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CHEMICAL COMPOUNDS AND THE MOLE
Chapter 7
Formula Mass
Mass of H2O? H 2(1.01)
+ O 16.00_
18.02 amu Formula Mass: mass of molecule, formula unit,
or ion is sum of masses of all atoms represented (amu)
Ca(NO3)2
Ca 40.08
N 2(14.01)
+ O 6 (16.00)
164.10 amu
Molar Mass
Definition: mass of 1 mole of compound – use molar masses of elements (g/mol)
MgCl224.31 g/mol + 2(35.45 g/mol) = 95.21 g/mol
(NH4)2CrO4
2(14.01 g/mol) + 8(1.01 g/mol) + 52.00 g/mol + 4(16.00 g/mol) = 152.10 g/mol
CuSO4 * 5H2O63.55 g/mol + 32.07 g/mol + 4(16.00 g/mol) + 5(18.02 g/mol) = 249.72 g/mol
Molar Mass in Conversions
Remember flow chart from chapter 3? What is mass (g) of 3.04 mol of ammonia
vapor, NH3? ? g = 3.04 mol X 17.04 g = 51.8 grams NH3
1 mol
How many moles of sodium chloride are present in 100.0 grams?
? Moles NaCl = 100.0 g X 1 mol = 1.711 mol NaCl
58.44 g
PRACTICE MOLAR MASS
Work as a group of 4. 1st group member: 1,5,9,13,17,21 GroupII:1 2nd group member: 2,6,10,14,18,22 Group
II:2 3rd group member: 3,7,11,15,19,23 Group II:3 4th group member:4,8,12,16,20,24 Group II:4
Show work for each problem you complete.Explain your work to the other group membersand write in their answers.HW: complete your set and 33-40.
“Super Mole” Conversions
How many molecules are in 4.15 x 10-5g C6H12O6?? Molec.= 4.15 x 10-5 g X 1 mol X 6.02 x 1023 molec. = 1.39 x 1017 molecules
180.18 g 1 mole
How many H atoms are in 7.1 moles of C6H12O6?
? Atoms = 7.1 mol X 12 mol H X NA = 5.1 x 1025 atoms H
1 mol C6H
12O
6 1 mol H
How many formula units are in 4.5 kg Ca(OH)2?
?f.un = 4.5 kg X 103 g X 1 mol X NA X 1 f. un = 3.7 x 1025 formula units
1 kg 74.10 g 1 mol 1 molecule
**NA = 6.02x1023
molecules
More “Super Mole” Conversions What is the mass of H2SO4, if you have 1.53 x
1023 sulfate ions your compound? 1.53x1023 ions x 1 mol SO4 x 1 mol H2SO4 x 98.09 g = 24.9 g
NA 1mol SO4 1 mol
How many water molecules are present in in a 5.00 g sample of copper (II) sulfate pentahydrate? 5.00 g CuSO4 * 5H2O x 1 mol x 5mol H2O x NA
249.72 g 1mol CuSO4 * 5H2O 1 mol H2O
= 6.03x1022 molecules H2O
Percent Composition
The percent by mass of each element in a compound.
% = mass due to 1 element x 100 mass of whole compound
Percent Composition
What is the percent composition by mass of each element in (NH4)2O? [MM (NH4)2O = 52.10 g/mol]
%N = 2(14.01) x 100 = 53.78% 52.10
% H = 8(1.01) x 100 = 15.5%52.10
% O = 16.00 x 100 = 30.71% 52.10 What percentage by mass of Al2(SO4)36H2O is water?
% H2O = 6(18.02) x 100 = 24.01%450.29
Given a 25.0 gram sample of aluminum sulfate hexahydrate, how much water (g) could be driven off?
g H2O= (%H2O) (total sample mass)(24.01%) (25.00 g) = 6.00g
Empirical Formula
Definition: formula showing smallest whole-number mole ratio of atoms in a compound
Ex: B2H6 Molecular Formula
BH3 Empirical Formula Given (CH2O)x as the empirical, determine
possible molecular formulas.x=2 x=1 x=6
C2H4O2 CH2O C6H12O6
Formaldehyde acetic acid glucose
(all have different molar masses)
Empirical Formula Calculation
Given the molecular formula: reduce Given % compostion data
Find grams of each element Find moles of each element Find mole ratio of atoms by dividing through
by the smallest number of moles
If the ratio yields a 0.33, 0.50 multiply the entire formula through to clear fractional mole amounts.
Finding Empirical Formula
Determine the empirical formula of the compound with 17.15% C, 1.44% H, and 81.41% F.
(CHF3)x
• Assume 100 g sample. C1.427H1.43F4.285
17.15g C x 1mol = 1.427mol C 1.427 1.427 1.427
12.01g
1.44g H x 1mol = 1.43mol H =(CHF3)x
1.01 g
81.41g F x 1mol = 4.285mol F
19.00g
Finding Empirical Formula
Find empirical formula of 26.56% K, 35.41% Cr, and rest O.
(K2Cr2O7)xAssume 100 g sample.
26.56 g K x 1 mol = .6793 moles K K.6793 Cr .6810 O 2.377
39.10 g .6793 .6793 .6793
35.41 g Cr x 1 mol = .6810 moles Cr
52.00 g = (KCrO3.5 ) x 2
38.03 g O x 1 mol = 2.377 moles O = (K2Cr2O7 )x
16.00 g
Finding Molecular Formula
x(empirical formula) = molecular formulax(emp.form mass) = molec.form mass
x = Molecular formula Mass Empirical formula Mass
Finding Molecular Formula
Determine molecular formula of compound with empirical formula CH and formula mass of 78.110 amu.x = 78.110 amu = 6 (CH)6 = C6H6
(12.01+1.01) amu
Finding Molecular Formula
Sample has formula mass of 34.00 amu has 0.44 g H and 6.92 g O. Find its molecular formula.%H = .44g x 100 = 6.0%
7.36g
%O = 6.92g x 100 = 94.0%
7.36 g
Assume 100 g sample.
6.0g H x 1 mol = 5.9 mol 94.0 g O x 1mol = 5.88 mol
1.01g 16.00g
H5.9 O5.88 = (HO)x x = 34.00 amu = 2 H2O2
5.88 5.88 (1.01 + 16.00) amu hydrogen peroxide
Combustion Analysis
A compound contains only carbon, hydrogen, and oxygen. Combustion of the compound yields .01068 grams of carbon dioxide and .00437 grams of water. The molar mass of the compound is 180.1 g/ mol. The sample has a total mass of .0100 grams. What are the empirical and molecular formulas of the compound?
CHO + O2 H20 + CO2
.01068g CO2 x 1 mol CO2 x 1 mol C = 2.427 x 10-4 mol C x 12.01 g = .002915 g C
44.01gCO2 1 mol CO2 1 mol
.00437g H20 x 1mol H20 x 2mol H = 4.85 x 10-4 mol H x 1.01 g = 4.90 x 10-4 g H
18.02 g 1mol H20 1 mol
C+H+O = CHO
.002915 g + 4.90 x 10-4 g + O = .0100 g
gO = .00660 g = 4.12 x 10-4 mol O
Combustion Analysis (continued)C 2.427 x 10-4 H 4.85 x 10-4 O 4.12 x 10-4 = (CH2O2)x
2.427 x 10 -4
X = molecular mass = 180.1 g = 4
empirical mass 46.03 g
(CH2O2)4 = C4H8O8