aqueous ionic substances dissociate. Gases, liquids, solids, and covalent aqueous species NEVER dissociate.

Scientists use the International System of Units (SI), which is based

on the metric system.

Average atomic mass =

(mass of element 1) (%abundance of element 1) + (mass of element 2)(% abundance of element 2)

Location of elements on Periodic Table:

*Metals* are located on the left and in the middle of the periodic table. ' *Nonmetals* are located on the right of the periodic table. *Metalloids* Elements that have some properties of metals and some properties of nonmetals. Also called *semimetals,* they are located on the "stairstep" boundary between metals and nonmetals.

Predicting Ionic Charges: Charge of Group 1A:

+1

Scientific law

-A brief statement that summarizes past observations and predict future ones -Example: The law of conservation of mass states, "In a chemical reaction, matter is neither created nor destroyed." -Laws are like hypotheses in that they are subject to experiments which can add support to them or prove them wrong.

Example: How many neutrons are there in an atom of the isotope 60Co?

-From the periodic table, the atomic number of Co is 27. -Mass number = 60 -Atomic number = protons = 27 -Neutrons = mass number - atomic number -Neutrons = 60 - 27 = 33 *Note: We often think of "isotope" in terms of radioactivity. Not all isotopes are radioactive, but 60Co happens to be. Its radioactivity is exploited in GammaKnifeTM surgery to treat brain tumors.

Experiments

-Highly controlled procedures designed to generate observations that can support or refute a hypothesis

John Dalton and the Atomic Theory

1)Each element is composed of tiny, indestructible particles called atoms. 2)All atoms of a given element have the same mass and other properties that distinguish them from the atoms of other elements. 3) Atoms combine in simple, whole-number ratios to form compounds. 4) Atoms of one element cannot change into atoms of another element. In a chemical reaction, atoms change they way they are bound together with other atoms to form a new substance

Predicting Ionic Charges: Charge of Group 7A

1-

Scientific theory

-A model for the way nature is that attempts to explain not merely what nature does, but why. -Often, theories predict behavior far beyond the observations or laws from which they were developed. -Example: Dalton's atomic theory proposed that matter is composed of small, indestructible particles (atoms) that rearrange during chemical changes such that the total amount of mass remains constant. Theories can be supported by experimental data, but they can never be conclusively proven. -Over time, poor theories and laws are eliminated or corrected and good theories and laws—those consistent with experimental results—remain.

Diatomic elements:

H2, O2, F2, Br2, I2, N2, Cl2

Ionic Compounds

In chemical reactions, metals tend to form cations (lose electrons) and nonmetals tend to form anions (gain electrons). The number of electrons lost or gained often is predictable, especially for main-group elements which tend to form ions that have the same number of electrons as the nearest noble gas.

There are several types of chemical bonds possible.

Ionic bonds Characterized by gain and loss of electrons (cation + anion; often a metal cation and a nonmetal anion) Covalent bonds Characterized by sharing of electrons between two nonmetals Metallic bonds Characterized by the "sea of electrons" model

Name the binary ionic compounds. MgCl2- CuI- NaNO3- Fe2(SO4)3-

Magnesium chloride Copper(I) iodide Sodium nitrate Iron(III) sulfate

Significant Figure Rules: 1)All nonzero digits are

Significant. *28*.0*3* 0.0*54*0

molar mass.

The mass of 1 mole of atoms of an element An element's molar mass in grams per mole is numerically equal to the element's atomic mass in atomic mass units. Molar mass allows conversion from mass to number of moles, much like a unit conversion. Avogadro's number functions much like a unit conversion between moles to number of particles

Groups

The vertical columns of the periodic table

Scientific notation

allows us to express very large or very small quantities in a compact way by using negative and positive exponents. The SI system uses prefix multipliers along with the standard units Scientific measurements are reported so that every digit is certain except the last, which is estimated. The number of digits reported in a measurement depends on the measuring device.

Chemical changes

alter the composition of matter. Atoms rearrange, transforming the original substances into different substances.

Isotopes

are atoms of the same element that have different numbers of neutrons. C-14 has a mass number of 14 while C-12 has a mass number of 12. C-14 has 14 - 6 = 8 neutrons while C-12 has 12 - 6 = 6 neutrons.

Molecules

are combinations of atoms bound together in specific geometric arrangements The properties of the substances around us depend on the atoms and molecules that compose them.

Derived units

are combinations of other units. Speed (m/s or km/hr) Volume (1 cm3 = 1 mL = 10-3 L) Density (mass / volume, g/cm3, etc.)

Mixtures

are composed of two or more different types of atoms or molecules that can be combined in variable proportions.

Polyatomic ions

are ions composed of two or more atoms that are covalently bonded. The charge is a property of the whole ion, not just of one of the atoms in it.

Atoms

are microscopic particles that represent the smallest unit of matter with the properties of an element -Atoms are the fundamental building blocks of ordinary matter. -Free atoms are rare in nature.

Physical properties

are properties that substances display without changing their compositions (odor, taste, color, melting point, density, etc.)

Significant figures

are the non-place-holding digits (those not simply marking the decimal place). The greater the number of significant figures, the greater the certainty of the measurement.

Extensive properties

are those at are dependent on the amount of substance Examples: volume, mass

Pure substances

are those composed of only a single type of atom or molecule. -Elements are substances that cannot be chemically broken down into simpler substances. -Compounds are composed of two or more elements in fixed, definite proportions.

Chemical properties

are those that substances display only by changing composition via chemical change (corrosiveness, flammability, acidity, toxicity)

Elemental analysis

can be used to determine the empirical formula of a compound. With the addition of the molar mass, you can also determine the molecular formula of the compound.

An atom is the

smallest identifiable unit of an element. About 91 naturally occurring elements Over 20 synthetic (man-made) elements

The basic unit of an ionic compound is

the formula unit (the smallest electrically neutral collection of ions).

Ionic compounds containing polyatomic ions are named

the same way as other ionic compounds, except that we use the name of the polyatomic ion whenever it occurs. Do not change the ending of the polyatomic anion's name. To name: Metals that form only one type of ion: name of cation(metal) + base name of anion (nonmetal) + -ide Metals that form more than one type of ion: name of cation(metal) (charge of cation(metal) in roman numerals in parentheses) + base name of anion (nonmetal) + -ide

Certain groups and periods have specific names. Name of Group 7A:

Halogens

Limiting reagent and theoretical yield

Assuming the hot dogs and buns combine in a one-to-one ratio, we will be limited by the number of hot dog buns we have since we will run out of buns first. In this less than ideal situation, we would call the hot dog buns the limiting reagent or limiting reactant. In a chemical reaction, the limiting reagent is the reactant that determines how much of the products are made. The other reactants are sometimes referred to as being in excess, since there will be some leftover after the limiting reagent is completely used up. The maximum amount of product that can be produced is called the theoretical yield. In the case of the hot dogs and hot dog buns, our theoretical yield is four complete hot dogs, since we have four hot dog buns. Enough about hot dogs, though! In the following example we will identify the limiting reagent and calculate the theoretical yield for an actual chemical reaction. Problem solving tip: The first and most important step for any stoichiometric calculation—such as finding the limiting reagent or theoretical yield—is to start with a balanced reaction! Since our calculations use ratios based on the stoichiometric coefficients, our answers will be incorrect if the stoichiometric coefficients are not right.

