Chapter 0211 With Answers

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Quizzes and Exams

• Please write in non-erasable blue or black pen, no pencil! (Will

lose 1 point for pencil or erasable ink)

• I will put periodic table on backs of quizzes, if you need them.

• BOX answers! If it’s not obvious where (or which) the answer is

that I’m supposed to grade, it will get no credit.

• Remember to bring your non-programmable calculators to class

on Wednesdays.

• Make sure your answers make sense (can there be 23 grams of

zinc in a 4.75 g sample of zinc oxide?).

• You may only use approved calculators on quizzes; others will

be confiscated for the duration of the quiz or exam.



Chapter 2: Atoms and Elements

1. Dalton’s Atomic Theory is based on empirical observations, formulated

as Laws of

Conservation of Mass

Definite Proportions

Multiple Proportions

No detectable gain or loss of mass occurs in chemical reactions.

In a compound, elements are always combined in the same proportion by mass.

Whenever two elements form more than one compound, the different masses of one element that combine with the same mass of the other element are in the ratio

of small whole numbers.

2. Summary of Dalton’s Theory

--Matter consists of tiny particles called atoms.

--Atoms are indestructible. In chemical reactions, the atoms rearrange but they do not

themselves break apart.

--In any sample of a pure element, all the atoms are identical in mass and other properties.

--The atoms of different elements differ in mass and other properties.

--In a given compound the constituent atoms are always present in the same fixednumerical ratio.



Sample Problems

1. A sample of caffeine contains 96.1 g of carbon for every 10.1 g of

hydrogen. If another sample of caffeine contains 30.0 g of carbon, howmany g of hydrogen does it contain?

2.* Phosphorus forms two compounds with chlorine. In 5.5 g of one of

these compounds, there were 1.24 g of P. In 5.5 g of the other compound,

there were 0.818 g of P. Explain how these compounds exhibit the law of

multiple proportions.



Sample Problems

1. A sample of caffeine contains 96.1 g of carbon for every 10.1 g of

hydrogen. If another sample of caffeine contains 30.0 g of carbon, howmany g of hydrogen does it contain?

Answer:

2.* Phosphorus forms two compounds with chlorine. In 5.5 g of one of

these compounds, there were 1.24 g of P. In 5.5 g of the other compound,

there were 0.818 g of P. Explain how these compounds exhibit the law of

multiple proportions.

Answer: According to the law of multiple proportions, if you compare the same

amounts of phosphorus, then the chlorine masses should be in a ratio of small

whole numbers. Since 1.24/0.818 = 1.516, then you should multiply the

compound B quantities by 1.516. In compound A, there is (5.5 - 1.24 =) 4.26 g Cl.

In compound B there is (5.5 - 0.818 =) 4.682 g Cl. Multiply that by 1.516 to get

7.10 g Cl for every 1.24 g P in compound B. Then you can check the Cl ratio:

4.26 g / 7.10 g = 0.6 = 3/5, a ratio of small whole numbers.

96.1

10.1=

30.0

x

x = 3.15 g



“Modern” View of an Atom

nucleus has protons(+) and neutrons

electrons(-) form a diffusecloud around the nucleus

Massproton neutron 1 amu

electron 10-4 amu

(amu = 1.66 x 10-24 g = 1/12 of mass of 12C)

~If there is a different number of protons, it is a different ___________?~If there is a different number of neutrons, it is a different __________?

Atomic number, Z = # of protons

Mass number, A = # of protons + number of neutrons

Isotopes = atoms of the same element with different mass numbers

Matter is neutral, so # of

electrons = # of protons

AEZ

~Where is most of the

mass of an atom?



Sample Problems

If a beryllium atom has 4 protons, then it should weigh 4 amu; but it actually

weighs 9 amu! Where is the extra mass coming from?

How many protons, electrons, and neutrons are in an atom of 52Cr?24



Sample Problems

If a beryllium atom has 4 protons, then it should weigh 4 amu; but it actually

weighs 9 amu! Where is the extra mass coming from?

