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CHAPTER 8: CHEMICAL COMPOSITION
Active Learning: 1-4, 6-8, 12, 18-25;
End-of-Chapter Problems: 3-4, 9-82, 84-85, 87-92, 94-104, 107-109, 111, 113, 119, 125-126
8.2 ATOMIC MASSES: COUNTING ATOMS BY WEIGHING
Atoms are too small to weigh directly
– e.g. one carbon atom has a mass of 1.99×10-23 g—too inconvenient an amount to use!
→ need more convenient unit for an atom’s mass
→ atomic mass unit (amu)
Carbon-12 was chosen as the reference and given a mass value of 12 amu
→ 1 amu = 1/12th the mass of a carbon-12 atom
→ Masses for all other elements are measured relative to mass of a carbon-12 atom
Weighted Average Atomic Mass of an Element
– Why is carbon’s mass on Periodic Table 12.01 amu, NOT 12.00 amu?!
– Atomic masses reported on the Periodic Table are weighted averages of the masses of all
the naturally occurring isotopes for each element based on their percent natural
abundance—i.e., the percentage of existing atoms that are a specific naturally occurring
isotopes.
Ex. 1 If 98.892% of carbon exists as carbon-12, which has a mass of 12.00000, while
1.108% exists as carbon-13, which has a mass of 13.00335, calculate the
average atomic mass for carbon.
average atomic mass =
(0.98892)(12.00000 amu)
+ (0.01108)(13.00335 amu)
Ex. 2 The atomic masses of the three naturally occurring isotopes or argon, Ar-36
(0.3365%), Ar-38 (0.0632%) and Ar-40 (99.6003%), are 35.96754552 amu,
37.9627325 amu, and 39.9623837 amu, respectively. Calculate the average
atomic mass for argon.
CHEM 139: Zumdahl Chapter 8 page 1 of 10 8.3 THE MOLE
Atomic Masses and Molar Masses
If 98.93% of C atoms are carbon-12 and 1.07% of C atoms are carbon-13,
then the (weighted average) atomic mass of carbon is calculated as follows:
(12.000000 amu)(0.9893) + (13.003354 amu)(0.0107) = 12.01 amu
So how many carbon atoms are present in 12.01 grams of carbon?
This number was determined experimentally to be 6.022×1023.
– It was named Avogadro’s number, to honor Italian scientist Amedeo Avogadro who first
proposed in 1811 that the volume of a gas depended only on the number (not type) of
atoms or molecules present at a given temperature and pressure
AVOGADRO'S NUMBER (NA)= 6.022×1023 (4 sig figs)
How big is this?
•
If 6.022×1023 hydrogen atoms were laid side by
side, the total length would encircle the
Earth about a million times.
•
The mass of 6.022×1023 Olympic shotput balls is
about equal to the mass of the Earth.
•
CHEM 139: Zumdahl Chapter 8 The volume of 6.022×1023 softballs is
about equal to the volume of the Earth.
page 2 of 10 THE MOLE CONCEPT
1 mole (abbreviated mol) = 6.022×1023 items
Similar to: 1 dozen = 12 items
1 dozen doughnuts = 12 doughnuts
1 mole of doughnuts = 6.022×1023 doughnuts
MOLE CALCULATIONS I
a. How many eggs are in 3 dozen eggs?
____________
b. How many eggs are in 3.00 moles of eggs?
c. How many moles of C atoms are present in a sample of 1.25×1024 C atoms?
8.5 MOLAR MASS
Thus, the mass of 1 mole of C atoms is 12.01 g (or 12.01 g/mol)
→ 1 mole (6.022×1023) is the amount of atoms of any element that has a mass in grams
equal to the mass of ONE atom in amu.
→ The atomic masses reported for each element in the Periodic Table give the average
atomic mass in amu and the molar mass in g/mol.
Molar mass (MM): Mass in grams of 1 mole of any element or compound
Example: Use the Periodic Table to provide the molar mass of each element below:
a. Al: ___________
CHEM 139: Zumdahl Chapter 8 b. Si: ___________
c. Ar: ___________
d. Sn: ___________
page 3 of 10 Molar mass (MM): Mass in grams of 1 mole of any element or compound
– To obtain the molar mass of a compound, multiply the molar mass of each element by the
number of each present, then add them all up.
Example: Determine the molar mass of each of the following compounds:
a. O2: 2 (molar mass of O) = 2 (16.00 g/mol) = 32.00 g/mol
b. NaCl:
c. CO2:
d. H2SO4:
e. Ca3(PO4)2:
Do not worry about rounding molar mass to the correct number of sig figs! Molar
mass will rarely be the measurement that limits the number of sig figs for a calculation.
CONVERSIONS AMONG MASS, NUMBER OF MOLES, AND NUMBER OF PARTICLES
Use the dimensional/unit analysis method:
1. Write the units of the final answer.
2. Write the given information related to the answer.
3. Determine the unit factors (Avogadro’s #, molar masses) necessary to solve the problem
using the given information
Ex. 1:
How many moles of helium are in 5.00 g of helium?
CHEM 139: Zumdahl Chapter 8 page 4 of 10 Ex. 2:
How many moles of neon are in 25.0 g of neon?
