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Chapter 8. Chemical Composition

- These glass bottles have something in them.

- The synthesis of new substances is very important.
- Polyvinylchloride is used in plastic water pipes, Teflon is the alloy that remembers its shape even after being severely distorted, and nylon is used in bulletproof vests and body parts of exotic cars.
- When a chemist makes a new substance, the first thing to do is identify it.

- We will learn how to determine a compound's formula in this chapter.
- We need to think about counting atoms before we can do that.
- It's not possible to count atoms individually.
- We usually count atoms by weighing them.
- The principle of counting by weighing should be considered first.

- The carbon-fiber material used in the body of the Enzo Ferrari is very strong.

- To understand the concept of average mass.

- You work in a candy store that sells gourmet Jelly beans.
- You have to count them out because people come in and ask for a lot of beans.
- A good problem solver tries to come up with a better system.
- It's possible to buy a scale and count the jelly beans by weighing them.

- All of the jelly beans have the same mass of 5 g. It takes a few seconds to weigh out the beans.

- Jelly beans are not the same.

- Imagine a plastic so smart that it can be used to sense a baby's breath, measure the force of a karate punch, or make a balloon that sings.
- There is a plastic film that can do these things.

- When it is processed in a way that makes it piezoelectric, it becomes pyroelectric.
- When a current is applied to a piezoelectric substance, it undergoes a change in shape.
- A material that develops an electrical potential in response to a change in temperature is called a pyroelectric material.

- A paper-thin microphone is possible because of the fact that it responds to sound by producing a current proportional to the sound waves.
- A ribbon of plastic one-quarter of an inch wide could be strung along a hallway and used to listen to conversations as people walk through.
- The PVDF film can be used to make a speaker.
- A balloon with a strip of film on it can play any song stored on a chip attached to the film.
- The PVDF film can be used to make a sleep apnea monitor that will set off an alarm if the breathing stops in an infant, which will help to prevent sudden infant death syndrome.
- The U.S. Olympic karate team uses the same type of film to measure the force of kicks and punches.
- A material that curls in response to a current can be created by glueing two strips of film together.
- It is useful for burglar alarm systems because it responds to the heat radiation emitted by a human as far away as 100 ft.
- It costs $10 per square foot to make the PVDF, but it's a small price to pay for its amazing properties.

- The average mass of the jelly beans is the most important piece of information we need.
- The average mass for the 10-bean sample is calculated.

- A jelly bean has an average mass of five grams.
- To count out 1000 beans, we need 5000 g of beans.
- A sample of beans with an average mass of 5.0 g can be treated like a sample where all of the beans are the same.
- It is not necessary for objects to have the same mass to be counted.
- The average mass of the objects is what we need to know.
- The objects behave as though they were all the same, as if they had the same mass.

- The problem needs to be solved quickly.

- You fill the scoop with beans and dump them on the scale, which reads 500 g.

- Put the 1500 g of mints in a bag.
- The customer leaves with your assurance that both the bag containing 500 g of jelly beans and the bag containing 1500 g of mints have the same number of candies.

- If the ratio of the sample mass is the same as the ratio of the individual components of A and B, both samples contain the same number of components.

- Let's use the example we just discussed to show this intimidating statement.
- The components have a mass of 5 g and 15 g. Consider a number of cases.

- The same number of components will be found in any two samples, one of mints and the other of jelly beans.
- In the next section, we will see how these same ideas apply to atoms.

- To understand the determination of atomic mass.

- If you have a pile of carbon, you want to know how many oxygen molecule are required to convert it into carbon dioxide.
- One oxygen molecule is required for each carbon atom according to the balanced equation.

- We need to know how many carbon atoms are in the pile of carbon to determine the number of oxygen molecule required.
- Individual atoms are too small to see.
- We need to learn how to count atoms by weighing samples.

- We can easily count things like mints and jelly beans by weighing them.
- The same principles can be used to count atoms.

- The gram and the kilogram are too large to be convenient because they are so small.
- The mass of a single carbon atom is.
- To avoid using terms like 10-23 when describing the mass of an atom, scientists have defined a much smaller unit of mass called the atomic mass unit.

- Sometimes the symbol U is used.

- We have a problem of counting carbon atoms.
- To count carbon atoms by weighing, we need to know the mass of individual atoms, just as we need to know the mass of individual jelly beans.
- The atoms of a given element are called isotopes.
- Every sample of carbon contains the same amount of these isotopes.
- There is a slightly different mass for each of these isotopes.
- We need to use an average mass for the carbon atoms.
- Carbon atoms have an average atomic mass of 12.01.
- Any sample of carbon from nature can be treated as if it were composed of identical carbon atoms, each with a mass of 12.01.
- We can count carbon atoms by weighing samples of natural carbon, since we know the average mass of the carbon atom.

- 1000 is an exact number.

- Let's assume that when we weigh the pile of natural carbon, the result is 3.00x1020 amu.