Molar Mass

Before applying stoichiometric factors to chemical equations, you need to understand molar mass. Molar mass is a useful chemical ratio between mass and moles. The atomic mass of each individual element as listed in the periodic table established this relationship for atoms or ions. For compounds or molecules, you have to take the sum of the atomic mass times the number of each atom in order to determine the molar mass

Limiting Reactant Problems

Chemistry Stoichiometry Stoichiometry: Fun to Say, Fun to Do! Simple Stoichiometry Calculations Limiting Reactant Problems Gas Stoichiometry Now that you're a pro at simple stoichiometry problems, let's try a more complex one. Using the recipe for ice water (1 glass of water + 4 ice cubes = 1 glass of ice water), determine how much ice water we can make if we have 10 glasses of water and 20 ice cubes. Hopefully, you didn't have too much trouble figuring out that we can make only five glasses of ice water. Let's go through this calculation carefully to see what we did (it'll be clear why we need to do this in a second). Molecular Meanings The limiting reactant in a stoichiometry problem is the one that runs out first, which limits the amount of product that can be formed. The other reactant is called the excess reactant. Using our recipe, we can make 10 glasses of ice water with 10 glasses of water. With the same recipe, we can make 5 glasses of ice water with 20 cubes of ice. Because we run out of ice before we run out of water, we can only make five glasses of ice water. There will be five glasses of warm water left over. In our example, we would say that ice is the limiting reactant. The ice is said to be "limiting" because it is the ingredient we would run out of first, which puts a limit on how much ice water we can make. The water is called the excess reactant because we had more of it than was needed. Limiting Reactants in Chemistry We can use this method in stoichiometry calculations. Again, if we're given a problem where we know the quantities of both reactants, all we need to do is figure out how much product will be formed from each. The smaller of these quantities will be the amount we can actually form. The reactant that resulted in the smallest amount of product is the limiting reactant. Let's see an example: Example: Using the equation 2 H2(g) + O2(g) ↔ 2 H2O(g), determine how many moles of water can be formed if I start with 1.75 moles of oxygen and 2.75 moles of hydrogen. You've Got Problems Problem 2: Using the following equation, determine how much lead iodide can be formed from 115 grams of lead nitrate and 265 grams of potassium iodide: Pb(NO3)2(aq) + 2 KI(aq) ⇔ PbI2(s) + 2 KNO3(aq) Solution: Do two stoichiometry calculations of the same sort we learned earlier. The first stoichiometry calculation will be performed using "1.75 mol O2" as our starting point, and the second will be performed using "2.75 mol H2" as our starting point. 1.75 mol O2 × 2 mol H2O⁄1 molO2 = 3.50 mol H2O 2.75 mol H2 × 2 mol H2O⁄2 mol H2 = 2.75 mol H2O Because "2.75 mol O2" is the smaller of these two answers, it is the amount of water that we can actually make. The limiting reactant is hydrogen because it is the reactant that limits the amount of water that can be formed since there is less of it than oxygen. How Much Excess Reactant Is Left Over? You've Got Problems Problem 3: Using your results from problem #2 in this section, determine the amount of excess reactant left over from the reaction. Once we've determined how much of each product can be formed, it's sometimes handy to figure out how much of the excess reactant is left over. This task can be accomplished by using the following formula:

A Guide to Combustion Analysis

Combustion occurs when fuel, most generally a fossil fuel, reacts with the oxygen in the air to produce heat. The heat generated by the burning of a fossil fuel is used in the operation of equipment such as boilers, furnaces, kilns and engines. Along with heat, CO2 (carbon dioxide) and H2O (water) are created as byproducts of the exothermic reaction. CH4 + 2O2 => CO2 + 2H2O Reactants => Products + Heat By monitoring and regulating some of the gases in the stack or exhaust, it is easy to improve combustion efficiency, which conserves fuel and lowers expenses. Combustion efficiency is the calculation of how effectively the combustion process runs. To achieve the highest levels of combustion efficiency, complete combustion should take place. Complete combustion occurs when all of the energy in the fuel being burned is extracted and none of the Carbon and Hydrogen compounds are left unburned. Complete combustion will occur when the proper amounts of fuel and air (fuel/air ratio) are mixed for the correct amount of time under the appropriate conditions of turbulence and temperature. Although theoretically stoichiometric combustion provides the perfect fuel to air ratio, which thus lowers losses and extracts all of the energy from the fuel; in reality, stoichiometric combustion is unattainable due to many varying factors. Heat losses are inevitable thus making 100% efficiency impossible. In practice, in order to achieve complete combustion, it is necessary to increase the amounts of air to the combustion process to ensure the burning of all of the fuel. The amount of air that must be added to make certain all energy is retrieved is known as excess air. In most combustion processes, some additional chemicals are formed during the combustion reactions. Some of the products created such as CO (carbon monoxide), NO (nitric oxide), NO2 (nitrogen dioxide), SO2 (sulfur dioxide), soot, and ash should be minimized and accurately measured. The EPA has set specific standards and regulations for emissions of some of these products, as they are harmful to the environment. Combustion analysis is a vital step to properly operate and control any combustion process in order to obtain the highest combustion efficiency with the lowest emissions of pollutants. Combustion analysis is part of a process intended to improve fuel economy, reduce undesirable exhaust emissions and improve the safety of fuel burning equipment. Combustion analysis begins with the measurement of flue gas concentrations and gas temperature, and may include the measurement of draft pressure and soot level.

Types of Reactions There are 6 basic types of reactions.

Combustion: Combustion is the formation of CO2 and H2O from the reaction of a chemical and O2 Combination (synthesis): Combination is the addition of 2 or more simple reactants to form a complex product. Decomposition: Decomposition is when complex reactants are broken down into simpler products. Single Displacement: Single displacement is when an element from on reactant switches with an element of the other to form two new reactants. Double Displacement: Double displacement is when two elements from on reactants switched with two elements of the other to form two new reactants. Acid-Base: Acid- base reactions are when two reactants form salts and water.

Problem #4: Quinone, which is used in the dye industry and in photography, is an organic compound containing only C, H, and O. What is the empirical formula of the compound if you find that 0.105 g of the compound gives 0.257 g of CO2 and 0.0350 g of H2O when burned completely? Given a molecular weight of approximately 108 g/mol, what is its molecular formula?

Comment: as a reminder, the following link goes to a discussion of how to calculate the molecular formula once you get the empirical formula. Solution: 1) mass of each element: carbon ⇒ 0.257 g x (12.011 / 44.0098) = 0.07014 g hydrogen ⇒ 0.0350 g x (2.016 / 18.015) = 0.00391674 g oxygen ⇒ 0.105 g minus (0.07014+0.00391674) = 0.03094326 g 2) moles of each element: carbon ⇒ 0.07014 g / 12.011 g/mol = 0.005840 mol hydrogen ⇒ 0.00391674 g / 1.008 g/mol = 0.0038856 mol oxygen ⇒ 0.03094326 g / 16.00 g/mol = 0.0019340 mol 3) Look for smallest whole-number ratio: carbon ⇒ 0.005840 / 0.0019340 = 3 hydrogen ⇒ 0.0038856 / 0.0019340 = 2 oxygen ⇒ 0.0019340 / 0.0019340 = 1 4) Empirical formula: C3H2O 5) Molecular formula: the weight of C3H2O is 54 108 / 54 = 2 C6H4O2

Solvent

Component present in greater amounts (usually much greater amounts) Water is the solvent in aqueous solutions.

