Moles and Compounds You need a calculator for use during these notes!

Moles and Compounds

You need a calculator for use during these notes!

Roses and eggs are conveniently packaged as a dozen. Sheets of paper are packaged as a ream. Small item are packaged in large amounts to make life easier. The same is true for atoms. However, because atoms are really, really, really small, the amounts of them that are packaged together are really, really, really big

A package of atoms or compounds that chemists use is called the Mole.

A package of atoms or compounds that chemists use is called the Mole. A mole (abbreviated “mol”) contains 602,200,000,000,000,000,000,000 of anything. This really large number is often called Avogadro’s Number and is abbreviated 6.022 × 1023.

Typical representative particles (things) measured in moles:

• atoms• molecules• compounds• formula units

If you know the number of moles, then the amount of atoms can be calculated.

How many atoms are in 2.00 moles of Al?



2.00 mols Al=



2.00 mols Al=

1 mol



2.00 mols Al 6.022 × 1023 atoms=

1 mol



In a calculator, you would put in 2 × 6.022 EE 23. Pushing the button that says EE takes the place of the “ × 10 ” and is the way calculators are meant to be used with scientific notation!

2.00 mols Al 6.022 × 1023 atoms=

1 mol



2.00 mols Al 6.022 × 1023 atoms= 1.20 × 1024 atoms Al

1 mol

The opposite can also be done.

How many moles is 9.03 × 1023 atoms of K?



9.03 × 1023 atoms K

=



9.03 × 1023 atoms K

=6.022 × 1023 atoms



9.03 × 1023 atoms K 1 mol

=6.022 × 1023 atoms



9.03 × 1023 atoms K 1 mol

= 1.50 mols K 6.022 × 1023 atoms

Notice, in both cases you use that fact that 1 mol = 6.022 × 1023. Whether the 6.022 × 1023 is on the top or bottom of the railroad tracks depends on what you start the problem with.

Also, this can be done with ANY representative particle, not just atoms. The only difference is the label!

Masses and the Mole

The mole is not only handy to counting a large amount of atoms or molecules, it is even more useful when measuring out amounts of elements or compounds. The masses on the periodic table have been designed to be the amount of grams in a mole of that element, thus it is called the “molar mass”

For example, look at sulfur.

The information shown for sulfur tells us that 1 mol of sulfur = 32.066 grams of sulfur. Likewise, 2 mols of sulfur = 2 × 32.066 or 64.132 grams of sulfur, etc. This allows for amounts of elements to be easily measured in the laboratory and turned into moles.

16S

32.066

A student measures out 12.0 g of Mg in lab. How many moles does this student have?


12.0 g Mg =


12.0 g Mg =

24.31 g


12.0 g Mg 1 mol =

24.31 g

A student measures out 12 g of Mg in lab. How many moles does this student have?

Notice the molar mass is unique for each element and must come from the periodic table.

12.0 g Mg 1 mol = 0.494 mols Mg

24.31 g


A student needs 2.30 mols of B for an experiment. How many grams should this student measure out?



2.30 mols B =



2.30 mols B =

1 mol



2.30 mols B 10.81 g =

1 mol



2.30 mols B 10.81 g = 24.9 g B

1 mol

Our previous work with representative particles and molar masses can be put together!

A student needs 2.60 × 1023 atoms of fluorine for an experiment. How many grams of fluorine should the student measure out?


2.60 × 1023 atoms F

=


2.60 × 1023 atoms F

=6.022 × 1023

atoms


2.60 × 1023 atoms F

1 mol =

6.022 × 1023

atoms


2.60 × 1023 atoms F

1 mol =

6.022 × 1023

atoms 1 mol


2.60 × 1023 atoms F

1 mol 18.99 g =

6.022 × 1023

atoms 1 mol


2.60 × 1023 atoms F

1 mol 18.99 g = 8.20 g F

6.022 × 1023

atoms 1 mol


A student has 56.7 grams of aluminum. How many atoms of aluminum does the student have?



56.7 g Al



56.7 g Al

26.98 g



56.7 g Al 1 mol

26.98 g



56.7 g Al 1 mol

26.98 g 1 mol



56.7 g Al 1 mol 6.022 × 1023 atoms

26.98 g 1 mol



56.7 g Al 1 mol 6.022 × 1023 atoms = 1.27 × 1024 atoms Al

26.98 g 1 mol

Moles of Compounds

A properly written compound shows the ratio of atoms in the compound. For example, sodium carbonate (Na2CO3) shows that for every 1 compound there are 2 atoms of Na, 1 atom of C, and 3 atoms of O.

This also works with the mole. For 1 mole of the compound, there are 2 moles of Na, 1 mole of C, and 3 moles of O. This information allows us to calculate the molar mass of the entire compound, often called the formula mass.