- The principles we have just discussed apply to all elements.
- The elements found in nature typically consist of a mixture of different elements.
- To count the atoms in a sample of a given element, we need to know the mass of the sample and the average mass for that element.
- Table 8.1 shows the average mass for common elements.

- The sample of aluminum has 75 atoms.

- The average mass for an aluminum atom is 26.98 amu.

- The number 75 is the number of atoms.

- The average mass of nitrogen is 14.01.

- Problems 8.5 and 8.8 can be seen.

- The opposite calculation can be done.
- We can determine the number of atoms by knowing the mass of the sample.
- This procedure is shown in an example.

- The number of sodium atoms in the sample is calculated.

- The problem can be solved by using the average atomic mass for sodium.

- The average mass of oxygen in the sample is 16.00 amu, which is equivalent to the number of oxygen atoms in the sample.
- There is a person named ]amu.

- Problems 8.6 and 8.7 can be seen.

- If we know the average atomic mass of the atom, we can count it by weighing it.
- In the next section, we will see how this is one of the fundamental operations in chemistry.

- The front cover of the book shows the average atomic mass for each element.
- The atomic weights for the elements are no longer used by chemists.

- To understand the mole concept.
- To learn to convert moles, mass, and number of atoms in a sample.

- In the previous section, we used atomic mass units for mass, but they are very small.
- The gram is the convenient unit for mass in the laboratory.
- In this section we will learn how to count the atoms in samples.

- A sample of aluminum has a mass of 26.98 g.

- To answer this question, we need to know the average atomic mass of both aluminum and copper.
- The answer is copper.
- We need more than 26.98 g of copper because each copper atom has a bigger mass than the aluminum atom.
- A given number of copper atoms will weigh more than an equal number of aluminum atoms.
- 26.98 g of aluminum and 63.55 g of copper have the same number of atoms.
- We need 63.55 g of copper.
- The ratio of the mass to the number of atoms in a sample is the same as the ratio of the mass to the number of atoms in the individual atoms.

- 26.98 g of aluminum has the same number of aluminum atoms as 63.55 g of copper.

- Carbon has an average atomic mass of 12.01 amu and helium has an average atomic mass of 4.003 amu.
- A sample of carbon has the same number of atoms as a sample of helium.
- The same number of atoms can be found in all the samples if we weigh them out with the average atomic mass in grams.
- The number of atoms in a sample is important in chemistry.
- The mole is a unit that all chemists use to describe numbers of atoms.
- One mole is equal to the number of carbon atoms in 12.01.
- This number has been determined using techniques for counting atoms very precisely.
- Avogadro's number is the number of atoms in 12 grams of pure C12 A mole of eggs is the same as a dozen eggs.

- The definition of the mole is slightly different from the SI definition, but it is easier to understand at this point.

- Avogadro's number is 6.022x10343.
- 6.022x1023 units of that substance is the mole of anything.

- The samples of pure elements all have the same number of atoms.

- It is difficult to imagine the magnitude of the number.
- A mole of atoms or molecule is a perfect quantity to use in a reaction because they are so tiny.

- One-mole samples of iron, iodine, liquid mercury, and powdered sulfur.

- Determine your hourly wage.

- Avogadro's number is defined by the fact that a sample of carbon contains 6.022x1013 atoms.
- Because the average atomic mass of hydrogen is 1.008 amu, 1.008 g of hydrogen contains 6.022x1013 hydrogen atoms.
- A sample of any element that weighs a number of grams equal to the average atomic mass of that element contains 6.022x1013 atoms.

- The table shows the mass of elements with a single mole of atoms.

- A sample of an element with a mass equal to that element's average atomic mass in grams contains 1 million atoms.

- Understanding what a mole means and how to determine the number of moles in a given mass of a substance is required for chemical calculations.
- Let's be sure that the process of counting by weighing is clear before we do any calculations.

- Consider the following "bag" of H atoms, which contains 1 mole of H atoms and has a mass of 1.008 g. The bag has no mass.

- The number of hydrogen atoms is unknown in another bag of hydrogen atoms.

- We want to know how many H atoms are present.
- The sample can be weighed.
- The mass of sample B is 0.500 g.

- A sample of carbon weighing 12.01 g is a 1-mole sample.

- The mass of 1 mole of H atoms is 1.008 g. Sample B has a mass of 0.500 g, which is half the mass of a mole of H atoms.

- We know the mass of 1 mole of H atoms, so we can determine the number of moles of H atoms in any other sample of pure hydrogen by weighing it and comparing it to the mass of 1 mole of H atoms.
- We know the mass of each element so we can follow the same process.

- We will often break the problem into smaller steps and report the answer to each step in the correct number of significant figures.
- It is a better idea to wait until the final step to round your answer to the correct number of significant figures.

- We can easily determine the number of atoms present if we know the moles of atoms present.
- We have a small amount of H atoms in sample B. H atoms are present in about 1/2 of 6x1013.

