LECTURE 3 CHM 151 ©slg

TOPICS:

1. DENSITY Calculations2. PERCENT Calculations3. The Nuclear Atom4. Atomic Weight, Isotopes

KOTZ: 1.7, 2.1-2.6

Common Density Values

Magnesium: 1.74 g/cm3

Aluminum: 2.70 g/cm3

Silver: 10.5 g/cm3

Gold: 19.3 g/cm3

Dry air: 1.2 g/L at 25 oC, 1 atm pressure

Water: .917 g/mL at 0 oC 1.00 g/ mL at 4.0 oC .997 g/mL at 25 oCSea Water: 1.025 g/mL at 15 oC Antifreeze: 1.1135 g/mL at 20oC

SOLVING DENSITY PROBLEMS

1. Density itself can be calculated from experimentalvalues: Density = mass of object, solution, substance volume occupied (mL, cm3, L)

Volume can be determined in several ways:

a) direct measurement, liquidb) liquid displacement, solidc) measurement of dimensions, calculation

Volume by Displacement:

20.0 mL

28.5 mLFluid displacement: 28.5 mL -20.0 mL

8.5 mL

8.5 mL = 8.5 cm3

A liquid in which the solid does not dissolve issuitable for this technique

Calculation, Volume:

V = l X w X ht (rectangle) V = e3 (cube) V = r2 ht (cylinder) radius = diameter / 2; all dimensions in same units

diameterheight

l

w

ht

2. In all other cases, where conversion of mass to volume or volume to mass is desired, density is used as a“conversion factor”:

SOLVING DENSITY PROBLEMS

2.70 g Al = 1 = 1 cm3 Al 1 cm3 Al 2.70 g Al

“conversion factors”

D, Al = 2.70 g /cm3 2.70 g Al = 1 cm3 Al

Sample, Obtaining Density Value

A “cup” is a volume used by cooks in US. One cup is equivalent to 237 mL. If one cup of olive oil has a mass of 205 g, what is the density of the oil in g/mL and inoz avoir / in3?

1. State the question: D, olive oil = ? g/ mL = ? oz avoir / in3

2. State formula: Density = mass, g volume, mL

3. Substitute values and solve:

D = mass = 205 g = .86497 g = .865 g volume 237 mL mL mL

Second question: What is this density factor expressedas oz avoir/ in3?

1. State Question: .865 g = ? oz avoir mL in3

2. State relationships:

453.6 g = 1 lb 1 lb = 16 oz avoir

1 mL = 1 cm3

2.540 cm = 1 in ( 2.540 cm)3 = (1 in)3

16.39 cm3 = 1 in3

.865 g = ? oz avoir mL in3

Pathway: g lb oz avoir; mL cm3 in3

Pathway: g lb oz avoir; mL cm3 in3

3. Setup and Solve:

.865 g X 1 lb X 16 oz avoir X 1 mL X 16.39 cm3 1 mL 453.6 g 1 lb 1 cm3 1 in3

= .50008 oz avoir in3

= .500 oz avoir in3

.865 g = ? oz avoir mL in3

Density as a Conversion Factor:

Peanut oil has a density of .92 g/mL. If the recipe callsfor 1 cup of oil (1 cup = 237 mL), what mass of oil, inlb, are you going to use?

State question: 237 mL oil =? lb oil

State relationship:

( D, oil= .92 g / mL) 1 mL oil = .92 g oil 453.6 g = 1 lb

Pathway: mL g lb

Solution:

237 mL oil =? lb oil

Pathway: mL g lb

237 ml oil X .92 g oil X 1 lb = .48068 lb oil 1 mL oil 453.6 g

= .48 lb oil

Density conversion factor

GROUP WORK: Setup and solve as Conversion Type Problem

A gold coin is 2.75 cm in diameter and .50 cm thick.If the density of gold is 19.3 g / cm3, what is the massof the coin in grams?

