Lesson StarterLesson Starter
Write the stock system Write the stock system names for MnO, PbOnames for MnO, PbO22
Manganese (II) OxideManganese (II) OxideLead (IV) OxideLead (IV) Oxide
How would you find the How would you find the mass of the system if mass of the system if
they could not measure they could not measure it as a whole ?it as a whole ?
Formula Weights and Formula Weights and Percentage CompositionPercentage Composition
Section 3.3Section 3.3
Three ConceptsThree Concepts
Formula MassFormula MassMolar MassMolar MassPercentage CompositionPercentage Composition
Formula MassFormula Mass
The The sum of the average atomic masses sum of the average atomic masses represented by the formula.represented by the formula.
HH220 as an example0 as an example
2 atoms of H
1 atom of O
1.01 amuxH atom
= 2.02 amu
16.00 amuxO atom
= 16.00 amu
18.02 amu
Formula MassFormula Mass
NaClNaCl HH22SOSO44
MgClMgCl22
HClHCl
58.44 amu58.44 amu 98.09 amu98.09 amu 95.21 amu95.21 amu 36.46 amu36.46 amu
Molar MassMolar Mass
The The mass in grams of 1 mole or 6.022 x 10mass in grams of 1 mole or 6.022 x 102323 particles. (same number as formula mass)particles. (same number as formula mass)
HH220 as an example0 as an example
2 moles of H
1 mole of O
1.01 gxH mole
= 2.02 g
16.00 gxO mole
= 16.00 g
18.02 g/mol
Molar MassMolar Mass
NaClNaCl HH22SOSO44
MgClMgCl22
HClHCl
58.44 g/mol58.44 g/mol 98.09 g/mol98.09 g/mol 95.21 g/mol95.21 g/mol 36.46 g/mol36.46 g/mol
Percentage Composition Percentage Composition (% Comp)(% Comp)
The The % by mass of each element in a % by mass of each element in a compound. compound. (FORMULA ON pg. 88)(FORMULA ON pg. 88)
Find the percentage composition of water.
2 moles of H
1 mole of O
1.01 gxH mole
= 2.02 g
16.00 gxO mole
= 16.00 g
18.02 g/mol
Percentage CompositionPercentage Composition
Find the percentage composition of water.
2.02 g H
16.00 g O
18.02 g H2Ox 100% = 11.21% H
18.02 g H2Ox 100% = 88.79 % O
PracticePracticeFind the percentage Find the percentage compositions of:compositions of:
Ba(NOBa(NO33))22
PbClPbCl22
Ba(NOBa(NO33))221 mol Ba X 137.33 g/ mol Ba = 137.33 g Ba
2 mol N X 14.00 g/ mol N = 28.00 g N
6 mol O X 16.00 g/ mol O = 96.00 g O
261.33 g total
137.33 g Ba
28.00 g N
96.00 g O
261.33 g total X 100 =
261.33 g total X 100 =
261.33 g total X 100 =
52.6 % Ba
10.7 % N
36.7 % O
PbClPbCl221 mol Pb X 207.2 g/ mol Pb = 207.2 g Pb
2 mol Cl X 35.45 g/ mol Cl = 70.9 g Cl
278.1 g total
X 100 =
X 100 =
74.5 % Pb
25.5 % Cl
278.1 g total
278.1 g total
207.2 g Pb
70.9 g Cl
HomeworkHomework
Page 110 Page 110 3.213.213.233.23
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