- The procedures are illustrated in an example.

- The mass of aluminum is 26.98 g. The sample we are considering has a mass greater than 10 g. The sample has less than 1 mole of aluminum atoms.

- A chip used in an integrated circuit of a computer has a mass of more than 5 grams.
- The average atomic mass is 28.09 amu.

- The chip is used in electronic equipment.

- Because the average mass of Silicon is 28.09 amu, we know that 1 mole of Si atoms weighs 28.09 g.

- Think about the reasonableness of your answers when you finish a problem.
- The final answer of 1.22x1020 atoms is the same as the one for 1 mole of Silicon, which has a mass of 28.09 g. 1.22x1020 atoms is a smaller number than 6.022x103 Make sure the correct units are obtained at the end by including the units in your calculations.
- If you pay attention to units and make this type of general check, you can detect errors such as an inverted conversion factor or a number that was entered into your calculator.

- The problems are getting more complex to solve.
- Strategies that will help you become a better problem solvers will be discussed in the next section.

- Problems 8.19, 8.20, 8.21, 8.22, 8.23 and 8.24 can be seen.

- To understand how to solve problems by asking and answering questions.

- Today is the first day of work for you.
- You don't know how to get there.
- As luck would have it, a friend knows how to drive you.
- If you want to get to work today, you might not pay attention to how to get there.
- You need to get there on your own tomorrow, so you need to pay attention to distances, signs, and turns.
- There is a difference between taking a passive role and learning how to do it on your own.
- In this section, we want you to be involved in reading the text and the solutions to the practice problems.

- One of the benefits of studying chemistry is that you can solve problems.
- Being able to solve complex problems is a talent that will serve you well.
- The purpose of this text is to help you learn to solve problems in a flexible, creative way by understanding the fundamental ideas of chemistry.
- This approach is called conceptual problem solving.
- To be able to solve new problems on your own is the ultimate goal.
- Instead of giving solutions for you to memorize, we will explain how to think about the solutions to the problems in this text.
- The thinking necessary to get to the answer is more important than the answers to these problems.
- We will solve the problem for you first.
- It's important that you don't take a passive role.
- While studying the solution, it is important that you interact with us so that eventually you can drive by yourself.
- Don't skip the discussion and jump to the answer.
- Asking a series of questions is usually the solution.
- You need to understand each step of the process.

- At some point, you will need to know how to think on your own, even if you are studying the solutions to the problems.
- You won't learn effectively if we help you too much as you solve the problems.
- You won't interact meaningfully with the material if we always drive.
- We will provide more help on the earlier problems as we progress in the later chapters.
- To learn how to solve a problem, you need to understand the main concepts and ideas of the problem.

- You now know how to get from home to work.
- You probably know from experience that not necessarily.
- If you don't understand fundamental principles such as "I traveled north to get to the workplace, so my house is south of the workplace", you may find yourself stranded.
- Understanding these fundamental principles is part of conceptual problem solving.

- From home to work and back there are many more places to go.
- Suppose you know how to get from your house to the library and back.
- If you don't have a big picture of where your house, your workplace, and the library are relative to one another, then probably not.
- The other part of conceptual problem solving is getting a real understanding of the situation.

- We let the problem guide us as we solve it.
- We use our knowledge of fundamental principles to answer the questions we are asked.
- The reward of learning this approach is that you will be able to solve any new problem that comes up in your work or daily life.

- The following principles will be useful to us as we solve the problem.

- We need to read the problem and make a decision on the final goal.
- We sort through the facts and then draw a diagram of the problem.
- In this part of the analysis, we need to state the problem clearly.
- We need to work backwards from the final goal in order to make a decision.
- We always start with the chemical reaction.
- Our understanding of the fundamental principles of chemistry will allow us to answer the simple questions and eventually lead us to the final solution.
- This is called the Reality Check.
- It's always a good idea to check your answer.

- A conceptual approach to problem solving will allow you to develop real confidence.
- When you see a problem that isn't the same as the one you have solved before, you won't be afraid.
- Although you might be frustrated at times as you learn this method, we guarantee that it will pay dividends later and will prepare you for any career you choose.

- A creative problem solver has an understanding of fundamental principles and a big picture of the situation.
- One of the main goals of this text is to help you solve problems.
- We will give you a lot of guidance on how to solve problems.
- We will gradually shift more responsibility to you as we move forward.
- As you gain confidence in letting the problem guide you, you will be amazed at how effective you can be at solving some really complex problems, just like the ones you will confront in real life.

- We will look at how conceptual problem solving works in practice.
- Let's consider a problem about driving because we used a driving analogy before.

- Estimate the amount of money you would spend on gas to drive from New York to Los Angeles.

- To understand the problem, we need to state it in words or as a diagram.

- We are trying to figure out how much we will spend on gasoline.
- We need to understand why we spend more or less money.

