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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
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

slide11

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
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
slide35

Mass

Periodic Table

Moles

slide36

Mass

PeriodicTable

Volume

Moles

slide37

Mass

Periodic Table

Volume

22.4 L

Moles

slide38

Mass

Periodic Table

Volume

22.4 L

Moles

Representative

Particles

slide39

Mass

Periodic Table

Volume

22.4 L

Moles

6.02 x 1023

Representative

Particles

slide40

Mass

Periodic Table

Volume

22.4 L

Moles

6.02 x 1023

Representative

Particles

Atoms

slide41

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

slide47

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
slide56

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

slide57

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

slide58

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
slide61

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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