(4.6/4.7) empirical and molecular formulas sch 3u
DESCRIPTION
(4.6/4.7) Empirical and Molecular Formulas SCH 3U. An empirical formula represents the simplest whole number ratio of the atoms in a compound. The molecular formula is the true or actual ratio of the atoms in a compound. Types of Formulas. - PowerPoint PPT PresentationTRANSCRIPT
(4.6/4.7) Empirical and Molecular Formulas
SCH 3U
• An empirical formula represents the simplest whole number ratio of the atoms in a compound.
• The molecular formula is the true or actual ratio of the atoms in a compound.
Types of Formulas
The formulas for compounds can be expressed as an empirical formula and as a molecular (true) formula.
eg. empirical formula = CH2O
Learning Check
A. What is the empirical formula for C4H8?
1) C2H4 2) CH2 3) CH
B. What is the empirical formula for C8H14?
1) C4H7 2) C6H12 3) C8H14
C. What is a molecular formula for CH2O?
1) CH2O 2) C2H4O2 3) C3H6O3
Finding the Empirical Formulaa) A compound is 71.65% Cl, 24.27% C, and 4.07% H. What is the empirical formula?
1. Assume a 100 g sample. Determine the mass, in grams, of each element present.
Cl 71.65 g C 24.27 g H 4.07 g
2. Calculate the number of moles of each element.
nCl = 71.65 g = 2.021 mol
35.45 g/mol
nC = 24.27 g = 2.021 mol
12.01 g/mol
nH = 4.07 g = 4.03 mol
1.01 g/mol
3. Divide each by the smallest number of moles to obtain the simplest whole number ratio.
**If whole numbers are not obtained, multiply subscripts by the smallest number that will give whole numbers**
Cl: 2.021 mol = 1.000 Cl (1) 2.021 mol
C: 2.021 mol = 1.000 C (1) 2.021 mol
H: 4.04 mol = 2.00 H (2) 2.021 mol4. Write the simplest or empirical formula.
CH2Cl
Learning Check
Aspirin is 60.0% C, 4.5 % H and 35.5% O. Calculate the empirical formula.
nC = 60.0 g = 5.00 mol
12.01 g/mol
nH = 4.5 g = 4.5 mol
1.01 g/mol
nO = 35.5 g = 2.22 mol
16.00 g/mol
5.00 mol C = 2.25 mol
2.22
4.5 mol H = 2.0 mol
2.22
2.22 mol O = 1.00 mol
2.22
Therefore, the Empirical Formula (EF) = C9H8O4
X 4 = 9 mol C
X 4 = 8 mol H
X 4 = 4 mol O
a) A compound is Cl 71.65%, C 24.27%, and H 4.07%. What is the empirical formula? (from yesterday, CH2Cl)
b) The molar mass is known to be 99.0 g/mol. What is the molecular formula?
1. Calculate EFM (empirical formula mass)
1(12.01g/mol) + 2(1.01 g/mol) + 1(35.45g/mol) = 49.48 g/mol
2. Calculate Multiplier: Molar mass (M) = 99.0 g/mol = 2.00 EFM 49.48 g/mol
3. Multiply the empirical formula subscripts by the multiplier
(CH2Cl) x 2 = C2H4Cl2
Finding the molecular formula
Learning Check
A compound is 27.4% S, 12.0% N and 60.6 % Cl. If the compound has a molar mass of 351 g/mol, what is the molecular formula?
nS = 27.4 g = 0.855 mol
32.06 g/mol
nN = 12.0 g = 0.857 mol
14.01 g/mol
nCl = 60.6 g = 1.71 mol
35.45 g/mol
Solution
0.855 mol S = 1.00 mol
0.855
0.857 mol N = 1.00 mol
0.855
1.71 mol Cl = 2.00 mol
0.855
Therefore, the Empirical Formula (EF) = SNCl2
Solution
Solution
empirical formula mass (EFM)= 32.06 g/mol + 14.01 g/mol + (2)(35.45 g/mol)= 116.97 g/molMolar Mass = 351 g/mol
Multiplier = 351 g/mol = 3.00 116.97 g/mol
so MF is S3N3Cl6