- Two people are in separate cars.
- Write down some ideas when you are done reading.

- We are not given any values but are asked to estimate the cost of gasoline.
- We have to estimate the required information.
- The distance between New York and Los Angeles is 3000 miles.
- $4.00 a gallon is a reasonable estimate for the cost of gasoline.
- We assume gas mileage is about 30 miles per gallon.

- We will solve the problem now that we have the necessary information.

- We need to understand how the information affects our answer to set up the solution.
- There is a relationship between the three factors we identified and our final answer.

- The price of gasoline is related.
- The more gasoline costs, the more we will spend.
- The distance is directly related.
- We will spend more on gasoline when we travel farther.
- It's related to gas mileage.
- The less we spend on gasoline the better.

- It makes sense to divide the distance and price by the gas mileage because they are related.
- We will use the same analysis that was used in Chapter 2.
- Let's figure out how much gasoline we will need.

- The distance is in the numerator and the gas mileage is in the denominator, just as we determined they should be.
- We will need 100 gallons of gasoline.

- The price of a gallon of gas is the same as predicted.
- We estimate the total cost of gasoline to be $400.
- If this answer is reasonable, the final step is to consider it.

- Our answer will depend on our familiarity with the situation, and this is always a good question to consider.
- When we are learning a new concept, we may not have a good feel for what the answer should be.
- We can claim that the answer seems reasonable if we have a rough idea, but we can't say it's right.
- If you are familiar with how much you spend on gasoline, this will be the case.
- The price to fill up the tank for an average car is between 40 and 80 dollars.
- If our answer is under $100, we should be suspicious.
- The answer in the thousands of dollars is too high.
- An answer in the hundreds of dollars seems reasonable.

- To understand what a mass is.
- To convert between moles and mass of a sample of a compound.

- A collection of atoms is a chemical compound.
- Each molecule of methane contains one carbon atom and four hydrogen atoms.
- One mole of CH4 molecule consists of 1 mole of carbon atoms and 4 mole of hydrogen atoms.

- The number of significant figures is limited by the number of decimal places.

- There are lots of methane molecule showing their atoms.

- When we say 1 mole of methane, we mean 1 mole of methane molecule.

- The mass of 1 mole of CH4 is called the molar mass.
- The mass in grams of 1 mole of a compound is the mass in grams of 1 mole of the substance.
- The mass of the component atoms is summed.

- The term "molecular weight" was once used.
- In this text, the term molar mass will be used.

- Sulfur dioxide is a gas produced when fuels are burned.
- Acid rain can be produced if sulfur dioxide reacts with water in the atmosphere.

- We want to know the mass of sulfur dioxide in g/mol.

- One mole of SO2 contains 1 mole of sulfur atoms and 2 mole of oxygen atoms.
- The atomic mass of sulfur is 32.07 g/mol and the atomic mass of oxygen is 16.00 g/mol.

- The molar mass of SO2 is 1 mole.

- The mass of SO2 is 64.07 g. It shows the mass of SO2 molecule.

- The answer is greater than the atomic mass of sulfur and oxygen.
- The units are correct, and the answer is reported to the correct number of significant figures.

- Polyvinylchloride, which is used for floor coverings and plastic pipes in plumbing systems, is made from a molecule with the formula C2H3Cl.

- There are Problems 8.27, 8.28, 8.29, and 8.30.

- Some substances are a collection of ion.
- Ordinary table salt is composed of an array of Na+ and Cl- ion.
- There is no NaCl molecule present.
- The term formula weight is used in some books.
- The term molar mass will be applied to both ionic and molecular substances in this book.

- One mole of NaCl contains 1 mole of Na+ ion and 1 mole of Cl- ion.

- The mass of 1 mole of sodium ion and the mass of 1 mole of chloride ion is represented by the molar mass.

- Na and Na have the same mass of the electron, even though Na has one less electron than Na.
- Even though it has one more electron, the mass of Cl is almost identical to the mass of Cl.

- The mass of NaCl is 58.44 g. The mass of the salt is 1 mole.

- CaCO3 is the principal mineral found in limestone, marble, chalk, pearls, and the shells of marine animals.

- Determine the mass of calcium carbonate.
- There is a sample of calcium carbonate.

- We want to know the mass of calcium carbonate in g/mol.

- CaCO3 is the formula for calcium carbonate.
- One mole of CaCO3 contains Ca, C, and O.
- The atomic mass of calcium is 40.08 g/mol, carbon is 12.01 g/mol, and oxygen is 16.00 g/mol.

- Calcium carbonate is an ionic compound.
- One mole of calcium carbonate contains both Ca2+ and CO32- ion.
- The mass of the components is summed up.

- The answer is greater than the atomic mass of calcium, carbon, and oxygen.
- The units are correct, and the answer is reported to the correct number of significant figures.

- The mass of CaCO3 needs to be determined.

