Ch 7 New CP 2013.notebook
October 24, 2013
THE MOLE!!!
Yes, math has returned to
Problem Solving
­ Techniques for solving word problems
Do you have one?
If not I have a suggestion!
Conversion Factors and Dimensional Analysis
our chemistry lives!
How many quarters are in 5.25 dollars?
13 quarters are how many dollars?
­ this technique is based on the fact that you can multiply any thing by one and not change it
­ these "ones" we are going to multiply by are called conversion factors
­ a conversion factor is two things which are equal made into a fraction.
Ex: 1 m = 100 cm therefore: 1 m = 1 and 100 cm = 1
1 m 100 cm
in dimensional analysis we use units (dimensions) which are part of the numbers to help solve (analyze) ­
the problems. We simply multiply by one (a conversion factor) any number of times until the units cancel correctly to get the desired units.
Oct 29 ­ 3:44 AM
Oct 2 ­ 8:57 AM
Broad outline of dimensional analysis:
1)
Write the starting number from the problem with units
Broad outline of dimensional analysis:
1)
Write the starting number from the problem with units
2)
2)
3)
4)
Find a conversion factor with units that cancel toward what you want (put the units down first, then add the numbers which make the fraction =1)(it may not be 1 step!)
Repeat step 2 until both the numerator and denominator are in the units you want
Then multiply across the top and bottom and divide putting your final answer in the correct units
3)
4)
Find a conversion factor with units that cancel toward what you want (put the units down first, then add the numbers which make the fraction =1)(it may not be 1 step!)
Repeat step 2 until both the numerator and denominator are in the units you want
Then multiply across the top and bottom and divide putting your final answer in the correct units
.
8373.4 ft x 1 mile = 8373.4 mile = 1.58587 mile = 1.5859 mile
(s.f. from #'s in prob)
1 5280.0 ft 5280.0
Oct 2 ­ 9:08 AM
Oct 2 ­ 9:08 AM
But what's a mole?
23 representative particles of • a mole (mol) of a substance represents 6.02 X 10
that substance.
chemical • The mole is a unit of measurement for the amount of substance or amount. It is one of the base units in the International System of Units, and has the unit symbol mol
1) One billion Pb atoms is how many mols of Pb?
1,000,000,000 Pb atoms
1 1 mol Pb
= 1.66 X 10 ­15 mol Pb
X 6.02 X 1023 atoms Pb
23
• What is means is that the mol is a word that stands for a number, like the word 23 dozen stands for the number 12. A mol of something has 6.02 X 10
representative particles of that substance.
Nov 2­12:25 PM
Nov 2­2:29 PM
1
Ch 7 New CP 2013.notebook
October 24, 2013
2) How many Hydrogen atoms are in 0.5 moles of sucrose (C12 H22 O11)
0.5 mol C12H22O11
1
x
6.02 X 1023 molecules C12H22O11
1 mol C12H22O11
22 atoms of H
x 1 molecule C
H O
Nov 2­2:29 PM
12
22
11
Nov 2­2:36 PM
EX: What is the molar mass of the following:
1)Pb
EX: What is the molar mass of the following:
1)Pb=207.2
2)NaCl
2)NaCl23+35.45=58.45
3)Water
3)WaterH2O 1+1+16=18
4)Calcium chloride
4)Calcium chloride CaCl2 40+2(35.45) =110.9
5)C12 H22 O11
6) Tin (IV) phosphate
5)C12 H22 O11 12(12)+22(1)+11(16)=342
6) Tin (IV) phosphate Sn3(PO4)4
3(118.7)+4(30.97)+16(16)=735.98
Nov 2­2:37 PM
Nov 2­2:37 PM
The molar mass of something is the mass of 1.0 mol of that stuff in grams; just like adding up the atomic masses gives us the mass of just one of them in amu’s. That makes the unit of molar mass = g/mol
(molar mass)
Oct 29 ­ 3:30 AM
Oct 29 ­ 3:30 AM
2
Ch 7 New CP 2013.notebook
October 24, 2013
Name each of the following chemical compounds and list their molar masses to the nearest g/mol:
Name each of the following chemical compounds and list their molar masses to the nearest g/mol:
1) PbSO4
1) PbSO4
lead (II) sulfate
2) N2O3
2) N2O3
dinitrogen trioxide 76 g/mol
3) CoCl2 . 4 H2O
3) CoCl2 . 4 H2O
4) B2F6
4) B2F6
diboron hexafluoride136 g/mol
5) Sn(CO3)2
5) Sn(CO3)2
tin (IV) carbonate 239 g/mol
Write the formulas of each of the following chemical compounds and list their molar masses to the nearest g/mol:
Write the formulas of each of the following chemical compounds and list their molar masses to the nearest g/mol:
6) copper (I) oxide
6) copper (I) oxide
Cu2O
143 g/mol
7) ammonium phosphate
7) ammonium phosphate
(NH4)3PO4
149 g/mol
8) vanadium (V) cyanide
8) vanadium (V) cyanide
V(CN)5
181 g/mol
9) nitrogen tribromide
9) nitrogen tribromide
NBr3
254 g/mol
10) iron (II) fluoride tetrahydrate
FeF2 . 4 H2O
166 g/mol
10) iron (II) fluoride tetrahydrate
Oct 24­7:39 AM
Review of Day 2
We now know two things:
1) There are 6.02 X 1023 particles in a mol
2) The molar mass gives us how many grams are in a mol
Oct 25­7:26 AM
So... the molar mass of tin (IV) phosphate is 735.98 g/mol.
