# MOLSREVIEWJune_3 ```3) What is the % by mass of oxygen in H2SO4 ? [ GFM=98]
% COMP (MASS) = PART X 100
WHOLE
ELEMENT
SUBSCRIPTS
MULTIPLY
ATOMIC
MASS
EQUALS
MASS
SUBTOTAL
H
2
X
=
2.01588 g
S
1
X
=
32.01 g
O
4
X
1.00794
g/mol
32.01
g/mol
15.99
g/mol
=
63.9976 g
% Oxygen =
63.9976
98
X 100 = 65.3 % ~ 65 %
#5) A hydrate is a compound that includes water molecules within its crystal
structure. During an experiment to determine the percent by mass of water in a
hydrated crystal, a student found the mass of the hydrated crystal to be 4.10
grams. After heating to constant mass, the mass was 3.70 grams. What is the
percent by mass of water in this crystal?
Mass water
% H2O =
Mass hydrate
x 100 =
4.10 g – 3.70 g
% H2O =
This is a modified
version of
part/whole x 100
from the reference
tables!
4.10 g
x 100 =
9.8 %
water
7) Given the reaction:
6 CO2 +
6 H2O  1 C6H12O6 +
6 O2
a) What is the total number of moles of water needed to make 2.5 moles of
C6H12O6?
H2O
.
.
C6H12O6 .
Theoretical
mole ratio
6
1
=
x
2.5
X = 15 moles of water
#13) A compound contains 0.5 moles of sodium, 0.5 moles of nitrogen, and
1.0 moles of hydrogen. The empirical formula of the compound is
Na0.5N0.5H1.0
Multiply subscripts by 2
Subscripts can not
be decimals, to get
rid of a 0.5 decimal
multiply all
subscripts by 2!
Na0.5x2N0.5x2H1.0x2
The new subscripts are:
Na1N1H2
or
NaNH2
#15) A compound contains 40% CALCIUM, 12% CARBON and 48%
OXYGEN by mass. What is the empirical formula of this compound?
ELEMEN
T
MASS
/ ATOMIC MASS
RAW
RATIO
Ca
40 g
/ 40.08g/mol
=0.998
/ 0.998
= 1.0
C
12 g
/ 12.01g/mol
=0.999
/ 0.998
= 1.0
O
48 g
/ 15.99g/mol
=3.00
/ 0.998
= 3.0
Assume 100g of the
sample, this will allow you
to assume 40% is 40
grams. Total mass does
NOT affect % composition.
DIVIDE BY
SMALLEST
Ca1C1O3
CaCO3
SUBSCRIPT
RATIO
```