Chemistry Workbook For Dummies with Online Practice book cover

Chemistry Workbook For Dummies with Online Practice

By: Chris Hren and Peter J. Mikulecky Published: 04-17-2017

Take the confusion out of chemistry with hundreds of practice problems

Chemistry Workbook For Dummies is your ultimate companion for introductory chemistry at the high school or college level. Packed with hundreds of practice problems, this workbook gives you the practice you need to internalize the essential concepts that form the foundations of chemistry. From matter and molecules to moles and measurements, these problems cover the full spectrum of topics you'll see in class—and each section includes key concept review and full explanations for every problem to quickly get you on the right track. This new third edition includes access to an online test bank, where you'll find bonus chapter quizzes to help you test your understanding and pinpoint areas in need of review. Whether you're preparing for an exam or seeking a start-to-finish study aid, this workbook is your ticket to acing basic chemistry.

Chemistry problems can look intimidating; it's a whole new language, with different rules, new symbols, and complex concepts. The good news is that practice makes perfect, and this book provides plenty of it—with easy-to-understand coaching every step of the way.

  • Delve deep into the parts of the periodic table
  • Get comfortable with units, scientific notation, and chemical equations
  • Work with states, phases, energy, and charges
  • Master nomenclature, acids, bases, titrations, redox reactions, and more

Understanding introductory chemistry is critical for your success in all science classes to follow; keeping up with the material now makes life much easier down the education road. Chemistry Workbook For Dummies gives you the practice you need to succeed!

Articles From Chemistry Workbook For Dummies with Online Practice

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20 results
How to Add and Subtract with Exponential Notation

Article / Updated 09-17-2021

In chemistry, you can add and subtract extreme numbers by using exponential notation, and expressing your numbers as coefficients of identical powers of 10. To wrestle your numbers into this form, you may need to use coefficients less than 1 or greater than 10. Adding with exponential notation To add two numbers by using exponential notation, you begin by expressing each number as a coefficient and a power of 10. In this example, you add these numbers, by following these steps: Convert both numbers to the same power of 10. Add the coefficients. Join your new coefficient to the shared power of 10. Subtracting with exponential notation To subtract numbers in exponential notation, you follow the same steps but subtract the coefficients. Here’s an example: 0.0743 – 0.0022 To perform the subtraction, follow these steps: Convert both numbers to the same power of 10. Subtract the coefficients. 7.43 – 0.22 = 7.21 Join your new coefficient to the shared power of 10. Now try a few practice questions. Practice questions Add the following: Use exponential notation to subtract the following: 9,352 – 431 Answers and Explanations The correct answer is Because the numbers are each already expressed with identical powers of 10 (in this case, 10–6), you can simply add the coefficients: 398 + 147 = 545 Then join the new coefficient with the original power of 10. The correct answer is (or an equivalent expression). First, convert the numbers so each uses the same power of 10: Here, you’ve picked 10², but any power is fine as long as the two numbers have the same power. Then subtract the coefficients: 93.52 – 4.31 = 89.21 Finally, join the new coefficient with the shared power of 10.

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Examining Equivalents and Normality

