Molecular Mass
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Concepts and Calculations in Molecular Mass and Formula Mass, Study notes of Chemistry
Various concepts and calculations related to molecular mass and formula mass, including determining molar mass, mass in grams, number of moles, and percent composition of compounds. It also discusses empirical and molecular formulas, and their relationship with elemental composition and molar mass.
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2021/2022
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Molecular Mass
synonymous with molar mass and
molecular weight
Molecular Mass
is the sum of the atomic masses of all
the atoms in a molecule
the mass in grams of one mole of a
compound
Determine the molar mass of Sr(NO 3 ) 2. pg. 322 problem 25 1 mol Sr(NO 3 ) 2 2 mol N = 2(14.00 g) 1 mol Sr = 87.62g 6 mol O = 6(16.00 g) = 211.62 g
pg. 323 problem 27 What is the mass in grams of 3.25 mol sulfuric acid ( H 2 SO 4 ). 3.25 mol H 2 SO 4 98.07 g H 2 SO 4 1 mol H 2 SO 4 x = 319 g H 2 SO 4 1 mol H 2 SO 4 2 mol H = 2(1.00 g) 1 mol S = 32.07g 4 mol O = 4(16.00 g) = 98.07 g
pg. 326 problem 31a A sample of silver chromate (Ag 2 CrO 4 ) has a mass of 25.8 g. 25.8 g Ag 2 CrO 4 1 mol Ag 2 CrO 4 331.74 g Ag 2 CrO 4 x = 9.36 x 10 22 ions Ag
1 mol Ag 2 CrO 4 1 mol Cr = 52.00 g 2 mol Ag = 107.87g 4 mol O = 4(16.00 g) = 331.74 g a. How many Ag
ions are present? 2 mol Ag
1 mol Ag 2 CrO 4 x 6.02 x 10 23 ions Ag
1 mol Ag
x
pg. 326 problem 31b A sample of silver chromate (Ag 2 CrO 4 ) has a mass of 25.8 g. 25.8 g Ag 2 CrO 4 1 mol Ag 2 CrO 4 331.74 g Ag 2 CrO 4 x = (^) 4.68 x 10^22 atoms CrO 4 2- 1 mol Ag 2 CrO 4 1 mol Cr = 52.00 g 2 mol Ag = 107.87g 4 mol O = 4(16.00 g) = 331.74 g b. How many CrO 4 2- ions are present? 1 mol CrO 4 2- 1 mol Ag 2 CrO 4 x 6.02 x 10 23 ions CrO 4 2- 1 mol CrO 4 2- x
Percent Composition
of Compounds
Percent composition is the percent by
mass of each element the compound
contains.
Obtained by dividing the mass of each
element in one mole of the compound by
the molar mass of the compound and
multiplying by 100%
Molar mass = 3(1.008g) + 30.97g + 4(16.00) = 97.
%H =^ x^ 100 % =
97.99g
3(1.008g)
%P = x 100% =
30.97g
97.99g
%O = 100% =
x
97.99g
Example
A sample of a compound containing carbon and
oxygen had a mass of 88g. Of this sample 24g
was carbon, 64g was oxygen. What is the
percent composition of this compound.
%oxygen = 100% =
88g
x
64g
%carbon = 100% =
88g
x
24g
Elemental Composition
Levels of Structure
Empirical Formula
Molecular Formula
Constitution
Configuration
Conformation
Examples:
Elemental Composition
Formaldehyde Glucose
C: 40.00% C: 40.00%
H: 6.73% H: 6.73%
O: 53.27% O: 53.27%
The empirical formula tells us which
elements are present and the simplest
whole-number ratio of their atoms.
Empirical Formula
Examples: Formaldehyde and Glucose
Empirical Formula