Chemical quantities the mole composition empirical and molecular formulas
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Chemical Quantities The Mole, % Composition, Empirical and Molecular Formulas. How you measure how much?. You can measure mass, or volume, or you can count pieces . We measure mass in grams. We measure volume in liters. We count pieces in MOLES. Moles.

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Chemical Quantities The Mole, % Composition, Empirical and Molecular Formulas

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Chemical quantities the mole composition empirical and molecular formulas

ChemicalQuantities

The Mole,

% Composition, Empirical and Molecular Formulas


How you measure how much

How you measure how much?

  • You can measure mass, or volume, or you can count pieces.

  • We measure mass in grams.

  • We measure volume in liters.

  • We count pieces in MOLES.


Moles

Moles

  • Defined as the number of carbon atoms in exactly

    12 grams of carbon-12.

  • 1 mole is 6.02 x 1023 particles.

  • Treat it like a very large dozen

  • 6.02 x 1023 is called Avagadro’s number.


Representative particles

Representative particles

  • The smallest pieces of a substance.

  • For a molecular compound it is a molecule.

  • For an ionic compound it is a formula unit.

  • For an element it is an atom.


Types of questions

Types of questions

  • How many oxygen atoms in the following?

    • CaCO3

    • Al2(SO4)3

  • How many ions in the following?

    • CaCl2

    • NaOH

    • Al2(SO4)3

3

12

3

2

5


Types of questions using the equality 1 mole 6 02 x 10 23

Types of questions using the equality;1 mole = 6.02 x 1023

  • How many molecules of CO2 are the in 4.56 moles of CO2?

  • 4.56 mole x 6.02x1023 mc =

    1 1 mole

  • How many moles of water is 5.87 x 1022 molecules?

  • 5.87 x 1022 mc x 1 mole =

    1 6.02x1023 mc

2.75x1024mc

0.0975 mole


Types of questions using the equality 1 mole 6 02 x 10 231

Types of questions using the equality;1 mole = 6.02 x 1023

4.44x1024 atoms

  • How many atoms of carbon are there in 1.23 moles of C6H12O6?

  • 1.23 moles x 6.02x1023 mc x 6 atoms =

    1 1 mole 1 mc

  • How many moles is 7.78 x 1024 formula units of MgCl2?

    7.78x1024 FU x 1 mole = 12.9 mole

    1 6.02x1024 FU


Measuring moles

Measuring Moles

  • Remember relative atomic mass?

  • The amu was one twelfth the mass of a carbon 12 atom.

  • Since the mole is the number of atoms in 12 grams of carbon-12,

  • the decimal number on the periodic table is also the mass of 1 mole of those atoms in grams.


Gram atomic mass

Gram Atomic Mass

  • The mass of 1 mole of an element in grams.

  • 12.01 grams of carbon has the same number of pieces as 1.008 grams of hydrogen and 55.85 grams of iron.

  • We can right this as 12.01 g C = 1 mole

  • We can count things by weighing them.


Examples

Examples

  • How much would 2.34 moles of carbon weigh?

  • 2.34 moles C x 12 g C

    1 1mole

  • How many moles of magnesium in 24.31 g of Mg?

    24.31 g Mg x 1 mole =

    1 24g Mg

= 28.08 g

1.013 mole


Chemical quantities the mole composition empirical and molecular formulas

8.60x1022 atoms

13.6 g

How many atoms of lithium in 1.00 g of Li?

1.00 g Li x 1 mole x 6.02x1023 atoms

1 7 g Li 1 mole

How much would 3.45 x 1022 atoms of U weigh?

3.45x1022 atoms U x 1 mole x 238 g U

1 6.02x1023atoms 1 mole


What about compounds

What about compounds?

  • in 1 mole of H2O molecules there are two moles of H atoms and 1 mole of O atoms

  • To find the mass of one mole of a compound

  • determine the moles of the elements they have

  • Find out how much they would weigh

  • add them up


What about compounds1

What about compounds?

  • What is the mass of one mole of CH4?

  • 1 mole of C = 12 g

  • 4 mole of H x 1 g = 4g

  • 1 mole CH4 = 12 + 4 = 16g

  • The Gram Molecular mass of CH4 is 16.05g

  • The mass of one mole of a molecular compound.