Article / Updated 07-26-2021

In the world of chemistry, not all acids and bases are created equally. Some have an innate ability to neutralize more effectively than others. Consider hydrochloric acid (HCl) and sulfuric acid (H2SO4), for example. If you mixed 1 M sodium hydroxide (NaOH) together with 1 M hydrochloric acid, you'd need to add equal amounts of each to create a neutral solution of sodium chloride (NaCl) and water (H2O). If you mixed sodium hydroxide with sulfuric acid, however, you'd need to add twice as much sodium hydroxide as sulfuric acid to create a solution of sodium sulfate (Na2SO4) and water. Why this blatant inequality of acids? The answer lies in the balanced neutralization reactions for both acid/base pairs: HCl + NaOH → NaCl + H2O H2SO4 + 2NaOH → Na2SO4 + 2H2O The coefficients in the balanced equations are the key to understanding this inequality. To balance the second equation, the coefficient 2 needs to be added to sodium hydroxide, indicating that 2 mol of it must be present to neutralize 1 mol of sulfuric acid. On a molecular level, this happens because sulfuric acid has two acidic hydrogen atoms to give up, and the single hydroxide in a molecule of sodium hydroxide can neutralize only one of those two acidic hydrogens to form water. Therefore, 2 mol of sodium hydroxide are needed for every 1 mol of sulfuric acid. Hydrochloric acid, on the other hand, has only one acidic hydrogen to contribute, so it can be neutralized by an equal amount of sodium hydroxide, which has only one hydroxide to contribute to neutralization. The number of moles of an acid or base multiplied by the number of hydrogens or hydroxides that a molecule has to contribute in a neutralization reaction is called the number of equivalents of that substance. Basically, the number of effective neutralizing moles available determines the ratio of acid to base in a neutralization reaction. In the hydrochloric acid example, there's 1 equivalent of acid (from HCl) present and 1 equivalent of hydroxide (from NaOH) present. In the second example, there are 2 equivalents of acid (from H2SO4) and 2 equivalents of hydroxide present (from NaOH and the coefficient of 2). In chemistry life, this idea can come in quite handy when you want to neutralize an acid or base for disposal or cleanup. In addition to dealing with the more common concept of molarity, you may encounter a concentration measure called normality, which is simply the number of equivalents divided by the volume in liters: As you can see from the formula, molarity and normality are very similar. Normality, however, takes equivalents into account. Mixing equal amounts of acidic and basic solutions of equal normality always results in a neutral solution, while the same can't be said of solutions of equal molarity.

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How to Work with Molality and Mole Fractions

Article / Updated 07-25-2021

Like the difference in their names, the practical difference between molarity and molality is subtle. Take a close look at their definitions, expressed next to one another in the following equations: The numerators in molarity and molality calculations are identical, but their denominators differ greatly. Molarity deals with liters of solution, while molality deals with kilograms of solvent. A solution is a mixture of solvent and solute; a solvent is the medium into which the solute is mixed. A further complication to the molarity/molality confusion is how to distinguish between their variables and units. To clarify this plethora of m-words and their abbreviations, check out the table. M Words Related to Concentration Name Variable Unit Abbreviation Molarity M M Molality m m Moles -- mol Occasionally, you may be asked to calculate the mole fraction of a solution, which is the ratio of the number of moles of either solute or solvent in a solution to the total number of moles of solute and solvent in the solution. Chemists defined this quantity using the variable X. Of course, chemists still need to distinguish between the mole fractions of the solute and the solvent, which unfortunately both start with the letter s. To avoid confusion, they decided to abbreviate solute and solvent as A and B, respectively, in the general formula. In practice, the chemical formulas of the solute and solvent are usually written as subscripts in place of A and B. For example, the mole fraction of sodium chloride in a solution would be written as XNaCl. In general, the mole ratio of the solute in a solution is expressed as where nA is the number of moles of solute and nB is the number of moles of solvent. The mole ratio of the solvent is then . These mole fractions are useful because they represent the ratio of solute to solution and solvent to solution very well and give you a general understanding of how much of your solution is solute and how much is solvent. Here's an example: How many grams of dihydrogen sulfide (H2S) must you add to 750 g of water to make a 0.35 m solution? This problem gives you molality and the mass of a solvent and asks you to solve for the mass of solute. Because molality involves moles and not grams of solute, you first need to solve for moles of solvent, and then you use the gram formula mass of sodium chloride to solve for the number of grams of solute. Before plugging the numbers into the molality equation, you must also note that the problem has given you the mass of the solvent in grams, but the formula calls for it to be in kilograms. Moving from grams to kilograms is equivalent to moving the decimal point three places to the left. Plugging everything you know into the equation for molality gives you the following: Solving for the unknown gives you 0.26 mol of H2S in solution. You then need to multiply this mole value by the molecular mass of H2S to determine the number of grams that need to be added: So, you require 8.9 grams of dihydrogen sulfide (H2S) to make a 0.35 m solution.

