Chapter 10 chem 2011

Document Sample
Chapter 10 chem 2011 Powered By Docstoc
					Chapter 10




    Chemical Quantities
Section 10.1 The Mole: A
Measurement of Matter

   When you measure things you can use three
    different methods.
    –   Count
    –   Mass
    –   Volume
Sample Problems 10.1
Section 10.1 Cont.

   A mole of any substance contains
    Avogadro’s number of representative
    particles, or 6.02 x 1023 representative
    particles.
   Converting number of particles to moles
    –   Moles = representative particles x 1 mole / 6.02 x 1023 representative particles
Sample problem 10.2
Section 10.1 cont.

   Converting Moles to number of particles
    –   Representative particles = moles x 6.02 x 1023 representative particles / 1 mole
Sample problem 10.3
Section 10.1 Cont.

   The atomic mass of an element expressed in grams
    is the mass of a mole of the element.
   The mass of a mole of an element is its molar mass.
    –   For Carbon the molar mass is 12.0 grams, for Hydrogen the
        molar mass is 1.0 grams.
   To calculate the molar mass of a compound, find the
    number of grams of each element in one mole of the
    compound. Then add the masses of the elements in
    the compound.
Sample problem 10.4
Homework 10.1

   Review Worksheet 10.1
   Section assessment page 296 (#9-15)
   Read 10.2
Section 10.2 Mole-Mass and Mole-
Volume relationships

   Use the Molar Mass of an element or
    compound to convert between the mass of a
    substance and the moles of a substance.
   Mass (grams) = number of moles x mass (grams)/ 1 mole

   Moles = mass (grams) x 1 mole / mass (grams)
Sample problem 10.5
The aluminum satellite dishes in Figure 10.8 are resistant to corrosion because the
aluminum reacts with oxygen in the air to form a coating of aluminum oxide (Al2O3). This
tough, resistant coating prevents any further corrosion. What is the mass of 9.45 moles
of aluminum oxide?
#16. Find the mass, in grams, of 4.52 x
10-3 mol C20H42.
#17 Calculate the mass, in grams, of
2.50 mol of Iron (II) hydroxide.
Sample problem 10.6
When iron is exposed to air, it corrodes to form red-brown rust. Rust
is Iron (III) oxide (Fe2O3). How many moles of iron (III) oxide are
contained in 92.2 grams of pure Fe2O3?
#18. Find the number of moles in 3.70
x 10-1 grams of boron.
#19. Calculate the number of moles in
75.0 grams of dinitrogen trioxide.
Section 10.2 Cont.

   Standard temperature and pressure (STP)
    means a temperature of 0oC and a pressure
    of 101.3 KPa, or 1 atmosphere (atm).
   At STP, 1 mol or 6.02 x 1023 representative
    particles, of any gas occupies a volume of
    22.4 Liters.
   The quantity, 22.4 Liters, is called the molar
    volume of a gas.
10.2 Cont.

   The molar volume is used to convert a
    known number of moles of gas to the volume
    of the gas at STP. The relationship 22.4 L =
    1 mol at STP provides the conversion factor.

   Volume of gas = moles of gas x 22.4 L / 1 mol
Sample problem 10.7
Sulfur dioxide (SO2) is a gas produced by burning coal. It is an air
pollutant and one of the causes of acid rain. Determine the volume, in
liters, of 0.60 mol SO2 gas at STP.
#20. What is the volume of these gases at STP?
A) 3.20 x 10-3 mol CO2
B) 3.70 mol N2
#21. At STP, what volume do these gases occupy?
A) 1.25 mol He
B) 0.335 mol C2H6
10.2 Cont.