That means 1 mol of tin (IV) phosphate has a mass of 735.98 g
...and if you have 735.98 g of tin (IV) phosphate you have 1 mol
In dimensional analysis terms:
735.98 g Sn3(PO4)4
1 mol Sn3(PO4)4
303 g/mol
cobalt (II) chloride tetrahydrate 202 g/mol
Oct 24­7:44 AM
Review of Day 2
We now know two things:
1) There are 6.02 X 1023 particles in a mol
2) The molar mass gives us how many grams are in a mol
So for water (for example):
6.02 X 1023 H2O molecules = 1 mol H2O and 1 mol H2O = 18 g H2O
Oct 25­7:26 AM
Review of Day 2
We now know two things:
1) There are 6.02 X 1023 particles in a mol
2) The molar mass gives us how many grams are in a mol
So for water (for example):
6.02 X 1023 H2O molecules = 1 mol and 1 mol H2O = 18 g H2O
both are legit conversion factors!
1 mol Sn3(PO4)4
735.98 g Sn3(PO4)4
Oct 22­10:31 AM
To that lets add one more tidbit: 1 mol of any gas at STP has a volume of 22.4 L
Oct 25­7:26 AM
3
Ch 7 New CP 2013.notebook
How many mols of water are in 47.9 g of water?
October 24, 2013
One mole of X is:
• 6.02 X 10 23 of X (in number)
• Has a mass in grams = the molar mass of X • If X is a gas at STP has a volume = 22.4 L
(L)
How many molecules of water is that?
How many molecules of water are in 47.9 g of water? (the above in one step)
• atoms
• molecules
• ions
• formula units
(grams)
Oct 29 ­ 5:31 AM
Convert the following to mols:
1) 12. 8 g of sodium acetate
Oct 25 ­ 1:47 AM
1) Convert the following to moles:
a) 1.6 L of dinitrogen pentoxide gas at STP
b) 43.8 g of copper metal
2) 8.0 L of dinitogen pentoxide gas at STP
c) 2.87 X 10 23 atoms of lead
3) 1.7 X1019 water molecules
d) 19.4 g of calcium carbonate
How many molecules and what is the mass of 150.0 mL of carbon dioxide at STP?
Nov 3­7:27 AM
2) How many carbon dioxide molecules are in 6.70 g of carbon dioxide?
3) How many L of volume would 5.6 g of nitrogen gas (N2) at STP occupy?
e) 45,235 ml of neon gas at STP
Nov 8 ­ 8:10 AM
5) How many ammonium ions are in .0000536 g of ammonium phosphate?
6) What is the mass of air (roughly 79% N2 and 21% O2 ) in a 55 gallon drum at STP? (1 gal = 3.785 L)(even though there is no “air” molecule air can be given an average molecular mass based on it’s percent composition)
4) If every cat in the world (11.2 billion of the beasts) was reduced to 1 carbon atom per cat (thus 11.2 billion C atoms) how many grams of carbon would they make?
Nov 8 ­ 8:11 AM
Nov 8 ­ 8:12 AM
4
Ch 7 New CP 2013.notebook
October 24, 2013
Percent Composition
Percent Composition
• Percent composition tells us the percent by mass of each element within an element
• Percent composition tells us the percent by mass of each element within an element
­We calculate this by: grams of the element X 100 % grams of the compound
if we’re not given grams we use 1 mol of the stuff so we can use the molar mass
­We calculate this by: grams of the element X 100 % grams of the compound
if we’re not given grams we use 1 mol of the stuff so we can use the molar mass
EX: What is the % composition of nitric acid?
1) What is nitric acid? HNO3
2) What is its molar mass? 1 + 14 + 3(16) = 63 g/mol
3) Calculate: % H = 1/63 = .0159 = 1.59 % H
% N = 14/63 = .222 = 22.2 % N
% O = [3(16)] / 63 = .762 = 76.2 % O
we could check by adding up and getting close to 100%
What is the % composition of calcium acetate?