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How to Calculate Partial Pressures Using Dalton's Law

Article / Updated 07-25-2021

When gases mix, each individual gas within the mixture contributes a partial pressure, and adding the partial pressures yields the total pressure. This relationship is summarized by Dalton's law of partial pressures for a mixture of individual gases: This relationship makes sense if you think about pressure in terms of kinetic molecular theory. Adding a gaseous sample into a particular volume that already contains other gases increases the number of particles in that space. Because pressure depends on the number of particles colliding with the container walls, increasing the number of particles increases the pressure proportionally. There's no one specific pressure unit you have to use when doing problems with Dalton's partial-pressures equation. As long as the pressure units for all the gases are the same, you're good to go. However, if all the pressures given aren't in the same units, then some conversion must take place! Here's an example: A chemist designs an experiment to study the chemistry of the atmosphere of the early Earth. She constructs an apparatus to combine pure samples of the primary volcanic gases that made up the atmosphere billions of years ago: carbon dioxide, ammonia, and water vapor. If the partial pressures of these gases are 50 kPa, 80 kPa, and 120 kPa, respectively, what's the pressure of the resulting mixture? However difficult early-Earth atmospheric chemistry may prove to be, this particular problem is a simple one. Dalton's law states that the total pressure is simply the sum of the partial pressures of the component gases: So, the pressure of the resulting mix is 250 kPa. Here's another problem: A chemist adds solid zinc powder to a solution of hydrochloric acid to initiate the following reaction: The chemist inverts a test tube and immerses the open mouth into the reaction beaker to collect the hydrogen gas that bubbles up from the solution. The reaction proceeds to equilibrium. At the end of the experiment, the water levels within the tube and outside the tube are equal. The pressure in the lab is 101.325 kPa, and the temperature of all components is 298 K. The vapor pressure of water at 298 K is 3.17 kPa. What is the partial pressure of hydrogen gas trapped in the tube? The system has come to equilibrium, so the interior of the tube contains a gaseous mixture of hydrogen gas and water vapor. Because the water levels inside and outside the tube are equal, you know that the total pressure inside the tube equals the ambient pressure of the lab, 101.325 kPa. The total pressure includes the partial pressure contributions from hydrogen gas and from water vapor. Set up an equation using Dalton's law, rearrange the equation to solve for the pressure of just the hydrogen gas, plug in your numbers, and solve: So, the partial pressure of hydrogen gas trapped in the tube is 98.1 kPa.

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How to Measure Enthalpy Change Using Hess's Law

Article / Updated 07-25-2021

For the chemist, Hess's law is a valuable tool for dissecting heat flow in complicated, multistep reactions. For the confused or disgruntled chemistry student, Hess's law is a breath of fresh air. In essence, the law confirms that heat behaves the way we'd like it to behave: predictably. Imagine that the product of one reaction serves as the reactant for another reaction. Now imagine that the product of the second reaction serves as the reactant for a third reaction. What you have is a set of coupled reactions, connected in series like the cars of a train: Therefore, You can think of these three reactions adding up to one big reaction What is the overall enthalpy change associated with this reaction Here's the good news: Enthalpy changes are additive. But the good news gets even better. Imagine that you're trying to figure out the total enthalpy change for the following multistep reaction: Here's a wrinkle: For technical reasons, you can't measure this enthalpy change directly but must calculate it from tabulated values for and No problem, right? You simply look up the tabulated values and add them. But here's another wrinkle: when you look up the tabulated values, you find the following: Gasp! You need but you're provided only Relax. The enthalpy change for a reaction has the same magnitude and opposite sign as the reverse reaction. So if then It really is that simple: Thanks be to Hess. Try an example. Calculate the reaction enthalpy for the following reaction: Use the following data: Reaction enthalpies are given for two reactions. Your task is to manipulate and add Reactions 1 and 2 so the sum is equivalent to the target reaction. First, reverse Reactions 1 and 2 to obtain Reactions and and add the two reactions. Identical species that appear on opposite sides of the equations cancel out (as occurs with species P4 and Cl2): Finally, divide the sum by 4 to yield the target reaction equation: So, the reaction enthalpy for the reaction is 83.8 kJ.