- The molar mass of CaCO3 is 100.09 g/mol.
- We have a lot of CaCO3.

- The mass of CaCO3 is determined by using the molar mass.

- We have less than 5 moles of CaCO3 which has a molar mass of 100 g/mol.
- Our answer makes sense because we should expect an answer less than 500 g. The number of significant figures in our answer is three, as required by the initial number of moles.

- The mass of Na2SO4 can be calculated.
- The formula for sodium sulfate is Na2SO4.
- A sample of sodium sulfate with a mass of 300.0 g is more than 1mol.

- Problems 8.35, 8.36, 8.37 and 8.38 can be seen.

- The molar mass of a substance can be obtained by summing the component atoms.
- The mole is the mass of the substance in grams.
- The number of moles present in a sample of known mass can be calculated once we know the molar mass of a compound.
- The reverse is also true, as shown in example 8.7.

- Black walnuts are used to make juglone, a dye known for centuries.
- It is a natural weed killer that does not affect grass or other noncompetitive plants.
- C10H6O3 is the formula for juglone.

- The mass of juglone can be calculated.
- A sample of pure juglone was taken from black walnuts.

- Black walnuts are growing on a tree.

- We want to know the mass of juglone in g/mol.

- C10H6O3 is the formula for juglone.
- One mole of juglone contains a lot of moles.
- We know the atomic mass of carbon, hydrogen, and oxygen.

- The mass of the component atoms is summed.
- There are 10 moles of carbon atoms, 6 moles of hydrogen atoms, and 3 moles of oxygen atoms in one mole of juglone.

- Ten moles of carbon would have a mass of about 120 g, and our answer is higher than that.
- The units are correct, and the answer is reported to the correct number of significant figures.

- We want to know the number of moles in a sample with a mass of 1.56 g.

- The mass of juglone is 174.1 g/mol.
- We have 1.56 g of juglone.

- The mass of this compound is 174.1 g, so 1.56 g is less than a mole.

- The mass of juglone is 174.1 g, so 1.56 g is less than 1 mole.
- Our answer has units of moles, and the number of significant figures in our answer is three, as required by the initial mass of 1.56 g.

- The compound responsible for the scent of bananas can be produced commercially.
- The bees join the attack.

- We want to know the number of moles and the number of molecule of isopentyl acetate in the sample.

- The formula for isopentyl acetate is C7H14O2.
- We know the atomic mass of carbon, hydrogen, and oxygen.
- There are 6.022x1023 molecules in 1 mole.

- We have a mass of isopentyl and want the number of molecules, so we need to calculate the molar mass.

- The mole of isopentyl acetate has a mass of 130.18 g.

- Each bee sting releases a large number of molecules.

- The number of molecules should be less than 5x1015.

- The number of significant figures in our answer is one, as required by the initial mass.

- Problems 8.39 and 8.40 can be seen.

- To find the mass percent of an element.

- We have talked about the composition of compounds in terms of the numbers of atoms.
- It is useful to know the mass of a compound.
- The formula of the compound allows us to compare the mass of the elements in 1 mole to the total mass of the compound.

- The mass fraction is converted to mass percent.

- We will use the compound alcohol obtained from the sugar in grapes, corn, and other fruits and grains to show this concept.
- Gasohol is a form of fuel that is created when gasoline is enriched with alcohol.
- The carbon monoxide in the exhaust of an automobile can be lowered with the addition of the added alcohol.

- Each mole contains 2 moles of carbon atoms, 6 moles of hydrogen atoms, and 1 mole of oxygen atoms.

- You might expect the formula to be written as C2H6O, but it is actually written as C2H5OH.

- The mass percent is the percent by mass of a component of a mixture or of a given element in a compound.

- The mass of carbon is 52.14%.
- The mass percents of hydrogen and oxygen are obtained in the same way.

- The sum of the mass percents in a compound can be rounded off.

- The mass percents of elements in a compound add up to 100%, although rounding-off effects may produce a small deviation.
- You can check the calculations by adding up the percentages.
- The sum of the mass percents is 100 percent.

- We want to know the mass percent of the elements.

- C10H14O is the formula for carvone.
- We know the atomic mass of carbon, hydrogen, and oxygen.
- There are 6.022x1023 molecules in 1 mole.

- The fraction of the total mass contributed by each element is converted to a percentage.

- They should total 100% within a small range due to rounding off.
- The percentages add up to 1000%.

- Alexander Fleming accidentally discovered penicillin in 1928, but he was never able to make it a pure compound.
- Millions of lives have been saved because of this and similar antibiotics.
- Penicillin is a large molecule with many atoms.
- The formula for penicillin F is C14H20N2SO4.

There are Problems 8.45, 8.46, 8.47, 8.48, and 8.49

- To understand the meaning of empirical formulas.

- If you mix two solutions, a solid product forms.
- There are several ways to answer these questions.
- If we know some facts about the solubilities of ionic compounds, we can usually predict the identity of a precipitate formed when two solutions are mixed in a reaction of this type.