Note:

Volume = r2 ht r = d / 2 ht = “thickness”

D, Au = 19.3 g/cm3 19.3 g Au = 1 cm3 Au

d = 2.75 cm, r = 2.75 cm / 2 = 1.375 cm ht = .50 cm

V=? = r2 ht = 3.1416 X (1.375 cm)2 X .50 cm V= 2.9698 cm3 = 3.0 cm3

First step: solve for volume

Second step: solve for mass

3.0 cm3 Au = ? g Au 19.3 g Au = 1 cm3 Au

3.0 cm3 Au X 19.3 g Au = 57.9 g Au = 58 g Au 1 cm3

Percent Composition by Mass

“Percent composition” is a convenient way to describe a mixture or solution or compound in terms of mass of the part contained in 100 mass units of the whole:

“This solution is 15% salt” means that for every 15 g of salt there is 100 g of solution or: 15 g salt = 100 g solution

“The brass alloy is 15 % tin and 45% copper”

100 g alloy = 15 g tin = 45 g copper

Like Density, Percent Composition is: calculated from experimental values used as a conversion factor.

Calculation of % by mass (g):

% Part = g part X 100% g whole

A brass alloy weighing 79.456 g was found to contain34.29 g of copper. What is the % of Cu in the alloy?

%Cu = 34.29 g Cu X 100% = 43.16% Cu 79.456 g alloy = 43.16 g Cu in 100 g alloy

Percent as a Conversion Factor

What mass in grams of a brass alloy would contain 25.00 g Cu if the alloy is 43.16% Cu?

Question: 25.00 g Cu = ? g brass

Relationship: 43.16 g Cu = 100 g brass

Setup and solve: 25.00 g Cu X 100 g brass = 57.92 g brass 43.16 g Cu

% factor

Density/Percent Solution Problems

Automobile batteries are filled with sulfuric acid, which is a solution of liquid H2SO4 in water.

What is the mass (in grams) of the H2SO4 in 500. mL of the battery acid solution, if the density of the solutionis 1.285 g /mL and the solution is 38.08% H2SO4 by mass?

Analysis of Problem

1.“Describing the Scene”:

Automobile batteries are filled with sulfuric acid, which is a solution of liquid H2SO4 in water.

3.“Giving the Relationships”:

if the density of the solution is 1.285 g /mL and the solution is 38.08% H2SO4 by mass?

2. “Stating the Question”:

What is the mass (in grams) of the H2SO4 in 500. mL of the battery acid solution

Automobile batteries are filled with sulfuric acid, which is a solution of liquid H2SO4 in water.

What is the mass (in grams) of the H2SO4 in 500. mL of the battery acid solution, if the density of the solutionis 1.285 g /mL and the solution is 38.08% H2SO4 by mass?

Question: 500. mL soltn = ? g H2SO4

Relationships: 1.285 g soltn = 1 mL soltn 38.08 g H2SO4 = 100 g soltn

Pathway: mL soltn g soltn g H2SO4

Question: 500. mL soltn = ? g H2SO4

Relationships: 1.285 g soltn = 1 mL soltn 38.08 g H2SO4 = 100 g soltn

Pathway: mL soltn g soltn g H2SO4

Setup and solve:

500. mL soltn X density factor X % factor = g H2SO4

500. mL soltn X 1.285 g soltn X 38.08 g H2SO4 1 mL soltn 100 g soltn

= 244.664 g H2SO4 = 245 g H2SO4

Group Work

Automobile batteries are filled with sulfuric acid, which is a solution of liquid H2SO4 in water.

How many mL of the acid solution would contain 15.00 gH2SO4, if the density of the solution is 1.285 g /mL and the solution is 38.08% H2SO4 by mass?

Question: 15.00 g H2SO4 = ? mL soltn

Relationships: 1.285 g soltn = 1 mL soltn 38.08 g H2SO4 = 100 g soltn

Pathway: g H2SO4 g soltn mL soltn

% D

Solution

Question: 15.00 g H2SO4 = ? mL soltn

Relationships: 1.285 g soltn = 1 mL soltn 38.08 g H2SO4 = 100 g soltn

Pathway: g H2SO4 g soltn mL soltn

Setup and Solve:

15.00 g H2SO4 X 100 g soltn X 1 mL soltn = 38.08 g H2SO4 1.285 g soltn

= 30.65 mL soltn

Chapter 2: Atoms and Elements

1. History of discovery of the atomic nature of all matter and the various particles found within the atom: Kotz, 2.1-2.2, excellent reading on all topics. Saunders CD-ROM: animated film clips

Reading, viewing assignment

Atomic Structure, Atomic Number, Mass

All matter (anything that has mass and occupiesvolume) can be classified as:

• an element (basic building blocks of nature: H, O, Au )

• a compound (made up of two or more elements)

• a mixture (any physical combination of the above)

Elements, the simplest forms of matter, are composed of unique tiny particles called atoms.

