Notes, Units 4 & 5

Elements, Compounds, Avogadro‘s

Hypothesis, Relative Mass, Molar Mass,

Balancing Equations, Mole Concept,

Counting by Weighing

Property water alcohol

appearance

odor

flammability

density

Melting point

Boiling point

Solubility of sugar (ability to dissolve)

Properties of Water and Alcohol

1.0 g/mL 0.79 g/mL

0°C -117°C

100°C 80°C

colorless colorless

none noticeable

non-flammable flammable

very soluble Insoluble (doesn’t dissolve)

Unit 4, Worksheet 1

A mixture of iron

and sulfur atoms

A compound of iron and

sulfur (FeS) in a 1:1 ratio

#3. Draw particle representations for :

Compound Elements

Pure

Compounds

Pure Elements

Mixtures: B & D

Temperature

and fixed

volume are

constant

n = 100

P = 1 atm

n = 200

P= 2 atm

n = 300

P=3 atm

Avogadro’s Hypothesis

of particles

Equal Volumes of Gases

Evidence for Avogadro’s Hypothesis The measured volume, pressure, and temperature are equal for these two volumes of gas molecules.

What would happen to the pressure if we increased the number of molecules by the same amount? The pressure would increase by the same amount due to more molecules hitting the walls. What does that tell you about the number of molecules in the above volume? The number of molecules in the volumes must be the same.

If the temperature and volume are constant

Equal volumes of gases have equal numbers of particles

This is known as Avogadro’s Hypothesis

Each rectangle is one volume of gas.

Each volume of gas has the same number of particles or molecules,

regardless of the identity of the gas

Equal volumes of gases have equal numbers of particles

Thus 2 volumes of hydrogen react with 1 volume of oxygen to

form 1 volume of water vaper .

The number of molecules in each volume is equal.

Equal volumes of gas have equal numbers of molecules

2 volumes of H + 1 volume of O

or:

2 H + 1 O 1 H2O

+

1 volume H2O

Avogadro’s Hypothesis The apparatus in the classroom breaks down water showing it is composed of 2 volumes of hydrogen and one volume of oxygen.

We end up with one volume of water vaper

1 1

H Cl

N H

1 volume HCl

1 volume

NH3

Using Avogadro’s Hypothesis to Deduce the

Composition of Compounds

Evidence for Avogadro’s Hypothesis The apparatus in the classroom breaks down water showing it is composed of 2 volumes of hydrogen and one volume of oxygen.

The conclusion that two molecules of hydrogen combine

with one molecule of oxygen to form water works only if

we assume that each volume of gas contains the same number of particles.

2 volumes H1 1 vol O1 + 1 volume H2O

Evidence for Avogadro’s Hypothesis But experiments showed two volumes of water was produced

How do we explain this and still have equal volumes of

gases with equal number of molecules?

`

2 volumes H2 1 vol O2 + 2 volumes H2O

Avogadro’s Hypothesis became Avogadro’s Law: Equal volumes of gases at the same pressure and temperature contain equal numbers of molecules.

The molecules of hydrogen and oxygen must contain two atoms and not 1 atom. In other words, its not H1, its H2. Its not O1, its O2.

These elements are diatomic: they are made up of two of

the same atom.

`

2 volumes H1 1 vol O1 + 1 volume H2O

When 1 volume of hydrogen and 1 volume of oxygen were mixed, it was

assumed that one volume of water vapor was produced. Experiments

showed two parts water vapor was produced.

The bottom diagram is correct (circled).

2 volumes H2 1 vol O2 + 2 volumes H2O

N H

N2 H2

On the Unit 4 Worksheet 2 homework you found the formula that 1

volume of nitrogen mixed with three volumes of hydrogen produce

one volume of ammonia:

Experiments showed that two volumes of ammonia were produced. Thus, nitrogen & hydrogen are not monatomic as shown above , but diatomic as shown below ( N2 instead of N1 & H2 instead of H1)

NH3

NH3

` 1 volume H1

1 vol Cl1 HCl 1 volume

Equal volumes of gasses have equal numbers of molecules

`

1 vol Cl1 1 volume

When 1 volume of hydrogen and 1 volume of chlorine were mixed, it was

assumed that one volume of hydrogen chloride was produced:

1 vol H1

1 vol H1

` 1 volume H1

1 vol Cl1 HCl 2 volumes

Equal volumes of gasses are supposed to have equal numbers of molecules:

`

This is not equal numbers of atoms on both sides of the arrow (not balanced).

