The MCAT Chemistry Book - Aryangat A. 2012

General Chemistry
Matter

A. INTRODUCTION

A firm grasp of the basic ideas of division of matter is important for the understanding of physical sciences. These basic ideas presented here are not only used in chemistry and physics, but in many diverse fields such as medicine, engineering, astronomy, geology, and so on. In this chapter, we will discuss ideas about atoms and molecules, and related aspects such as moles, Avogadro number, percentage composition, atomic mass, atomic weight, and subatomic particles.

B. ATOMS

Atoms are the basic units of elements and compounds. In normal chemical reactions, atoms retain their identity. In this section, we will present a quick review of some of the basic terms and concepts such as elements, compounds, and mixtures.

Elements

An element is defined as matter that is made of only one type of atom. Elements are the basic building blocks of more complex matter. Some examples of elements include hydrogen (H), helium (He), potassium (K), carbon (C), and mercury (Hg).

Compounds

A compound is matter formed by the combination of two or more elements in fixed ratios. Let’s consider an example. Hydrogen peroxide (H2O2) is a compound composed of two elements, hydrogen and oxygen, in a fixed ratio.

Mixtures

A combination of different elements, or a combination of elements and compounds, or a combination of different compounds is called a mixture. For example, an aqueous solution of potassium hydroxide (KOH + H2O). In this example, the two components are potassium hydroxide and water.

Though these definitions illustrate basic ideas, you need to understand them fully; otherwise it will be almost impossible to decipher higher concepts that are based on these simple ideas. The MCAT tests your understanding of basic concepts by incorporating simple ideas into passages. So in order to succeed on the test, you need to thoroughly understand the basics.

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C. DALTON’S ATOMIC THEORY

In 1803, John Dalton proposed the atomic theory of matter. The main postulates of his atomic theory can be summarized as follows:

1) Matter is composed of indivisible particles - atoms.

2) An element is composed of only one kind of atom. These atoms in a particular element have the same properties such as mass, size, or even shape.

3) A compound is composed of two or more elements combined in fixed ratios or proportions.

4) In a chemical reaction, the atoms in the reactants recombine, resulting in products which represent the combination of atoms present in the reactants. In the process, atoms are neither created, nor destroyed. So a chemical reaction is essentially a rearrangement of atoms.

Ramifications of Dalton’s Theory

The atomic theory put forward by Dalton is consistent with the law of conservation of mass. As the fourth postulate says, chemical reaction is just a rearrangement of atoms, and thus the total mass remains constant during a chemical reaction.

The postulates also account for the law of definite proportions. Compounds are made of elements in fixed or definite proportions. Since the atoms have fixed mass, compounds should have elements in a fixed ratio with respect to mass. Finally, these postulates predict what is known as the law of multiple proportions. According to this law, if two elements form two or more different compounds, the ratio of the mass of one element of these compounds to a fixed mass of the other element is a simple whole number.

D. THE GENERAL STRUCTURE OF THE ATOM

During the early twentieth century, scientists discovered that atoms can be divided into more basic particles. Their findings made it clear that atoms contain a central portion called the nucleus. The nucleus contains protons and neutrons. Protons are positively charged, and neutrons are neutral. Whirling about the nucleus are particles called electrons which are negatively charged. The electrons are relatively small in mass. Take a look at Table 1-1 for a size comparison.

Table 1-1

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E. ELECTRONS

As mentioned above, the late nineteenth century scientists conducted several experiments, and found that atoms are divisible. They conducted experiments with gas discharge tubes.

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Figure 1-1 Gas discharge tube

A gas discharge tube is shown in Figure 1-1. The gas discharge tube is an evacuated glass tube and has two electrodes, a cathode (negative electrode) and an anode (positive electrode). The electrodes are connected to a high voltage source. Inside the tube, an electric discharge occurs between the electrodes. The discharge or `rays’ originate from the cathode and move toward the anode, and hence are called cathode rays. Using luminescent techniques, the cathode rays are made visible and it was found that these rays are deflected away from negatively charged plates. The scientist J. J. Thompson concluded that the cathode ray consists of negatively charged particles (electrons).

