Math with Chemical Formulas Honors Chemistry Unit 5.
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Transcript of Math with Chemical Formulas Honors Chemistry Unit 5.
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Math with Chemical Formulas
Honors Chemistry Unit 5
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Unit ObjectivesBe able to perform math
functions with and without your calculator using correct scientific notation.
Be able to find molar/molecular/formula mass using the periodic table.
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Unit Objectives, cont.Be able to calculate Molarity.Be able to calculate percent
composition.Be able to determine empirical
and molecular formulas using lab data.
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Unit Objectives, cont.Understand the mole and
Avogadro’s number.Be able to convert to/from
atoms, ions, molecules, moles and grams
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What is a “Mole” (mol)?
A mole is a counting unitjust like a dozenother examples…..
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What is a Mole, cont.?“Official Definition”
the amount of a substance that contains as many particles as there are atoms in exactly 12g of carbon-12
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Avogadro’s NumberConstantthe number of particles in
exactly one mole of a pure substance
6.02 X 1023
Memorize this NumberMemorize this Number
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Conversion Factors – 1 mol
1 mol = 6.02 X1023 of anything
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Conversion Factors – 1 mol Examples…
1 mol 6.02 X 1023 atoms
1 mol 6.02 X 1023 ions
1 mol 6.02 X 1023 molecules
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Molar Mass
Mass in g of 1 mole of anything
For elements, the molar mass is equal to the atomic mass
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Molar Mass Examples…..
1 mol atomic wt. (g)
1 mol C 12.01 g C
1 mol Li 6.94 g Li
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You try……
1 mol Ca ? g Ca
1 mol Fe ? g Fe
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Now you have Two conversion Factors for a mole…..
1 mol 6.02 X 1023 atoms, ions or molecules
andand
1 mol ______(g)
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Example 1How many g in 2.0 mol of He?
2.0 mol He 4.0g He = 8.0g He 1.0 mol He
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Oct. 16-22, 2005
Example 2How many moles in 3.01 X 1023
atoms Ag?
3.01 X1023 atoms Ag 1mol Ag =
6.02 X 1023 atoms Ag
0.500 mol Ag
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Example 3 What is the mass of 1.20 X 108 atoms of Cu?
1.20 X108 atoms Cu 1mol Cu 63.6g Cu =
6.02 X 1023 atoms Cu 1 mol Cu
1.27 X 10-14 g Cu
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You try:Convert 11.5 g B to moles B
1.06 mol BConvert 8.0 X 1019 atoms of
Ag to g
0.014 g Ag
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Formula Mass/Molecular Mass/Molar Mass
Sum of masses in a compoundMolar Mass of sodium chloride
NaClNa 1 mol X 22.99g/mol = 22.99g
Cl 1 mol X 35.45g/mol = 35.45g
Total: 58.44g
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Example 2Molar Mass of magnesium chloride
MgCl2Mg 1 mol X 24.31g/mol = 24.31g
Cl 2 mol X 35.45g/mol = 70.90g
Total: 95.21g
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Example 3 - Calcium Nitrate
Molar Mass of Ca(NO3)2
Ca 1 mol X 40.08 g/mol = 40.08g
N 2 mol X 14.01 g/mol = 28.02g
O 6 mol X 16.00 g/mol = 96.00g
Total: 164.10g
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You try:Calculate the molar mass of
sodium phosphate
Na3PO4
Na 3 mol X 22.99 g/mol = 68.97g
P 1 mol X 30.97 g/mol = 30.97g
O 4 mol X 16.00 g/mol = 64.00g
Total: 163.94g
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Conversion Factors using Formula Mass, Molecular Mass/Molar Mass
1 mol NaCl58.44g NaCl
1 mol MgCl295.21g MgCl2
1 mol Ca(NO3)2
164.1g Ca(NO3)2
can be used as a conversion factors
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You try….How many mol in 127g barium
chloride?
(set up on board)
Answer: 0.610 mol BaCl2
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Percent CompositionPercent composition is the percent by
mass of each element in a compound.
Percent composition is the same, regardless of the size of the sample.