2 mols Na 22.9897 g = 45.9794 g Na

1 mol

1 mol C 12.0107 g = 12.0107 g C

1 mol

3 mols O 15.9994 g= 47.9982 g O

1 mol

The total molar mass of Na2CO3 is 105.9883 g.

Once it is known that 1 mol of Na2CO3 = 105.9883 g, conversion factors can be made of this information:

These conversion factors can then be used in other calculations.

105.9883 g Na2CO3or

1 mol

1 mol 105.9883 g Na2CO3

What is the molar mass for potassium oxide (K2O)?


K2O = 2 x K + 1 x O


K2O = 2 x K + 1 x O

K2O = (2 x 39.0983 g) + (15.9994 g)


K2O = 2 x K + 1 x O

K2O = (2 x 39.0983 g) + (15.9994 g)

K2O = 78.1966 g + 15.9994 g = 94.1960 g

If you have 560.0 grams of K2O, how many moles do you have?


560.0 g K2O =


560.0 g K2O 1 mol =

94.196 g


560.0 g K2O 1 mol = 5.945 mol K2O

94.196 g

If you have 3.4 moles of K2O, how many grams do you have?


3.4 mols K2O =


3.4 mols K2O 94.2 g =

1 mol


3.4 mols K2O 94.2 g = 320 g K2O

1 mol

If you have 54.3 grams of cobalt(II) chloride, how many formula units (compounds) and how many atoms is this?

54.3 g CoCl2

=


54.3 g CoCl2

=129.8 g

58.93g + (2 x 35.45 g)


54.3 g CoCl2 1 mol CoCl2

=129.8 g

58.93g + (2 x 35.45 g)


54.3 g CoCl2 1 mol CoCl2

=129.8 g 1 mol CoCl2


54.3 g CoCl2 1 mol CoCl26.022 x 1023

formula units =129.8 g 1 mol CoCl2


54.3 g CoCl2 1 mol CoCl26.022 x 1023

formula units = 2.52 x 1023

formula units129.8 g 1 mol CoCl2


2.52 x 1023 formula units CoCl2



1 formula unit CoCl2



3 atoms




3 atoms= 7.56 x 1023 atoms


Percent Composition and Empirical Formulas

Percent Composition

Percent Composition

The percent composition shows the relative percent (by mass) of each element in a compound.

Percent Composition

The percent composition shows the relative percent (by mass) of each element in a compound.

The percent composition is determined by dividing the mass of the individual elements in a compound by the entire formula mass of the compound.

Percent Composition =

Mass of individual element (g)

× 100 %

= % of element

Formula Mass of Compound (g)

For example, when the correct percent composition for HF is determined, the process is as follows:

For example, when the correct percent composition for HF is determined, the process is as follows:

Find the total formula molar mass: 1 mol H = 1.0079 g/mol1 mol F = + 18.9884 g/mol total = 19.9963 g/mol

Take the individual molar mass of each element and divide by the total formula mass, and turn it into a percent:

Take the individual molar mass of each element and divide by the total formula mass, and turn it into a percent:

for H1.0079 g/mol H

× 100 % = 5.04% H19.9963 g/mol HF

for F18.9884 g/mol F

× 100 % = 94.96% F19.9963 g/mol HF

A good way to quickly check the answers is to sum the percentages, which should equal 100% (or 1). There will be cases where the percentages might not equal exactly 100% because of rounding, but the total should always be VERY close to 100%.

As another example, consider sulfuric acid:H2SO4 :


2 mol H × 1.008 g/mol = 2.016 g/mol

1 mol S × 32.066 g/mol = 32.066 g/mol

4 mol O × 15.999 g/mol = 63.996 g/mol

total = 98.078 g/mol


for H2.016 g/mol H

× 100% = 2.06% H98.078 g/mol H2SO4

for S 32.066 g/mol S× 100% = 32.69% S

98.078 g/mol H2SO4

for O63.996 g/mol O

× 100% = 65.25% O98.078 g/mol H2SO4

Empirical Formula

Empirical FormulaOnce the percent composition of a compound is known, the empirical formula of the compound can be determined. An empirical formula shows the lowest whole-number ratio of the elements in a compound

1. Turn the percent composition information into mass. This is made simple by assuming a theoretical amount of 100 grams.

Thus 50% composition is turned into 50 grams, and 36.8% composition is turned into 36.8 grams, etc.

2. Calculate the number of moles for each element that would contain the amount of mass from step 1.

This involves dividing the mass from step 1 by the molar mass shown for the element on the periodic table.

3. The simplest whole-number ratio of each element needs to be found.

One of the ways to get a good start on this is to divide each number of moles from step 2 by the smallest amount of moles.

This will guarantee at least one whole number to start with (a “1” amount).

3. a. If the other molar amounts are within 0.15 of a whole number, it is usually safe to round up or down to that whole number.