Experiments showed that when 1 volume of hydrogen and 1 volume of chlorine were mixed, two volumes of hydrogen chloride were produced:

1 vol H1

H = 3 Cl = 3 H = 6

Cl = 6

`

1 volume H1 1 vol Cl1 HCl

2 HCl

Diatomic molecule:

1 volumes

Equal volumes of gasses have equal numbers of molecules

3atoms 3 atoms

3 atoms

3 atoms

6 atoms

hydrogen's

6 atoms

chlorines

6 atoms 6atoms

6 atoms

hydrogen's

6 atoms

chlorines

`

1 volume H 1 vol Cl HCl

2 HCl Diatomic molecule

2 volumes

Equal volumes of gasses have equal numbers of molecules

3 3 3 3

3 3

6

6

Total H atoms:

Total Cl atoms: + +

+

Copyright © by Holt, Rinehart and Winston. All rights reserved.

Resources Chapter menu

Interpreting a Chemical Reaction

Section 1 Describing Chemical Reactions,

Balancing Chemical Equations Unit 7

This 2 is called a coefficient

This 2 is called a subscript

A balanced equation shows that mass is conserved

This illustrates the Law of Conservation of Mass

Copyright © Houghton Mifflin Company. All rights reserved. 3–22

C?O? + O2 CO2 + H2O

• The products can be trapped in the

device shown below.

•The increase in the mass of the H2O absorber and the CO2

absorber are a direct result of the mass of the C and the O from

C?O? ending up in the absorber.

•The result of such analysis provides the masses of

each type of element in the compound.

Two 100 g samples of unknown compounds each made up of only carbon and oxygen give us the following mass data. Compound A: 57.1 g O / 42.9 g C

Compound B: 72.7 g O and 27.3 C

a. Determine the value of the ratio of oxygen to carbon in

each compound.

Compound A: 1.33

b. How does the mass ratio for compound B compare to that

in compound A?

c. Express these ratios as improper fractions.

A: 4/3, B: 8/3

d. sketch particle diagrams for the compounds of A and B that

account for these mass ratios. Write the formula for the compound in

each diagram.

Relative Mass Notes

Compound B: __2.66___

double

Atoms of C and O have the same mass

Atoms of O are heavier than C atoms by the ratio in compound A.

A 4/3:

C3O4

A ratio of O to C=1.33

CO

B 8/3:

C3O8

B ratio of O to C = 2.66

CO2

Mole Concept

Count by Weighing, Relative Mass,

Molar mass, moles, Avogadro’s

number, Mole Calculations

What is the best way to

figure out how many

“peanuts” are in the

bag?

Weigh some of the peanuts to get an

average mass for a peanut.

Then weigh the big bag of peanuts.

Divide the mass of bag by the mass

the average peanut to determine

the number of peanuts in the bag.

Count by weighing

Relative Mass Activity Copy this data

Go to the Relative Mass Activity sheet, problem #1, and

calculate the mass of a box of washers and the mass of a

box of hex nuts using the data above.

**Be careful. The masses above include the bag & tape.**

Hardware Mass (g)

Empty bag & tape

Bag/ tape + Washers

Bag/ tape + Hex Nuts

Bag/ tape + Bolts

Relative Mass Activity Copy this data

Go to the Relative Mass Activity sheet, problem #1, and

calculate the mass of a box of washers and the mass of a

box of hex nuts using the data above.

**Be careful. The masses above include the bag & tape.**

Hardware Mass (g)

Empty bag & tape 5.8

Bag/ tape + Washers 73.5

Bag/ tape + Hex Nuts 133.2

Bag/ tape + Bolts 215.5

1. A box of hardware contains 100 pieces. Assuming

there are 25 pieces in each vial, calculate the mass of a

box of each kind of hardware. Express these values in

units of g/box. Calculate the mass of one piece assuming the mass in your data table is the mass of 25 pieces. Now multiply this mass x 100.

2. If you had 1.00 kg of each kind of hardware, how

many boxes of each would you have?

1000 g ÷ mass of a box of 100

3. You learned that a barrel of the 1” bolts had a

mass of 65.2 kg. The mass of the barrel was 9.6 kg.

How many boxes of bolts are in the barrel?

(62.2 kg – 9.6 kg) x 1000 g/kg = 52600g bolts 52600 g bolts ÷ mass in g of 1 box bolts = # boxes of bolts

4. Someone at the Home Depot tells you that a 2” bolt is

6.75 times as heavy as a washer. What would be the

mass of a box of such bolts? Mass of a box of 100 washers x 6.75 = mass of box 2” bolts 5. Suppose that you were given the job of shipping 25,000 hex

nuts to a customer. All you have is a hanging scale and a barrel of hex nuts. Describe how you could determine the proper

number of pieces without physically counting them out. – Determine how many boxes. 25000/ 100 = 250 boxes – Multiply 250 boxes x the mass of one box of hex nuts. – This equals the mass of 25,000 hex nuts

Count By Weighing • Let say we are using a kitchen scale to weigh hex nus, bolts &

washers. • A standard amount of bolts, nuts and washers (hardware) is a

hundred pieces (which can be more accurately weighed than 1 piece).

