Chris Hren

Article / Updated 03-20-2024

Chemists aren’t satisfied with measuring length, mass, temperature, and time alone. On the contrary, chemistry often deals in calculated quantities. These kinds of quantities are expressed with derived units, which are built from combinations of base units. Here are some examples: Area (for example, catalytic surface). and area has units of length squared (square meters, or m2, for example). Volume (of a reaction vessel, for example). You calculate volume by using the familiar formula Because length, width, and height are all length units, you end up with or a length cubed (for example, cubic meters, or m³). Density (of an unidentified substance). Density, arguably the most important derived unit to a chemist, is built by using the basic formula Density = Mass / Volume. Pressure (of gaseous reactants, for example): Pressure units are derived using the formula Pressure = Force / Area. The SI units for force and area are newtons (N) and square meters (m²), so the SI unit of pressure, the pascal (Pa), can be expressed as N/m². Let’s try an example. A physicist measures the density of a substance to be 20 kg/m³. His chemist colleague, appalled with the excessively large units, decides to change the units of the measurement to the more familiar grams per cubic centimeter. What is the new expression of the density? The answer is 0.02 g/cm³. A kilogram contains 1,000 (10³) grams, so 20 kg equals 20,000 g. Well, 100 cm = 1 m; therefore, (100 cm)³= (1 m)³. In other words, there are 100³ (or 106) cubic centimeters in 1 cubic meter. Doing the division gives you 0.02 g/cm³. You can write out the conversion as follows:

Article / Updated 04-14-2023

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. It is important to know some common conversions of temperature, size, and pressure as well as metric prefixes. 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.

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.

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.

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.

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.

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.

Article / Updated 07-25-2021

Increasing temperature magnifies the effects of entropy on a system. Because the entropy of a solute is usually increased when it dissolves, increasing temperature usually increases solubility — for solid and liquid solutes, anyway. Another way to understand the effect of temperature on solubility is to think about heat as a reactant in the dissolution reaction: Heat is usually absorbed into solute particles when a solute dissolves. Increasing temperature corresponds to added heat. So by increasing temperature, you supply a needed reactant in the dissolution reaction. (In those rare cases where dissolution releases heat, increasing temperature can decrease solubility.) Gaseous solutes behave differently from solid or liquid solutes with respect to temperature. Increasing the temperature tends to decrease the solubility of gas in liquid. To understand this pattern, check out the concept of vapor pressure. Increasing temperature increases vapor pressure because added heat increases the kinetic energy of the particles in solution. With added energy, these particles stand a greater chance of breaking free from the intermolecular forces that hold them in solution. A classic, real-life example of temperature's effect on gas solubility is carbonated soda. Which goes flat (loses its dissolved carbon dioxide gas) more quickly — warm soda or cold soda? Comparing gas solubility in liquids with the concept of vapor pressure highlights another important pattern: Increasing pressure increases the solubility of a gas in liquid. Just as high pressures make it more difficult for surface-dwelling liquid molecules to escape into vapor phase, high pressures inhibit the escape of gases dissolved in solvent. Henry's law summarizes this relationship between pressure and gas solubility: Solubility = Constant x Pressure The "constant" in Henry's law is Henry's constant, and its value depends on the gas, solvent, and temperature. A particularly useful form of Henry's law relates the solubility (S) and pressure (P) between two different states: According to this relationship, tripling the pressure triples the gas solubility. Try an example. Henry's constant for nitrogen gas (N2) in water at 293 Kelvin (K) is The partial pressure of nitrogen in air at sea level is 0.78 atm. What is the solubility of N2 in a glass of water at 20 degrees Celsius sitting on a coffee table within a beach house? This problem requires the direct application of Henry's law. The glass of water is at 20 degrees Celsius, which is equivalent to 293 K (just add 273 to any Celsius temperature to get the Kelvin equivalent). Because the glass sits within a beach house, you can assume the glass is at sea level. So you can use the provided values for Henry's constant and the partial pressure of N2: So, the solubility of N2 is

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.

Article / Updated 07-23-2021

By calculating the enthalpy change in a chemical reaction, you can determine whether the reaction is endothermic or exothermic. Chemical reactions transform both matter and energy. Though chemical equations usually list only the matter components of a reaction, you can also consider heat energy as a reactant or product. When chemists are interested in heat flow during a reaction (and when the reaction is run at constant pressure), they may list an enthalpy change to the right of the reaction equation. At constant pressure, heat flow equals enthalpy change: If the enthalpy change listed for a reaction is negative, then that reaction releases heat as it proceeds — the reaction is exothermic (exo- = out). If the enthalpy change listed for the reaction is positive, then that reaction absorbs heat as it proceeds — the reaction is endothermic (endo- = in). In other words, exothermic reactions release heat as a product, and endothermic reactions consume heat as a reactant. The sign of the tells you the direction of heat flow, but what about the magnitude? The coefficients of a chemical reaction represent molar equivalents, so the value listed for the refers to the enthalpy change for one mole equivalent of the reaction. Here's an example: This reaction equation describes the combustion of methane, a reaction you might expect to release heat. The enthalpy change listed for the reaction confirms this expectation: For each mole of methane that combusts, 802 kJ of heat is released. The reaction is highly exothermic. Based on the stoichiometry of the equation, you can also say that 802 kJ of heat is released for every 2 mol of water produced. So reaction enthalpy changes (or reaction "heats") are a useful way to measure or predict chemical change. But they're just as useful in dealing with physical changes, like freezing and melting, evaporating and condensing, and others. For example, water (like most substances) absorbs heat as it melts (or fuses) and as it evaporates. Here are the molar enthalpies for such changes: Molar enthalpy of fusion: Molar enthalpy of vaporization: The same sorts of rules apply to enthalpy changes listed for chemical changes and physical changes. Here's a summary of the rules that apply to both: The heat absorbed or released by a process is proportional to the moles of substance that undergo that process. For example, 2 mol of combusting methane release twice as much heat as 1 mol of combusting methane. Running a process in reverse produces heat flow of the same magnitude but of opposite sign as running the forward process. For example, freezing 1 mol of water releases the same amount of heat that is absorbed when 1 mol of water melts. Try an example: here is a balanced chemical equation for the oxidation of hydrogen gas to form liquid water, along with the corresponding enthalpy change: How much electrical energy must be expended to perform electrolysis of 3.76 mol of liquid water, converting that water into hydrogen gas and oxygen gas? First, recognize that the given enthalpy change is for the reverse of the electrolysis reaction, so you must reverse its sign from –572 kJ to 572 kJ. Second, recall that heats of reaction are proportional to the amount of substance reacting (2 mol of H2O in this case), so the calculation is