Determining Molecular Formulas Calculate the molecular formula of a compound whose molar mass is 600 gmol and has an empirical formula of CH4N What you need Empirical Formula Molar Mass Calculate the molar mass of the empirical formula ID: 917094
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Slide1
Molecular Formulas
A molecular formula is either the same as an empirical formula or it is a simple whole number multiple of its empirical formula.
Slide2Determining Molecular Formulas
Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and has an empirical formula of CH4N.
What you need:
Empirical Formula
Molar Mass
Calculate the molar mass of the empirical formula
Molecular Formula Molar Mass
Empirical Formula Molar Mass
=
Multiple
Empirical Formula
X
Multiple
= Molecular Formula
C: 12.0g x 1H: 1.0g x 4N: 14.0g x 1
= 12.0g= 4.0g= 14.0g
30.0 g/mol
60.0 g/mol
30.0 g/mol
=
2
CH4N x 2
= C2H8N2
Slide3Determining Molecular Formulas
Find the molecular formula of ethylene glycol which has a molar mass of 62 g/mol and an empirical formula of CH3O.
What you need:
Empirical Formula
Molar Mass
Calculate the molar mass of the empirical formula
Molecular Formula Molar Mass
Empirical Formula Molar Mass
=
Multiple
Empirical Formula
X
Multiple
= Molecular Formula
C: 12.0g x 1H: 1.0g x 3O: 16.0g x 1
= 12.0g= 4.0g= 16.0g
31.0 g/mol
62 g/mol
31.0 g/mol
=
2
CH3O x 2
= C2H6O2
Slide4Determining Molecular Formulas
The compound methyl
butanoate
has a percent composition of
58.8% C, 9.8% H, and 31.4% O and has a molar mass
of 102 g/mol
. What is its molecular formula?
What you need:
Empirical Formula
Molar Mass
58.8g C
12.0g C
1 mol C
9.8g H
1.0g H
1 mol H
31.4g O
16.0g O
1 mol O
=
4.90 mol C
=
9.8 mol H
=
1.96 mol O
1.96 mol
1.96 mol
1.96 mol
=
2.50
=
5.0
=
1.00
x 2
x 2
x 2
= 5
= 10
= 2
Empirical Formula = C5H10O2
C: 12.0g x 5
H: 1.0g x 10
O: 16.0g x 2
= 60.0g
= 10.0g
= 32.0g
102.0 g/mol
Calculate the molar mass of the empirical formula
102 g/mol
102.0 g/mol
=
1
C5H10O2 x 1
= C5H10O2
Slide5Determining Molecular Formulas
What is the molecular formula of a compound that has a percent composition of
94.1% O and 5.9% H
and has a
molar
mass of 34 g/mol.
What you need:
Empirical Formula
Molar Mass
5.9g H
1.0g H
1 mol H
94.1g O
16.0g O
1 mol C
=
5.9 mol H
=
5.88 mol O
5.88 mol
5.88 mol
=
1.0
=
1.00
Empirical Formula = HO or OH
H: 1.0g
O: 16.0g
17.0 g/mol
Calculate the molar mass of the empirical formula
34 g/mol
17.0 g/mol
=
2
HO
x
2
= H2O2 or O2H2