Solute

Component present in lesser amounts; the minor component In all solutions, solute dissolves in solvent.

Breathing: Putting Pressure to Work

Every day, without even thinking about it, you move approximately 8500 L of air into and out of your lungs. The total mass of this air is about 11 kg (or 24 lb). How do you do it? The simple answer is pressure. You rely on your body's ability to create pressure differences to move air into and out of your lungs. Pressure is the force exerted per unit area by gas particles (molecules or atoms) as they strike the surfaces around them (Figure 5.1▶ on the next page). Just as a ball exerts a force when it bounces against a wall, so a gaseous molecule exerts a force when it collides with a surface. The result of all these collisions is pressure—a constant force on surfaces exposed to any gas. The total pressure exerted by a gas depends on several factors, including the concentration of gas particles in the sample; the higher the concentration, the greater the pressure. When you inhale, the muscles that surround your chest cavity expand the volume of your lungs. The expanded volume results in a lower concentration of gas molecules (the number of molecules does not change, but since the volume increases, the concentration goes down). This in turn results in fewer molecular collisions, which results in lower pressure. The external pressure (the pressure outside of your lungs) remains relatively constant and once you inhale, it is now higher than the pressure within your lungs. As a result, gaseous molecules flow into your lungs (moving from higher pressure to lower pressure). When you exhale, the process is reversed. The chest cavity muscles relax, which decreases your lung volume, increasing the pressure within the lungs and forcing air back out. In this way, over your lifetime, you will take about half a billion breaths and move about 250 million L of air through your lungs. With each breath you create pressure differences that allow you to obtain the oxygen that you need to live.

When doing stoichiometric calculations with solutions, the point is still to get to moles of one substance so you can use mole ratios.

For reactions occurring in solution, the concentration and volume of reactants and products are often used instead of mass to solve solution stoichiometry problems.

Molarity

Molarity (moles/L) establishes a relationship between moles and liters. Given volume and molarity, it is possible to calculate mole or use moles and molarity to calculate volume. This is useful in chemical equations and dilutions.

Certain groups and periods have specific names. Name of Group A:

Nobel gases or rare gases

Compounds containing the following ions are generally insouble

OH⁻ and S²⁻ = When these ions pair with Li⁺, Na⁺, K⁺, or NH₄⁺, THE resulting compounds are soluble When S²⁻ pairs with Ca²⁺, Sr²⁺, or Ba²⁺, the resulting compound is soluble When OH⁻ pairs with Ca²⁺, Sr²⁺, or Ba²⁺, the resulting compound is soluble CO₃²⁻ and PO₄³⁻ = When these ions pair with Li⁺, Na⁺, K⁺, or NH₄⁺, THE COMPOUND is soluble

Percent Mass

Percents establish a relationship as well. A percent mass states how many grams of a mixture are of a certain element or molecule. The percent X% states that of every 100 grams of a mixture, X grams are of the stated element or compound. This is useful in determining mass of a desired substance in a molecule.

Stoichiometry (Rules):

Rules for determining amount of substance reacting or formed in chemical equations: Balance the equation. Decide what quantity of each substance is given. Write it below the substance (one line down if a mole quantity, two lines down if a laboratory quantity). Decide what quantity you want to calculate. Represent it with a question mark directly under the substance (one line down if a mole quantity, two lines down if a laboratory quantity). Complete the diagram with arrows by going from the known laboratory quantity to moles, then to moles of unknown quantity, and finally to laboratory quantity of the unknown. (Step(s) can be omitted if either or both the given quantity and the unknown quantity are moles.) Calculate each quantity following the steps in the diagram.

How can we "see" atoms?

Scanning Tunneling microscopy (STM) Movement of tip is used to create an image with atomic resolution. Tunneling current is extremely sensitive to distance. Tip is scanned across surface and moved up an down to maintain constant tunneling format. STM can also be used to move atoms.

Significant Figure Rules: 2) Interior zeroes (zeroes between two digits) are

Significant 4*0*8 7.*0*3*0*1

Net ionic equations

Simplify the total ionic equation to only the components that undergo direct change in the reaction by omitting the spectator ions or species that appear exactly the same on both sides of the chemical equation.

Avogadro's Law: Volume and Amount (in Moles)

So far, we have discussed the relationships between volume and pressure, and volume and temperature, but we have considered only a constant amount of a gas. What happens when the amount of gas changes? The volume of a gas sample (at constant temperature and pressure) as a function of the amount of gas (in moles) in the sample is shown in Figure 5.12▶. We can see that the relationship between volume and amount is linear. As we might expect, extrapolation to zero moles shows zero volume. This relationship, first stated formally by Amedeo Avogadro, is Avogadro's law: Avogadro's law V ∝ n (constant T and P) When the amount of gas in a sample is increased at constant temperature and pressure, its volume increases in direct proportion because the greater number of gas particles fill more space. You experience Avogadro's law when you inflate a balloon. With each exhaled breath, you add more gas particles to the inside of the balloon, increasing its volume. Avogadro's law can be used to calculate the volume of a gas following a change in the amount of the gas as long as the pressure and temperature of the gas are constant. For these types of calcu- lations, we express Avogadro's law as V1/n1 = V2/n2 where V1 and n1 are the initial volume and number of moles of the gas and V2 and n2 are the final volume and number of moles. In calculations, we use Avogadro's law in a manner similar to the other gas laws, as shown in

What is the empirical formulate for isopropyl alcohol (which contains only C, H and O) if the combustion of a 0.255 grams isopropyl alcohol sample produces 0.561 grams of CO2 and 0.306 grams of H2O?

Solution From this information quantitate the amount of C and H in the sample. (0.561gCO2)⎛⎝1molCO244.0gCO2⎞⎠=0.0128molCO2(3.5.11) Since one mole of CO2 is made up of one mole of C and two moles of O, if we have 0.0128 moles of CO2 in our sample, then we know we have 0.0128 moles of C in the sample. How many grams of C is this? (0.0128molC)(12.011gC1molC)=0.154gC(3.5.12) How about the hydrogen? (0.306gH2O)⎛⎝1molH2O18.0gH2O⎞⎠=0.017molH2O(3.5.13) Since one mole of H2O is made up of one mole of oxygen and two moles of hydrogen, if we have 0.017 moles of H2O, then we have 2*(0.017) = 0.034 moles of hydrogen. Since hydrogen is about 1 gram/mole, we must have 0.034 grams of hydrogen in our original sample. When we add our carbon and hydrogen together we get: 0.154 grams (C) + 0.034 grams (H) = 0.188 grams But we know we combusted 0.255 grams of isopropyl alcohol. The 'missing' mass must be from the oxygen atoms in the isopropyl alcohol: 0.255 grams - 0.188 grams = 0.067 grams oxygen This much oxygen is how many moles? (0.067gO)⎛⎝1molO15.994gO⎞⎠=0.0042molO(3.5.14) Overall therefore, we have: 0.0128 moles Carbon 0.0340 moles Hydrogen 0.0042 moles Oxygen Divide by the smallest molar amount to normalize: C = 3.05 atoms H = 8.1 atoms O = 1 atom Within experimental error, the most likely empirical formula for propanol would be C3H8O

Problem #3: A 0.2500 g sample of a compound known to contain carbon, hydrogen and oxygen undergoes complete combustion to produce 0.3664 g of CO2 and 0.1500 g of H2O. What is the empirical formula of this compound?