- Although an experienced chemist can often predict the product expected in a chemical reaction, the only sure way to identify the product is to perform experiments.
- The product's physical properties are compared to those of known compounds.

- A chemical reaction can give a product that has never been obtained before.
- A chemist determines what compound has been formed by determining which elements are present and how much of each.
- The formula of the compound can be obtained with these data.
- We used the formula of the compound to determine the mass of each element in a mole of the compound.
- We do the opposite of obtaining the formula of an unknown compound.
- We use the measured mass of the elements to determine the formula.

- The formula of a compound shows the relative numbers of different types of atoms.
- The formula CO2 tells us that there are two oxygen atoms in each molecule of carbon dioxide.
- To determine the formula of a substance, we need to count the atoms.
- We can do this by weighing.
- If we know that a compound only contains carbon, hydrogen, and oxygen, we can use a 0.2015-g sample for analysis.
- The 0.2015-g sample of compound contains 0.0806 g of carbon, 0.01353 g of hydrogen, and 0.1074 g of oxygen.
- We have learned how to use the atomic mass of the elements to convert them to numbers of atoms.
- We start by converting to moles.

- The compound has a mole of C atoms, a mole of H atoms, and a mole of O atoms.
- The quantities can be converted to actual numbers of atoms.

- These are the numbers of the different types of atoms.

- The compound has the same number of atoms.
- There are more H atoms than C atoms.

- The formula CH2O expresses the relative numbers of C, H, and O atoms present.
- It could be made up of C2H4O2 molecule, C3H6O3 molecule, C4H8O4 molecule, C5H10O5 molecule, and so on.
- The required ratio of carbon to hydrogen to oxygen is shown in the experiment.

- We only learn the ratio of atoms when we break a compound down into its separate elements.
- The simplest whole-number ratio of atoms in a compound is called the empirical formula.
- A compound that contains the molecule C4H8O4 has the same formula as a compound that contains the molecule C6H12O6 The empirical formula is CH2O.
- The exact formula of a molecule, giving the types of atoms and the number of each type, is called the molecular formula.
- The molecule C6H12O6 is used to make the sugar called glucose.
- The empirical formula is CH2O.

- The formula can be verified by counting the atoms.
- The empirical formula is CH2O.

- In the next section, we will look at how to calculate the empirical formula for a compound from the relative mass of the elements present.
- We must know the mass of a compound to determine its formula.

- The formula for a compound is given in each case.
- Determine the formula for each compound.

- The formula for benzene is used in industry as a starting material for many important products.
- Dioxin is a powerful poison that sometimes occurs as a by-product of the production of other chemicals.
- This is the formula for one of the reactants.

- CH is the empirical formula.
- The empirical formula uses 6 subscript to get the molecular formula.
- The empirical formula is C6H2Cl2O.
- The empirical formula uses 2 subscript to get the molecular formula.
- The empirical formula is C6H16N2.
- The empirical formula uses 2 subscript to get the molecular formula.

- To learn how to calculate formulas.

- One of the most important things we can learn about a new compound is its chemical formula.
- To calculate the empirical formula of a compound, we first determine the relative mass of the various elements.

- The mass of elements that react to form a compound can be measured.
- If we heat the nickel in the air to form a nickel oxide compound, it will react with oxygen to form a nickel oxide compound.
- After the sample has cooled, we weighed it again and found its mass to be less than one gram.
- Oxygen reacts with nickel to form oxide, which leads to a gain in mass.

- The nickel metal originally weighed out is the mass of nickel present in the compound.
- We know that the nickel oxide contains nickel and oxygen.

- There were four significant figures allowed.

- There were three significant figures allowed.

- The mole quantities represent the number of atoms.
- The formula is NiO because the compound contains an equal number of Ni and O atoms.

- This compound has the same number of nickel and oxygen atoms.
- The ratio of nickel atoms to oxygen atoms is 1 to 1.

- We want to figure out the formula for aluminum oxide.
- We want to find a solution for x and y.

- The compound contains aluminum and oxygen.
- The atomic mass of aluminum is 26.98 g/mol and the atomic mass of oxygen is 16.00 g/mol.

- To get the empirical formula, we need to know the relative numbers of each atom, so we have to convert the mass of the atom to the mole of the atom.
- The conversion is carried out using the atomic mass of the elements.

- Because chemical formulas only use whole numbers, we find the whole-number ratio of the atoms.
- The first thing we need to do is divide the two numbers by the smallest of them.
- The smallest number is converted to one.

- The relative numbers of oxygen and aluminum atoms don't change when you divide the moles of atoms by the same number.

- We need a set of whole numbers to express the empirical formula because only whole atoms combine to form compounds.
- We get the numbers we need when we divide 1.000 and 1.500 by 2.