Gram formula mass

Gram Formula Mass

  • The mass of one mole of an ionic compound.

  • Calculated the same way.

  • What is the GFM of Fe2O3?

  • 2 moles of Fe x 56 g = 112 g

  • 3 moles of O x 16 g = 48 g

  • The GFM = 112 g + 48 g = 160g


Molar mass

Molar Mass

  • The generic term for the mass of one mole.

  • The same as gram molecular mass, gram formula mass, and gram atomic mass.


Examples1

Examples

  • Calculate the molar mass of the following and indicate what type it is.

  • Na2S

  • 2 (23) + 32 =

  • N2O4

  • 2(14) + 4(16) =

  • C

46 + 32 = 78g

1mole Na2S = 78g

Gram Formula Mass

28 + 64 = 92g

1 mole N2O4 = 92g

Gram Molecular Mass

12g

1 mole C = 12 g

Gram Atomic Mass


Molar mass cont

Molar Mass Cont.

Gram Formula Mass

  • Ca(NO3)2

  • 40 + 2(14) + 6(16) = 40 + 28 + 96 = 164g

  • 1 mole Ca(NO3)2 = 164g

  • C6H12O6

  • 6(12) + 12(1) + 6(16) = 70 + 12 + 96 = 180g

  • 1 mole C6H12)6 = 180g Gram Molecular Mass

  • (NH4)3PO4

  • 3(14) + 12(1) + 31 + 4(16) = 42+12+31+64 = 149g

  • 1 mole (NH4)3PO4 = 149g Gram Formula Mass


Using molar mass

Using Molar Mass

Finding moles of compounds

Counting pieces by weighing


Molar mass1

Molar Mass

  • The number of grams of 1 mole of atoms, ions, or molecules.

  • We can make conversion factors from these. To change grams of a compound to moles of a compound.


For example

For example

  • How many moles is 5.69 g of NaOH?


For example1

For example

  • How many moles is 5.69 g of NaOH?


For example2

For example

  • How many moles is 5.69 g of NaOH?

  • need to change grams to moles


For example3

For example

  • How many moles is 5.69 g of NaOH?

  • need to change grams to moles

  • for NaOH


For example4

For example

  • How many moles is 5.69 g of NaOH?

  • need to change grams to moles

  • for NaOH

  • 1mole Na = 23g 1 mol O = 16.00 g 1 mole of H = 1 g


For example5

For example

  • How many moles is 5.69 g of NaOH?

  • need to change grams to moles

  • for NaOH

  • 1mole Na = 23g 1 mol O = 16 g 1 mole of H = 1 g

  • 1 mole NaOH = 40 g


For example6

For example

  • How many moles is 5.69 g of NaOH?

  • need to change grams to moles

  • for NaOH

  • 1mole Na = 23g 1 mol O = 16 g 1 mole of H = 1 g

  • 1 mole NaOH = 40 g


For example7

For example

  • How many moles is 5.69 g of NaOH?

  • need to change grams to moles

  • for NaOH

  • 1mole Na = 23g 1 mol O = 16 g 1 mole of H = 1 g

  • 1 mole NaOH = 40 g


Examples2

Examples

  • How many moles is 4.56 g of CO2?

  • 4.56g CO2 x 1 mole

    1 44gCO2

  • How many grams is 9.87 moles of H2O?

  • 9.87 moles x 18g H2O

    1 1 mole

= 0.104 moles

= 178g


Examples3

Examples

= 2.56x1023 mc

  • How many molecules in 6.8 g of CH4?

  • 6.8g CH4 x 1 mole x 6.02x1023 mc

    1 16g CH4 1mole

  • 49 molecules of C6H12O6 weighs how much?

  • 49 mc x 1 mole x 180g C6H12O6 =

    1 6.02x1023 mc 1 mole

  • 8820 g____= 1.5x10-20 g

    6.02x1023


Gases and the mole

Gases and the Mole


Gases

Gases

  • Many of the chemicals we deal with are gases.

  • They are difficult to weigh.

  • Need to know how many moles of gas we have.

