the mole
DESCRIPTION
The Mole. Molar mass, percent composition, empirical and molecular formulas. How many atoms are in here? What about weight?. Atomic Mass. A single atom has a very small mass. on the order of 10 -23 grams per atom 1 atom has a mass of about 0.000 000 000 000 000 000 000 01g - PowerPoint PPT PresentationTRANSCRIPT
The Mole
Molar mass, percent composition, empirical and molecular formulas
How many atoms are in here? What about weight?
.
Atomic Mass
A single atom has a very small mass. on the order of 10-23 grams per atom 1 atom has a mass of about
0.000 000 000 000 000 000 000 01g Because this mass is so small, we use a unit called
amu to describe the mass of a single atom. 1 atom of carbon has a mass of 12.011 amu 1 atom of hydrogen has a mass of 1.00794 amu 1 atom of gold has a mass of 196.967 amu
Formula Mass
Formula Mass – the mass, in amu, of one formula unit of a compound.
To calculate the formula mass of a compound; Count and record the number of each atom
in the formula. Multiply the number of atoms by the mass of
that atom from the periodic table. Add these products together.
Definitions:
Atom: The smallest particle of an element (ex: He).
Molecule: The smallest particle of a compound (ex: H2O)
Atomic Weight: The decimal number on the periodic chart (Units: amu)
Formula Mass: add up the total atomic weights of the elements in a compound (Units: amu).
Formula Mass
Calculate the formula mass of Ca
1. Ca = 1(Ca)2. Ca = 1(40.08) = 40.083. 40.08 amu4. Units = amu
Ca
Cl
Cl
Formula Mass
Calculate the formula mass of CaCl2
1. CaCl2 = 1(Ca) + 2(Cl)
2. CaCl2 = 1(40.08) + 2(35.45)= 110.981. 110.982. Units = amu
Ca
The Mole
The mass of a single atom is too small a quantity for use in lab.
We need to come up with a way to take atomic masses from the periodic table and turn them into a mass that is useable in lab.
In 1811, Amedeo Avogadro determined the number of atoms in one mole of a substance.
The Mole
The mole establishes a relationship between the atomic mass unit (on periodic table) and the gram (used in lab).
Mole (mol) - The amount of a pure substance that contains 6.02 x 1023 particles of that substance.
The mole is used to describe a huge amount of any extremely small particle. A mole of gold, a mole of salt and a mole of water each contain 6.02 x 1023 individual units.
The Mole
One mole of oxygen, one mole of salt and one mole of water.
Each sample contains one mole, 6.02 x 1023, particles.
Do you think that one mole of gold atoms is going to weigh the same as one mole of aluminum atoms? Why?
It doesn’t make much sense to weigh atoms in amu, because the units are so small……
• If you express atomic mass (amu) in GRAMS instead of amu (like we did yesterday), than you have one MOLE of atoms.
Molar Mass
A molar mass is calculated the same way that you calculated formula mass (HUP, TWO, THREE, FOUR).
The only difference is the unit that will be placed on the final answer because you are calculating the mass of one mole of a substance.
If the formula mass of CaCl2 is 110.98 amu, then the molar mass of CaCl2 is 110.98 grams.
Formula vs. Molecular Mass
The Mole
There are 6.02 x 1023 carbon atoms in 12 grams, one mole, of carbon.
There are 6.02 x 1023 gold atoms in 197 grams, one mole, of gold.
There are 6.02 x 1023 formula units of calcium chloride in 110.98 grams, one mole, of CaCl2.
So, how do we use the mole in Chemistry?
We use the mole to convert between the mass, volume and number of particles in a substance.
We can make conversions from:Moles to particles and particles to moles.Moles to grams and grams to moles.Moles to volume of a gas and volume of
a gas to moles.Any combination of the above categories!
Molar Conversions
number ofrepresentative
particles
mass
1 mol22.4L
22.4L1 mol
6.02 x 10 particles1 mol
23
6.02 x 10 particles231 mol
1 molmolar mass
molar mass1 mol
moles
volumeat STP
Converting from Moles to Particles.
How many particles of Sodium Chloride are there in 2.50 moles of Sodium Chloride (NaCl)?
2.50 mol NaCl x 6.02 x 1023part. = 1.51 x 1024
1 mole part. NaCl
Converting from Moles to Particles
How many particles of gold are there in 4.25 moles of Au?
