Unit 5Counting Atoms

1, 2, 3, 4, …..

10, 11, ………

1000, 1001, 1002…………

Counting Suppose I asked you

to count all the peanuts in this bag?

But what if I knew that 100 peanuts had a mass of 5 grams – can you think of another way to estimate the number of peanuts in the sample?

Connecting Mass to Number of Particles

One of the greatest challenges early chemists faced was trying to find a way to connect the mass of a substance to the number of particles in the sample.

Recall it was determined that particles combined in fixed ratios by weight.

Connecting Mass to Number of Particles

This led Dalton to the “atomic model” of matterExample: The mass ratio of oxygen to

hydrogen in water is 8:1This does not tell us how many atoms of each

element are involvedIt could tell us this if we knew the relative mass of

each kind of atom

Relative Mass

To assign relative masses to elements it is necessary to know that the samples being compared have the same number of particles

Relative Mass

Which has more balls in it: a bucket of golf balls or an identical bucket of baseballs?

If this is true in the macroscopic world, why wouldn’t it be true in the sub-microscopic one?

Relative Mass

Was iron more dense than aluminum because iron had more particles per given volume than aluminum or because iron’s individual particles were more massive than aluminum’s?

Could it be some combination of both?

Relative Mass

The truth is, based on the experiments we conducted earlier in the year, we couldn’t say which was true.

Dalton did not know what was true during his time either.Since the mass of individual atoms could not

be determined, a system of atomic masses had to be determined by comparison.

Relative Mass

To determine a system of masses by comparison, one element would have to be chosen as the basis of comparison for all others

Dalton chose hydrogen and assigned it a mass of 1.

Relative Mass

To find the mass of another element like oxygen:Compare the masses of equal number of

oxygen and hydrogen atoms ORFind the combining masses of oxygen and

hydrogen in water

Avogadro’s Hypothesis

Remember..Avogadro assumed:Equal volumes of gases have equal numbers

of moleculesThese molecules can be split during chemical

reactionsThat molecules of elemental gases could

contain more than a single atom

Avogadro’s Hypothesis

If we accept Avogadro’s Hypothesis, we can compare the mass of various gases and deduce the relative mass of the molecules

To do this, we pick a weighable amount of the lightest element (how about 1.0 g?) then use mass ratios to assign atomic masses to the other elements

Implications

If two volumes of hydrogen combine with one volume of oxygen gas, it is reasonable to assume that two molecules of hydrogen are reacting with each molecule of oxygen

Relative mass activity

vials containing several different items pretend that the items in the vials are

enlarged particles of various gases. Since the vials are the same size, we will

assume that each contains the same number of particles.

Your job is to determine the relative mass of each kind of item in the vial, without opening the vial to examine single items.

Comparison of Counting Methods P vs. n Lab to Avogadro’s Hypothesis

Our Previous Way of Counting Particles

Avogadro’s Way of Counting Particles

Assumptions Equal volumes of air in an open syringe contain equal numbers of particles (based on density – equal vol – equal mass, then equal # of particles)

Equal volumes of any gas contain equal numbers of gas molecules. Comparing masses of gas with the same volume (therefore, count) gives us relative masses of substances.

Unit Definition

Volume of 1 syringe-ful at constant density = 1 ‘puff’’ Actual count unknown; must be measured indirectly using volume

2 grams of hydrogen gas = 1 ‘mole’ of gas molecules (mole = ‘lump’)Actual count unknown; must be measured indirectly using volume and mass.

The Mole

The word chosen to represent the standard weighable amount of stuff, the mole, comes from the Latin “mole cula” or little lump

Mole

A mole of any element is defined as:

the amount of the element that contains as many atoms as there are in exactly 12 g of the carbon-12 isotope.

Avogadro’s Number

A mole was found experimentally to be equal to 6.022 X 1023 atoms of C-12, which is called Avogadro’s Number (NA).12 g Carbon-12 = 1 mol Carbon-12 atoms = 6.022 X 1023 Carbon-12 atoms

The MOLE (mol)The molar mass of any substance is the

mass of one mole of that substance.Molar mass is numerically equal to the atomic

mass of the substance.A mole of any substance contains Avogadro’s

number of units of that substance (6.022 X 1023 units).