EX: What is the % composition of nitric acid?
1) What is nitric acid? HNO3
2) What is its molar mass? 1 + 14 + 3(16) = 63 g/mol
3) Calculate: % H = 1/63 = .0159 = 1.59 % H
% N = 14/63 = .222 = 22.2 % N
% O = [3(16)] / 63 = .762 = 76.2 % O
we could check by adding up and getting close to 100%
What is the % composition of calcium acetate?
% Ca = 40/158 = .253 = 25.3%
% C = 48/158 = .304 = 30.4%
% H = 6/158 = .038 = 3.8%
% O = 64/158 = .405 = 40.5%
Nov 1 ­ 11:16 AM
• Another way we find % composition is by experiment
In an experiment 0.500 g of an acid was analyzed. It was found to be 0.010 g H, 0.163 g S, and 0.327 g O. What is the acid’s % composition?
Nov 1 ­ 11:16 AM
• Another way we find % composition is by experiment
In an experiment 0.500 g of an acid was analyzed. It was found to be 0.010 g H, 0.163 g S, and 0.327 g O. What is the acid’s % composition?
% H = .01/.5 = .02 = 2%
% S = .163/.5 = .326 = 32.6%
% O = .327/.5 = .654 = 65.4%
Nov 1 ­ 11:18 AM
Nov 1 ­ 11:18 AM
Determining Empirical and Molecular formulas
­We can use % composition data to find empirical formulas
• Before we go into the math of this we have to define these two things:
• Empirical formula: the lowest whole number ratio of the atoms in a compound
• Molecular formula: the actual number of atoms in the compound
1) assume you have 100 g of the compound
that means you would have 79.8 g of C and 20.2 g of H
EX: What is the empirical formula of a compound that is 79.8 % C, and 20.2 % H 2) convert these to moles
79.8 g of C 1 mole of C = 6.65 mol
1 X 12 g of C of C
20.2 g of H 1 mole of H = 20.2 mol
1 X 1 g of H of H
3) divide by the smallest of the mole numbers
6.65 mol of C = 1 mol of C
6.65
20.2 mol of H = 3 mol of H
6.65
therefore the EF is: CH3
Nov 1 ­ 11:18 AM
Nov 1 ­ 11:20 AM
5
Ch 7 New CP 2013.notebook
October 24, 2013
what if we don’t get whole numbers when we divide?
• if it’s close to a whole number it probably is that number (4.06 = 4)
• all the numbers may need multiplied by a whole number to become whole.
EX : C = 2.3, H = 3, N = 1 X 3 to get C 7H9N3
what if we don’t get whole numbers when we divide?
• if it’s close to a whole number it probably is that number (4.06 = 4)
• all the numbers may need multiplied by a whole number to become whole.
EX : C = 2.3, H = 3, N = 1 X 3 to get C 7H9N3
Try finding the empirical formula for the acid in the previous part
Try finding the empirical formula for the acid in the previous part
H = 2%, S = 32.6%, O = 65.4%
2 g H 1 mol H 1 1 g H 2 mol H / 1.017 = 1.96 2
32.6 g S 1 mol S 1.017 mol S / 1.017 = 1
1 32.06 g S
65.4 g O 1 mol O 4.088 mol O / 1.017 = 4.02
1 16 g O
Nov 1 ­ 11:22 AM
Finding molecular formulas:
• To find molecular formulas(MF) from empirical formulas(EF) we need another piece of information. The compounds molecular mass(MM)
• If we found a compounds EF to be CH 3 and we also know its MM to be 30
Add up the weight of the EF CH 3 = 15
Find how many times that goes into the compounds MM 30/15 = 2
Multiply the EF through by that number to get the MF CH3 X 2 = C2H6
Nov 1 ­ 11:23 AM
4
Nov 1 ­ 11:22 AM
Summary Problem:
1,6 diaminohexane is used to make nylon. Analysis of this compound finds its percent composition to be 62.1% C, 13.8% H, and 24.1% N. Its molecular mass is also found to be 116 g/mol. What are the empirical and molecular formulas of this compound?
Nov 1 ­ 11:43 AM
Summary Problem:
1,6 diaminohexane is used to make nylon. Analysis of this compound finds its percent composition to be 62.1% C, 13.8% H, and 24.1% N. Its molecular mass is also found to be 116 g/mol. What are the empirical and molecular formulas of this compound?
Nov 1 ­ 11:43 AM
Nov 2­2:54 PM
6
Attachments
solid_atoms.swf