ch3notes_08_1.notebook
October 20, 2008
Objective: To convert from grams to moles and number of atoms.
1 mol = 6.02 x 1023 atoms or molecules = Molar mass in grams
Convert 1.55 x 109 Rh atoms to grams.
Oct 6­9:23 AM
1
ch3notes_08_1.notebook
October 20, 2008
C
Oct 16­9:52 AM
2
ch3notes_08_1.notebook
October 20, 2008
Oct 16­9:57 AM
3
ch3notes_08_1.notebook
October 20, 2008
Convert
3.45 x 1022 atoms of gold to mol
2.27 mol of Zr to grams
4.48 x 10­2 g Pb to mmol
20.5 g water to molecules
0.767 mol CO2 to grams
6.47 g lead (II) nitrate to mol
How many atoms of oxygen are there in 6.12 g of aluminum nitrate?
How many grams of oxygen are there in 0.100 mol of copper (II) sulfate pentahydrate?
Oct 7­10:41 AM
4
ch3notes_08_1.notebook
October 20, 2008
Convert
3.45 x 1022 atoms of gold to mol
2.27 mol of Zr to grams
4.48 x 10­2 g Pb to mmol
20.5 g water to molecules
0.767 mol CO2 to grams
6.47 g lead (II) nitrate to mol
How many atoms of oxygen are there in 6.12 g of aluminum nitrate?
How many grams of oxygen are there in 0.100 mol of copper (II) sulfate pentahydrate?
Oct 7­8:22 AM
5
ch3notes_08_1.notebook
October 20, 2008
Mass crucible: Mass crucible + Mg: Mass Mg:
Mass crucible and product:
Mass combustion product:
18.902 g
19.111 g
0.209 g
19.203 g
0.301 g
Oct 10­9:31 AM
6
ch3notes_08_1.notebook
October 20, 2008
Mass crucible: Mass crucible + Mg: Mass Mg:
Mass crucible and product:
Mass combustion product:
18.902 g
19.111 g
0.209 g
19.203 g
0.301 g
Oct 10­9:11 AM
7
ch3notes_08_1.notebook
October 20, 2008
Objective: To find empirical formula of a compound.
To find empirical formula
1. % to mass
2. mass to moles
3. divide by smallest
4. multiply 'till whole
Oct 10­9:31 AM
8
ch3notes_08_1.notebook
October 20, 2008
Find the empirical formula of a compound that is 2.1 %H, 65.3 % O, 32.6 %S by mass.
Oct 14­9:24 AM
9
ch3notes_08_1.notebook
October 20, 2008
Find the empirical formula of a compound that is 2.1 %H, 65.3 % O, 32.6 %S by mass.
Oct 10­9:23 AM
10
ch3notes_08_1.notebook
October 20, 2008
assume that all the C from lysine gets converted to C in CO2
assume that all the H from lysine gets converted to H in H2O
%C=1.08gC/2.175g lysine
%H = 0.212 g H/ 2.175 g lysine
%N = 0.358 g N/1.873 g lysine
Oct 15­8:07 AM
11
ch3notes_08_1.notebook
5000 mL blood
October 20, 2008
5.0 x109 eryth
2.8 x 108 hemo
65332 g hemo
1.0 mL
1 eryth
6.02 x 1023 molecules
Oct 16­9:05 AM
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ch3notes_08_1.notebook
8.00 g Ca3(PO4)2
1 mol Ca3(PO4)2
October 20, 2008
3 mol Ca2+
6.02 x1023 ions
4.22 x 1022 Ca2+ ions
8.00 g Ca3(PO4)2
310.18 g Ca3(PO4)2
1 mol Ca3(PO4)2
1 mol
1 mol Ca3(PO4)2
2 mol PO43­
6.02 x1023 ions
2.81 x 1022 PO43­ ions
310.18 g Ca3(PO4)2
1 mol Ca3(PO4)2
1 mol
Oct 16­9:13 AM
13
ch3notes_08_1.notebook
October 20, 2008
Oct 16­8:23 AM
14
ch3notes_08_1.notebook
October 20, 2008
Oct 16­8:26 AM
15
ch3notes_08_1.notebook
October 20, 2008
Oct 16­9:27 AM
16
ch3notes_08_1.notebook
October 20, 2008
Oct 16­8:46 AM
17
ch3notes_08_1.notebook
October 20, 2008
find empirical formula of a compound that is 40.1 % C, 6.6 % H and 53.3 % O by mass
Oct 10­10:48 AM
18
ch3notes_08_1.notebook
October 20, 2008
Objective­ To balance equations and to perform stoichiometric calculations.
10.0 grams of sodium react with excess water to form sodium hydroxide and hydrogen gas. How many grams of hydrogen are produced?
Na + H2O NaOH + H2
Oct 17­9:59 AM
19
ch3notes_08_1.notebook
October 20, 2008
Objective­ To balance equations and to perform stoichiometric calculations.
10.0 grams of sodium react with excess water to form sodium hydroxide and hydrogen gas. How many grams of hydrogen are produced?
Na + H2O NaOH + H2
0.439 g H2
Oct 17­8:36 AM
20
ch3notes_08_1.notebook
October 20, 2008
NaHCO3 CO2 + H2O +Na2CO3
Oct 20­9:35 AM
21
ch3notes_08_1.notebook
October 20, 2008
14.0 g Al react with 35.0 g iron (III) oxide to form aluminum oxide and iron metal.
How many grams iron is produced?
What is the limiting reactant?
How many grams of the excess reactant is left over?
Oct 20­10:05 AM
22
ch3notes_08_1.notebook
October 20, 2008
14.0 g Al react with 35.0 g iron (III) oxide to form aluminum oxide and iron metal.
2 Al + Fe2O3 Al2O3 + 2 Fe
How many grams iron is produced?
What is the limiting reactant?
How many grams of the excess reactant is left over?
Oct 20­9:49 AM
23
ch3notes_08_1.notebook
October 20, 2008
Theoetical yield is the maximum amount that can be obtained in a chemical reaction. It is the amount obtained if all of the limiting reactant is converted to product.
Actual yield is how much of the product you actually put on the balance after the reaction is done and the product is isolated.
If 22.0 g of Fe were isolated in the reaction on the previous slide, what is the yield?
Oct 20­10:23 AM
24
ch3notes_08_1.notebook
October 20, 2008
The reaction of silver nitrate and copper metal to form silver metal and copper (II) nitrate has a 78.2 % yield. If you wish to attain 13.0 g silver from this reaction, how much copper metal should you start with?
Oct 20­10:32 AM
25
ch3notes_08_1.notebook
October 20, 2008
The reaction of silver nitrate and copper metal to form silver metal and copper (II) nitrate has a 78.2 % yield. If you wish to attain 13.0 g silver from this reaction, how much copper metal should you start with?
Oct 20­10:47 AM
26