4.25 mol Au x 6.02 x 1023 part.= 2.56x1024 part 1 mole Au
Converting from Particles to Moles.
How many moles of gold are there in 1.5 x 1024
atoms of gold?
1.5 x 1024 atoms Au x 1 mole = 2.5 moles Au
6.02 x 1023 atoms
Converting from Particles to Moles
How many moles of calcium chloride are there in 3.6 x 1024 particles of CaCl2?
3.6 x 1024 units x 1 mole CaCl2 = 6.0 mol CaCl26.02 x 1023 part.
Converting from Mass to Moles
How many moles of gold are there in 25.0 grams of gold?
25.0 g Au x 1 mole Au = 0.127 mol Au
196.97 g Au
Converting from Mass to Moles
How many moles of calcium chloride are there in 75.5 grams of CaCl2?
75.5 g CaCl2 x 1 mole CaCl2 = 0.680mol CaCl2
110.98 g CaCl2
Converting from Moles to Mass
How many grams of Carbon are there in 4.2 moles of Carbon?
4.2 mole C x 12.01 g C = 50. g C
1 mole C
Converting from Moles to Mass
How many grams of water are there in 2.75 moles of water?
The molar mass of water is: 16.00 + 1.01 + 1.01 = 18.02 g/mol
2.75 moles H2O x 18.02 g H2O = 49.6 g H2O
1 mole H2O
Volume of Gases
One mole of gas (6.02 x 1023 particles of gas) occupies a volume of 22.4 liters at standard temperature and pressure.
This is true if the gas is monatomic like Helium (He), diatomic like Oxygen (O2) or a compound like Carbon dioxide (CO2).
Converting from Volume to Moles of a gas
How many moles of oxygen are there in 10.0L of oxygen?
10.0L O2 x 1 mole O2 = 0.446 mole O2
22.4 L O2
Converting from Volume to Moles of a gas
How many moles of Carbon dioxide (CO2) are there in 2.50 L of CO2 gas?
2.50 L CO2 x 1 mole CO2 = 0.112 mole CO2
22.4 L CO2
Converting from Moles to Volume of a gas
How many liters of Helium are there in 3.6 moles of Helium?
3.6 moles He x 22.4 L He = 81 L He
1 mole He
Converting from Moles to Volume of a gas
What volume will 8.24 moles of CO2 occupy?
8.24 mol CO2 x 22.4 L CO2 = 185 L CO2
1 mole CO2
Multi-Step Conversions
If you need to convert from: Mass to particles or particles to mass, Mass to volume or volume to mass, Volume to particles or particles to volume,
then you need to perform a 2-step conversion.
Notice the mole is not present in the starting or ending quantities. You will ALWAYS use the mole when converting between these units!
Multi-Step Conversions
How many formula units of calcium chloride are there in 10.0 grams of CaCl2?
10.0 g CaCl2 x 1 mole CaCl2 x 6.02 x 1023 units =
110.98 g CaCl2 1 mole CaCl2
5.42 x 1022 formula units of CaCl2
Multi-Step Conversions
What volume would 2.25 grams of hydrogen gas (H2) occupy at standard temperature and pressure?
2.25 g H2 x 1 mole H2 x 22.4 L = 25.0 L H2
2.02 g H2 1 mole
Multi-Step Conversions
How many atoms of Argon gas are there in a typical light bulb (0.20 liters) at standard temperature and pressure?
0.20 L Ar x 1 mole x 6.02 x 1023 atoms = 5.4 x 1021
22.4 L 1 mole atoms Ar
Percent Composition
The Percent Composition – the percent by mass of each element in a compound.
To calculate %Comp:
mass of element x 100 = % by mass
mass of compound
Percent Composition
Determine the percent by mass of each element in calcium chloride (CaCl2).
Ca 1 x 40.08 = 40.08 40.08 x 100 = 36.11% Ca
110.98
Cl 2 x 35.45 = 70.90 70.90 x 100 = 63.89% Cl
110.98
40.08 + 70.90 =110.98
(molar mass)
Percent Composition
Calculate the percent by mass of each element in Calcium sulfate, CaSO4.
Ca 1 x 40.08 = 40.08 40.08 x 100 = 29.44% Ca 136.15
S 1 x 32.07 = 32.07 32.07 x 100 = 23.55% S 136.15
O 4 x 16.00 = 64.00 64.00 x 100 = 47.01% O
136.15
40.08 + 32.07 + 64.00 = 136.15 molar mass
Percent Composition
You can determine the %Comp of a substance experimentally.Measure the mass of a sampleDecompose the sample (usually by heating) to
separate the component substances.Measure the mass of the substance that
remains.Calculate %Comp as before
Empirical Formula
Empirical Formula – the smallest whole number mole ratio of elements in a compound.