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How to Convert between Units Using Conversion Factors

Article / Updated 07-24-2021

A conversion factor uses your knowledge of the relationships between units to convert from one unit to another. For example, if you know that there are 2.54 centimeters in every inch (or 2.2 pounds in every kilogram or 101.3 kilopascals in every atmosphere), then converting between those units becomes simple algebra. Conversion factor table The following table includes some useful conversion factors. Using conversion factors example The following example shows how to use a basic conversion factor to fix non-SI units. Dr. Geekmajor absentmindedly measures the mass of a sample to be 0.75 lb and records his measurement in his lab notebook. His astute lab assistant, who wants to save the doctor some embarrassment, knows that there are 2.2 lbs in every kilogram. The assistant quickly converts the doctor’s measurement to SI units. What does she get? The answer is 0.34 kg. Let’s try another example. A chemistry student, daydreaming during lab, suddenly looks down to find that he’s measured the volume of his sample to be 1.5 cubic inches. What does he get when he converts this quantity to cubic centimeters? The answer is 25 cm3. Rookie chemists often mistakenly assume that if there are 2.54 centimeters in every inch, then there are 2.54 cubic centimeters in every cubic inch. No! Although this assumption seems logical at first glance, it leads to catastrophically wrong answers. Remember that cubic units are units of volume and that the formula for volume is Imagine 1 cubic inch as a cube with 1-inch sides. The cube’s volume is Now consider the dimensions of the cube in centimeters: Calculate the volume using these measurements, and you get This volume is much greater than 2.54 cm3! To convert units of area or volume using length measurements, square or cube everything in your conversion factor, not just the units, and everything works out just fine.

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How to Elevate and Calculate Boiling Points of Solvents

Article / Updated 07-23-2021

Boiling point elevation refers to the tendency of a solvent's boiling point to increase when an impurity (a solute) is added to it. In fact, the more solute that is added, the greater the change in the boiling point. Boiling point elevations are directly proportional to the molality of a solution, but chemists have found that some solvents are more susceptible to this change than others. The formula for the change in the boiling point of a solution, therefore, contains a proportionality constant, abbreviated Kb, which is a property determined experimentally and must be read from a table such as the one below. Common Kb Values Solvent Kb in Degrees C/m Boiling Point in Degrees C Acetic acid 3.07 118.1 Benzene 2.53 80.1 Camphor 5.95 204.0 Carbon tetrachloride 4.95 76.7 Cyclohexane 2.79 80.7 Ethanol 1.19 78.4 Phenol 3.56 181.7 Water 0.512 100.0 Here's the formula The formula for the boiling point elevation is where m is molality. Note the use of the Greek letter delta in the formula to indicate that you're calculating a change in the boiling point, not the boiling point itself. You need to add this number to the boiling point of the pure solvent to get the boiling point of the solution. The units of Kb are typically given in degrees Celsius per molality. What causes boiling points to rise? Boiling point elevations are a result of the attraction between solvent and solute particles in a solution. Colligative properties such as boiling point elevation depend on only the number of particles in solution. Adding solute particles increases these intermolecular attractions because more particles are around to attract one another. To boil, solvent particles must therefore achieve a greater kinetic energy to overcome this extra attractive force, which translates into a higher boiling point. Here's a sample problem Try this example: What is the boiling point of a solution containing 45.2 g of menthol (C10H20O) dissolved in 350 g of acetic acid? The problem asks for the boiling point of the solution, so you know that first you have to calculate the boiling point elevation. This means you need to know the molality of the solution and the Kb value of the solvent (acetic acid). The table tells you that the Kb of acetic acid is 3.07 degrees C/m. To calculate the molality, you must convert 45.2 g of menthol to moles: You can now calculate the molality of the solution, taking care to convert grams of acetic acid to kilograms: Now that you have molality, you can plug it and your Kb value into the formula to find the change in boiling point: You're not quite done, because the problem asks for the boiling point of the solution, not the change in the boiling point. Luckily, the last step is just simple arithmetic. You must add your to the boiling point of pure acetic acid, which, according to the table, is 118.1 degrees Celsius. This gives you a final boiling point of for the solution.