Charge of Electrons

R. A. Millikan conducted the famous oil drop experiments and came to several conclusions: The charge of an electron is —1.602 x 10—19 C. From the charge-to-mass ratio, the mass of an electron was also calculated.

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F. PROTONS

Protons are positively charged nuclear particles. The charge of a proton is (positive electronic charge) +1.6 x 10—19 C. The net positive charge of the nucleus is due to the presence of the protons. A proton is about 1800 times more massive than an electron.

G. NEUTRONS

Neutrons have mass comparable to that of protons, but neutrons are devoid of any electric charge. We will talk more about neutrons and their whereabouts when we study radioactivity.

Now a natural question is whether electrons, protons, and neutrons are the most fundamental particles. The answer is no. These fundamental particles are made of more fundamental particles called quarks. But, we don’t have to go that far for the MCAT. Just be aware that such sub-fundamental particles exist, fundamental particles being electrons, protons, and neutrons.

H. ATOMIC NUMBER AND MASS NUMBER

The atomic number denotes the number of protons in an atom’s nucleus. The mass number denotes the total number of protons and neutrons. Protons and neutrons are often called nucleons. By convention, the atomic number is usually written to the left of the elemental notation, and the mass number to the left above the elemental notation as represented by the example below. The element shown is aluminum.

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Some atoms have the same atomic number, but different mass numbers. This means different number of neutrons. Such atoms are called isotopes.

I. ATOMIC WEIGHT

The atomic weight of an element is the average weight of all the isotopic masses of the element, calculated on the basis of their relative abundance in nature. The atomic weights are set on a “carbon-12” scale. This is the standard weight scale that is used worldwide to express atomic weights. Exploring this further, we can say that 12 atomic mass units (amu) make up the mass of one atom of Image isotope. In other words, one amu is equal to 1/12 the mass of one carbon-12 atom. We can also say that the atomic weight of carbon-12 is 12 amu. Even though it is popular to use the term atomic weight, atomic mass is a more appropriate term since we are really talking about the mass rather than the weight.

J. MOLECULES

A molecule is a set or group of atoms which are chemically bonded. It can be represented by a molecular formula. A molecular formula represents the different kinds of atoms that are present in a molecule, along with the actual ratio of its atomic combination in forming that molecule.

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A molecule of H2O (water) contains two hydrogen atoms bonded to one atom of oxygen.

NaOH Sodium Hydroxide

A molecule of sodium hydroxide contains one sodium atom, one oxygen atom, and one hydrogen atom. The point is that molecular formula represents the molecules and the actual ratio of the atoms present in them.

Molecular Weight

Molecular weight represents the sum of the atomic weights of all the atoms in that molecule. Molecular weight is also known as formula weight.

Example 1-1

Calculate the molecular weight of sulfuric acid (H2SO4).

2 hydrogens 2 x 1 = 2

1 sulfur 1 x 32 = 32

4 oxygens 4 x 16 = 64

     98 g/mol

Example 1-2

Calculate the molecular weight of carbon dioxide (CO2).

1 carbon 1 x 12 = 12

2 oxygens 2 x 16 = 32

    44 g/mol

Empirical Formula

Empirical formula of a molecule represents the simplest ratio of the atoms present in the molecule. For example, acetylene has the molecular formula C2H2. The empirical formula of acetylene is CH. In essence, empirical formula gives the simplest ratio of atoms in a molecule.

Problem 1-1

Write the empirical formula of the following molecules.

1. H2O2

2) C2H6

3) H2O

Answers:

1) HO

2) CH3

3) H2O

Notice that sometimes the empirical formula is the same as the molecular formula as in the case of water (H2O).

K. THE CONCEPT OF MOLE

The quantity of a given substance that contains as many units or molecules as the number of atoms in 12 grams of carbon-12 is called a mole. For example, one mole of glucose contains the same number of glucose molecules as in 12 grams of carbon-12. The mass of one mole of a substance is called its molar mass. The number of atoms in 12 grams of carbon-12 is represented by the Avogadro number (6.023 x 1023). So one mole of any substance contains Avogadro number of units in them.

