TEST II MOLES (10/25/05) PART ONE GRADING RUBRIC
Question 2
a) NO, coefficients do NOT affect molar mass, the molar mass is a constant not
affected by how many moles you have.( 3 points)
b) Coefficients are THEORETICAL MOLE RATIOS in a balanced reaction EX,
H2O/ CO2 IS 6/4 RATIO. ( 7 points)
c) The molar mass of CO2 is, (round atomic molar mass to 3 significant figures.
Element Subscript sum
Atomic molar mass Mass subtotal for element
C
1 mol
X 12.0 g/mol
=
12.0g
O
2 mol
X 16.0 g/mol
=
32.0g
+
molar mass is sum of subtotal mass 44.0 g/mol CO2
d) You were given 88.0 grams of carbon dioxide to convert to moles.
2 moles CO2
=
Answer(10 points)
88.0 g
MOLES CO2 = MASS CO2
/ 44.0 g/mol CO2
/ MOLAR MASS CO2
e) You are asked to calculate the moles of C2H6 that would be react with 2 moles of
CO2. To solve set up a mole ratio between your known 2 moles of CO2 and your
objective C2H6, using the 4 to 2 ratio from reaction coefficients.
CO2 =
C2H6
4
2
= 2 moles CO2
X
where 4X = 4 therefore X= 1 mol C2H6 answer(10pts)
f) You are asked to convert the 1 mol C2H6 from question e) into grams.
1 mol C2H6
=
X
MOLES C2H6 = MASS C2H6
/ 30.0g/mol C2H6 where X = 30.0g C2H6 answer(10)
/ MOLAR MASS C2H6
Element Sum of subscripts Atomic Molar Mass
Mass subtotal for element
C
2 mol
X 12.0 g/mol
= 24.0 g
H
6 mol
X 1.00 g/mol
=
6.00 g
+
30.0g/mol C2H6
Question 3
a) You are asked to calculate the % by mass composition of CH3COOH from this
given formula.
You use the equation: The whole is the molar mass of the CH3COOH, the part is the
subtotal of the oxygen.
% Composition by Mass = Part / Whole
= 32.0g O/60.0 grams /mol C2H6 = 53.3% answer(10pts)
.
Element
C
H
O
Sum of subscripts Atomic molar mass
2 mol
x 12.0 g/mol
=
4 mol
x
1.00 g/mol =
2 mol
x 16.0 g/mol
=
+
Molar mass of C2H6 is:
Mass subtotal for element
24.0g
4.00g
32.0g
60.0 grams /mol C2H6
b) You are asked how many moles of C are in one mole of CH3COOH. Use the sum
of the subscripts for this. 2 mol answer(7pts)
c) In one mol of CH3COOH there are 2 mol of O, therefore in a half mole of
CH3COOH there will be ONE mole of O. answer(8pts).
Question 4
Element Given
mass
F
O
Atomic
mass
Raw
mole
ratio
48.0g / 19.0g/mol =2.5mol
16.0g / 16.0g/mol =1mol
Double all moles to
make 2.5 an integer
5mol
2mol
F5O2 answer(10pts). F3O is acceptable for part credit.
b) You are asked to calculate the molar mass of F5O2.
Element Subscript sum Atomic molar mass Element mass subtotal
F
5
x 19.0
= 95.0
O
2
x 16.0
= 32.0
+
127g/mol answer(8pts).
c) You are asked to calculate the molecular (true) formula if the molecular mass is 128g
Molecular Mass
Empirical Mass
= 128g/mol = 1, therefore F5O2 answer(7pts).
127g/mol