3. b. If the other molar amounts cannot be rounded, it will be necessary to multiply ALL the molar amounts by a whole number to obtain a whole number (or a number close to a whole number.) Thus, if a molar amount had the decimal value of 0.20, it would be necessary to multiply by 5. If the decimal value is 0.25,it would be necessary to multiply by 4, and it would if the decimal value is 0.33, it would be necessary to multiply by 3, etc.

Example:White gold is 75.0% gold, 10.0% palladium, 10.0% nickel,

and 5.00% zinc. What would be the empirical formula of white gold?

75.0% Au → 75.0 g Au 1 mole Au

= 0.3807 moles Au

197.0 g

10.0% Pd → 10.0 g Pd 1 mole Pd= 0.09398 moles Pd

106.4 g

10.0% Ni → 10.0 g Ni 1 mole Ni= 0.1704 moles Ni

58.69 g

5.00% Zn → 5.00 g Zn 1 mole Zn= 0.07646 moles Zn

65.39 g

Dividing by the lowest amount of moles from above (0.07646 mol): 0.3807 moles Au

= 4.979 moles Au ≈ 5 moles Au0.07646

0.09398 moles Pd = 1.229 moles Pd

0.07646

0.1704 moles Ni = 2.229 moles Ni

0.07646

0.07646 moles Zn = 1 moles Zn

0.07646

The gold and zinc are already expressed in a whole number, but to express the palladium and nickel as a whole number, it will be necessary to multiply everything by 4.

This would make the palladium and nickel 4.916 moles and 8.916 moles (respectively), which are now close enough to round. Do not forget to multiply everything, even the ones that are already whole numbers!

The gold and zinc are already expressed in a whole number, but to express the palladium and nickel as a whole number, it will be necessary to multiply everything by 4.

This would make the palladium and nickel 4.916 moles and 8.916 moles (respectively), which are now close enough to round. Do not forget to multiply everything, even the ones that are already whole numbers!

Thus the final relative amount of moles is 20 Au, 5 Pd, 9 Ni, 4 Zn. The empirical formula is Au20Pd5Ni9Zn4.

Practice Exercise:Find the empirical formula for purple gold, Purple Gold =

80% Au, 20% Al

mass 1 mol

=

answer

= answer × factor = whole #P.T. mass lowest #

mass 1 mol

=

answer

= answer × factor = whole #P.T. mass lowest #

80 g 1 mol=

0.4062 mol

= answer × factor = whole #

196.97 g lowest #

20 g 1 mol=

0.7413 mol

= answer × factor = whole #

26.98 g lowest #

80 g 1 mol=

0.4062 mol

= 1 × factor = whole #

196.97 g 0.4062 mol

20 g 1 mol=

0.7413 mol

= 1.825 × factor = whole #

26.98 g 0.4062 mol

80 g 1 mol=

0.4062 mol

= 1 × 5 = 5196.97 g 0.4062

mol

20 g 1 mol=

0.7413 mol

= 1.825 × 5 = 926.98 g 0.4062

mol

80 g 1 mol=

0.4062 mol

= 1 × 5 = 5196.97 g 0.4062

mol

20 g 1 mol=

0.7413 mol

= 1.825 × 5 = 926.98 g 0.4062

mol

Empirical Formula = Au5Al9

Molecular Formulas

An empirical formula shows the lowest whole-number ratio of the elements in a compound, but may not be the actual formula for the molecular compound, called the molecular formula. The molecular formula is always some whole number multiple of the empirical formula.

For Example

Compound E.F. M.F. MultipleWater H2O H2O 1

Hydrogen Peroxide HO H2O2 2

Glucose CH2O C6H12O6 6

To find the molecular formula it is necessary to know the empirical formula and the multiple.

Often the empirical formula is calculated from the percent composition, and then the multiple is calculated by knowing the actual molar mass of the molecular compound.

For example:A compound has a formula mass

of 78.11 g/mol and is 92.24% carbon and 7.76% hydrogen. What is the molecular formula?

Step 1: Determine the empirical formula:


92.24 g C= 7.68 mols

12.01 g/mol

7.76 g H= 7.68 mols

1.01 g/mol


Dividing both by 7.68 moles gives an empirical formula of CH.

92.24 g C= 7.68 mols

12.01 g/mol

7.76 g H= 7.68 mols

1.01 g/mol

Step 2: Determine the molar mass of the empirical formula:

Step 2: Determine the molar mass of the empirical formula:

CH = 13.02 g/mol

Step 3: Determine the factor:


molecular formula mass = 78.11empirical formula mass =

13.02= 5.999 or 6


Thus the actual molecular formula is 6 times more than the empirical formula, or C6H6.

molecular formula mass = 78.11empirical formula mass =

13.02= 5.999 or 6

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Moles and Compounds You need a calculator for use during these notes!