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Transcript
Chapter 7: The Mole and Chemical Composition
Notes
Section 1: Avogadro’s Number and Mole Conversions
Mole – a number
-
also the SI unit for amount of particles
-
6.02 x 1023 things, like atoms or molecules
-
An Avogadro’s number of things (6.02 x 1023 is Avogadro’s
Number)
A mole is used in chemistry, like a dozen is used for donuts. Let’s say you buy
5 dozen donuts. How many donuts have you purchased?
12 donuts
5 dozen donuts X
-------------------- =
60 donuts
1 dozen donuts
Just like the above example, you can use Avogadro’s number to convert from
moles to particles or from particles to moles. For example, if you have 0.50 moles of
carbon atoms, how many atoms is that?
6.02 x 1023 atoms Carbon
0.50 mol Carbon X --------------------------------------
= 3.01 x 1023 atoms
Carbon
1 mol Carbon
You can also convert from atoms to moles. For example, if you have 1.2 x 1024
atoms of Helium, how many moles do you have?
1 mol He
1.2 x 1024 atoms He X ------------------------------- = 2.0 mol Helium
6.02 x 1023 atoms He
Notice that the unit of mole is abbreviated (mol), we like to abbreviate whenever
possible in chemistry, even if it is only 1 letter…
You can also use this same technique to convert from moles of an element or
compound to grams of that element or compound. See…we use a balance to measure
out grams of substances, not moles. So it would be useful to be able to convert from
grams to moles and vice versa. You use the molar mass (the number located below the
symbol on the periodic table). Here is how it is done. Let’s say you have 0.50 moles of
Carbon. How many grams is that?
12.01 grams Carbon
0.50 mol Carbon X --------------------------- = 6.0 grams of Carbon
1 mol Carbon
This method only works if you label the units of the numbers you are working with. You
must label your units with these kinds of problems!
Chapter 7: The Mole and Chemical Composition
Notes
Section 2: Relative Atomic Mass and Chemical Formula
Elements that are found in nature contain a mixture of different isotopes (an
element with the same number of protons, but a different number of neutrons) of the
element. The periodic table reflects this by showing the mass number as the average
atomic mass. The average atomic mass is the weighted average of the atomic mass of
all of the different isotopes of an element.
Practice – If there are 2 isotopes of Copper, use their mass and relative abundance to
calculate the average atomic mass of copper.
Isotope Abundance
Cu-62.94
69.17%
Cu-64.93
30.83%
Calculation:
Average Atomic Mass = Sum of (Isotope x Decimal abundance)
Average Atomic Mass = (62.94 amu x 0.6917) + (64.93 amu x 0.3083)
Average Atomic Mass = (43.53 amu) + (20.02 amu) = 63.55 amu
Second Practice – If there are 2 isotopes of Gallium, use their mass and relative
abundance to calculate the average atomic mass of copper.
Isotope Abundance
Ga-68.926
60.00%
Ga-70.925
40.00%
Calculation:
Average Atomic Mass = Sum of (Isotope x Decimal abundance)
Average Atomic Mass = (68.926 amu x 0.6000) + (70.925 amu x 0.4000)
Average Atomic Mass = (41.36 amu) + (28.37 amu) = 69.73 amu
Formula Mass is defined as the molar mass of the elements in an element or
compound’s formula. It is calculated by adding up the atomic mass of all of the atoms
in a formula.
Example: Find the formula mass of Zinc chloride
Step 1 – Write the formula for zinc chloride = ZnCl2
Step 2 – Look up the atomic mass of elements = Zn (65.39 amu), Cl (35.45 amu)
Step 3 – Add up the mass of all of the elements in the formula, using the subscripts to
guide you = 65.39 amu + 2 x 35.45 amu = 126.29 amu
Second Example: Find the formula mass of Zinc sulfate
Step 1 – Write the formula for zinc sulfate = ZnSO4
Step 2 – Look up the atomic mass of elements =
Zn (65.39 amu), S (32.07 amu), O (16.00 amu)
Step 3 – Add up the mass of all of the elements in the formula, using the subscripts to
guide you =
65.39 amu + 32.07 amu + 4 x 16.00 amu = 161.46 amu
Third Example: Find the formula mass of Ammonium sulfate
Step 1 – Write the formula for ammonium sulfate = (NH4)2SO4
Step 2 – Look up the atomic mass of elements =
N (14.01 amu), H (1.01 amu), S (32.07 amu), O (16.00 amu)
Step 3 – Add up the mass of all of the elements in the formula, using the subscripts to
guide you =
2 x 14.01 amu 8 x 1.01 amu + 32.07 amu + 4 x 16.00 amu = 132.17 amu
Chapter 7: The Mole and Chemical Composition
Notes
Section 3: Formulas and Percentage Composition
When chemists analyze the composition of a substance, the result is reported as
a percentage of the elements that comprise the substance. This information is used to
calculate the empirical formula for a compound. The empirical formula is the simplest
whole number ratio of atoms in a molecule.
Example: Chemical analysis of a liquid shows that it is 60.0% Carbon, 13.4%
Hydrogen, and 26.6% Oxygen by mass. Calculate the empirical formula of the
substance.
Step 1 – Assume you have 100 grams of the sample
Step 2 – Convert from grams of each element to moles
1 mol C
60.0 g C x
------------ = 5.00 mol C
12.01 g C
1 mol H
13.4 g H x
------------ = 13.3 mol H
1.01 g H
1 mol O
26.6 g O x
------------ = 1.66 mol O
16.00 g O
Step 3 – In order to convert the ratios to whole numbers, divide the moles of each
element by the lowest value of moles from step 2.
5.00 mol C
---------------
=
3.01 mol C
1.66
1.3 mol H
------------- =
8.01 mol H
1.66
1.66 mol O
-------------- =
1.00 mol O
1.66
Step 4 – Round the ratios to the closest whole number and write the formula for the
compound
C3H8O
Molecular formulas are whole number multiples of the empirical formula.
Sometimes the empirical and molecular formula for a compound are the same, like
water for example (H2O).
Molecular Formula = n (Empirical Formula)
Where n equals some whole number
Example Problem: The empirical formula for a compound is P2O5. The molar mass of
the compound is 284 g/mol. Determine the molecular compound.
Step 1 – Calculate the formula mass of the empirical formula = 141.94 g/mol
Step 2 – Solve the Equation for n
Experimental molar mass of compound
n = ---------------------------------------------------------Molar mass of the empirical formula
284 g/mol
n = ---------------------
= 2.00 = 2
141.04 g/mol
Step 3 – Multiply the subscripts of the empirical formula by n to find the molecular
formula of the compound
P2O5  P(2x2)O(5x2) = P2O10
Calculating percent composition from empirical formula
Example: calculate the percent composition of Cu2S
Step 1 – Calculate the formula mass of each component of the compound
Cu – 2 x 63.55 g/mol = 127.10 g/mol
S – 32.07 g/mol = 32.07 g/mol
Step 2 – Add them up to find the molar mass
127.10 g/mol + 32.07 g/mol = 159.17 g/mol
Step 3 – Calculate the fraction that each component contributes to the molecule
127.10 g/mol
Mass % Cu = ------------------
x 100 = 79.85% Cu
159.17 g/mol
32.07 g/mol
Mass % S =
------------------ x 100 = 20.15% S
159.17 g/mol