The empirical formula for a class of molecules is often the same for each sample.Examples: C5H10O5 = CH2O
C6H12O6 = CH2O
C11H22O11 = CH2O
Calculating Empirical Formula
1) Assume you are working with a 100.0g sample so the mass of each element will be the same as the percent of that element (for simplicity).
2) Convert each mass in grams to moles using the molar mass of the element.
3) Find the whole number ratio of these calculated amounts by dividing each mole value by the smallest number of moles calculated.
4) Round to whole numbers and use the ratio to determine the empirical formula.
Calculating Empirical Formula
A compound was found to contain 29.6% Calcium, 23.7% Sulfur and 46.8% Oxygen. What is the empirical formula for the compound?
If you had a 100 gram sample, you would have 29.6g Ca, 23.7g S and 46.8g O.
Calculating Empirical Formula
Ca 29.6g x 1 mole = 0.739 mol Ca 40.08g
S 23.7g x 1 mole = 0.739 mole S 32.07g
O 46.8 x 1 mole = 2.93 mole O 16.00g
0.739 = 1 0.739 = 1 2.93 = 3.960.739 0.739 0.739
Round to the nearest whole numbers, 1 : 1 : 4 Therefore the empirical formula is CaSO4
Molecular Formulas
Molecular Formula – the actual number of atoms of each element in one molecule of a compound.
The ratio of molecular mass to empirical mass is the same as the ratio of molecular formula to empirical formula.
Calculating Molecular Formula
Example:Molecular formula Molecular Mass
C6H12O6 180.18
Empirical formula Empirical Mass
CH2O 30.03
Ratio C6H12O6 = 6 Ratio 180.18 = 6
CH2O 30.03
Calculating Molecular Formula
Ribose has a molar mass of 150g/mol and a chemical composition of 40.0%C, 6.67% hydrogen and 53.3% oxygen. What is the molecular formula for ribose?Begin by calculating the empirical
formula for Ribose.
Calculating the Molecular Formula for Ribose
40.0g x 1 mole = 3.33 mole C 12.01g
6.67g x 1 mole = 6.60 mole H 1.01g
53.3g x 1 mole = 3.33 mole O 16.00g
3.33 = 1 6.60 = 1.98 3.33 = 13.33 3.33 3.33
Empirical Formula for RiboseCH2O
Calculating Molecular Formula
Once you know the empirical formula, you can calculate the molecular formula.
Covalent bonds can form in many different ratios.
Use the empirical formula and the ratio of molar mass to empirical formula mass to determine the molecular formula.
Calculating the Molecular Formula for Ribose
The molecular mass of Ribose is 150 g/mol The empirical mass of Ribose is 30 g/mol
(CH2O = 12.01 + (2 x 1.01) + 16.00 = 30.03 g/mol)
150 = 5 Therefore, the Molecular
30 formula is 5 times the
Empirical Formula
C5H10O5
The Formula of a Hydrate
Hydrate – a compound that has a specific number of water molecules bound to its atoms.
Examples:
CaCl2 . 2H2O
FePO4 . 4H2O
MgSO4 . 7H2O
Determining the Formula for a Hydrate
To determine the formula of a hydrate, the sample must be heated to remove all of the water.
The mass of the anhydrous sample is subtracted from the mass of the hydrated sample to determine the mass of water removed.
The mass of the anhydrous sample and the mass of the water are used to determine the empirical formula and molecular formula as before.
Determining the Molecular Formula of a Hydrate
Cobalt (II) chloride is a hydrated salt. If 11.75g of this hydrate is heated, 9.25g
of anhydrous cobalt chloride remains. What is the formula for this hydrate?
11.75g hydrated salt- 9.25g anhydrous salt
2.50g water
Determining the Molecular Formula of a Hydrate
9.25g CoCl2 x 1 mole = 0.0712 mol CoCl2
129.83g
2.50g H2O x 1 mole = 0.139 mol H2O 18.02g
0.0712 = 1 0.139 = 1.95
0.0712 0.0712
1:2 ratio Therefore the formula for the hydrate is
CoCl2 . 2H2O