Solution: 1a) Determine the grams of carbon in 0.3664 g CO2 and the grams of hydrogen in 0.1500 g H2O. carbon: 0.3664 g x (12.011 g / 44.0098 g) = 0.1000 g hydrogen: 0.1500 g x (2.0158 g / 18.0152 g) = 0.01678 g 1b) Determine the grams of oxygen in the sample by subtraction. 0.2500 − (0.1000 g + 0.01678) = 0.1332 g Notice that the subtraction is the mass of the sample minus the sum of the carbon and hydrogen in the sample. Also, it is quite typical of these problems to specify that only C, H and O are involved. A warning: sometimes the problem will give the CO2 and H2O values, but FAIL to say that C, H, and O are involved. Make sure you add the C and H values (or sometimes the C, H, and N values) and check against the mass of the sample. Any difference would be an amount of oxygen present (or it might be a mistake!! Keep double checking your work as you do each calculation.) 2) Convert grams of C, H and O to their respective amount of moles. carbon: 0.1000 g / 12.011 g / mol = 0.008325 mol hydrogen: 0.01678 g / 1.0079 g/mol = 0.01665 mol oxygen: 0.1332 g / 15.9994 g/mol = 0.008327 mol 3) Divide each molar amount by the lowest value, seeking to modify the molar amounts into small, whole numbers. carbon: 0.008325 mol / 0.008325 mol = 1 hydrogen: 0.01665 mol / 0.008325 mol = 2 oxygen: 0.008327 mol / 0.008325 mol = 1 We have now arrived at the answer: the empirical formula of the substance is CH2O

Ions

Species with a net charge. The numbers of protons and electrons in the species are not equal. *Anions* Have more electrons than protons *Cations* Have more protons than electrons Ions can be either monatomic (derived from a single atom) or polyatomic (a group of atoms with an overall charge).

Soluble

Substances dissolve easily and to a large extent in a specified solvent.

Insoluble

Substances do not dissolve to any measurable extent in a specified solvent

Electrolytes

Substances that dissolve in water to produce a conductive solution. Weak electrolytes Dissociate only a little bit (weak acids and weak bases) Nonelectrolytes Do not dissociate (covalent molecules) Strong electrolytes Dissociate completely (soluble ionic compounds, strong acids, and strong bases) Weak electrolytes Dissociate only a little bit (weak acids and weak bases) Nonelectrolytes Do not dissociate (covalent molecules)

Charles's Law: Volume and Temperature

Suppose we keep the pressure of a gas sample constant and measure its volume at a number of different temperatures. The results of several such measurements are shown in Figure 5.9▶. From the plot we can see the relationship between volume and temperature: The volume of a gas increases with increasing temperature. Looking at the plot more closely, however, reveals more—volume and temperature are linearly related. If two variables are linearly related, then plotting one against the other produces a straight line. Another interesting feature emerges if we extend or extrapolate the line in the plot backward from the lowest measured temperature. The extrap- olated line shows that the gas should have a zero volume at -273.15 °C. Recall from Chapter 1 that -273.15 °C corresponds to 0 K (zero on the Kelvin scale), the coldest possible temperature. The extrapolated line indicates that below -273.15 °C, the gas would have a negative volume, which is physically impossible. For this reason, we refer to 0 K as absolute zero—colder temperatures do not exist. The first person to carefully quantify the rela- tionship between the volume of a gas and its temperature was J. A. C. Charles (1746-1823), a French mathematician and physicist. Charles was interested in gases and was among the first people to ascend in a hydrogen-filled balloon. The direct proportionality between volume and temperature is named Charles's law after him. Charles's law: V ∝ T (constant P and n) When the temperature of a gas sample is increased, the gas particles move faster; collisions with the walls are more frequent, and the force exerted with each collision is greater. The only way for the pressure (the force per unit area) to remain constant is for the gas to occupy a larger volume so that collisions become less frequent and occur over a larger area (Figure 5.10▲). Charles's law explains why the second floor of a house is usually a bit warmer than the ground floor. According to Charles's law, when air is heated, its volume increases, resulting in a lower density. The warm, less dense air tends to rise in a room filled with colder, denser air. Similarly, Charles's law explains why a hot-air balloon can take flight. The gas that fills a hot-air balloon is warmed with a burner, increasing its volume and lowering its density, and causing it to float in the colder, denser surrounding air. You can experience Charles's law directly by holding a partially inflated balloon over a warm toaster. As the air in the balloon warms, you can feel the balloon expanding. Alternatively, you can put an inflated balloon into liquid nitrogen and watch it become smaller as it cools (Figure 5.11◀). We can use Charles's law to calculate the volume of a gas following a temperature change or the temperature of a gas following a volume change as long as the pressure and the amount of gas are constant. For these types of calculations, we rearrange Charles's law as follows: Since V ∝ T, then V = constant * T If we divide both sides by T, we get: V/T = constant If the temperature increases, the volume increases in direct proportion so that the quo- tient, V>T, is always equal to the same constant. So, for two different measurements, we can say that: V1/T1 = constant = V2/T2, or V1/T1 = V2/T2 [5.3] where V1 and T1 are the initial volume and temperature of the gas and V2 and T2 are the final volume and temperature. The temperatures must always be expressed in kelvins (K), because, as we can see from Figure 5.9, the volume of a gas is directly proportional to its absolute temperature, not its temperature in °C. For example, doubling the temperature of a gas sample from 1 °C to 2 °C does not double its volume, but doubling the temperature from 200 K to 400 K does.

Many factors determine the amount of desired product actually produced in a reaction.

Temperature of the reaction The possibility of side reactions Further reaction of the initial product to produce new product

Actual yield

The amount the reaction produces in the laboratory.

When ionic compounds dissolve, they dissociate or break apart into their constituent ions.

The associated reaction is called a dissociation reaction. Ionic compounds (strong electrolytes) NaCl(s) ---> Na+(aq) + Cl-(aq) Uneven distribution of electrons within the water molecule casue the O side to have a partial negative charge and the H side to have a parital positive charge.

