CHEMICAL FOUNDATIONS

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CHAPTER 1

CHEMICAL FOUNDATIONS

Questions

• A law summarizes what happens, e.g., law of conservation of mass in a chemical reaction or

the ideal gas law, PV = nRT. A theory (model) is an attempt to explain why something happens. Dalton’s atomic theory explains why mass is conserved in a chemical reaction. The kinetic molecular theory explains why pressure and volume are inversely related at constant temperature and moles of gas present, as well as explaining the other mathematical relationships summarized in PV = nRT.

• a. At 8 a.m., approximately 57 cars pass through the intersection per hour.
• At 12 a.m. (midnight), only 1 or 2 cars pass through the intersection per hour.
• Traffic at the intersection is limited to less than 10 cars per hour from 8 p.m. to 5 a.m.

Starting at 6 a.m., there is a steady increase in traffic through the intersection, peaking at

• a.m. when approximately 57 cars pass per hour. Past 8 a.m. traffic moderates to about

40 cars through the intersection per hour until noon, and then decreases to 21 cars per hour by 3 p.m. Past 3 p.m. traffic steadily increases to a peak of 52 cars per hour at 5 p.m., and then steadily decreases to the overnight level of less than 10 cars through the intersection per hour.

• The traffic pattern through the intersection is directly related to the work schedules of the

general population as well as to the store hours of the businesses in downtown.

• Run the same experiment on a Sunday, when most of the general population doesn’t

work and when a significant number of downtown stores are closed in the morning.

• The fundamental steps are

(1) making observations; (2) formulating hypotheses; (3) performing experiments to test the hypotheses.

The key to the scientific method is performing experiments to test hypotheses. If after the test of time the hypotheses seem to account satisfactorily for some aspect of natural behavior, then the set of tested hypotheses turns into a theory (model). However, scientists continue to perform experiments to refine or replace existing theories. Hence, science is a dynamic or active process, not a static one.

• A compound will always contain the same numbers (and types) of atoms. A given amount of

hydrogen will react only with a specific amount of oxygen. Any excess oxygen will remain unreacted.

• Law of conservation of mass: Mass is neither created nor destroyed. The total mass before a

chemical reaction always equals the total mass after a chemical reaction.(Chemistry An Atoms First Approach 2e Steven Zumdahl Susan Zumdahl) (Solution Manual, For Complete File, Download link at the end of this File) 1 / 4

CHAPTER 1 CHEMICAL FOUNDATIONS 25

Law of definite proportion: A given compound always contains exactly the same proportion of elements by mass. For example, water is always 1 g H for every 8 g oxygen.

Law of multiple proportions: When two elements form a series of compounds, the ratios of the mass of the second element that combine with 1 g of the first element always can be reduced to small whole numbers. For CO2 and CO discussed in Section 1.4, the mass ratios

of oxygen that react with 1 g carbon in each compound are in a 2 : 1 ratio.

• Yes, 1.0 g H would react with 37.0 g

37 Cl, and 1.0 g H would react with 35.0 g 35 Cl.

No, the mass ratio of H/Cl would always be 1 g H/37 g Cl for 37 Cl and 1 g H/35 g Cl for 35 Cl.As long as we had pure 37 Cl or pure 35 Cl, the ratios will always hold. If we have a mixture (such as the natural abundance of chlorine), the ratio will also be constant as long as the composition of the mixture of the two isotopes does not change.

• Natural niacin and commercially produced niacin have the exact same formula of C6H5NO2.

Therefore, both sources produce niacin having an identical nutritional value. There may be other compounds present in natural niacin that would increase the nutritional value, but the nutritional value due to just niacin is identical to the commercially produced niacin.

• a. The smaller parts are electrons and the nucleus. The nucleus is broken down into protons

and neutrons, which can be broken down into quarks. For our purpose, electrons, neutrons, and protons are the key smaller parts of an atom.

• All atoms of hydrogen have 1 proton in the nucleus. Different isotopes of hydrogen have

0, 1, or 2 neutrons in the nucleus. Because we are talking about atoms, this implies a neutral charge, which dictates 1 electron present for all hydrogen atoms. If charged ions were included, then different ions/atoms of H could have different numbers of electrons.

• Hydrogen atoms always have 1 proton in the nucleus, and helium atoms always have 2

protons in the nucleus. The number of neutrons can be the same for a hydrogen atom and a helium atom. Tritium ( 3

• and

4 He both have 2 neutrons. Assuming neutral atoms, then the number of electrons will be 1 for hydrogen and 2 for helium.

• Water (H2O) is always 1 g hydrogen for every 8 g of O present, whereas H2O2 is always 1

g hydrogen for every 16 g of O present. These are distinctly different compounds, each with its own unique relative number and types of atoms present.

• A chemical equation involves a reorganization of the atoms. Bonds are broken between

atoms in the reactants, and new bonds are formed in the products. The number and types of atoms between reactants and products do not change. Because atoms are conserved in a chemical reaction, mass is also conserved.