To understand this concept thoroughly, try different possible scenarios where the Avogadro number can be used. Here are some. One mole of hydrogen atoms contains Avogadro number of hydrogen atoms. One mole of hydrogen molecules contains Avogadro number of hydrogen molecules. A mole of water contains 6.023 x 1023 water molecules. These are all different ways of expressing the same concept.

The molar mass of a substance is equal to the molecular weight of that substance. The molecular weight (formula weight) of water is 18 amu. Since this is the molar mass, we can express it as 18 grams/mol.

Example 1-3

Calculate the mass of one molecule of sodium hydroxide (NaOH).

Ans: We know that the formula weight of sodium hydroxide is 40 g/mol.

Sodium 23

Oxygen 16

Hydrogen 1

   40 g/mol

We also know that one mole of NaOH contains Avogadro number of molecules. So the mass of the NaOH molecule can be found by the following method:

Mass of one molecule of sodium hydroxide

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Example 1-4

Calculate the number of moles in 109.5 grams of hydrogen chloride.

Ans: Just like the last example, we have to find the molar mass of the molecule. The molar mass is 35.5 + 1 = 36.5 g/mol.

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So, 109.5 grams of HCl correspond to 3 moles of HCl.

You should be able to do these types of conversions back and forth, from grams to moles and moles to grams.

Try the next problem to see whether you have mastered the idea.

Problem 1-2

Calculate the number of grams in 8 moles of sulfur dioxide.

Ans: If your answer is close to 512.8 g, you solved the problem correctly.

L. COMPOSITION BY PERCENTAGE

The MCAT often contains percentage composition problems. Percentage composition is the percentage contribution (by weight) of each element to the total mass. Let’s explore this idea by looking at some examples.

Example 1-5

Calcium carbonate (CaCO3), commonly known as limestone is used in the preparation of a variety of compounds. Calculate the percentage composition of each element in calcium carbonate.

Solution:

# of atoms per molecule

molar weight of the atoms

total mass of the element per mol

1 Calcium

1 Carbon

3 Oxygen

1 x 40.1 g

1 x 12.0 g

3 x 16.0 g

40.1 g

12.0 g

48.0 g



100.1 g

The percentage composition of each element can be found as follows:

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Predicting Formulas from Percentage Compositions

You should be able to predict the formula of a compound on the basis of a given data of percentage compositions. Study the next example to understand how it is done.

Example 1-6

A carbon compound contains 27.27% carbon and 72.73% oxygen by mass. Predict the simplest ratio or formula of the compound.

Solution:

The best way to approach this problem is to consider that we have 100 grams of this compound. Logically it should contain 27.27 grams of carbon and 72.73 grams of oxygen. With that in mind, we can calculate the number of moles of each element or atom. After that we can obtain the simple ratio.

Step 1

# of moles of carbon atoms equals 27.27/12 = 2.275 moles of carbon atoms

# of moles of oxygen atoms equals 72.73/16 = 4.546 moles of oxygen atoms

Step 2

Divide every number of moles with the smallest number of moles that you got in Step 1. Here the smaller one is 2.2725. So divide the number of moles of carbon atoms and the number of moles of oxygen atoms by 2.2725. That will give you the simplest ratio between them.

Carbon: 2.2725/2.2725 = 1

Oxygen: 4.546/2.2725 ≈ 2

Since the ratio of carbon to oxygen is 1:2, the compound is CO2.

Example 1-7

Calculate the mass of sulfur in 150 grams of H2SO4.

Solution:

The easiest way to calculate this is to find the percentage composition of sulfur. Then, use that percentage to find the mass of sulfur in the given amount of substance.

Step 1

% of sulfur = 32.1/98 x 100, roughly 33%

Step 2

The mass of sulfur present in 150 grams of sulfuric acid is

150 x 33%/100 = 49.5 g

M. DENSITY

Density is defined as the mass per unit volume.

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This property can be used to identify a compound or an element, since the density of a pure substance is a constant. Since density relates mass and volume, it can be used to find the volume occupied by a given mass, or if the volume is known, we can find the mass.