Mass

2 ways to measure mass:Atomic Mass Units (amu)The atomic mass of an element is the

mass (in amus) as shown on the periodic table for an atom.

It is a average relative mass.

C = 12.01 amu

Average Atomic Mass of Elements

ELEMENT AVERAGE ATOMIC MASS (amu)

AVERAGE ATOMIC MASS, rounded (amu)

Hydrogen 1.00794 1.01

Carbon 12.0111 12.01

Oxygen 15.9994 16.00

Chlorine 35.453 35.45

Iron 55.847 55.85

Formula MassFormula Mass: the sum of the atomic masses

of all atoms in one compound.Water (H2O) has 2 Hydrogen atoms and 1

Oxygen atom.Formula Mass of H2O is

H 2 X 1.01 amu = 2.02 amu

O 1 X 16.00 amu = 16.00 amu

18.02 amu

Your turn…..Formula MassWhat is the formula mass of methane, CH4?

Of NaCl?

Of ammonia, NH3?

Of glucose, C6H12O6?

Molar Mass

Molar Mass (grams per mole) The mass of 1 mol of a compound. SO3 = 1S + 3O =

32.10 g + 3(16.00 g) = 80.10 g CaI2 = 1 Ca + 2 I =

40.10 g + 2(126.90 g) = 293.00 g

Molar mass and formula mass

1 atom of Carbon weighs 12.01 amus (formula mass)

1 mole of Carbon weights 12.01 grams (molar mass)

1 molecule of H20 weighs 18.02 amus (formula mass)

1 mole of H20 weighs 18.02 grams (molar mass)

Same numerically but different units!!!!!

What’s in a Mole? Atoms – one mole of an element contains 6.022 X 1023

atoms of the element. Molecules – one mole of a molecular (covalent) compound

contains 6.022 X 1023 molecules. Gizmos – one mole of gizmos contains 6.022 X 1023

gizmos. Anything – one mole of anything contains 6.022 X 1023

anythings! NA (6.022 X 1023) is very practical for counting small

particles, especially things like atoms, ions and molecules.

EXAMPLESHow many pens in 1 mole of pens?

How many atoms in 63.546g of Cu?How many atoms in 6.3546g of Cu?

EXAMPLES

How many molecules in 1 mole of sugar (C6H12O6)?

How many molecules in 10 moles of sugar?

How many carbon atoms in 1 mole of sugar? How many oxygen atoms?

Solving ‘Mole Problems”

MOLES

MASS

Molar mass

Mole Conversions by Factor Label Method

Mass and Moles Use the molar mass of the substance.

1 Mole = n grams of the substance, so have 2 conversion factors:

1 mole or n grams

n grams 1 mole

Mole Conversions by Factor Label

Example A: How many moles in 75.0 g of iron?How many moles in 75.0 g of iron?

Example B: How many grams in 0.250 mol Na?How many grams in 0.250 mol Na?

75.0g Fe X1 mol Fe

55.85g Fe= 1.34 mol Fe

0.250 mol Na X 22.99g Na

1 mol Na= 5.75g Na

Solving ‘Mole Problems”

MOLES

PARTICLES

Number of Particles in1 mole(6.02 X 10(6.02 X 102323))

Mole Conversions by Factor LabelParticles and Moles

Use Avogadro’s Number (6.022 X 1023) of particles.

(6.022 X 1023)particles 1 mol

1 mole (6.022 X 1023)particles Again, set up Factor Labels to cancel units!

Mole Conversions by Factor Label Method

= 0.697 mol = 0.697 mol COCO22

Example 1: How many atoms in 0.25 mol Na?How many atoms in 0.25 mol Na?

0.250 mol Na X0.250 mol Na X 6.022 X106.022 X102323 atoms Na atoms Na

1 mol Namol Na= 1.51 X 10= 1.51 X 102323 atoms Na atoms Na

Example DExample D: How many moles in 4.20 X 10: How many moles in 4.20 X 102323 molecules of CO molecules of CO22??

