Inclass worksheet
Empirical and Molecular Formulas
3/7/2014
1.) Define molar mass. Mass of 1 mole of a substance
2.) What is the molar mass of carbon? CO2 ? In2(SO4)3 ?
C- 12.01
CO2 12.01 + (2*16)= 44.01
In2(SO4)3- (2*114.8)+(32.07*3)+(16*12)= 517.81
3.) What is the mass in grams of

4.31 moles H2SO4
4.31 * ( (2*1.01) + (1*32.07) +(4*16)) = 422.77 g

7.2 x 103 moles NaHCO3
7.2 x 103 * ( ( 22.99)+ (1.01) + (12.01) + (16*3)) = 6.049X105 g

.013 moles NaCl
.013 * ( ( 22.99) + (35.45)) = .760 g
4.) How many moles is
 125 g Fe 125 / 55.85 = 2.24 moles
 2.35 x 10-5 g C10H6O3
( ( 12.01*10) + (1.01*6) + (16*3))= 174.16g/mole
2.35 x 10-5 / 174.16 = 1.349 x 10-7 moles

1.69 g AgCl
1.69 / ( 107.87 + 35.45) = .0118 moles
See also page 246 #116a,b and #117b
Inclass worksheet
Empirical and Molecular Formulas
3/7/2014
5. ) Calculate the % composition by mass for each element in these
compounds:

K3PO4 K 55.23 % P 14.50% O 30.15%

NaNO3 Na 27.05% N 16.47 %

NH4NO3 N 34.98% H 5.05% O 59.97%
6.) Define empirical formula - Page 288 in text in bold
molar mass of empirical formula - sum of molar masses of elements in
empirical formula [CH] 12.01 + 1.01 = 13.02
molecular mass – sum of molar masses in elements in the molecule
C6H6 (12.01*6) + (1.01*6) = 78.12
What is n in the relationship between the empirical formula and the
molecular formula? You multiply the empirical form by this to get
molecule formula
What is n in the relationship between the molar mass of the empirical
formula and the molecular mass? You multiply the molar mass of the
empirical form by this to get the molar mass of the molecular formula
7. ) What is the empirical formula for each of these compounds :
C8H6O4 , C12H12N6O3 , C12H10O ?
[C4H3O2]
[C4H4N2O]
[C12H10O]
See also page 246 #116a,b and #117b
Inclass worksheet
Empirical and Molecular Formulas
3/7/2014
8.) Find the empirical formula for a compound that is
Cu 33.88%
O 51.18%
N 14.94%
Cu 33.88 / 63.55 = .5331/.5331 = 1
O 51.15/ 16 = 3.1968/ .5331
N 14.94 / 14 = 1.067/ .5331
= 6
=2
Empirical formula is [ CuO6N2]
9.) A binary compound of magnesium and oxygen has the following
percentage composition: 60.31% Mg 39.69% O . If the molar mass of
the compound is determined by experiment to be between 39 g and 41
g, what are the empirical and molecular formulas?
Mg 60.31 / 24.31 = 2.48/ 2.48 = 1 mole Mg
O 39.69/16 = 2.4806/ 2.48 = 1 mole O
empirical formula [MgO]
molar mass of emp. form =(24.31+16)= 40.31
n = molar mass of molecular form
= 39 to 41
molar mass of emp. form
if n =1 the molecular formula is MgO
See also page 246 #116a,b and #117b
40.31
=
1
Inclass worksheet
Empirical and Molecular Formulas
3/7/2014
10.) A 0.7221g sample of a new compound has been analyzed and
found to contain the following masses of elements:
C 0.2990g
H 0.05849g
N 0.2318 g
O 0.1328 g
10a.) What is the empirical formula of this compound?
C 0.2990g / 12.01 = .0248959 /.0083 = 3
H 0.05849g / 1.01 = .057840/.0083 = 7
N 0.2318/ 14 = .016557/.0083 = 2
O 0.1328/ 16 = .0083/.0083 = 1
empirical formula [C3H7N2O]
molar mass of empirical formula
C 3 * 12.01 = 36.03
H 7 * 1.01 = 7.07
N 2 *14 = 28
O 1*16 = 16
Total 87.1 g/mole molar mass of empirical formula
10b.) If the molecular mass is 261.3 g/mole what is the molecular
formula?
n = molar mass of molecular form
= 261.3
molar mass of emp. form
=
3
87.1
if n =3 the molecular formula is [C3H7N2O] *3 = C9H21N6O3
See also page 246 #116a,b and #117b