Ex. 3:
How many neon atoms are in 25.0 g of neon?
Ex. 4:
How many moles of H2O are in 50.0 g of H2O?
Ex. 5:
How many H2O molecules are in 50.0 g of H2O?
Ex. 6:
How many H atoms are in 50.0 g of H2O?
CHEM 139: Zumdahl Chapter 8 page 5 of 10 MOLAR VOLUME: The volume occupied by 1 mole of any gas
Avogadro's Law: At the same temperature and pressure, equal volumes of gases contain the
same number of molecules.
Standard temperature and pressure
(STP): T=0˚C and P=1.00 atm
At STP, 1 mole of gas occupies 22.4 L!
3 sig figs
Write 2 unit factors using this information:
Mole Calculations with Molar Volume:
Ex. 1
What mass of CO2 occupies a volume of 25.0 L at STP?
Ex. 2
What is the volume occupied by 10.0 g of propane gas (C3H8) at STP?
Ex. 3
How many He atoms are present in a 10.0 L balloon filled with He gas at STP?
CHEM 139: Zumdahl Chapter 8 page 6 of 10 8.6 PERCENT COMPOSITION OF COMPOUNDS
Mass Percent Composition (or Mass Percent): The mass of each element in a compound
divided by the mass of the entire compound.
Mass Percent of Element =
Mass of Element
× 100%
Total Mass of Compound
Steps to determine percentage composition:
1. Calculate the total mass of each individual element in the compound
2. Add up all the masses of each element to get the total mass of the compound
3. Divide the mass of each individual element with the total mass of compound
Ex. 1
What is the percent composition by mass of each element in H2O?
____________% H
____________% O
Ex. 2
What is the percent composition by mass of each element in K2SO4?
____________% K
____________% S
____________% O
Ex. 3
What is the percent composition by mass of each element in Al2(CO3)3?
____________% Al
____________% C
____________% O
CHEM 139: Zumdahl Chapter 8 page 7 of 10 Ex. 4:
Calculate the mass of iron in 50.0 g of rust, iron(III) oxide.
Ex. 5:
Calculate the mass of silver present in 125 g of silver carbonate.
Ex. 6
What mass of nitrogen is present in 75.0 g of ammonium nitrate?
8.7 FORMULAS OF COMPOUNDS
Empirical Formula: Simplest whole-number ratio of atoms in a compound
– where the term “empirical” means “derived from experiment”
Molecular Formula:
Chemical formula of a compound that expresses the actual number of
atoms present in one molecule.
– The molecular formula will either be exactly the same or some
multiple of the empirical formula!
Fill in the table below:
glucose, C6H12O6
caffeine, C6H10N4O2
acrylonitrile, C3H3N
Empirical Formula
Molecular Formula
CHEM 139: Zumdahl Chapter 8 page 8 of 10 8.8 CALCULATION OF EMPIRICAL FORMULAS
Guidelines for Determining the Empirical Formula of a Compound
1. Find the # of moles of each element in the compound.
2. Divide each # of moles of each element by smallest # of moles to get ratio of atoms.
3. Get a whole number ratio for all atoms in the compound:
– If within 0.1 of a whole number, round to that whole number
– If any ratio ends close to 0.5 → multiply ALL subscripts by 2
– If any ratio ends close to 0.33 or 0.66 → multiply ALL subscripts by 3
Ex. 1: Determine the empirical formula for nickel oxide if a sample of nickel oxide consists of
17.74 g of nickel and 7.26 g of oxygen?
empirical formula: ______________
name: __________________________
Ex. 2: Ascorbic acid, known more commonly as Vitamin C, is a hydrocarbon derivative, a
compound that consists of carbon, hydrogen, and oxygen. If a 25.00 g sample of
ascorbic acid contains 10.23 g of carbon and 1.14 g of hydrogen, determine the
empirical formula for ascorbic acid.
Empirical Formula of ascorbic acid: ______________
CHEM 139: Zumdahl Chapter 8 page 9 of 10 8.9 CALCULATION OF MOLECULAR FORMULAS
Determining the Molecular Formula of a Compound
1. For the Molecular Formula, you will be given the molar mass of the compound, and you need
to calculate the molar mass of the empirical formula.
2. Divide the molar mass of the compound by the molar mass of the empirical formula to get the
factor by which to multiply each subscript in the empirical formula.
Ex. 3: If the ascorbic acid in Ex. 2 above has a molar mass of 176.1 g/mol, determine its
molecular formula.
Empirical and Molecular Formulas from Percent Composition:
1. Assume 100.0 g of the compound is present, so change percent units to grams.
2. Follow the same steps for determining empirical formula and molecular formula.
Ex. 4: Quinine is used as an antimalarial drug. An analysis of quinine indicates the compound
consists of 74.03% carbon, 7.47% hydrogen, 8.64% nitrogen, and 9.86% oxygen. If
quinine's molar mass is 325 g/mol, determine the empirical and molecular formulas for
quinine.
Empirical Formula of Quinine: __________________
Molecular Formula of Quinine: __________________
CHEM 139: Zumdahl Chapter 8 page 10 of 10