MOLES REVIEW, MAY 28
2 Al2O3  3 O2 + 2
4 Al
O2 =
Al =
6
2
4
6
Al = 4
2
O2 =
2 Al2O3  3 O2 +
4 Al
a) What is the mole ratio of O2 to Al?
O2
Al
=
3
4
THE MOLE RATIOS OF
SUBSTACES IN A
REACTION ARE THE
BALANCED REACTION
COEFFICIENTS.
2 Al2O3  3 O2 +
4 Al
b) If 6 mol of O2 is produced, how many moles of Al are
produced?
O2 =
Al
3 = 6
4
X
X = 8 MOL of Al
2 Al2O3  3 O2 +
4 Al
c) If 6 mol of O2 is produced, how many grams of Al are
produced?
Moles = mass
8 = mass = 215.2g Al
GFM
26.9
2 Al2O3  3 O2 +
4 Al
d) What
is the % by mass composition of Al2O3 ?
NEVER USE
COEFFICENTS
FOR MOLAR
MASS
CALCULATION
S SUBSCRIPTS
ELEMENT
% COMP (MASS) = PART X 100
WHOLE
MULTIPLY
ATOMIC
MASS
26.98g/ =
mol
15.99g/ =
mol
Al
2
X
O
3
X
First you need to find the molar
mass (gfm) of the substance
EQUALS
MASS
SUBTOTAL
53.96 g
47.97 g
+
101.93
g/mol
% COMP (MASS) = PART X 100
WHOLE
% COMP Al = 53.96 g X 100 = 52.93 %
101.93 g
ELEMENT
SUBSCRIPTS
MULTIPLY
ATOMIC
MASS
Al
2
X
O
3
X
26.98g/ =
mol
15.99g/ =
mol
EQUALS
MASS
SUBTOTAL
53.96 g
47.97 g
101.93
g/mol
REVIEW CLASS MULTIPLE CHOICE QUESTIONS,
FRIDAY JUNE 1 2007
#7) CALCULATE THE MOLECULAR FORMULA OF A COMPOUND THAT IS
CH, GIVEN MOLECULAR MASS IS 78 G/MOL.
ELEMENT
SUBSCRIPTS
MULTIPLY
ATOMIC
MASS
C
1
X
H
1
X
12.01g/ =
mol
1.007g/ =
mol
THIS IS THE MASS (GFM)OF
THE EMPIRICAL FORMULA
EQUALS
MASS
SUBTOTAL
12.01 g
1.007 g
+
13.017
g/mol
REVIEW CLASS MULTIPLE CHOICE QUESTIONS,
FRIDAY JUNE 1 2007
GIVEN IN PROBLEM
#7)CONTINUED
MOLAR
MASS
78. g/mol
=
EMPIRICAL
MASS
CH X 6 = C6H6
=
13.017
g/mol
5.99 = 6
THIS MEANS THAT THE
MOLECULAR(TRUE)
FROMULA IS 6 TIMES AS
LARGE AS THE
EMPIRICAL(SIMPLEST)
FORMULA. MULTIPLY THE
SUBSCRIPTS OF THE
EMPIRICAL FORMULA BY
6.
REVIEW CLASS MULTIPLE CHOICE QUESTIONS,
FRIDAY JUNE 1 2007
#10) CALCULATE THE FORMULA OF A COMPOUND THAT IS 85% SILVER
and 15% FLUORINE BY MASS.
ELEMENT
MASS
Ag
85 g / 107.86g/mol =0.7880
F
15 g / 18.99 g/mol =0.7898 / 0.7880 = 1.0
/ ATOMIC MASS
Assume 100g of the
sample, this will allow you
to assume 85% is 85
grams. Total mass does
NOT affect % composition.
RAW
RATIO
DIVIDE BY
SMALLEST
SUBSCRIPT
RATIO
/ 0.7880 = 1.0
Ag1F1
AgF
#23) CALCULATE THE LITERS OF AMMONIA (NH3) GAS FORMED FROM 20
LITERS OF N2 GAS REACTED in the reaction given below.
3 H2 (g) + N2 (g) 2 NH3 (g)
NH3 (g)
N2 (g)
=
2
1
X liters
=
20
X = 40 Liters of ammonia gas
GAS VOLUMES CAN EXIST IN A RATIO AS
DEFINED BY THE COEFFICIENTS OF THE
BALANCED EQUATION, RATIO AS YOU
WOULD RATIO MOLES! THIS ASSUMES P
AND T ARE CONSTANT.
#28) WHAT IS THE MOLARITY OF A SOLUTION OF KNO3 (MOLECULAR
MASS=101) THAT CONTAINS 404 GRAMS OF KNO3 IN 2.00 LITERS OF
SOLUTION?
MOLES = MASS
GFM
MOLES =
404. g
= 4mol KNO3
101. g/mol
MOLARITY = MOLES
VOLUME
MOLARITY = 4 mol
2.00L
= 2.00M
#15) The density of a gas is 3.00 g/L at STP, WHAT IS THE GFM OF THE GAS?
3.00 g
1 LITER
DENSITY RATIO,
GIVEN
X
=
22.4
LITERS
1 MOLE = 22.4L
THE MASS OF
22.4 L IS THE
MOLAR MASS
AT STP
= 67.2 g/mol