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How to Use Empirical Formulas to Find Molecular Formulas

Article / Updated 07-21-2021

Many compounds in nature are composed of atoms that occur in numbers that are multiples of their empirical formula. In other words, their empirical formulas don't reflect the actual numbers of atoms within them; instead, they reflect only the ratios of those atoms. What a pain! Fortunately, this is an old pain, so chemists have devised a means to deal with it. To account for these annoying types of compounds, chemists are careful to differentiate between an empirical formula and a molecular formula. A molecular formula uses subscripts that report the actual number of each type of atom in a molecule of the compound (a formula unit accomplishes the same thing for ionic compounds). Molecular formulas are associated with gram molecular masses that are simple whole-number multiples of the corresponding empirical formula mass. For example, a molecule with the empirical formula CH2O has an empirical formula mass of about 30 g/mol (12 for the carbon + 2 for the two hydrogens + 16 for the oxygen). The molecule may have a molecular formula of CH2O, C2H4O2, C3H6O3, or the like. As a result, the compound may have a gram molecular mass of 30 g/mol, 60 g/mol, 90 g/mol, or another multiple of 30 g/mol. You can't calculate a molecular formula based on percent composition alone. If you attempt to do so, the famous chemists (who named the amount of particles in a mole) Avogadro and Perrin will rise from their graves, find you, and slap you 6.02 x 1023 times per cheek. You can clearly see the folly of such an approach by comparing formaldehyde with glucose. The two compounds have the same empirical formula, CH2O, but different molecular formulas, CH2O and C6H12O6, respectively. Glucose is a simple sugar, the one made by photosynthesis and the one broken down during cellular respiration. You can dissolve it in your coffee with pleasant results. Formaldehyde is a carcinogenic component of smog. Solutions of formaldehyde have historically been used to embalm dead bodies. (Dissolving formaldehyde in your coffee is not advised.) In other words, molecular formulas differ from empirical formulas, and the difference is important in the real world. Determining a molecular formula To determine a molecular formula, you must know the gram formula mass of the compound as well as the empirical formula (or enough information to calculate it yourself from the percent composition). With these tools in hand, calculating the molecular formula involves three steps: Calculate the empirical formula mass. Divide the gram molecular mass by the empirical formula mass. Multiply each of the subscripts within the empirical formula by the number calculated in Step 2. Here's an example: What is the molecular formula of a compound that has a gram molecular mass of 34 g/mol and the empirical formula HO? Calculate the empirical formula mass. You determine this number by finding the mass of HO (1 hydrogen atom and 1 oxygen atom). So, the empirical formula mass is 17.01 g/mol. Divide the gram molecular mass by the empirical formula mass. Dividing the gram molecular mass by this value yields the following: Multiply each of the subscripts within the empirical formula by the number calculated in Step 2. Multiplying the subscripts within the empirical formula by this number gives you the molecular formula H2O2. This formula corresponds to the compound hydrogen peroxide.