4.20 X 104.20 X 102323 Molecules CO Molecules CO22 X X1 mol CO2

6.022 X 106.022 X 102323 molecules CO molecules CO22

Mole is the bridge…

“How much” “How many”

Mass Number of

particles

MOLEMOLE

BridgeBridge

Summary: Solving ‘Mole Problems” using the Mole Map

MOLES

MASS

Molar mass

PARTICLES

Number of Particles in1 mole(6.02 X 10(6.02 X 102323))

Summary of Mole Conversions by Factor Label Method

Mass and Moles Use the molar mass of the substance.

Particles and Moles Use Avogadro’s Number (6.02 X 1023) of particles.

Set up Factor Labels to cancel units!DON’T GET LAZY! Include labels to ensure

that ALL units cancel correctly. Multi-step conversions are easily done if

you are careful with the labels!

The mass of each element in a compound compared to the entire mass of the compound and multiplied by 100%.

Ex: CH4

C 1 x 12.00g = 12.00 gH 4 x 1.01g= 4.04 g

total 16.04 g

%C = 12.00g/16.04g X 100 = 74.8%

%H = 4.04g/16.04g x 100 = 25.2%

Percentage Composition

Example 2 2.45 g aluminum oxide decomposes into 1.30 g

aluminum & 1.15 g oxygen. What is the percentage composition?

%O = 1.15g O X 100% = 46.9% O

2.45g Al Oxide note: Oxygen (O, not O2)

%Al = 1.30g Al X 100% = 53.1% Al 2.45g Al Oxide

As a check, note that 46.9% + 53.1% = 100.0%.

Percentage Composition

Determining Empirical FormulasEmpirical Formula

The formula that gives the simplest whole number ratio of the atoms of the elements in the formula.

Example 1 What is the empirical formula of a compound

containing 2.644g of gold and 0.476g of chlorine?

0.476g Cl X 1 mol Cl = 0.0134 mol Cl 35.45g Cl 2.644g of Au X 1 mol Au = 0.0134 mol Au

196.97g Au Empirical formula = Au0.0134 Cl0.0134 or AuCl

Determining Empirical FormulasExample 2

What is the empirical formula of a compound with 5.75 g Na, 3.50 g N & 12.00 g O?

First, find the mole amounts. 5.75g Na X (1 mol Na/22.99g Na) = 0.250 mol Na 3.50 g N X (1 mol N/14.01g N) = 0.250 mol N 12.00g O X (1 mol O/16.00g O) = 0.750 mol O Empirical formula = Na0.250 N0.250 O0.750

Divide each mole quantity by the smallest to get whole numbers. (0.250 in this case)

Empirical formula = NaNO3

Determining Molecular Formulas

Molecular FormulaThe formula that gives the actual number

of atoms of each element in a molecular compound.

Example Hydrogen peroxide has a molar mass

of 34.00 g/mol and a chemical composition of 5.90% H & 94.1% O. What is its molecular formula?

Determining Molecular Formulas

1. First, find the empirical formula, assuming the percents are mass.

5.90gH X (1 mol H/1.01g H) = 5.84 mol H

94.1g O X (1 mol O/16.00g O) = 5.88 mol O

Empirical formula = H5.84O5.88 or HO.

2. Find the molar mass of the empirical formula = 17.01g/mol

3. Divide by molar mass of the empirical formula

34.00/17.01 = 2

4. Multiply the empirical formula subscripts by this number: = 2(HO) or H2O2

The story so far….

From Avogadro’s Hypothesis we are able to count molecules by weighing macroscopic samples.

For gases at the same temperature and pressure we can deduce the following:1. From combining volumes we can

determine the ratio in which molecules react.2. From masses of these gases we can

determine the relative mass of individual molecules.

The story so far….

From these results it is possible to determine: the molar masses of the elements; using these masses and formulas of compounds, one

can determine molar masses of compounds. One can also determine empirical formulas and

molecular formulas

These tools allow one to relate

“how much stuff” to “how many particles”.