Title: Chapter 8: Atomic Masses: counting by weighing
1Chapter 8 Atomic Masses counting by weighing
- Objective To understand atomic mass and its
experimental determination
2Atomic 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 unit12.01
3So we can calculate the mass of any of the
following
- Calculate the mass (in amu) of a sample of Carbon
that contains 62 atoms. - Calculate the mass (in amu) of a sample of iron
that contains 15 atoms. - Calculate the mass (in amu) of a sample of
aluminum that contains 75 atoms.
4Calculate 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.
5Chapter 8.3 The Mole
- Objectives To understand the mole concept and
Avogadros number. - To learn to convert among moles, mass, and number
of atoms in a given sample.
6The 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. - Avogadros number is this number6.022 x 1023
Average atomic mass contains 1 mol of atoms.
Just like a dozen 12 eggs.. A mole of water
6.022x 1023
7The mole Revisited
Sample A Mass1.008 g
Sample B Mass0.500 g
? Number of H atoms.
1 mol of H atoms
8Calculating Moles and Numbers of Atoms
- Calculate the number of moles of atoms and the
number of atoms in a 25.0g sample of calcium. - Calculate the number of moles of atoms and the
number of atoms in a 57.7 g sample of sulfur.
0.624 mol Ca 3.76 x 1023 Ca atoms
1.80 mol S 1.08 x 1024 S atoms
9Calculating the Number of Atoms
- Calculate the number of atoms in a 23.6 mg sample
of Zinc. - Calculate the number of atoms in a 128.3 mg
sample of silver.
2.17 x 1020 Zinc atoms
7.16 x 1020 Ag atoms
108.4 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.
11Mass of 1 mol of C 1 x 12.01 g12.01 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.
12Calculate 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
13Calculating 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.
14Calculating Moles from Mass
- Formaldehyde has the formula H2CO. How many
moles of formaldehyde does a 7.55 g sample
represent? - How many moles of tetraphosphorous decoxide does
a 250.0 g sample represent?
0.251 mol
0.8805 mol
15Calculating Number of Molecules
- How many water molecules are in a 10.0g sample of
water? - Sucrose or table sugar has the formula C12H22O11.
How many molecules of sugar are in a 5.00 lb bag
of sugar?
3.34 x 1023
3.99 x 1024
168.5 Percent Composition of Compounds
- Objective To learn to find the mass percent of
an element in a given compound.
17Percent 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
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
C2H5OH
18Compounds containing H2O
- The crystals of a hydrate are made of solid
substances combined chemically w/ water. - Sodium carbonate decahydrate
- Na(CO3). 10 H2O
- First find the weight of NaCO3
- Find the weight of 10 molecules of water
- (Mass of water/ mass of hydrate) x 100
19Percent CompositionMass Percent
- Determine the mass percent of each element in
sulfuric acid (H2SO4) - 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.
208.6 Formulas for Compounds
- Objective To understand the meaning of the
empirical formula of compounds.
21Empirical 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
22Calculation of Empirical Formulas
- To learn to calculate empirical formulas
23Calculation of Empirical Formulas
Total mass of nickel oxide
Mass of nickel originally present
Mass of oxygen that reacted with the nickel
0.3354 g - 0.2636
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
24Calculation 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.
25Calculation 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. 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.
CH3 N2O
26Calculation of Empirical Formulas for Binary
Compounds
- 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. - 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.
FeO iron (II) oxide CoCl2
cobalt(II)chloride
278.8 Calculation of Molecular Formula
- Objective To learn to calculate the molecular
formula of a compound, given its empirical
formula and molar mass.
28Molecular Formula
- Molecular formula (empirical formula)n
- Molar mass n x empirical formula
- So to solve for n molar mass
- Empirical formula
mass
298.8 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
30EXAMPLES
- 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
31Example 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