• J. J. Thomson’s study of cathode-ray tubes led him to postulate the existence of negatively

charged particles that we now call electrons. Ernest Rutherford and his alpha bombardment of metal foil experiments led him to postulate the nuclear atoman atom with a tiny dense center of positive charge (the nucleus) with electrons moving about the nucleus at relatively large distances away; the distance is so large that an atom is mostly empty space.

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CHAPTER 1 CHEMICAL FOUNDATIONS

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• The atom is composed of a tiny dense nucleus containing most of the mass of the atom. The

nucleus itself is composed of neutrons and protons. Neutrons have a mass slightly larger than that of a proton and have no charge. Protons, on the other hand, have a 1+ relative charge as compared to the 1 charged electrons; the electrons move about the nucleus at relatively large distances. The volume of space that the electrons move about is so large, as compared to the nucleus, that we say an atom is mostly empty space.

• The number and arrangement of electrons in an atom determine how the atom will react with

other atoms. The electrons determine the chemical properties of an atom. The number of neutrons present determines the isotope identity.

• Density = mass/volume; if the volumes are assumed equal, then the much more massive

proton would have a much larger density than the relatively light electron.

• For lighter, stable isotopes, the number of protons in the nucleus is about equal to the number

of neutrons. When the number of protons and neutrons is equal to each other, the mass number (protons + neutrons) will be twice the atomic number (protons). Therefore, for lighter isotopes, the ratio of the mass number to the atomic number is close to 2. For example, consider 28 Si, which has 14 protons and (28 – 14 =) 14 neutrons. Here, the mass number to atomic number ratio is 28/14 = 2.0. For heavier isotopes, there are more neutrons than protons in the nucleus. Therefore, the ratio of the mass number to the atomic number increases steadily upward from 2 as the isotopes get heavier and heavier. For example, 238 U has 92 protons and (238 – 92 =) 146 neutrons. The ratio of the mass number to the atomic number for 238 U is 238/92 = 2.6.

• Some elements exist as molecular substances. That is, hydrogen normally exists as H2

molecules, not single hydrogen atoms. The same is true for N2, O2, F2, Cl2, Br2, and I2.

Exercises

Development of the Atomic Theory

• a. The composition of a substance depends on the numbers of atoms of each element

making up the compound (depends on the formula of the compound) and not on the composition of the mixture from which it was formed.

• Avogadro’s hypothesis (law) implies that volume ratios are equal to molecule ratios at

constant temperature and pressure. H2(g) + Cl2(g) → 2 HCl(g). From the balanced equation, the volume of HCl produced will be twice the volume of H2 (or Cl2) reacted.

• Avogadro’s hypothesis (law) implies that volume ratios are equal to molecule ratios at

constant temperature and pressure. Here, 1 volume of N2 reacts with 3 volumes of H2 to

produce 2 volumes of the gaseous product or in terms of molecule ratios:

• N2 + 3 H2  2 product

In order for the equation to be balanced, the product must be NH3. 3 / 4

CHAPTER 1 CHEMICAL FOUNDATIONS 27

• From the law of definite proportions, a given compound always contains exactly the same

proportion of elements by mass. The first sample of chloroform has a total mass of 12.0 g C

• 106.4 g Cl + 1.01 g H = 119.41 g (carrying extra significant figures). The mass percent of

carbon in this sample of chloroform is:

totalg41.119 Cg0.12 × 100 = 10.05% C by mass From the law of definite proportions, the second sample of chloroform must also contain

10.05% C by mass. Let x = mass of chloroform in the second sample:

x Cg0.30 × 100 = 10.05, x = 299 g chloroform

• A compound will always have a constant composition by mass. From the initial data given,

the mass ratio of H : S : O in sulfuric acid (H2SO4) is: 02.2 00.64

:

02.2 07.32

:

02.2

02.2 = 1 : 15.9 : 31.7

If we have 7.27 g H, then we will have 7.27 × 15.9 = 116 g S and 7.27 × 31.7 = 230. g O in the second sample of H2SO4.

29. Hydrazine: 1.44 × 1

10

 g H/g N; ammonia: 2.16 × 1

10

 g H/g N; hydrogen azide:

2.40 × 2

10

 g H/g N. Let's try all of the ratios:

0240.0

144.0 = 6.00; 0240.0

216.0 = 9.00; 0240.0

0240.0 = 1.00; 144.0

216.0 = 1.50 = 2

3 All the masses of hydrogen in these three compounds can be expressed as simple whole- number ratios. The g H/g N in hydrazine, ammonia, and hydrogen azide are in the ratios

6 : 9 : 1.

• The law of multiple proportions does not involve looking at the ratio of the mass of one

element with the total mass of the compounds. To illustrate the law of multiple proportions, we compare the mass of carbon that combines with 1.0 g of oxygen in each compound:

compound 1: 27.2 g C and 72.8 g O (100.0  27.2 = mass O)

compound 2: 42.9 g C and 57.1 g O (100.0  42.9 = mass O)

The mass of carbon that combines with 1.0 g of oxygen is:

compound 1: Og8.72

Cg2.27 = 0.374 g C/g O

compound 2: Og1.57

Cg9.42 = 0.751 g C/g O 1 2 374.0 751.0  ; this supports the law of multiple proportions because this carbon ratio is a whole number.

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