Name
Date
.'
PI!It
Class
PERCENT COMPOSITION
.
.,
,
,
,
,
Part 0 Problems
AND
Solve the fallowing
CHEMICAL FORMULAS
problems
14. What is the percent
in the space provided. Shaw your work.
composition
of each of the [oil owing'
a. Cr,03
c.
HgS
Section Revgew
Objectives
0. Mn,P~OT
$
Calculate the percent
by mass of all element
•
Interpret an empirical
d.
in a compound
CarNO"h
formula
• Compare and contrast
empirical
and molecular
formulas
15.
Vocabulary
Determine the empirical formula of the compound
compo.ition
of 29.1 % Na. 40.5% S. and 30.4% O.
with the percent
• percent composition
III
empirical
formula
16.
Key Equation
mass of element
= mass of compound
• % mass of element
]
~
~
]
~~
~
I
~
.sa
j-
~
g
~
element
compound
ill a compound.
compound
is the number
of the compound,
is the percent by mass of each
The percent
ofgrarns
multiplied
by mass of an element
1.
_
2. Find the percent composition of a compound
containing
18.35 g of the compound contains 5.74 g of tin,
2.
g
3.
by
4.
divide the mass of the
5.
6. How many grams of iron are in 21.6 g of iron(IJl) oxide (FeoOs)?
6.
7.
by 100%. To calculate
per __ 2__
the percent
3. If 3.907 g of carbon combines completely with 0.874 g of hydrogen
compound,
what is the percent composition
of this compound?
4. Frorn the formula for culciurn acetate, Ca(C,H30,)"
calculate
carbon that can be obtained frorn 65.3 g 01" the compound.
5. How many grams of aluminum
in one mole by the __ 3__
Aln) _4 __
elements
and multiply by 100%.
formula represents
in a compound.
percent composition.
the lowest __ 5__
lt can be calculated
The __ 6_
ratio of the
formuJa of a compound
multiple
formula,
or it is some whole-number
of it.
245
if
to form a
the mass of
oxide (A120~)?
formula of each of the following compounds
b. 10.0% C, 0.30% H, 89.1 % C!
is either
Chapter 70 ChemicalQuantities
Determine the empirical
percent composition:
are in 25.0 g of aluminum
a. 7.8% carbon and 92.2% chlorine
from a compound's
@
the same as its empirical
tin and chlorine
in a
of the element
mass of on element in a known compound.
element
of the
1. i..sample of a compound analyzed in a chemistry laboratory consists of 5.34 g
of carbon, 0.42 g of hydrogen, and 47.08 g of chlorine. What is the percent
composition
of this compound?
I"~'
..of a
from 639 kilograms
10.3 PERCENT COMPOSiTION AND CHEMICAL FORMULAS
Use this completion exercise to check your kllowledge of the terms and your
understanding
of the concepts introduced in
section. Each blank can be
completed with CL term. short phrase, or number.
The __..L
of iron can be recovered
X 100%
Part A Completion
:g.
'"
How many kilograms
ore Fe203?
from the
Section Review 10.3
Hi. molar mass FeZ03
vf,
Part A Completion
L percent
2 rnoiFe x,
=
55.8 g Fe
1 IT~e
composition
=
2. 100
3. molar mass
%Fe
4. empirical
=
112 g Fe
a po
0 -
16.0 gO
~ ~O
1 1.y>11 ,
v
. / + 3 ~O",
X
v
g Fe203
°
+ 48 g
112 g
= ---
100
100
X
160 g
5. whole-number
70.0% Fe
=
6. molecular
=
70.0 leg Fe
x 100.kg-FezCJ3
639~
160 g Fe203
=
447 kg Fe
Part D Questions and Problems
,
14.
104 g Cr
1.
_
II
0"
X -,-00 - 68.",,% 01
152 g Cr203
48 z O
0
X 100 = 31.6%
152 g CrZ03
110 g Mn
b.
X 100 = 38.7% Mn
284 g Mn2P20,
a.
°
62
ap
__
---"'-0 __
284 g Mn2P207
112 z O
X
Section 10.3
X
b
284 g Mn2P :P7
201 a Hg
c.
b
X 100
233 g HgS
32.1
S X 100
233 g hg
as
~T
100 = 39.4% 0
=
86.3% Ha
b
=
13.8% S
X
')' ,100 = ",'±.J% Ca
X
100
=
17.1%N
96 gO
X 100
164 g Ca(NOa}2
_ T,X::
1.00 mol Na
15. 29.l2A:\la x
_ "-""
'"
23.0 ~a
1.27 mol Na/l.27 = 1 x 2
=
58.5%
=
')
, ._
1.~7mal Na
d.
40.1 g Ca
264 g Ca(N03)2
28 aN
100 = 21.8% P
b
164 g Ca(N03)2
°
=
U
1.00 malO
X
aA'''')
16.0 g-'-/
=
1.90 mol 0/1.27 = 1.5 x 2
Empirical
formula
_
1.90 mol
=
= NazS203
5.34
=
Percent
H=
Percent
Cl
3
°
a
C
b
d x 100
=
10.1% C
0
d x 100
=
0.79% H
=
89.1 % Cl
52.84 g cp
0.42 aH
52.84 g cp
47.08 a
=.
CI
b
52.84 g cp
d
X
100
2. Mass of Cl
= total mass of compound - mass of Sn
= 18.35 g of compound - 5.74 g Sn
= 12.61 g Cl
e
=
=
=
or Sn
Percent
5.74
a
Sn
0
18.35 g cp
d
X
of CI
=
12.61 a Cl
0
d X 100
18.35 g cp
68.7% Cl
3. Percent
C=
3.907 a
C
d x 100
4.781 g cp
0.874 g H
Percent H =
d x 100
4.781 g cp
~
48.0 IT C
4. Percent C =
b
158.1 g Ca(C2H302)2
=
100
31.3% Sn
2
1.00 mol S
_
.
40.5 ~
/, ')~
A
= 1.21 mol S
,J~.O ~
1.27 mol S/1.27 = 1 X 2 = 2
gD
a
C
Percent
...(v
30.4
1. Percent
0
=
81.7% C
=
18.3% H
X
30.4% C
Mass C = 30.4% C x 65.3 g
5. 13.2 g Al
6. 15.11 g Fe
7. a. Get!
b. CHCI3
=
19.8 g
100