How many protons, electrons, and neutrons are in an atom of 52Cr?24

neutrons

Atomic number = 24, so 24 protonsNeutral, so 24 electrons

52 – 24 = 27, so 27 neutrons

Answer:

Answer:



Some Notes on Charge

• Two kinds of charge called + and –

• Opposite charges attract

– + attracted to –

• Like charges repel

– + repels +

– – repels –

• To be neutral, something must have

no charge or equal amounts of opposite charges



The Periodic Table

Average among

all isotopes

Decides ordering of elements

Inner transition elements

Transition metals



Types of Elements

Metals:Shiny, malleable, ductile solids with

high mp and bp

Good electrical conductors

Nonmetals:Gases, liquids, or low-melting solids

Non-conductors of electricity

Metalloids:

Intermediate properties, often semiconductors

Diatomic elements: H2, O2, N2, F2, Cl2, Br 2, I2



Predicting Ionic Charges

Li+1

Na+1

K+1

Rb+1

Cs+1

Mg+2

Ca+2

Sr +2

Ba+2

Al+3

O-2

S-2

Se-2

Te-2

F-1

Cl-1

Br -1

I-1

N-3

1A

2A 3A 7A6A5A He

Ne

Ar

Kr

Xe

Rn

F?

Rb?

N?

Noble gases have very favorable electron configurations. The other elements are

jealous…

If an ion forms, it will commonly match the nearest noble gas.



Calculating Atomic Mass (Weighted Avg)

Atomic Mass =∑ (fraction of isotope n) x (mass of isotope n)

= (fraction1∙mass1) + (fraction2 ∙ mass2) + (fraction3 ∙ mass3) + … n

Example

63Cu has mass = 62.9396, natural abundance 69.17%,

and 65Cu has mass = 64.9278 and natural abundance 30.83%. What is the

atomic mass?

Answer: Atomic mass = (62.9396)(0.6917) + (64.9273)(0.3083)

= 63.55 amu (4 sig fig!)



Sample Problems

1. In a class A is worth 4 points and B is worth 3. Out of 20 grades, if you

have 4 A’s (20%) and 16 B’s (80%), what is your average?

If an element has two isotopes, A (Z = 4 amu) and B (Z = 3 amu), and their natural abundance is 20% A and 80% B, what is the average molecular

weight?

2. Naturally occurring boron (Z = 10.8110 amu) is composed of two isotopes,10B and 11B. Atoms of 10B have a mass of 10.0129 amu and those of 11B have a mass of 11.0093 amu. Calculate the percentages by mass

of the individual isotopes, 10B and 11B.



Sample Problems

1. In a class A is worth 4 points and B is worth 3. Out of 20 grades, if you

have 4 A’s (20%) and 16 B’s (80%), what is your average?

If an element has two isotopes, A (Z = 4 amu) and B (Z = 3 amu), and their natural abundance is 20% A and 80% B, what is the average molecular

weight?

Answer: Grade average = 3.2 points

Average molecular weight = 3.2 amu

2. Naturally occurring boron (Z = 10.8110 amu) is composed of two isotopes,10B and 11B. Atoms of 10B have a mass of 10.0129 amu and those of 11B have a mass of 11.0093 amu. Calculate the percentages by mass

of the individual isotopes, 10B and 11B.

Answer:

Mass percent of 10B = 19.90%

Mass percent of 11B = 80.10%



The Mole; How Chemists “Count”

1. Avogadro’s Number -- The Chemists’ “Dozen”

N0 = number of atoms in exactly 12 g of carbon-12

= 6.022 x 1023 “things” ( a very large number!)

This is a conversion factor, just like 12 things per dozen, e.g.

Mass of one atom of carbon-12 = (12.0 g)/6.022 x 1023 atoms)

= 1.99 x 10-23 g/atom

2. The Mole

One Mole of a substance contains an Avogadro’s

number of formula units.

e.g. 1 mole of H2O = 6.022 x 1023 H2O

molecules



Sample Problems

• How many atoms of gallium are there in 0.2398 moles of gallium?

• How many molecules of sugar are there in 0.2398 moles of

sugar?

• How many marbles are there in 0.2398 moles of marbles?



Sample Problems

• How many atoms of gallium are there in 0.2398 moles of gallium?

• Answer: 1.444 x 1023 atoms

• How many molecules of sugar are there in 0.2398 moles of

sugar?

• Answer: 1.444 x 1023 molecules

• How many marbles are there in 0.2398 moles of marbles?

• Answer: 1.444 x 1023 marbles

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Chapter 0211 With Answers