Transcript Chapter 6: Atomic Masses: counting by weighing

Chapter 8: Atomic Masses:
counting by weighing
Objective: To understand atomic
mass and its experimental
determination
Atomic Mass: Atomic Mass Unit
C(s) + O2 (g)
CO2 (g)
Atomic mass unit = unit of mass
1 amu = 1.66 x 10-24
Remember C has several isotopes: 12C, 13C, and
14C.
Average atomic mass unit=12.01
So we can calculate the mass of
any of the following:
1) Calculate the mass (in amu) of a sample
of Carbon that contains 62 atoms.
2) Calculate the mass (in amu) of a sample
of iron that contains 15 atoms.
3) Calculate the mass (in amu) of a sample
of aluminum that contains 75 atoms.
Calculate the number of atoms in
• Calculate the number of copper atoms
present in a sample that has a mass of
1779.4 amu.
• Calculate the number of argon atoms
present in a sample that has a mass of
3755.3 amu.
• Calculate the number of sodium atoms
present in a sample that has a mass of
1172.49 amu.
Chapter 8.3: The Mole
• Objectives: To understand the mole
concept and Avogadro’s number.
• To learn to convert among moles, mass,
and number of atoms in a given sample.
The Mole: Revisited
• A mole (mol) is the unit a chemist uses to
describe the numbers of atoms. It is
defined as the number = to the number of
carbon atoms in 12.01 g of carbon.
• Avogadro’s number is this number=6.022
x 1023
Just like a dozen = 12 eggs…..
A mole of water = 6.022x 1023
Average atomic mass contains 1
mol of atoms.
The mole: Revisited
Sample A
Mass=1.008 g
1 mol of H atoms
Sample B
Mass=0.500 g
? Number of H atoms.
Calculating Moles and Numbers of
Atoms
• Calculate the number of moles of atoms
and the number of atoms in a 25.0g
sample of calcium. 0.624 mol Ca; 3.76 x 10 Ca atoms
• Calculate the number of moles of atoms
and the number of atoms in a 57.7 g
sample of sulfur. 1.80 mol S; 1.08 x 10 S atoms
23
24
Calculating the Number of Atoms
• Calculate the number of atoms in a 23.6
mg sample of Zinc. 2.17 x 10 Zinc atoms
• Calculate the number of atoms in a 128.3
mg sample of silver. 7.16 x 10 Ag atoms
20
20
8.4: Molar Mass
• To understand the definition of molar mass
• To learn to convert between moles and
mass of a given sample of a chemical
compound.
Mass of 1 mol of C= 1 x 12.01 g=12.01 g
Mass of 4 mol of H= 4 x 1.008 g= 4.032 g
Mass of 1 mol of CH4= 16.04 g.
Molar Mass of any substance
Is the mass ( in grams) of
1 mol of the substance.
Calculate the molar mass
Water H2O
Ammonia NH3
Propane C3H8
Glucose C6H12O6
18.02 g/mol
17.03 g/mol
44.09 g/mol
180.2 g/mol
Calculating Mass from Moles
Calculate the mass of 1.48 mol of potassium
oxide (K2O)
Calculate the mass of 4.85 mol of acetic
acid, HC2H3O2.
Calculating Moles from Mass
• Formaldehyde has the formula H2CO.
How many moles of formaldehyde does a
0.251 mol
7.55 g sample represent?
• How many moles of tetraphosphorous
decoxide does a 250.0 g sample
represent?
0.8805 mol
Calculating Number of Molecules
• How many water molecules are in a 10.0g
3.34 x 10
sample of water?
23
• Sucrose or table sugar has the formula
C12H22O11. How many molecules of sugar
are in a 5.00 lb bag of sugar? 3.99 x 10
24
8.5: Percent Composition of
Compounds
• Objective: To learn to find the mass
percent of an element in a given
compound.
Percent Composition:
• Mass fraction
for a given element = mass of the element present in 1 mol
of compound
Mass of 1 mol of compound
Mass percent = mass fraction x 100
C2H5OH
Mass of C= 2 mol x 12.01 g/mol = 24.02
Mass of H= 6 mol x 1.008 g/mol = 6.048
Mass of O= 1 mol x 16.00 g/mol= 16.00 g
Compounds containing H2O
•
The crystals of a hydrate are made of solid
substances combined chemically w/ water.