Molecular formula

The complete inventory of atoms

Intensive properties

are those that are independent of the amount of substance Example: density

How to Write a Net Ionic Equation

There are three steps to writing a net ionic equation: Balance the chemical equation. Write the equation in terms of all of the ions in the solution. In other words, break all of the strong electrolytes into the ions they form in aqueous solution. Make sure to indicate the formula and charge of each ion, use coefficients (numbers in front of a species) to indicate the quantity of each ion, and write (aq) after each ion to indicate it's in aqueous solution. In the net ionic equation, all species with (s), (l), and (g) will be unchanged. Any (aq) that remain on both sides of the equation (reactants and products) can be cancelled out. These are called "spectator ions" and they don't participate in the reaction. Tips for Writing the Net Ionic Equation The key to knowing which species dissociate into ions and which form solids (precipitates) is to be able to recognize molecular and ionic compounds, know the strong acids and bases, and predict the solubility of compounds. Molecular compounds, like sucrose or sugar, don't dissociate in water. Ionic compounds, like sodium chloride, dissociate according to solubility rules. Strong acids and bases completely dissociate into ions, while weak acids and bases only partially dissociate. For the ionic compounds, it helps to consult the solubility rules. Follow the rules in order: All alkali metal salts are soluble. (e.g., salts of Li, Na, K, etc. - consult a periodic table if you're unsure) All NH4+ salts are soluble. All NO3-, C2H3O2-, ClO3-, and ClO4- salts are soluble. All Ag+, Pb2+, and Hg22+ salts are insoluble. All Cl-, Br-, and I- salts are soluble. All CO32-, O2-, S2-, OH-, PO43-, CrO42-, Cr2O72-, and SO32- salts are insoluble (with exceptions). All SO42- salts are soluble (with exceptions). For example, following these rules you know sodium sulfate is soluble, while iron sulfate is not. The six strong acids that completely dissociate are HCl, HBr, HI, HNO3, H2SO4, HClO4. The oxides and hydroxides of alkali (group 1A) and alkaline earth (group 2A) metals are strong bases that completely dissociate.

Why do we care about limiting reactants and excess reactants?

Think about industry. In manufacturing, you'd want to make sure the chemical process used makes any scarce or expensive reactants the limiting reactant so that they don't go to waste. In the design of rocket engines, total mass of the rocket is a majorconsideration, so the aim is to use the minimum amount of fuel possible (no excess reactants). (see Example 4.7)

Determining Molecular Formulas

To determine a molecular formula, first determine the empirical formula for the compound as shown in the section above and then determine the molecular mass experimentally. Next, divide the molecular mass by the molar mass of the empirical formula (calculated by finding the sum the total atomic masses of all the elements in the empirical formula). Multiply the subscripts of the molecular formula by this answer to get the molecular formula.

SI Base Unit: Quantity: Electric Current Give Unit and Symbol

Unit: Ampere Symbol: A

Problem #5: A 1.000 g sample of a compound is combusted in excess oxygen and the products are 2.492 g of CO2 and 0.6495 g of H2O.

a) Determine the empirical formula of the compound. b) Given that its molar mass is 388.46 g/mol, determine the compound's molecular formula. Solution: 1) mass of each element: carbon ⇒ 2.492 g x (12.011 / 44.0098) = 0.68011 g hydrogen ⇒ 0.6495 g x (2.016 / 18.015) = 0.07268343 g oxygen ⇒ 1.000 minus (0.68011+0.07268343) = 0.24720657 g Notice that there was oxygen in the compound and that the problem did not tell you that. 2) moles of each element: carbon ⇒ 0.68011 g / 12.011 g/mol = 0.0566240 mol hydrogen ⇒ 0.07268343 g / 1.008 g/mol = 0.0721066 mol oxygen ⇒ 0.24720657 g / 16.00 g/mol = 0.01545 mol 3) Look for smallest whole-number ratio: carbon ⇒ 0.0566240 / 0.01545 = 3.665 hydrogen ⇒ 0.0721066 / 0.01545 = 4.667 oxygen ⇒ 0.01545 / 0.01545 = 1 Do NOT round these off. You should only round off with numbers very close to a whole number. How close? Something like 2.995 goes to 3. 4) I want to change the numbers to improper fractions carbon ⇒ 3.665 = 11/3 hydrogen ⇒ 4.667 = 14/3 oxygen ⇒ 1 = 3/3 Sometimes, a textbook will "magically" tell you what factor to multiply by and you will wonder why. Notice that 3.665 is three-and-two-thirds, 4.667 is four-and-two-thirds. I changed everything to fractions to try and highlight why three is used. 5) Multiply by three to get the whole-number ratio: 11 : 14 : 3 empirical formula = C11H14O3 6) The weight of the empirical formula is 194: 388 / 194 = 2 the molecular formula is C22H28O6

Physical changes

alter only state or appearance but not composition

Solubility rules

are a quick way to see what substances are soluble or insoluble.

Mass number

is the number of protons plus the number of neutrons in an atom

Predicting Ionic Charges: Charge of Aluminum:

3+

Hypothesis

A tentative interpretation or explanation of observations -Should be falsifiable—it makes predictions that can be supported or refuted by further observation

