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The Mole

1 atom or 1 molecule is a very small entity not convenient to operate withThe masses we usually encounter in chemical experiments vary from milligrams to kilograms

Just like one dozen = 12 thingsOne mole = 6.022 x 1023 things

Avogadro’s number:

NA = 6.022 x 1023

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The Mole

Why 6.022 x 1023 ?This is the number of carbon atoms found in 12 g of the carbon-12 isotope

Molar mass – mass of one mole of atoms, molecules, ions, etc.Numerically equal to the atomic or molecular weight of the substance in grams:

m (1 mole H2) = Mr(H2) =

m (1 mole Fe) = Mr(Fe) =

NA = 6.022 x 1023

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The Mole: Example 1Example: Calculate the mass of a single Mg atom in grams to 3 significant figures.

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The Mole: Example 2Example: How many C6H14 molecules are contained in 55 ml of hexane (d = 0.78 g/ml).

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The Mole: Example 3Example: Calculate the number of O atoms in 26.5 g of lithium carbonate, Li2CO3.

Solution:1) Calculate the molar mass of Li2CO3:

Mr(Li2CO3) = 73.89 g/mol

2) Calculate the amount of Li2CO3 in moles:n (Li2CO3) = 0.359 mol

3) Calculate the amount of O atoms in moles:n (O) = (0.359x3) mol = 1.077 mol

4) Multiply by the Avogadro’s number to obtain the answer:N (O) = 1.077 mol x 6.022·1023 atom/mol =

= 6.50·1023 atoms

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Percent Composition and Formulas of Compounds

If the formula of a compound is known, its chemical composition can be expressed as the mass percent of each element in the compound (percent composition), and vice versa.

When solving this kind of problems, we can use masses expressed in a.m.u. or in g/mol

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What is the percent composition of each element in sodium chloride, NaCl?

Percent Composition: Example 1

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Percent Composition: Example 2Calculate the percent composition of iron(III) sulfate, Fe2(SO4)3, to 3 significant figures

Solution:1) Calculate the molar mass of Fe2(SO4)3:

Mr(Fe2(SO4)3) = 399.88 g/mol

2) Calculate the percent composition for the elements:

27.93% 100%g/mol 88399g/mol 85552 100%

))(SO(Fe(Fe)2 = ω(Fe)

342r

r =××

=××

..

MM

24.06% 100%g/mol 88399g/mol 07323 100%

))(SO(Fe(S)3 = ω(S)

342r

r =××

=××

..

MM

48.01% 100%g/mol 88399g/mol 001612 100%

))(SO(Fe(O)12 = ω(O)

342r

r =××

=××

..

MM

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The smallest whole-number ratio of atoms present in the compound

Molecular formula, on the other hand, indicates the actual number of atoms present in a molecule of the compound

Simplest (Empirical) Formula

water hydrogen peroxide

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The first high-temperature superconductor, prepared by Bednorz and Müller in 1986, contained 68.54% lanthanum, 15.68% copper, and 15.79% oxygen by mass. What was the simplest formula of this compound?

Empirical Formula: Example 1

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A sample of a compound contains 6.541g of Co and 2.368g of O. What is its empirical formula?

Empirical Formula: Example 2

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Elemental CompositionA combustion train for carbon-hydrogen analysis

percent composition is determined experimentally

magnesium perchlorate

sodium hydroxide

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0.1172 g of a pure hydrocarbon was burned in a C-H combustion train to produce 0.3509 g of CO2 and 0.1915 g of H2O. Determine the masses of C and H in the sample, the percentage of these elements in this hydrocarbon, and the empirical formula of the compound.

Empirical Formula: Example 3

Solution:1) First we calculate the molar weights of CO2 and H2O:

Mr(CO2) = 44.01 g/mol; Mr(H2O) = 18.02 g/mol

2) Now we can calculate the amount of both CO2 and H2O in moles:

mol 107.973 mol 0.007973 g/mol 0144

g0.3509 )(CO)(CO = )(CO 3

2r

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−⋅===.M

mn

mol 101.063 mol 0.01063 g/mol 8.021

g0.1915 O)(HO)(H = O)(H 2

2r

22

−⋅===Mmn

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Example 3 (continued)3) Notice that 1 mole of CO2 molecules contains 1 mole of C atoms

and 1 mole of H2O molecules contain 2 moles of H atomsTherefore,n (C) = 7.973·10-3 mol; n (H) = 2 x 1.063·10-2 mol = 2.126·10-2 mol

4) Now we can answer the first question – determine the masses of carbon and hydrogen in the starting compound – using the formula:

r = Mmn r= Mnm ×

m (C) = 7.973·10-3 mol x 12.01 g/mol = 0.09576 g

m (H) = 2.126·10-2 mol x 1.008 g/mol = 0.02143 g

Now, if you add these masses, they should give you the mass of the starting compound: 0.1172 g

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Example 3 (continued)5) Now we can easily find the mass percentage of the elements

in the hydrocarbon by dividing the mass of the element by the total mass of the hydrocarbon and multiplying by 100%:

We can also find the percentage of hydrogen by subtracting the percentage of carbon from 100%:

The slight difference between two results in caused by rounding errors.

81.71% 100%g 11720g0.09576 = ω(C) =×

.

18.28% 100%g 11720g0.02143 = ω(H) =×

.

18.29%ω(C)-100%=ω(H) =

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Example 3 (continued)6) Finally, to find the empirical formula we use the amounts

of C and H in moles found in step 3 and divide them by the smallest of both numbers:

We know that the fraction like .666 is in fact 2/3. Therefore, to convert both results to whole numbers, we need to multiply them by 3:

C: 3.000 H: 7.998

These result tells us that the empirical formula for the hydrocarbon in question is

C3H8

1.000 mol 107.973mol 107.973 :C 3-

-3

=⋅⋅ 2.666

mol107.973mol 101262 :H 3-

-2

=⋅⋅.

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0.1014 g sample of purified glucose was burned in a C-H combustion train to produce 0.1486 g of CO2 and 0.0609 g of H2O. An elemental analysis showed that glucose contains only carbon, hydrogen, and oxygen. Determine the empirical formula of the compound.

Empirical Formula: Example 4

This example is worked out in your textbook

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Indicates the actual number of atoms present in a molecule of the compound

To determine the molecular formula for a molecular compound, both its empirical formula and its molecular weight must be known

The molecular formula for a compound is either the same as, or an integer of, the empirical formula

Molecular Formula

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A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula?

Molecular Formula: Example

I’ll work on this example in the beginning of the next lecture.

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What mass of ammonium phosphate, (NH4)3PO4, would contain 15.0 g of N?

Other Examples

You do it !

The answer is: 53.2 g

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Reading Assignment

Go through Lecture 3 notes and learn again the problems we solved in class

Read Chapter 2 to the end

Learn Key Terms (p. 82)

Take a look at Lecture 4 notes (already posted on the web)

If you have time, read Chapter 3

Homework #1 due by 9/13