Unit III Chemical Composition (i.e. The Mole) Atomic Masses Atomic masses use Carbon 12 ( 12 C) as...
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Transcript of Unit III Chemical Composition (i.e. The Mole) Atomic Masses Atomic masses use Carbon 12 ( 12 C) as...
Unit IIIChemical
Composition(i.e. The Mole)
Atomic MassesAtomic masses use Carbon 12 (12C) as the standardCalculated with the aide of a mass
spectrometer
This process is also known as gas chromatography
As gas chromatograph counts the number of particles present in a given sample
Produces a graph like the one aboveWith this information percentages can be
calculated and formulas can be determined
Atomic mass is the average of all the naturally occurring isotopes of that elementCarbon = 12.011
Isotope Symbol Composition of the nucleus
% in nature
Carbon-12
12C 6 protons
6 neutrons
98.89%
Carbon-13
13C 6 protons
7 neutrons
1.11%
Carbon-14
14C 6 protons
8 neutrons
<0.01%
The MoleAbbrev. mol1 mol = # C atoms in 12 g of pure
12C
Avogadro’s numberEqual to 6.022 x 1023 atoms in
1 mol CNamed in honor of the Italian
chemist Amadeo Avogradro (1776-1855)
I didn’t discover it. Its just named
after me!
1 mol = 6.022 x 1023 atoms = molar mass (g) = 22.4 L
To covert amongst mol, atoms, grams, & liters, use the following equivalencies:1 mol = 6.022 x 1023 = molar mass = 22.4 L
atoms (grams)Or use ½ of the following chart:
Keys to Use:1. Only use 1 side of the chart
2. Proceed from one shaded box to another shaded box
3. When following the arrows, perform the indicated function
4. When going against the arrows, perform the opposite function
Molar MassThe molar mass is determined by summing the masses of the component atomsExample: What is the molar mass of MgCO3
24.31 g + 12.01 g + 3(16.00 g) =24.31 g + 12.01 g + 3(16.00 g) =84.32 84.32 gg
Molar CalculationsExample 1: How many grams of lithium are in 3.50 mol of lithium?3.50 mol Li → g Li
(3.50 • 6.94) / 1 = 24.29(3 SF’s in original problem) → Round24.3 → SSN → 2.43 x 101
Don’t forget the units → g Li
3.50 mol Li
1 mol Li
6.94 g Li= 2.43 x 101 g Li
Example 2: How many mol are in 98.2 g of NaCl?98.2 g NaCl → mol NaCl
(98.2 • 1) / 58.44 = 1.680355921(3 SF’s in original problem) → Round1.68 → SSN → 1.68 x 100
Don’t forget the units → mol NaCl
98.2 g NaCl
58.44 g NaCl
1 mol NaCl= 1.68 x 100 mol
NaCl
Example 3: How many atoms are in 411 g of calcium phosphate?411 g Ca3(PO4)2 → atoms Ca3(PO4)2
(411 • 6.022 x 1023) / 310.18 = 7.9793 x 1023
(3 SF’s in original problem) → Round7.98 x 1023 → SSN → 7.98 x 1023
Don’t forget the units → atoms Ca3(PO4)2
411 g Ca3(PO4)2
310.18 g Ca3(PO4)2
6.022 x 1023 atoms Ca3(PO4)2 =7.98 x 1023 atoms
Ca3(PO4)2
Example 4: How many liters of dioxygen heptioide gas does 7.5 x1024 atoms occupy?7.5 x 1024 atoms O2I7 → L O2I7
(7.5 x 1024 • 22.4) / 6.022 x 1023 = 278.977084(2 SF’s in original problem) → Round280 → SSN → 2.8 x 102
Don’t forget the units → L I2O7
7.5 x 1024 atoms O2I7
6.022 x 1023 atoms O2I7
22.4 L I2O7
= 2.8 x 102 L O2I7
Percent Composition of Compounds
Mass Percent = (Part / Whole) 100%Calculate the percent composition of
magnesium carbonate (MgCO3)Molar Mass = 84.32 g
24.31 g + 12.01 g + 3 (16.00)
Mg = (24.31 / 84.32) 100 = 28.83 %
C = (12.01 / 84.32) 100 = 14.24 %
O = (48.00 / 84.32) 100 = 56.93 %
100 %
Determining the Formula of a Compound
There are 2 basic formulas:1. Empirical
Simple – Can not be simplifiedExamples: H20, MgCl2
2. MolecularComplex – Can be simplifiedExamples: H8O4, Mg3Cl6Molecular Formula = (Empirical Formula )n
Example: (H20)4 = H8O4
n = molar mass / Empirical Formula Mass
Note: % are grams (if based on 100)Adipic acid contains 49.32% C, 6.85% H, &
43.84% O by mass. What is the E.F.?
C: (49.32/12.01)= 4.10 / 2.74 = 1.50 2 = 3
H: (6.85 /1.01) = 6.78 / 2.74 = 2.47 2 = 5
O: (43.84 /16.00)= 2.74 / 2.74 = 1 2 = 2
Answer:
C3H5O2
Moles Divide by smallest #
of mol
Mole Ratio X by
integer to obtain a whole #
# of atoms
Empirical Formula Determination
Empirical Formula Determination
Sample ProblemsSample Problem 1:.6884 g of lead combined with .2356 g of Cl
to form a binary compound. Calculate the empirical formula.
Sample Problem 2:A compound’s percent by mass is as follows:
Copper = 33.88%, Nitrogen = 14.94%, and Oxygen = 51.18%. Determine the empirical formula of the compound.
Problem:The E.F. for adipic acid is C3H5O2
The molar mass is 146 g/ molSteps:
1. Determine the E.F. mass of C3H5O2
(3 12.01) + (5 1.01) + (2 16.00) = 73.08 g
2. Find the n factorn = molar mass / E.F. massn = 146 / 73.08 = 2
3. M.F. = (E.F.)n = (C3H5O2)2
= C6H10O4
Molecular Formula Determination
Molecular Formula Determination
Sample ProblemsSample Problem 1:Calculate the M.F. of a compound with the
E.F. CH2O and a molar mass of 150 g/mol.
Sample Problem 2:A gas is composed of 71.75% Cl, 24.27% C,
and 4.07% H (by mass) & its molar mass is 98.96 g/mol. What is the M.F.?
Formulas of HydratesA hydrate is a compound that has a specific number of water molecules bound to its atomsWritten with each formula unit following a dotExamples:
1. Calcium chloride dihydrate → CaCl2 • 2 H2O
2. Magnesium sulfate heptahydrate→ MgSO4 • 7 H2O
Samples:1. Sodium cabonate decahydrate (molar mass ?)
2. Barium hydroxide octahydrate (molar mass ?)