Stoichiometric Calculations

Applying Conversion Factors to Stoichiometry Now you're ready to use what you know about conversion factors to solve some stoichiometric problems in chemistry. Almost all stoichiometric problems can be solved in just four simple steps: Balance the equation. Convert units of a given substance to moles. Using the mole ratio, calculate the moles of substance yielded by the reaction. Convert moles of wanted substance to desired units. These "simple" steps probably look complicated at first glance, but relax, they will all become clear. Let's begin our tour of stoichiometry by looking at the equation for how iron rusts: Fe + O2→Fe2O3 Step 1. Balancing the Equation The constituent parts of a chemical equation are never destroyed or lost: the yield of a reaction must exactly correspond to the original reagents. This fact holds not just for the type of elements in the yield, but also the number. Given our unbalanced equation: Fe + O2→Fe2O3 This equation states that 1 iron (Fe) atom will react with two oxygen (O) atoms to yield 2 iron atoms and 3 oxygen atoms. (The subscript number, such as the two in O2 describe how many atoms of an element are in a molecule.) This unbalanced reaction can't possibly represent a real reaction because it describes a reaction in which one Fe atom magically becomes two Fe atoms. Therefore, we must balance the equation by placing coefficients before the various molecules and atoms to ensure that the number of atoms on the left side of the arrow corresponds exactly to the number of elements on the right. 4Fe +3O2→2Fe2O3 Let's count up the atoms in this new, balanced version of the reaction. On the left of the arrow we have 4 atoms of iron and 6 atoms of oxygen (since 3×2 = 6). On the right we also have 4 iron (since 2×2 = 4) and 6 oxygen (2×3 = 6). The atoms on both sides of the equation match. The process of balancing an equation is basically trial and error. It gets easier and easier with practice. You will likely start to balance equations almost automatically in your mind. Step 2. Converting Given Units of a Substance to Moles The process of converting given units into moles involves conversion factors. Below we will provide the most common and important conversion factors to convert between moles and grams, moles and volumes of gases, moles and molecules, and moles and solutions. These conversion factors function in the same way as those discussed in the previous section Note also that though these conversion factors focus on converting from some other unit to moles, they can also be turned around, allowing you to convert from moles to some other unit. ←Converting from Grams to Moles The gram formula mass of a compound (or element) can be defined as the mass of one mole of the compound. As the definition suggests, it is measured in grams/mole and is found by summing the atomic weights of every atom in the compound. Atomic weights on the periodic table are given in terms of amu (atomic mass units), but, by design, amu correspond to the gram formula mass. In other words, a mole of a 12 amu carbon atom will weigh 12 grams. The gram formula mass can be used as a conversion factor in stoichiometric calculations through the following equation: Moles = Gram formula mass is also known as GFM. You may also see the term gram molecular mass, abbreviated GMM. This term is often used instead of GFM when the substance is molecular and not ionic. However, only the terminology is different, GMM is used in the same way as GFM. Therefore, I will use the catch- all term GFM in this study guide. Converting between Volume of a Gas and Moles The Ideal Gas law, discussed at length in the Sparknote on Gases, provides a handy means of converting between moles and a gas, provided you know certain qualities of that gas. The Ideal Gas Law is PV = nRT, with n representing the number of moles. If we rearrange the equation to solve for n, we get: n = with P representing pressure in atm, V representing volume in liters, T representing temperature in Kelvins, and R the gas constant, which equals .0821 L-atm/mol-K. Given P, V, and T, you can calculate the number of moles of substance in a gas. In those instances when a problem specifies that the calculations are to be made at STP (Standard Temperature and Pressure; P = 1 atm, T = 273 K)), the problem becomes even simpler. At STP, a mole of gas will always occupy 22.4 L of volume. If you are given a volume of a gas at STP, you can calculate the moles in that gas by calculating the volume you are given as a fraction of 22.4 L. At STP, 11.2 L of a gas will be .5 moles; 89.6 L of gas will be 4 moles.'' Stoichiometric Calculations Converting between Individual Particles and Moles Avogadro's Number provides the conversion factor for moving from number of particles to moles. There are 6.02×1023 formula units of particles in every mole of substance, with formula unit describing the substance we are looking at, whether it is a compound, molecule, atom, or ion. A formula unit is the smallest unit of a substance that still retains that substance's properties and is the simplest way to write the formula of the substance without coefficients. Some representative formula units are listed below. Compounds: Cu2S, NaCl Molecules: N2, H2 Atoms: Fe, Na Ions: Na+(aq), Cl-(aq) Since 1 mole = 6.02×1023 formula units, the conversion from formula units to moles is simple: Moles = Converting between Solutions and Moles Solutions are discussed in much greater detail in the series of Solutions SparkNotes. But it is possible, and fairly easy to convert between the measures of solution (molarity and molality) and moles. Molarity is defined as the number of moles of solute divided by the number of liters of solvent. Rearranging the equation to solve for moles yields: Moles = molarity × liters of solution MolaLity is defined as the number of moles of solute divided by the number of kilograms of solvent. Rearranging the equation to solve for moles yields: Moles = molality × kilograms of solution Using the Mole Ratio to Calulate Yield Before demonstrating how to calculate how much yield a reaction will produce, we must first explain what the mole ratio is. The Mole Ratio Let's look once again at our balanced demonstration reaction: 4Fe +3O2→2Fe2O3 The coefficients in front of iron, oxygen, and iron (III) oxide are ratios that govern the reaction; in other words, these numbers do not demand that the reaction can only take place with the presence of exactly 4 moles of iron and 3 moles of oxygen, producing 2 moles of iron (III) oxide. Instead, these numbers state the ratio of the reaction: the amount of iron and oxygen reaction together will follow a ratio of 4 to 3. The mole ratio describes exactly what its name suggests, the molar ratio at which a reaction will proceed. For example, 2 moles of Fe will react with 1.5 moles of O2 to yield 1 mole of Fe2O3. Alternatively, 20 moles of Fe will react with 15 moles of O2 to yield 10 moles of Fe2O3. Each of these examples of the reaction follow the 4:3:2 ratio described by the coefficients. Now, with a balanced equation, the given units converted to moles, and our understanding of the mole ration, which will allow us to see the ratio of reactants to each other and to their product, we can calculate the yield of a reaction in moles. Step 4 demands that we be able to convert from moles to back to the units requested in a specific problem, but that only involves turning backwards the specific converstion factors described above. Sample Problems Problem: Given the following equation at STP: N2(g) + H2(g)→NH3(g) Determine what volume of H2(g) is needed to produce 224 L of NH3(g). Solution: br> Step 1: Balance the equation. N2(g) + 3H2(g)→2NH3(g) Step 2: Convert the given quantity to moles. Note in this step, 22.4 L is on the denominator of the conversion factors since we want to convert from liters to moles. Remember your conversion factors must always be arranged so that the units cancel. = 10 moles of NH3(g) Step 3: mole ratio. = 15 moles H2(g) Step 4, convert to desired units: = 336 L H2(g) Now for a more challenging problem: Given the following reaction: 2H2S(g) + O2(g)→SO2(g) + 2H2O(s) How many atoms of oxygen do I need in order to get 18 g of ice? Solution Step 1. The equation is partially balanced already, but let's finish the job. 2H2S(g) +3O2(g)→2SO2(g) + 2H2O(s) Step 2, convert to moles: 1 formula unit of H2O has 2 atoms of H and 1 atom of O The atomic mass of H is 1 gram/mole Atomic mass of O = 16 grams/mole GFM of H2O(s) = + = 18 grams / mole ×1 mole = 1 mole of H2O(s) Step 3, mole ratio: ×3 moles O2(g) = 1.5 moles O2(g) Step 4, convert to desired units: = 9.03×1023 molecules O2(g) Is this the answer? No. The question asks for ATOMS of oxygen. There are two atoms of oxygen in each molecule of O2(g). ×2 atoms O = 1.806×1024 atoms O Now we're done. Note how important it was to write out not only your units, but what substance you're currently working with throughout the problem. Only a brief check was needed to ascertain if we were really answering the given question. Always check to make sure you have answered the correct question.

Example: How many protons and electrons are there in a chlorine atom? How many of each in a chloride ion (Cl -)?

Chlorine atom = Cl From the periodic table, Cl has 17 protons. In the neutral atom, there are also 17 electrons. Chloride ion = Cl- The chlorine atom gained one electron to form an anion. The chloride ion thus has 17 protons and 18 electrons.

Chemical formula

Describes compounds in terms of their atomic substituents Empirical formula The lowest whole number ratio of atoms in a compound

Example: Name the binary covalent compounds: N2O6 CO ICl4

Dinitrogen hexoxide Carbon monoxide Iodine tetrachloride

Molecular elements

Do NOT normally exist with single atoms at their basic units Two or more atoms of the element bond together, and they exist as molecules (O2, P4, S8) Diatomic elements: H2, O2, F2, Br2, I2, N2, Cl2

Atomic elements

Exist in nature with single atoms at their basic units (Examples: He, Ne, Ar)

Example: A given compound is 41.39% C and 3.47% H by mass. The rest of the compound is composed of O. The molecular weight of the compound is 116.28 g/mol. What is its molecular formula?

