5.5 molecular formula percent composition. review of empirical formula
TRANSCRIPT
5.5 Molecular FormulaPercent Composition
Review of
Empirical Formula
A compound is found to contain 77.9% I, 22.1 % O . Calculate the empirical formula.
=
=
= mol77.9 g I x 1 mole
g
= mol22.1 g O x 1 mole
g
Molecular Formula
• Actual number of atoms of each element in a molecule.
• Compare it to empirical formula, which is the simplest whole number ratio of atoms in a molecule.
Empirical Formula Molar Mass Molecular Formula Molecular Mass
C4H8NO 86.08 g/mol C12H24N3O3 258.24 g/mol
192.24 g/mol C14H24 96.12 g/molC7H12
360.4 g/molC20H40O572.08 g/mol C4H8O
78.06 g/molC6H613.01 g/molCH
192.24 96.12
= 2360.472.08
= 578.06
13.01= 6
A compound is found to contain 30.8 % C, 4.27 % H, 23.9 % N, and 41.0 % O. The molecular mass is 468.0 g/mole. Determine the empirical and molecular formulas.
= 2
= 1.504
= 5
117.0 g/mole 468.0 g/moleC3H5N2O3 C12H20N8O12
30.8 g C x 1 mol = 2.567 mol
12.0 g4.27 g H x 1 mol = 4.227 mol
1.01 g
23.9 g N x 1 mol = 1.707 mol14.0 g
41.0 g O x 1 mol = 2.563 mol 16.0 g
1.707 mol
1.707 mol
1.707 mol
1.707 mol
= 2.476
= 1
= 1.501
x 4 =
x 2 =
= 3
= 3
How to find the molecular formula
1- Find the empirical formula as you always do.
2- Calculate the molar mass of the empirical formula you got.
3- Divide the molecular mass by the molar mass:Molecular mass
Molar mass
4- Multiply the subscripts in the EF by this number in step(3)
This new formula will be your molecular formula!
TRY:A compound is found to contain 66.35 g C, 17.54 g H, 33.17 g N, and 63.22 g O. The molecular mass is 456.44 g/mole. Determine the empirical and molecular formulas. Hint: use 1.01 g/mol for H.
= 3
= 2.334
= 22
228.22 g/mole 456.44 g/moleC7H22N3O5C14H44N6O10
66.35 g C x 1 mol = 5.529 mol
12.0 g17.54 g H x 1 mol = 17.37 mol
1.01 g
33.17 g N x 1 mol = 2.369 mol14.0 g
63.22 g O x 1 mol = 3.9513 mol 16.0 g
2.369 mol
2.369 mol
2.369 mol
2.369 mol
= 7.332
= 1
= 1.668
x 2 =
x 3 =
= 7
= 5
Another type of questions
• You are not given the molecular mass.
• What to do?!
• Usually these questions are only about gases at STP.
A gas has the empirical formula CH2. If 0.85 L of the gas at STP has a mass of 1.59 g, what is the molecular formula?
What did we just learn?
• We use 22.4 L/mol to help us find the molecular mass
– For gases only
– At STP
• We use the density (g/L) to find molecular mass (g/mole)
Homework
Try in this order!
Page 95
#54, 50, 52.