  • Two things effect the volume of a gas

  • Temperature and pressure

  • Scientists compare gases at Standard Temperature and Pressure


Standard temperature and pressure

Standard Temperature and Pressure

  • 0ºC and 1 atm pressure

  • abbreviated STP

  • At STP 1 mole of gas occupies 22.4 L

  • Called the molar volume

  • Avogadro’s Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles.


Examples4

Examples

  • What is the volume of 4.59 mole of CO2 gas at STP?

  • 4.59mole x 22.4L

    1 1mole

  • How many moles is 5.67 L of O2 at STP?

  • 5.67L x 1 mole

    1 22.4L

  • What is the volume of 8.80g of CH4 gas at STP?

  • 8.80g CH4x 1mole x 22.4L

    1 16g CH4 1mole

= 102.816 = 103L

= .523moles

= 12.32 = 12.3L


We have learned how to

We have learned how to

  • change moles to grams

  • moles to atoms

  • moles to formula units

  • moles to molecules

  • moles to liters

  • molecules to atoms

  • formula units to atoms

  • formula units to ions


Chemical quantities the mole composition empirical and molecular formulas

Mass

Periodic Table

Moles


Chemical quantities the mole composition empirical and molecular formulas

Mass

PeriodicTable

Volume

Moles


Chemical quantities the mole composition empirical and molecular formulas

Mass

Periodic Table

Volume

22.4 L

Moles


Chemical quantities the mole composition empirical and molecular formulas

Mass

Periodic Table

Volume

22.4 L

Moles

Representative

Particles


Chemical quantities the mole composition empirical and molecular formulas

Mass

Periodic Table

Volume

22.4 L

Moles

6.02 x 1023

Representative

Particles


Chemical quantities the mole composition empirical and molecular formulas

Mass

Periodic Table

Volume

22.4 L

Moles

6.02 x 1023

Representative

Particles

Atoms


Chemical quantities the mole composition empirical and molecular formulas

Mass

Periodic Table

Volume

22.4 L

Moles

6.02 x 1023

Representative

Particles

Ions

Atoms


Percent composition

Percent Composition

  • Like all percents

  • Part x 100 % whole

  • Find the mass of each component,

  • divide by the total mass.


Example

Example

  • Calculate the percent composition of each element in a compound that is 29.0 g of Ag with 4.30 g of S.

  • Ag 29.0g

  • S + 4.30g

  • 33.3g

/33.3

= .8709 x 100 = 87.09%

= .1291 x 100 = 12.91%

/33.3


Getting from the formula

Getting % from the formula

  • If we know the formula, assume you have 1 mole.

  • Then you know the pieces and the whole.


Examples5

Examples

  • Calculate the percent composition of C2H4?

  • C 2(12g)=24

  • H 4(1g) = +4

  • 28g

/28

= .8571 x 100 = 85.71%

/28

= .1429 x 100 = 14.29%


Example1

Example

/234

= .2308 x 100 = 23.08%

/234

= .15.38 x 100 = 15.38%

/234

= .6154 x 100 = 61.54%

Calculate the percent composition of

Aluminum carbonate.

Al2(CO3)3

Al 2(27g)= 54

C 3(12g)= 36

O 9(16)= 144

234g


Chemical quantities the mole composition empirical and molecular formulas

You can also calculate the mass of an element in a given amount of a compound using % composition.

•Step 1: calculate the % comp. only of the element you want to find the mass of.

•Step 2: Multiply the elements %, by the mass of the compound given.

Example: Calculate the mass of sulfur in 3.54g of H2S.

MM of H2S = H 2 (1) = 2

S 1(32) = +32

34g H2S

% S = 32/34 x 100 = 94.1% S

94.1% x 3.54g = 3.33g S


Calculate the mass of nitrogen in 25g of nh 4 2 co 3

Calculate the mass of nitrogen in 25g of (NH4)2CO3.

Calculate the mass of nitrogen in 25g of (NH4)2CO3.