• A “box” is defined as a count of one hundred pieces. • The mass of a box (box mass) = a weighable amount of

hardware that always has the same count (100 pieces). • One hex nut, one bolts and one washer all have different masses. • Thus a box of hex nuts, washers, and bolts all have different box

masses. • If you know the “box mass” of each type of hardware you know

the count. • Thus you can count by weighing.

• The standard “lump” of bolts, nuts and washers (hardware) is called a box.

• The standard “lump” of elements or compounds is called a mole.

• The number of particles in a box = 100 • The number of particles in a mole = 6.022 x 1023 , Avogadro’s number. • The mass of 1 box of nuts, bolts or washers is called box mass. • The mass of 1 mole of elements or compounds is called the

molar mass. • Nuts, bolts and washer all have different masses. Thus a box of

hex nuts, washers, and bolts all have different box masses. • But a box mass always = 1 box of pieces = 100 pieces • Different elements & compounds have different masses and

thus have different molar masses. The molar mass of elements is found on the periodic table.

• The molar mass always = mass 1 mole = 6.022 x 1023 particles.

Bell Work about Moles

4. What is molar mass for elements. • It is the relative mass in grams of one standard lump of

elements. A standard lump of an element has 1 mole of atoms. • Scientist choose carbon to compare relative masses. • By definition, something called the carbon-12 isotope (a carbon

atom with 6 protons and six neutrons) has molar mass of exactly 12 grams.

5. What is the relative mass and the molar mass of an element that is ½ the mass of C-12?

½ of 12 = 6, so the relative mass = 6, the molar mass = 6 grams 6. What is the relative mass of an element that is 2 x the mass of C-12? 2 x 12 = 24, so the relative mass = 24, the molar mass = 24 grams 7. What is a mole?

The number of particles in a mole = 6.022 x 1023

The molar mass of any element contains 6.022 x 1023 atoms

the molar mass in grams of any element or compound = 1 mole

Atomic number

Molar Mass (mass in grams of one mole of the element).

Chemical symbol, begins with a capital letter

Atomic masses are relative masses

(comparisons to the mass of one carbon

atom). Example, carbon (#6) is 12 times

more massive than hydrogen (#1).

Atomic mass

2. The molar mass or formula mass of KNO3 is:

Molar mass of K = ________ Molar mass of N = ______

Molar mass of O = ______

Molar mass of 1 K is 2 x 22.99 g =

Molar mass of 1 N is 1 x 14.01 g =

Molar mass of 3 O’s is 3 x 16.0 g =

Molar mass of KNO3

14.01 g

16.00 g

14.01 g

48.00 g

101.02 g

1. The number nitrogen, potassium , nitrogen , & oxygen atoms represented by the formula KNO3

_____ potassium (K), ____ nitrogen (N) , ______ oxygen (O) atoms.

3 1

39.10 g

39.01 g

1

Molar Mass of a Compound

Molar (Formula) Mass Examples

a. H2ClO3 = 2 H, 1 Cl, 3 O

(2 *1.008) + (1*35.453) + (3 *16) = 84.469 g/mol

b. Ag2CrO4 = __2_ Ag, 2 Cr, __4_ O

(2 *107.87) + (1* 51.99) + (4 *16) = 331.73g/mol

Ag is # 47 , Cr is # 24

3. Convert from g moles or from moles grams.

_____ x = ______ (given) times factor = (answer)

a. 12.0 g Fe 12.0 g Fe x = ______

mm of Fe = 55.8 g/ mol

55.8 g Fe = 1 mol Fe

= 0.215 mol Fe

b. 0.0280 moles NO2 grams

mm NO2 = 46.0 g/mol

0.0280 mol NO2 x = _________

46 g NO2 = 1 mol NO2

1.29 g

1 mol Fe

55.8 g Fe

? mol Fe

1 mol NO2

46.0 g NO2 ? g NO2

4. Grams moles # of atoms

or molecules

A 4.07 g sample of NaI contains how many atoms of

Na?

Molar mass of Na =

1 mole of NaI = 149.9 g (= molar mass Na)

1 mole of NaI = 6.022 x 1023 atoms of NaI 1 molecule of NaI contains 1 atom of Na (and 1 aom of I)

Divide by the molar mass Multiply by Avogadro’s #

5.#of atoms moles Grams or molecules

What is the mass of 100 million atoms (1 x 108 atoms) of

gold (Au)?

Avogadro’s number = 6.022 x 1023

1 mole of Au = 6.022 x 1023 atoms of Au

1 mole of Au = 197.0 g of Au ( = the molar mass of Au)

Multiply by the molar mass Divide by Avogadro’s #