math with chemical formulas honors chemistry unit 5
TRANSCRIPT
Math with Chemical Formulas
Honors Chemistry Unit 5
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.
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.
Unit Objectives, cont.Understand the mole and
Avogadro’s number.Be able to convert to/from
atoms, ions, molecules, moles and grams
What is a “Mole” (mol)?
A mole is a counting unitjust like a dozenother examples…..
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
Avogadro’s NumberConstantthe number of particles in
exactly one mole of a pure substance
6.02 X 1023
Memorize this NumberMemorize this Number
Conversion Factors – 1 mol
1 mol = 6.02 X1023 of anything
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
Molar Mass
Mass in g of 1 mole of anything
For elements, the molar mass is equal to the atomic mass
Molar Mass Examples…..
1 mol atomic wt. (g)
1 mol C 12.01 g C
1 mol Li 6.94 g Li
You try……
1 mol Ca ? g Ca
1 mol Fe ? g Fe
Now you have Two conversion Factors for a mole…..
1 mol 6.02 X 1023 atoms, ions or molecules
andand
1 mol ______(g)
Example 1How many g in 2.0 mol of He?
2.0 mol He 4.0g He = 8.0g He 1.0 mol He
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
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
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
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
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
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
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
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
You try….How many mol in 127g barium
chloride?
(set up on board)
Answer: 0.610 mol BaCl2
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.
% Composition Calculations
% comp = mass of element X 100% molar mass of cpd
= % element in the compound
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
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?
You try:barium choride
sodium phosphate
Answers:barium chloride
barium 65.90%
chloride 34.10%sodium phosphate
sodium 42.07%
phosphorus 18.89%
oxygen 39.04%
Back to Objectives
Determining Formulas
Empirical Formula = Simplest Formula
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.
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
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
Example, cont. Finally, use these whole numbers as the
number of each individual element. They are the subscripts.
Empirical Formula = BH3
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
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
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
So, if you have: multiply all by:
.25 or .75 4
.33 or .66 3
.50 2
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
You try:What is the empirical formula if we have
a sample containing 66.0% Ca and 34.0% P?
Answer: Ca3P2
You try:Find the empirical formula of a
compound with 32.38% Na; 22.65% S; and 44.99% O.
Answer: Na2SO4
Molecular FormulaMolecular Formula = Actual Formula
Example:
C2H6 CH3
molecular empirical
MF = (EF)x where X = Molecular mass Empirical mass
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
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)
Example What is the molarity of a 0.5L solution
containing 2 moles of NaCl?
M = 2moles NaCl = 4 M NaCl0.5 L
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
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
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