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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ChemicalQuantities

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

  • 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

  • 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

  • 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 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 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

  • 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

  • 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

  • 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


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?

  • 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 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

  • 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

  • The generic term for the mass of one mole.

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


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.

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

Finding moles of compounds

Counting pieces by weighing


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

  • How many moles is 5.69 g of NaOH?


For example

  • How many moles is 5.69 g of NaOH?


For example

  • How many moles is 5.69 g of NaOH?

  • need to change grams to moles


For example

  • How many moles is 5.69 g of NaOH?

  • need to change grams to moles

  • for NaOH


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 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 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 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


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


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

  • 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

  • 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.


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

  • 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


Mass

Periodic Table

Moles


Mass

PeriodicTable

Volume

Moles


Mass

Periodic Table

Volume

22.4 L

Moles


Mass

Periodic Table

Volume

22.4 L

Moles

Representative

Particles


Mass

Periodic Table

Volume

22.4 L

Moles

6.02 x 1023

Representative

Particles


Mass

Periodic Table

Volume

22.4 L

Moles

6.02 x 1023

Representative

Particles

Atoms


Mass

Periodic Table

Volume

22.4 L

Moles

6.02 x 1023

Representative

Particles

Ions

Atoms


Percent Composition

  • Like all percents

  • Part x 100 % whole

  • Find the mass of each component,

  • divide by the total mass.


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

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

  • Then you know the pieces and the whole.


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%


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


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 (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.

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

From percentage to 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

  • 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 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.


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


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


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


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


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

  • 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


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


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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