Atomic Structure

The atom itself is composed of three types of “subatomic particles”, the proton (p), the neutron (n),and the electron (e).

Each element has its own unique pairings of thesethree particles.

It is the number and the placement of these particles which gives rise to the different properties exhibited by each element.

proton neutron electron

mass, g 1.673 E-24 g 1.675 E-24 g 9.11 E-28 g

mass, amu 1.007 amu 1.009 amu 0.000549 amu

comparative mass 1 1 0

relative charge 1 0 -1

Comparative Mass, Charge: Nuclear Particles

Nuclear Particle Location Within the Atom

1. Protons and Neutrons: “Nucleus of Atom” compact positive mass in center of atom

“all” of the mass, negligible volume (10-2 pm)

2. Electrons: “Outside the Nucleus” cloud of negative charge

“all” of the volume (103 pm), negligible mass

THE “NUCLEAR” ATOM

Nucleus, C atom

Electron cloud

.

ATOMIC SYMBOLS

Each atom of an element can be represented by a symbol that describes how many protons, neutronsand electrons are contained in this basic unit:

XA

Z

Mass number, A:total #, p + n

Atomic Number, Z: #p (equals #e)

Elementalsymbol

Accordingly, A, the mass number of the atom, representsboth the total number of nuclear particles (p + n) and the approximate mass of the atom (in amu’s)

Because atoms (and the subatomic particles) are so tiny,a relative mass scale was setup to describe atomicweights in a convenient numerical range.

The atom of Carbon which contains 6 protons and 6 neutrons in its nucleus is assigned the weight of 12.00 amu (atomic weight units), which essentiallymakes the mass of each proton and neutron 1.00 amu.

The Mass Number, A

Z, The Atomic Number

All known elements are listed in the familiar “PeriodicTable of the Elements” in order of increasing atomicnumber, found usually in the upper right hand cornerabove each element’s symbol.

The atomic number represents the number of protonsin the nucleus of every atom of that element.

Since every atom is electrically neutral, the number ofprotons in the nucleus represents the total positivecharge of the nucleus, which is exactly balanced by the total number of negatively charged electrons outside the nucleus.

Group Work: Atomic Symbols

Br35

80

Pb82

207

Cr52

24

??40

20

p's n's e's

Br35

80

Pb82

207

Cr52

24

Ca40

20

p's n's e's

35 80-35

45

35

82207-82

12582

24 52-24

2824

20 40-20

20

20

Isotopes of the Elements

For a given element, the number of protons and electrons is a fixed value and determines which element is being described.

The number of neutrons in the atom of a givenelement is not fixed, and several different atoms ofa given element are generally found, varying in atomic mass due to differing numbers of neutrons.

The different atoms of a given element which vary byneutron count and by mass are described as “isotopes”of that element.

Isotopic Symbols

Isotopic symbols represent specific isotopic forms of an element. Consider hydrogen:

H H H1 1 1

1 2 3

1 p0 n

1 p1 n

1 p2 n

"protium" "deuterium" "tritium"

Atomic Mass, revisited:

The atomic mass of any isotope of an element can beapproximated by a simple addition of the number ofp’s and n’s in the nucleus.

However, naturally occurring samples of any element generally include several different isotopes of different atomic masses.

The atomic mass value given for each element (asfound in your Periodic Table) is a weighted average of all the isotopes.

The given atomic mass for any element will beclosest to the most abundant element in most cases. Consider below the calculation of theatomic mass of magnesium, 24.305 amu:

Average Mass Element, amu =

(mass, isotope A X % abundance A)

+ (mass, isotope B X % abundance B)

+ (mass, isotope C X % abundance C)......

General Method of Calculation:

Calculation of average atomic mass of Magnesium:

Mg Mg12 12 12

24 25 26Mg

23.9850 amu

78.99%

24.9858 amu

10.00 %

25.9826 amu

11.01%

(23.9850 x .7899) + (24.9858 x .1000) + (25.9826 x .1101)

= (18.95) + (2.499) + (2.861) = 24.31 amu