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% Composition Calculations
% comp = mass of element X 100% molar mass of cpd
= % element in the compound
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ExamplesFind the % composition of Cu2SFirst, find the molar mass:
2 mol Cu = 2 X 63.55g/mol = 127.1g Cu
1 mol S = 1 X 32.06g/mol = 32.06g S
Total: 159.15g/mol
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Example, cont. Next, find the % of each elementFor Cu:For Cu:% Cu = 127.1g X 100% = 79.86% Cu
159.15gFor S:For S:% S = 32.06g X 100% = 20.14% S
159.15gNext, check your work – do the %s add up to 100?
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You try:barium choride
sodium phosphate
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Answers:barium chloride
barium 65.90%
chloride 34.10%sodium phosphate
sodium 42.07%
phosphorus 18.89%
oxygen 39.04%
Back to Objectives
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Determining Formulas
Empirical Formula = Simplest Formula
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To find the empirical formula from data:
1. Assume 100% sample; change % to grams for each element
2. Find moles from the grams of each element
3. Find the smallest whole # ratio by dividing by the smallest number of moles
4. If necessary, multiply to get rid of fractions.
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Example A compound is 78% B and 22% H. What is
the empirical formula?
FirstFirst, change % to grams and find moles:
78g B 1mol B = 7.22 mol10.81g B
22g H 1mol H = 21.78 mol1.01g H
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Example, cont.NextNext, divide all mole numbers by the smallest
number of moles:
B: 7.22 mol = 17.22 mol
H: 21.78 mol = 3.02 = 37.22 mol
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Example, cont. Finally, use these whole numbers as the
number of each individual element. They are the subscripts.
Empirical Formula = BH3
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Example 2 Analysis shows a compound to contain
26.56% K, 35.41% Cr, and 38.03% O. Find the empirical formula of this compound:
FirstFirst (always!) assume 100g sample, convert % to g and then find moles of each element
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Example 2 cont. Next,Next, Conversion to moles:
26.56g K 1mol K = 0.6793 mol K39.10g K
35.41g Cr 1mol Cr = 0.6810 mol Cr52.00g Cr
38.03g O 1mol O = 2.377 mol O16.00g O
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Next,Next, divide all numbers by the smallest whole number to find the smallest whole number ratios:
0.6793 mol K = 1.00 mol K0.6793
0.6810 mol Cr = 1.003 mol Cr ~ 1.00 mol Cr0.6793
2.377 mol O = 3.499 mol O – can’t be rounded
0.6793
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So, if you have: multiply all by:
.25 or .75 4
.33 or .66 3
.50 2
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For our example:
2 X 1.00 mol K = 2 mol K
2 X 1.00 mol Cr = 2 mol Cr
2 X 3.499 mol O= 7 mol O
Empirical Formula = K2Cr2O7
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You try:What is the empirical formula if we have
a sample containing 66.0% Ca and 34.0% P?
Answer: Ca3P2
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You try:Find the empirical formula of a
compound with 32.38% Na; 22.65% S; and 44.99% O.
Answer: Na2SO4
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Molecular FormulaMolecular Formula = Actual Formula
Example:
C2H6 CH3
molecular empirical
MF = (EF)x where X = Molecular mass Empirical mass
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Example The empirical formula of a compound was
found to be P2O5. Experimentation shows that the molar mass of this compound is 283.89 g/mol. What is the compound’s molecular formula?
Back to Objectives
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Moles in Solution Molarity is the term used for moles dissolved
in solution Symbol for Molarity = M Definition – moles of solute per liter of
solution Formula
M = moles solute (mol)liter solution (L)
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Example What is the molarity of a 0.5L solution
containing 2 moles of NaCl?
M = 2moles NaCl = 4 M NaCl0.5 L
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Example 2 What is the molarity of a 250 mL solution
containing 12.7 g of lithium bromide?
M = moles = 12.7g LiBr 1mol LiBr 1000mL
L 86.8g LiBr 250mL 1L
= 0.59 M LiBr
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Example 3 How would you make 500mL of a 0.32M
solution of LiBr and water? Given: M=0.32M, L=0.500 M= #moles/liters .32M= #moles .500L #moles= 0.16 molesLiBr
0.16 molesLiBr X 86.841g = 13.894gLiBr in .500L of Water 1 mole
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You try: Calculate the M of a 700. mL solution of
23.2g calcium chloride
How would you make a 0.2 L solution of 0.50 M CaCl2 solution?
Back to Objectives