Sodium carbonate decahydrate
Na(CO3). 10 H2O
1) First find the weight of NaCO3
2) Find the weight of 10 molecules of water
3) (Mass of water/ mass of hydrate) x 100
Percent Composition:Mass Percent
1) Determine the mass percent of each
element in sulfuric acid (H2SO4)
2) Rubbing alcohol is an aqueous solution
of isopropyl alcohol. Isopropyl alcohol
has the formula C3H7OH. Determine the
mass percent of each element in
isopropyl alcohol.
8.6: Formulas for Compounds
Objective: To understand the meaning of the
empirical formula of compounds.
Empirical Formulas
Empirical formula: smallest whole number
ratio of the atoms. (simplest formula)
Molecular formula: actual formula
(CH2O)6= C6H12O6
Give the empirical formula for the following:
H2O2 hydrogen peroxide
C4H10 butane
Calculation of Empirical Formulas
• To learn to calculate empirical formulas
Calculation of Empirical Formulas
Total mass
of nickel
oxide
0.3354 g
Mass of
nickel
originally
present
-
0.2636
=
=
Mass of oxygen
that reacted
with the
nickel
0.0718 g
Calculate the amount of moles
0.2636 g Ni x 1mol Ni atoms = 0.00491
58.69 g Ni
0.0718 g O x 1 mol O atoms = 0.00449 mol O atoms
16.00 g O
Calculation of Empirical Formulas
Steps for Determining the Empirical Formula of a Compound
STEP 1: Obtain the mass of each element present (in grams).
STEP 2: Determine the number of moles of each type of atoms present.
STEP 3: Divide the # of moles of each element by the smallest # of moles to convert
the smalles # to 1. If all of the #s so obtained are integers, they are the
subscripts in the empirical formula. If one or more are not integers, go to
STEP 4: Multiply the #s you derived in step 3 by the smallest integer that will
convert all of them to whole numbers. This set of whole numbers
represents the subscripts in the empirical formula.
Calculation of Empirical Formulas
A 1.500 g sample of a compound containing only carbon and
hydrogen is found to contain 1.198 g of carbon.
CH3
Determine the empirical formula for this compound.
2) A 3.450 g sample of Nitrogen reacts w/ 1.970 g of oxygen.
Determine the empirical formula for this compound.
N2O
Calculation of Empirical Formulas
for Binary Compounds
1) When a 2.000 g sample of iron metal is heated in air,
it reacts with oxygen to achieve a final mass of 2.573 g.
Determine the empirical formula for this Iron oxide.
FeO [iron (II) oxide]
2) A 4.550 g sample of cobalt reacts with 5.475 g chlorine to
form a binary compound.
Determine the empirical formula for this compound.
CoCl2 [cobalt(II)chloride]
8.8 Calculation of Molecular
Formula
Objective: To learn to calculate the
molecular formula of a compound, given
its empirical formula and molar mass.
Molecular Formula
• Molecular formula = (empirical formula)n
Molar mass= n x empirical formula
So to solve for n= molar mass
Empirical formula mass
8.8: Molecular Formula
Example: A white powder is analyzed and found to
have an empirical formula of P2O5. The
compound has a molar mass of 283.88 g. What
is the compound molecular formula?
Empirical formula weight:
Molar Mass
2 P: 2 x 30.97 g= 61.94 g
283.88 g =2
5 O: 5 x 16.00 g= 80.00 g
141.94 g
141.94 g
MOLECULAR FORMULA is P4O10
EXAMPLES
A compound containing carbon, hydrogen,
and oxygen is found to be 40.00% carbon
and 6.700% hydrogen by mass. The
molar mass of this compound is between
115 g/mol and 125 g/mol. Determine the
molecular formula for this compound.
C4H8O4
Example #2
• Caffeine is a compound containing carbon,
hydrogen, nitrogen, oxygen. The mass
percent composition of caffeine is 49.47%
carbon, 5.191% hydrogen, 28.86%
nitrogen and 16.48% oxygen. The molar
mass of caffeine is approximately 194
g/mol. Determine the molecular formula
for caffeine.
C8H10N4O2