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How to Calculate the Empirical Formula of a Compound

Article / Updated 07-21-2021

If you don't know the empirical formula of a compound, you can analyze samples of the unknown compound to identify the percent composition. From there, you calculate the ratios of different types of atoms in the compound. You express these ratios as the empirical formula. An empirical formula represents the lowest whole-number ratio of elements in a compound. Here's how to find an empirical formula when given percent composition: Assume that you have 100 g of the unknown compound. The beauty of this little trick is that you conveniently gift yourself with the same number of grams of each elemental component as its contribution to the percent composition. For example, if you assume that you have 100 g of a compound composed of 60.3% magnesium and 39.7% oxygen, you know that you have 60.3 g of magnesium and 39.7 g of oxygen. (The only time you don't do this is if the problem specifically gives you the masses of each element present in the unknown compound.) Convert the masses from Step 1 into moles using the molar mass. Determine which element has the smallest mole value. Then divide all the mole values you calculated in Step 2 by this smallest value. This division yields the mole ratios of the elements of the compound. If any of your mole ratios aren't whole numbers, multiply all numbers by the smallest possible factor that produces whole-number mole ratios for all the elements. For example, if you have 1 nitrogen atom for every 0.5 oxygen atoms in a compound, the empirical formula is not N1O0.5. Such a formula casually suggests that an oxygen atom has been split, something that would create a small-scale nuclear explosion. (Kaboom!) Though impressive sounding, this scenario is almost certainly false. Far more likely is that the atoms of nitrogen and oxygen are combining in a 1 : 0.5 ratio but do so in a larger but equivalent ratio of 2 : 1. The empirical formula is thus N2O. Because the original percent composition data is typically experimental, expect to see a bit of error in the numbers. For example, 2.03 is probably within experimental error of 2, 2.99 is probably 3, and so on. Write the empirical formula by attaching these whole-number mole ratios as subscripts to the chemical symbol of each element. Order the elements according to the general rules for naming ionic and molecular compounds. Here's an example: What is the empirical formula of a substance that is 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass? For the sake of simplicity, assume that you have a total of 100 g of this mystery compound. Therefore, you have 40.0 g of carbon, 6.7 g of hydrogen, and 53.3 g of oxygen. Convert each of these masses to moles by using the gram atomic masses of C, H, and O: Notice that the carbon and oxygen mole numbers are the same, so you know the ratio of these two elements is 1:1 within the compound. Next, divide all the mole numbers by the smallest among them, which is 3.33. This division yields The compound has the empirical formula CH2O. The actual number of atoms within each particle of the compound is some multiple of the numbers expressed in this formula.

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How to Do Arithmetic with Significant Figures

Article / Updated 07-21-2021

When doing math in chemistry, you need to follow some rules to make sure that your sums, differences, products, and quotients honestly reflect the amount of precision present in the original measurements. That’s where significant figures come in. You can be honest (and avoid the skeptical jeers of surly chemists) by taking things one calculation at a time, following a few simple rules. One rule applies to addition and subtraction, and another rule applies to multiplication and division. Adding or subtracting: Round the sum or difference to the same number of decimal places as the measurement with the fewest decimal places. Rounding like this is honest, because you’re acknowledging that your answer can’t be any more precise than the least-precise measurement that went into it. Multiplying or dividing: Round the product or quotient so that it has the same number of significant figures as the least-precise measurement—the measurement with the fewest significant figures. Notice the difference between the two rules. When you add or subtract, you assign significant figures in the answer based on the number of decimal places in each original measurement. When you multiply or divide, you assign significant figures in the answer based on the smallest number of significant figures from your original set of measurements. Here’s an example. Express the following sum with the proper number of significant figures: 35.7 miles + 634.38 miles + 0.97 miles = ? The answer is 671.1 miles. Adding the three values yields a raw sum of 671.05 miles. However, the 35.7 miles measurement extends only to the tenths place. Therefore, you round the answer to the tenths place, from 671.05 to 671.1 miles. Now let’s try an example with multiplication. Express the following product with the proper number of significant figures: 27 feet * 13.45 feet = ? Of the two measurements, one has two significant figures (27 feet) and the other has four significant figures (13.45 feet). The answer is therefore limited to two significant figures. You need to round the raw product, 363.15 feet². You could write 360 feet², but doing so may imply that the final 0 is significant and not just a placeholder. For clarity, express the product in scientific notation, as 3.6 * 10² feet².

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