Mass of O: 100.0% - 41.39% C - 3.47% H = 55.14% O Assume 100.0 g of substance: 41. 39 g C, 3.47 g H, and 55.14 g O Use molar mass to convert g of each substance to moles Divide each number of moles by the smallest number of moles Calculate the molar mass of CHO. 29.02 g/mol Divide the molar mass of the molecular formula by the molar mass of the empirical formula. Multiply the empirical formula by the result to get the molecular formula. 116.28 g/mole / 29.02 g/mol = 4 CHO C4H4O4

Significant Figure Rules: 3) Leading zeroes (zeroes to the left of the first nonzero digit) are

Not significant. They only serve to locate the decimal spot *0.00*32 *0.0000*6

Chemical Bonds

Sodium chloride (NaCl) is a chemical compound, a pure substance made up of atoms of two or more elements. In a chemical compound, atoms are joined together by chemical bonds (connections between atoms) . In any compound, the atoms combine in fixed whole number ratios. The resultant substance behaves differently from the atoms alone.

Problem #6: A carbohydrate is a compound composed solely of carbon, hydrogen and oxygen. When 10.7695 g of an unknown carbohydrate (MW = 128.2080 g/mol) was subjected to combustion analysis with excess oxygen, it produced 29.5747 g CO2 and 12.1068 g H2O. What is its molecular formula?

Solution: 1) Carbon: mass of CO2 = 29.5747 g ⇒ 0.6722 moles of CO2 ⇒ 0.6722 moles of C ⇒ 8.066 g of C 2) Hydrogen: mass of H2O = 12.107 g ⇒ 0.67260 moles of H2O ⇒ 1.3452 moles of H ⇒ 1.3558 g of H 3) Oxygen: mass of compound burnt = 10.770 g mass of C + H = 9.422 g 10.770 g - 9.422 g = 1.348 g of O ⇒ 0.08424 mol of O 4) Determine empirical and molecular formula: molar ratio of C : H : O ⇒ 0.6722 : 1.3452 : 0.08424 after dividing by the smallest molar ratio of C : H : O ⇒ 7.98 : 15.97 : 1.00 empirical formula is C8H16O "empirical formula weight" = 96+16+16 = 128 which is the molecular weight so the molecular formula is also C8H16O Note: I use "EFW" for the term "empirical formula weight." Neither is in standard usage in the world of chemistry.

Caffeine, a stimulant found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 1.000 mg of caffeine produces 1.813 mg CO2, 0.4639 mg H2O, and 0.2885 mg N2. Estimate the molar mass of caffeine, which lies between 150 and 200 g/mol.

Solution: What we must determine is the molecular formula and the only way to get it with the data given is to determine the empirical formula. One we have that, we can get the "empirical formula weight" and multiply it by the proper scaling factor to get the molar mass. Step One: determine the mass of each element present. Carbon: 1.813 mg x (12.011 / 44.0098) = 0.4948 mg Hydrogen: 0.4639 mg x (2.016 / 18.0152) = 0.0519 mg Nitrogen: 0.2885 mg is given in problem Oxygen: 1.000 mg minus (0.4948 + 0.519 + 0.2885) = 0.1648 mg If the mass of N was given in terms of NH3, we would use the factor (14.007 / 17.031) to get mass of N. If the mass of N was given as N2O5, we would use the factor (28.014 / 108.009) to get the mass of N. Step Two: Convert mass of each element to moles. Carbon: 0.4948 mg ÷ 12.011 mg/mmol = 0.0412 mmol Hydrogen: 0.0519 mg ÷ 1.008 mg/mmol = 0.0515 mmol Nitrogen: 0.2885 mg ÷ 14.007 mg/mmol = 0.0206 mmol Oxygen: 0.1648 mg ÷ 15.999 mg/mmol = 0.0103 mmol Comment #1: notice that the mass for nitrogen is is given as mass of N2, but 14.007 (the atomic weight) is used. Why? Answer: we are interested in moles of N atoms involved, NOT moles of N2 molecules. Comment #2: notice that I used the unit mg/mmol (milligrams per millimole) for the molar mass. When milligram amounts of substance are used, using mg/mmol as the unit on the molar mass (rather than g/mol) allows us to use the mass unit of mg rather than g. Note also that the numerical value is not changed in using mg/mmol. That is because both the numerator (grams) and the denominator (mol) of the original unit have been divided by 1000 to obtain the "milli" prefix. Step Three: find the ratio of molar amounts, expressed in smallest, whole numbers. Carbon: 0.0412 mmol ÷ 0.0103 mmol = 4 Hydrogen: 0.0515 mmol ÷ 0.0103 mmol = 5 Nitrogen: 0.0206 mmol ÷ 0.0103 mmol = 2 Oxygen: 0.0103 mmol ÷ 0.0103 mmol = 1 The empirical formula is C4H5N2O. Since the "EFW" of this is about 97, let's multiply by two. This gives us a molar mass for caffeine of about 194. The value given in reference books is 194.19.

A 1.50 gram sample of cadaverine gives 3.23 g of CO2, 1.58 g of N2O5, and 1.865 g of H2O. Its molar mass is 102.2 g/mol. Determine the empirical and molecular formulas.

Step One: determine the mass of each element present. Carbon: 3.23 g x (12.011 / 44.0098) = 0.8815 g Hydrogen: 1.865 g x (2.016 / 18.0152) = 0.2087 g Nitrogen: 1.58 g x (28.014 / 108.009) = 0.4098 g Oxygen: 1.50 g minus (0.8815 + 0.2087 + 0.4098) = zero Although the problem does not specifically mention oxygen, there may be some in the compound. Consequently, make sure you check by adding the masses of the other elements and subtracting the answer from the total mass of the starting compound. Any difference would be the oxygen (or a mistake in the calculation to this point!!). If the masses add up exactly to the starting total, as they do here, that's a nice indicator you've done the calculation correctly (or committed two mistakes that canceled each other out!!). Step Two: Convert mass of each element to moles. Carbon: 0.8815 g ÷ 12.011 g/mol = 0.0734 mol Hydrogen: 0.2087 g ÷ 1.008 g/mol = 0.207 mol Nitrogen: 0.4098 g ÷ 14.007 g/mol = 0.02926 mol Step Three: find the ratio of molar amounts, expressed in smallest, whole numbers. Carbon: 0.0734 mol ÷ 0.02926 mol = 2.51 Hydrogen: 0.207 mol ÷ 0.02926 mol = 7.07 Nitrogen: 0.02926 mol ÷ 0.02926 mol = 1 Doubling each value gives C = 5, H = 14.14, N = 2, so the empirical formula is C5H14N2. Since the "EFW" of this is about 102, we know that it is also the molecular formula. Notice that the hydrogen value is off by about 10%, but I went ahead and rounded off to the nearest whole number. This raises the question "How far off can a value be and be rounded off?" Another way to express this would be "When can I round off, rather than continuing to search for a set of smallest whole-number values?" Frankly, this can be a difficult question to answer, since the borderline cases require a bit of a judgement call, based on experience. However this is one fact in the example above that justifies our decision. The problem gave the molar mass as about 102 and we got 102 with our formula of C5H14N2. This validates our decision to round off. You might think that having the molar mass so easily available is just a bit too easy, too magical. However, I assure you that determining molar masses in this modern area is a perfectly straightforward process, with very exact values easily obtainable. This was not the case in the 1800's, but that is a different story. In the empirical formula section, there was a practice problem, numbered 6. It shows a problem where you might be tempted to round off too early. Go to answer for problem number 6

Lysine is an amino acid which has the following elemental composition: C, H, O, N. In one experiment, 2.175 g of lysine was combusted to produce 3.94 g of CO2 and 1.89 g H2O. In a separate experiment, 1.873 g of lysine was burned to produce 0.436 g of NH2. The molar mass of lysine is 150 g/mol. Determine the empirical and molecular formula of lysine.

