Transcript Empirical and Molecular Formulas

Empirical and Molecular Formulas
Empirical Formula
Empirical Formula
A formula that gives the simplest whole-number ratio of
the atoms of each element in a compound.
Molecular Formula
Empirical Formula
H2O2
HO
C6H12O6
CH2O
CH3O
CH3O
CH3OOCH = C2H4O2
CH2O
Determine the empirical formula for a
compound containing 2.128 g Cl and
1.203 g Ca.
Steps
1. Find mole amounts.
2. Divide each mole by the smallest
mole.
1. Find mole amounts.
2.128 g Cl x 1 mol Cl = 0.0600 mol Cl
35.45 g Cl
1.203 g Ca x 1 mol Ca = 0.0300 mol Ca
40.08 g Ca
2. Divide each mole by the smallest
mole.
Cl = 0.0600 mol Cl = 2.00 mol Cl
0.0300
Ca = 0.0300 mol Ca = 1.00 mol Ca
0.0300
Ratio – 1 Ca: 2 Cl
Empirical Formula = CaCl2
A compound weighing 298.12 g consists of
72.2% magnesium and 27.8% nitrogen by
mass. What is the empirical formula?
The percentages must add up to 100%
If grams are not given assume 100 g
Hint
“Percent to mass
Mass to mole
Divide by small
Multiply ‘til whole”
A compound weighing 298.12 g consists of 72.2%
magnesium and 27.8% nitrogen by mass. What is the
empirical formula?
Percent to mass: Mg – (72.2%/100)*298.12 g = 215.24 g
N – (27.8%/100)*298.12 g = 82.88 g
Mass to mole: Mg – 215.24 g * ( 1 mole ) = 8.86 mole
24.3 g
N – 82.88 g * ( 1 mole ) = 5.92 mole
14.01 g
Divide by small: Mg - 8.86 mole/5.92 mole = 1.50???
N - 5.92 mole/5.92 mole = 1.00 mole
Multiply ‘til whole: Mg – 1.50 x 2 = 3.00
N – 1.00 x 2 = 2.00
Mg3N2
Empirical formula practice
Fat makes up a major portion of all soaps. A fat used in
many soaps is 76.5% carbon, 12.2% hydrogen, and 11.3%
oxygen
A 10 gram sample of a compound contains 7.22 grams of
magnesium and 2.78 grams of nitrogen. What is its
empirical formula?
Strychnine, a deadly poison, has a molecular mass of 334
amu and a percentage composition of 75.42% carbon,
6.63% hydrogen, 8.38% nitrogen, and the balance
oxygen. What is the empirical formula of strychnine?
Molecular Formula
The molecular formula gives the actual number of atoms of
each element in a molecular compound.
1.
2.
3.
4.
Steps
Find the empirical formula.
Calculate the Empirical Formula Mass.
Divide the molar mass by the “EFM”.
Multiply empirical formula by factor.
Find the molecular formula for a compound whose molar mass is
~124.06 and empirical formula is CH2O3.
2.
“EFM” = 62.03 g
3.
124.06/62.03 = 2
4.
2(CH2O3) = C2H4O6
Find the molecular formula for a compound
that contains 4.90 g N and 11.2 g O. The
molar mass of the compound is 92.0 g/mol.
Steps
1. Find the empirical formula.
2. Calculate the Empirical Formula Mass.
3. Divide the molar mass by the “EFM”.
4. Multiply empirical formula by factor.
Empirical formula.
A. Find mole amounts.
4.90 g N x 1 mol N = 0.350 mol N
14.01 g N
11.2 g O x 1 mol O = 0.700 mol O
16.00 g O
B.
Divide each mole by the smallest mole.
N = 0.350 = 1.00 mol N
0.350
O = 0.700 = 2.00 mol O
0.350
Empirical Formula = NO2
Empirical Formula Mass = 46.01 g/mol
Molecular formula
Molar Mass
=
Emp. Formula Mass
92.0 g/mol = 2.00
46.01 g/mol
Molecular Formula = 2 x Emp. Formula =
N2O4
A 528.39 g compound containing only
carbon, hydrogen, and oxygen is found to
be 48.38% carbon and 8.12% hydrogen by
mass. The molar mass of this compound is
known to be ~222.25 g/mol. What is its
molecular formula?
A 528.39 g compound containing only carbon, hydrogen, and oxygen is found
to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this
compound is known to be ~222.25 g/mol. What is its molecular formula?
g C – (48.38/100)*528.39 g = 255.64 g
g H – (8.12/100)*528.39 g = 42.91 g
g O – (43.5/100)*528.39 g = 229.85 g
mole C - 255.64 g * ( 1 mole ) = 21.29 mol
12.01 g
mole H – 42.91 g * ( 1 mole ) = 42.49 mol
1.01 g
mole O – 229.85 g * ( 1 mole ) = 14.37 mol
16.00 g
A 528.39 g compound containing only carbon, hydrogen, and oxygen is
found to be 48.38% carbon and 8.12% hydrogen by mass. The molar
mass of this compound is known to be ~222.25 g/mol. What is its
molecular formula?
From last slide: 21.29 mol C, 42.49 mol H, 14.27 mol O
C – 21.29/14.27 = 1.49
H – 42.49/14.27 = 2.98 (esentially 3)
O – 14.27/14.27 = 1.00
C – 1.49 x 2 = 3
H–3x2=6
O–1x2=2
C3H6O2
A 528.39 g compound containing only carbon, hydrogen, and oxygen is found
to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this
compound is known to be ~222.25 g/mol. What is its molecular formula?
From last slide: Empirical formula = C3H6O2
“EFM” = 74.09
Molar mass = 222.24 = ~3
EFM
74.09
3(C3H6O2) = C9H18O6