N 2(14g) = 28

H 8(1g) = 8

C 1(12g) = 12

O 8(16g) = +128

176g (NH4)2CO3

%N = 28/176 x 100 = 15.91%

15.91% x 25g = 4.0g N


Calculate the mass of magnesium in 97 4g of mg oh 2

Calculate the mass of magnesium in 97.4g of Mg(OH)2.

Mg 1 (24g) = 24

O 2(16g) = 32

H 2 ( 1g) = +2

58g Mg(OH)2

%Mg = 24/58 x 100 = 41.38%

41.38% x 97.4g = 40.3g Mg


Empirical formula

Empirical Formula

From percentage to formula


The empirical formula

The Empirical Formula

  • The lowest whole number ratio of elements in a compound.

  • The molecular formula the actual ration of elements in a compound.

  • The two can be the same.

  • CH2 empirical formula

  • C2H4 molecular formula

  • C3H6 molecular formula

  • H2O both


Calculating empirical

Calculating Empirical

  • Just find the lowest whole number ratio

  • C6H12O6

  • CH4N

  • It is not just the ratio of atoms, it is also the ratio of moles of atoms.

  • In 1 mole of CO2there is 1 mole of carbon and 2 moles of oxygen.

  • In one molecule of CO2 there is 1 atom of C and 2 atoms of O.


Calculating empirical1

Calculating Empirical

  • Means we can get ratio from percent composition.

  • Assume you have a 100 g.

  • The percentages become grams.

  • Can turn grams to moles.

  • Find lowest whole number ratio by dividing by the smallest moles.


Example2

Example

  • Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.

  • Assume 100 g so

  • 38.67 g C x 1mol C = 3.223 mole C 12 g C

  • 16.22 g H x 1mol H = 16.22 mole H 1 g H

  • 45.11 g N x 1mol N = 3.222 mole N 14 g N


Example3

Example

  • The ratio is 3.223 mol C = 1 mol C 3.222 mol N 1 mol N

  • The ratio is 16.00 mol H = 5 mol H 3.222 mol N 1 mol N

  • C1H5N1


Chemical quantities the mole composition empirical and molecular formulas

A compound is 43.64 % P and 56.36 % O. What is the empirical formula?

43.64 g P x 1mol P= 1.408 mole P 31 g P

56.36 g O x 1mol O= 3.523 mole O 16 g O

The ratio is 3.523 mol O= 2.5 mol O 1.408 mol P 1 mol P

Can not have 2.5 atoms! Double

P2O5


Chemical quantities the mole composition empirical and molecular formulas

Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?

49.48 g C x 1mol C = 4.123 mole C 12 g C

5.15 g H x 1mol H = 5.15 mole H 1 g H

28.87 g N x 1mol N = 2.062 mole N 14 g N

16.49 g O x 1mol O = 1.031 mole O 16 g O


Chemical quantities the mole composition empirical and molecular formulas

The ratio is 4.123 mol C = 4 mol C 1.031 mol O 1 mol O

The ratio is 5.15 mol H = 5 mol H 1.031 mol O 1 mol O

The ratio is 2.062 mol N = 2 mol N 1.031 mol O 1 mol O

C4H5N2O1


Empirical to molecular

Empirical to molecular

  • Since the empirical formula is the lowest ratio the actual molecule would weigh more.

  • By a whole number multiple.

  • Divide the actual molar mass by the the mass of one mole of the empirical formula.

  • Caffeine has a molar mass of 194 g. what is its molecular formula?

  • C4H5N2O1 = 97g

  • 194/97 = 2

  • Molecular Formula = C8H10N4O2


Example4

Example

  • A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mas is known (from gas density) is known to be 98.96 g. What is its molecular formula?

  • 71.65 g Cl x 1mol Cl = 2.047 mole Cl35 g Cl

  • 24.27 g C x 1mol C = 2.023 mole C 12 g C

  • 4.07 g H x 1mol H = 4.07 mole H 1 g H


Chemical quantities the mole composition empirical and molecular formulas

The ratio is 2.047 mol Cl = 1 mol Cl2.023 mol C 1 mol C

The ratio is 4.07 mol H = 4 mol H 1.031 mol C 1 mol C

Empirical Formula = CClH4

12+35+4 = 51g

98.96/51 = 2

Molecular Formula = C2Cl2H8


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