Density of water is 1.0 g/ml.

Let’s explore some calculations involving density. You’ll see that these calculations have tremendous laboratory significance.

Example 1-8

The density of carbon tetrachloride is about 1.6 g/ml at 20°C. Calculate the volume occupied by 320 g of CCl4.

Solution:

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So, 320 grams of carbon tetrachloride will occupy a volume of 200 ml.

Example 1-9

A 20 ml sample of mercury has a mass of 271 g. Calculate the density of mercury.

Solution:

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This question tests your knowledge of a basic equation. Though the equation is simple, you should be able to manipulate this equation so that you can connect this piece of information with other facts and formulas that are given in your test question or passage.

CHAPTER 1 PRACTICE QUESTIONS

1. A student preparing a solution for an experiment measured the weight of the sample solute to be used. If she is supposed to use 2 moles of calcium hydroxide, she must use:

A. 57.1 grams.

B. 74.1 grams.

C. 114.2 grams.

D. 148.2 grams.

2. Which of the following best represents the total number of ions present in a sample of NaCl weighing 102 g?

A. 6.0 x 1023 ions

B. 10.5 x 1023 ions

C. 12.0 x 1023 ions

D. 21 x 1023 ions

3. Experiments conducted with gas discharge tubes during the late 19th century resulted in many important conclusions in atomic chemistry. Among those, one of the most important was the discovery of rays known as cathode rays. Cathode rays when passed near negative plates will most likely:

A. bend toward the negative plate.

B. bend away from the negative plate.

C. will not bend, because they are uncharged.

D. will not bend, because they are high energy radiations.

4. Which of the following is true regarding a typical atom?

A. Neutrons and electrons have the same mass.

B. The mass of neutrons is much less than that of electrons.

C. Neutrons and protons together make the nucleus electrically neutral.

D. Protons are more massive than electrons.

5. The empirical formula of butane is:

A. CH3

B. C2H5

C. C4H10

D. CH2

6. The mass of one mol of a substance is numerically equal to:

A. mass number.

B. Avogadro number.

C. molecular weight.

D. 22.4.

7. If m represents the number of moles of a substance, M represents the molar mass of the substance, and d represents the density of the substance, which of the following expressions equals to the volume of the sample substance?

A. mM/D

B. m/dM

C. d/mM

D. m/d

8. The mass of oxygen in 96 grams of sulfur dioxide is closest to:

A. 16 g.

B. 24 g.

C. 32 g.

D. 47 g.

9. Choose the best value that corresponds to the percentage composition of chlorine in carbon tetrachloride?

A. 92%

B. 90%

C. 86%

D. None of the above

10. A student researcher analyzing the identity of the by-product of a reaction found that the compound contained 63.6% nitrogen and 36.4% oxygen. What is the most likely formula of this compound?

A. NO

B. NO2

C. N2O

D. N2O3

Questions 11-14 are based on the following passage.

Passage 1

The following reactions were conducted in a lab for studies related to reaction kinetics. Consider the reactions shown below:

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11. Reaction 2 is best described by which of the following?

A. A metathesis reaction

B. A combustion reaction

C. An endothermic reaction

D. A decomposition reaction

12. In Reaction 2, if 54 g of water was formed, how much ethane and oxygen must have reacted?

A. 30 g of ethane and 224 g of oxygen

B. 60 g of ethane and 224 g of oxygen

C. 30 g of ethane and 112 g of oxygen

D. 60 g of ethane and 112 g of oxygen

13. Which of the following is the actual formula of dextrose, if the empirical proportion is 1:2:1, where the proportion number 2 is that of hydrogen. Dextrose has a molecular weight of 180 g/mol? (Dextrose molecule is composed of carbon, hydrogen, and oxygen)

A. CHO

B. C6H12O6

C. C12H6O3

D. CH2O

14. Which of the following equals the number of hydrogen atoms in 40 grams of methane?

A. 1.5 x 1024

B. 2.4 x 1024

C. 6.0 x 1024

D. 6.023 x 1023