Step One: determine the mass of each element present. Carbon: 3.94 g x (12.011 / 44.0098) = 1.0753 g Hydrogen: 1.89 g x (2.016 / 18.0152) = 0.2115 g Nitrogen: 0.436 g x (14.007 / 16.023) = 0.38114 g Oxygen: cannot yet be done Why can't the oxygen be determined yet? It is because our C, H, and N data come from TWO different sources. Step Two: Convert mass of each element to percentages. Carbon: 1.0753 g ÷ 2.175 g = 49.44 % Hydrogen: 0.2115 g ÷ 2.175 g = 9.72 % Nitrogen: 0.38114 g ÷ 1.873 g = 19.17 % Oxygen: 100 - (49.44 + 9.7 + 19.17) = 21.67 % Step Three: Determine the moles of each element present in 100 g of lysine. Carbon: 49.44 g ÷ 12.011 g/mol = 4.116 mol Hydrogen: 9.72 g ÷ 1.008 g/mol = 9.643 mol Nitrogen: 19.17 g ÷ 14.007 g/mol = 1.3686 mol Oxygen: 21.67 g ÷ 15.9994 g/mol = 1.3544 mol Step Four: find the ratio of molar amounts, expressed in smallest, whole numbers. Carbon: 4.115 mol ÷ 1.3544 mol = 3.04 Hydrogen: 9.643 mol ÷ 1.3544 mol = 7.12 Nitrogen: 1.3686 mol ÷ 1.3544 mol = 1.01 Oxygen: 1.3544 mol ÷ 1.3544 mol = 1 The empirical formula is C3H7NO. In order to determine the molecular formula, we need to know the "empirical formula weight." This value is 73.1. We see that the approximate molecular weight is just about double this value, leading to the molecular formula of C6H14N2O2

Combustion analysis

is a good way to obtain empirical formulas for unknown compounds, especially those containing carbon and hydrogen. CO2 and H2O produced are weighed and mole ratios used to determine moles of C and H in the original sample.

Combustion Analysis

This technique requires that you burn a sample of the unknown substance in a large excess of oxygen gas. The combustion products will be trapped separately from each other and the weight of each combustion product will be determined. From this, you will be able to calculate the empirical formula of the substance. This technique has been most often applied to organic compounds. A brief discussion (yet quite informative) of the history of this technique can be found here. This technique is also called "elemental analysis" Some points to make about combustion analysis: 1) The elements making up the unknown substance almost always include carbon and hydrogen. Oxygen is often involved and nitrogen is involved sometimes. Other elements can be involved, but problems with C and H tend to predominate followed by C, H and O and then by C, H, O and N. 2) We must know the mass of the unknown substance before burning it. 3) All the carbon in the sample winds up as CO2 and all the hydrogen in the sample winds up as H2O. 4) If oxygen is part of the unknown compound, then its oxygen winds up incorporated into the oxides. The mass of oxygen in the sample will almost always be determined by subtraction. 5) Often the N is determined via a second experiment and this introduces a bit of complexity to the problem. Nitrogen dioxide is the usual product when nitrogen is involved. Sometimes the nitrogen product is N2, sometimes NH3. 6) Sometimes the problem asks you for the empirical formula and sometimes for the molecular formula (or both). Two points: (a) you have to know the molar mass to get to the molecular formula and (b) you have to calculate the empirical formla first, even if the question doesn't ask for it. A few lines below is a link that goes to a file that discusses how to go from empirical to molecular formulas. Here is a brief overview of the solution steps before doing the example problems: 1) Determine the grams of each element present in the original compound. Carbon is always in CO2 in the ratio (12.011 g / 44.0098 g), hydrogen is always in H2O in the ratio (2.0158 g / 18.0152 g), etc. 2) Convert grams of each elment to the number of moles. You do this by dividing the grams by the atomic weight of the element. Many times students will want to use 2.016 for hydrogen, thinking that it is H2. This is wrong, use 1.008 for H. 3) Divide each molar amount by the lowest value, seeking to modify the molar amounts into small, whole numbers. Steps 2 and 3 are the technique for determining the empirical formula. Step one is required because you have all your carbon, for example, in the form of CO2 instead of a simpler problem where it tells you how much carbon is present. Finally, a common component of this type of problem is to provide the molecular weight of the substance and ask for the molecular formula. For example, the empirical formula of benzene is CH while the molecular formula is C6H6. Several of the problems below include this question and you can go here for a discussion about calculating the molecular formula once you know the empirical formula.

Significant Figure Rules: 4) Trailing zeroes (zeroes at the end of a number are categorized as follows:

Trailing zeroes after a decimal point are always significant. Ex) 45.*000* 3.56*00* Trailing zeroes before a decimal point(and after a nonzero number) are always significant. Ex) 14*0.00* 25.*0505* Trailing zeroes before an implied decimal point are ambiguous and should by using scientific notation. 1200 = ambiguous 1.2 x 10³ = 2 Significant figures

SI Base Unit: Quantity: Luminous Intensity Give Unit and Symbol

Unit: Candela Symbol: cd

SI Base Unit: Quantity: Temperature Give Unit and Symbol

Unit: Kelvin Symbol: K

SI Base Unit: Quantity: Mass Give Unit and Symbol

Unit: Kilogram Symbol: kg

SI Base Unit: Quantity: Length Give Unit and Symbol

Unit: Meter, Symbol: m

SI Base Unit: Quantity: Amount of Substance Give Unit and Symbol

Unit: Mole Symbol: mol

SI Base Unit: Quantity: Time Give Unit and Symbol

Unit: Second, Symbol: s

Dmitri Mendeleev developed the periodic law:

When the elements are arranged in order of increasing mass, certain sets of properties recur periodically. Mendeleev's periodic table contained gaps but allowed him to predict the existence of elements that had not yet been discovered. Ge is an example

Matter

is anything that occupies space and has mass. -Matter is classified according to its state (solid, liquid, or gas) and according to its composition (the kinds and amounts of substances that compose it).

Exact numbers

has no uncertainty, and thus do not limit the number of significant figures in any calculations. They originate from three sources: Accurate counting of discrete objects Defined quantities Integral numbers that are part of an equation

Heterogeneous mixtures

have compositions that vary from one region to another. -Example: wet sand

Homogeneous mixtures

have the same composition throughout -Example: tea with sugar

Chemical formulas also tell us

how much of a an element is in a compound by providing atom and mole ratios.

Ionic bonding occurs when

ions assemble into an extended array called a lattice and are held together by the attraction between oppositely charged ions.

Atomic number

is the number of protons in the nucleus of an atom. All atoms of a given element have the same number of protons (the atomic number identifies the element).

Remember that molecular (or covalent) compounds form between two or more nonmetals. To name a molecular compound:

prefix + name of first element + prefix + name of second element + ide. Prefixes: mono = 1 di = 2 tri = 3 tetra = 4 penta = 5 hexa = 6 hepta = 7 octa = 8 nona = 9 deca = 10