   Different gases have different densities.
    Usually the density of a gas is measured in
    grams per liter (g/L) and at a specific
    temperature.
   Molar mass = density at STP x molar volume at STP
   Grams/mole = (grams/ liter) x (22.4L/1mole)
Sample problem 10.8
The density of a gaseous compound containing carbon and oxygen is
found to be 1.964 g/L at STP. What is the molar mass of the
compound?
#22. A gaseous compound composed of sulfur and oxygen,
which is linked to the formation of acid rain, has a density of
3.58 g/L at STP. What is the molar mass of this gas?
#23. What is the density of krypton
gas at STP?
Homework 10.2

   Review worksheet 4 pages
   Section Assessment page 303 (#24-31)
   Read 10.3
Section 10.3 Percent Composition and
Chemical Formulas

   The percent composition by mass of an
    element in a compound is the number of
    grams of the element divided by the mass in
    grams of the compound, multiplied by 100%.
   % mass of element = (mass of element / mass of compound) x 100%
Sample Problem 10.9
When a 13.60 gram sample of a compound containing only magnesium and
oxygen is decomposed, 5.40 grams of oxygen is obtained. What is the
percent composition of this compound?
Practice Problem #32
A compound is formed when 9.03 grams Mg combines completely with 3.48
grams N. What is the percent composition of this compound?
Practice Problem #33
When a 14.2 gram sample of mercury (II) oxide is decomposed into its
elements by heating, 13.2 grams of Hg is obtained. What is the percent
composition of the compound?
Section 10.3 Cont.

   You can also calculate the percent
    composition of a compound is you know only
    its chemical formula.
   %mass = (mass of element in 1 mol compound / molar mass of compound) x 100%
Sample Problem 10.10
Propane (C3H8), the fuel commonly used in gas grills, is one of the compounds
obtained from petroleum. Calculate the percent composition of propane?
Practice Problem #34
Calculate the percent composition of these compounds.
A) Ethane (C2H6)
B) Sodium Hydrogen Sulfate (NaHSO4)
Practice Problem #35
Calculate the percent nitrogen in these common fertilizers.
A)NH3
B)NH4NO3
Section 10.3 Cont.

   You can use percent composition to
    calculate the number of grams of any
    element in a specific mass of a compound.
   For example if you said that 81.8% of
    Propane was Carbon it would be the same
    as saying 81.8 grams of Carbon/100 grams
    of Propane.
Example
How many grams of Carbon and Hydrogen are in 82.0 grams of
Propane?



   Known
    –   82.0 grams of C3H8
    –   81.8% Carbon
    –   18% Hydrogen
Section 10.3 Cont.

   The empirical formula of a compound shows
    the smallest whole-number ration of the
    atoms in the compounds.
   The molecular formula tells the actual
    number of each kind of atom present in a
    molecule of the compound.
Sample Problem 10.11
A compound is analyzed and found to contain 25.9% nitrogen and
74.1% oxygen. What is the empirical formula of the compound?
Practice Problem#36
Calculate the empirical formula of each compound.
A)94.1% O, 5.9% H
B)67.6% Hg, 10.8% S, 21.6% O
Practice Problem #37
1,6 Diaminohexane is used to make nylon. What is the empirical formula of
this compound if it is 62.1% C, 13.8% H, and 24.1 % N?
Section 10.3 Cont.

   The molecular formula of a compound is
    either the same as its experimentally
    determined empirical formula, or it is a
    simple whole-number multiple of its empirical
    formula.
   To determine the molecular formula you
    divide the molecular mass by the empirical
    mass. Take that number and multiply the
    subscripts by that number.
Sample Problem 10.12
Calculate the molecular formula of a compound whose molar mass is 60.0g
per mole and empirical formula is CH4N.
Practice Problem #38
Find the molecular formula of ethylene glycol, which is used as antifreeze.
The molar mass is 62.0 g/mol and the empirical formula is CH3O.
Practice Problem #39
Which pair of molecules has the same empirical formula?
A)C2H4O2, C6H12O6
B)NaCrO4, Na2Cr2O7
Homework 10.3

   Section assessment page 312 (#40-46)
   Review worksheet

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:70
posted:9/21/2012
language:English
pages:54