- The empirical formula is Al2O3 because this compound contains two Al atoms for every three O atoms.

- The values for x and y are whole numbers.

- In example 8.11, the relative numbers of moles you get when you calculate an empirical formula will turn out to be nonintegers.
- You have to convert to the correct whole numbers when this happens.
- The numbers can be found by trial and error by using the same small number.
- The multipliers are usually between 1 and 6.
- What we have learned about calculating empirical formulas will be summarized.

- Determine the number of moles for each atom.

- To convert the smallest number to one, divide the number of moles by the smallest number of moles.
- The subscripts in the empirical formula are the numbers that are integers.
- If more than one of these numbers are not integers, go to step 4.

- If you want to convert all of them to whole numbers, use the smallest number you got in step 3.
- The subscripts in the empirical formula are represented by this set of whole numbers.

- We want to figure out the formula for VxOy.
- We want to find a solution for x and y.

- The total mass of the compound is 0.6330 g.

- We need to know how much oxygen is in the sample.

- We divide the moles by the smaller number.

- We go on to step 4 because one of these numbers is not an integer.

- 2x2.500 is equivalent to 5.000 and 2x1.000 is equivalent to 2.

- V2O5 becomes V1.000O2.500.

- The empirical formula is V2O5 and the compound contains 2V atoms for every 5O atoms.

- The values for x and y are whole numbers.

- A lab experiment showed that 0.6884 g of lead combined with 0.2356 g of chlorine to form a compound.
- The formula is empirical.

- There are problems 8.61, 8.63, and 8.65.

- The same procedures are used for compounds with three or more elements.

- We want to figure out the formula for lead arsenate.
- We want to find a solution for a, b, c, and d.

- The compound contains 1.3813 g of Pb.
- We know the atomic mass of lead (207.2 g/mol), hydrogen (1.008 g/mol), arsenic (74.92 g/mol), and oxygen (16.00 g/mol).

- The compound has 1.3813 g Pb, 0.00672 g H, 0.4995 g As, and 0.4267 g O.

- The moles of the elements are calculated using the atomic mass of the elements.

- We divide by the smallest number of moles.

- The empirical formula is PbHAsO4 because the numbers of moles are all whole numbers.

- The values for a, b, c, and d are whole numbers.

- The commercial name for an insecticidal agent used to protect crops such as cotton, vegetables, and fruit is sevi.
- A sample of carbon, nitrogen, hydrogen, and oxygen is found by a chemist.

- Problems 8.57 and 8.59 can be seen.

- When a compound is analyzed to determine the relative amounts of elements present, the results are usually given in terms of percentages by mass.
- Section 8.6 shows us how to calculate the percent composition of a compound.
- We will do the opposite now.
- The empirical formula will be calculated based on the percent composition.

- The meaning of percent is what you need to understand this procedure.
- It's important to remember that percent means parts of a given component per 100 parts of the total mixture.
- If a compound is 15% carbon by mass, it has 15 g of carbon per 100 g of compound.

- The grams of that element in 100 g of the compound is the percent by mass.

The empirical formula of a compound is shown in example 8.14

- We want to figure out the formula for cisplatin.
- We want to find a solution for a, b, c, and d.

- We know that the atomic mass of chlorine is 35.45 g/mol.

- To get to the number of moles, we need to know the mass of each element in the sample.

- Determine the number of grams of each element in 100 g of compound.
- There is 65.02 g of Platinum (Pt) per 100.00 g of compound.

- Determine the number of moles for each atom.
- The atomic mass is used to calculate moles.

- PtN2H6Cl2 is the empirical formula for cisplatin.
- The number for hydrogen is slightly greater than 6 because of rounding-off effects.

- The values for a, b, c, and d are whole numbers.

- The 100 g of compound is used to base the calculation for empirical formula determination.

- The most common form of nylon is carbon, nitrogen, hydrogen, and oxygen.

- There are Problems 8.67, 8.68, 8.69, 8.70, 8.71, 8.72, and 8.74.

- This example is the same as earlier examples in which the mass was given directly.

- To learn how to calculate the formula of a compound.

- The empirical formula can be calculated if we know the composition of a compound in terms of the mass of the elements present.
- For reasons that will become clear as we consider example 8.15 we must know the molar mass.
- In this section, we will look at compounds where the percent composition and the molar mass are known.

- An empirical formula of P2O5 is found in a white powder.
- The compound's mass is 283.88 g.

- We want to know the formula for the compound.
- We want to find a solution for x and y.

- The formula for the compound is P2O5.
- The compound has a mass of 283.88 g/mol.
- The atomic mass of oxygen is 16.00 g/mol and the atomic mass of phosphorus is 30.97 g/mol.
- There are a lot of empirical formula units in the formula.
- The formula will be P2xO5y.

- We need to know the formula mass.

- We need to compare the empirical formula mass to the molar mass to get the formula.
- The mass of P2O5 units is the empirical formula mass.

- The formula contains a lot of empirical formula units.

- The empirical formula mass is the first thing we divide by to determine the formula.
- This tells us how many empirical formula mass there are.

- The result shows that the formula consists of two empirical formula units and the molecule is P4O10 The structure of this compound is shown.

- This compound is used as a drying agent and has a great affinity for water.

- The values for x and y are whole numbers.
- The ratio of P:O in the formula is 2:1.

- The empirical formula mass is known to be 98.96.
- We know thatMolar mass is a formula that can be used to get the value of n.

- Problems 8.81 and 8.82 can be seen.

- The empirical formula is always an multiple of the molecular formula.
- The empirical formula CH2O and the molecular formula C6H12O6 are used for the sugar.

- The empirical formula is the same as the molecular formula.
- The empirical formula for carbon dioxide is the same as the molecular formula.
- The empirical formula is P2O5 and the molecular formula is P4O10

- It is not necessary for objects to have the same mass to be counted.
- The average mass of the objects is all we need to know.
- To count the atoms in a sample of a given element, we need to know the mass of the sample and the average mass for that element.
- The ratio of the mass to the number of atoms in a sample is the same as the ratio of the mass to individual atoms.
- 6.022x1023 units of that substance are contained in one mole of anything.
- A sample of an element with a mass equal to that element's average atomic mass contains 1 mole of atoms.
- The mass in grams of 1 mole of a compound is called the molar mass.
- The sum of the component atoms in a compound is called the molar mass.
- The percent of the compound can be used to find the empirical formula.
- The formula for the molecule present in the substance is called the molecular formula.
- The empirical formula has a whole-number multiple.
- There are different ways of expressing the same information.

- Group of students in class will be asked these questions.
- For introducing a topic in class, these questions work well.

- The Student Solutions Guide has full solutions for the questions and problems below.

- The average washer weighs 0.110 g. The simplest whole-number ratio of atoms in a compound is an empirical formula.
- The empirical formula is the lowest whole-number ratio of atoms in the compound.
- The empirical formula is CHF because the graphic shows four different types of atom.

- The Student Solutions Guide has full solutions for the questions and problems below.

- The average atomic mass takes into account the various isotopes of an element and the relative abundances in which those elements are found.

- The mass of copper is 63.55 amu.
- A sample containing 75Cu atoms would weigh 4766 amu and represent 96Cu atoms.

- The Student Solutions Guide has full solutions for the questions and problems below.

- 55.85 g of iron has iron atoms.
- There are zinc atoms in 65.38 g.

- The Student Solutions Guide has full solutions for the questions and problems below.

- The mass of a substance is 1 mole.

- The compound's molar mass can be calculated by summing the individual atomic mass of the atoms in the formula.
- A molar mass of 16.042 g/mol is given for the atomic mass of carbon and four hydrogens in the compound CH4.

- The Student Solutions Guide has full solutions for the questions and problems below.

- The mass fraction of an element can be obtained by comparing the mass of the element present in 1 mole of the compound to the mass of the compound.

- The percent composition of each element in the compound does not change because a compound consists of the same percent composition by mass regardless of the starting amount in the sample.

- The Student Solutions Guide has full solutions for the questions and problems below.

- The empirical formula indicates the smallest whole-number ratio of the number and type of atoms present in a molecule.
- NO2 and N2O4 have the same empirical formula because they both have two oxygen atoms for every nitrogen atom.

- The Student Solutions Guide has full solutions for the questions and problems below.

- The Student Solutions Guide has full solutions for the questions and problems below.

- A compound of hydrogen and boron has a percentage composition of 21.86%.
- The compound with the empirical formula C2H5O was found to have a 90 g molar mass.
- Determine the formulas of the compound.
- The mass of the tooth is 184 g/mol.

- You have a sample of both copper and aluminum.
- Each sample has the same number of atoms.
- The mass of the copper sample is more than the mass of the aluminum sample, but it is not twice as great.

- The number of grams of iron and cobalt are the same number of atoms.
- The empirical formula of the compound was found to contain the following percentages of elements: hydrogen, 20.11%, and carbon,79.89%.

- If you have equal mole samples of NO2 and F2, which must be true?
- The number of atoms in each sample is the same.

- Determine the formula for the compound.

- A transition metal ion forms a compound with oxygen.
- The compound's mass is 250.2 g/mol.

- The mass is equivalent to one.

- The average mass of an element's atoms is used in most chemical calculations.
- The average mass takes into account the relative abundance of the elements in nature as well as the exact mass of the elements.

- A species with only nitrogen and oxygen has a percent by mass of nitrogen of 46.7%.

- The same type of assistance a student would get from an instructor can be found in these multiconcept problems.

- Determine the molar mass for the compounds.
- The formula is C63H88CoN14O14P.
- Anemia can be caused by a lack of this vitamins in the diet.

- The formula for aspirin is C9H8O4.

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