SCHOOL OF CHEMISTRY & PHYSICS
UNIVERSITY OF KWAZULU-NATAL, WESTVILLE CAMPUS
CHEM 110 – General Principles of Chemistry
TUTORIAL 3
05th & 07st March 2014
1. Determine the empirical formula of each of the following compounds:
(a) 5.28 g of Sn and 3.37 g of F
Calculate moles of each element present, then the simplest ratio of moles.
5.28 g Sn
= 0.04448 mol Sn;
118.7 g mol-1
3.37 g F
= 0.1774 mol F;
19.00 g mol-1
0.04448 / 0.04448 = 1
0.1774 / 0.04448  4
The integer ratio is 1 Sn : 4 F; the empirical formula is SnF4.
(b) 87.5% N and 12.5% H by mass
Assume 100 g sample, calculate moles of each element, find the simplest ratio of
moles.
87.5 g N
= 6.25 mol N;
14.01 g mol-1
12.5 g H
12.5% H =
= 12.4 mol H;
1.008 g mol-1
87.5% N =
6.25 / 6.25 = 1
12.4 / 6.25  2
The empirical formula is NH 2 .
(c) Combustion analysis of menthol is composed of C, H and O. A 0.1005 g sample of
menthol is combusted, producing 0.2829 g of CO2 and 0.1159 g of H2O. What is the
empirical formula of menthol? If the compound has a molar mass of 156 g mol-1,
what is the molecular formula?
no. of mol C and mol H in the sample.
0.2829 g CO2
1 mol C
×
= 0.0064281 mol C = 0.006428 mol C
-1
1 mol CO2
44.01 g mol
0.1159 g H2O
2 mol H
×
= 0.012863 mol H = 0.01286 mol H
-1
1 mol H2O
18.02 g mol
Mass of C, H and get mass of O by subtraction.
0.0064281 mol C × 12.01 g C mol-1 = 0.07720 g C
0.012863 mol H × 1.008 g H mol-1 = 0.01297 g H
mass O = 0.1005 g sample – (0.07720 g C + 0.01297 g H) = 0.01033 g O
Page 1 of 4
SCHOOL OF CHEMISTRY & PHYSICS
UNIVERSITY OF KWAZULU-NATAL, WESTVILLE CAMPUS
no. of mol O:
0.01033 g O
= 6.456 × 10-4 mol O
16.00 g mol-1
-4
Find integer ratio of mol C: mol H: mol O by dividing moles by 6.456  10 .
C:
0.006428
 10;
6.456 × 10-4
H:
0.01286
 20;
6.456 × 10-4
O:
6.456 × 10-4
=1
6.456 × 10-4
The empirical formula is C 1 0H 2 0O.
Empirical Formula mass (FM)
= 10(12.01 g mol-1) + 20(1.008 g mol-1)) + 16.00 g mol-1 = 156.26 g mol-1
=
=1
The molecular formula is the same as the empirical formula, C 1 0H 2 0O.
2. Car air bags inflate when sodium azide, NaN3, rapidly decompose to its component
elements.
2 NaN3(s) → 2 Na(s) + 3N2(g)
(a) How many moles of N2 are produced by the decomposition of 1.50 mol of NaN3?
(b) How many grams of NaN3 are required to form 10.0 g of nitrogen gas?
(a)
Use mole ratio from balanced equation.
no. of mol of NaN3 = no. of mol of N2
2
3
1.50 mol NaN3 ×
3 mol N2
= 2.25 mol N2
2 mol NaN3
(b)
Given: mass of N2. Find: mass of NaN3.
Use molar masses to get from and to grams and mol ratio to relate moles of the two
substances.
no. of mol of NaN3 = no. of mol of N2
2
3
2 mol NaN3
10.0 g N2
×
× 65.01 g mol-1 = 15.5 g NaN3
3 mol N2
28.01 g mol-1
Page 2 of 4
SCHOOL OF CHEMISTRY & PHYSICS
UNIVERSITY OF KWAZULU-NATAL, WESTVILLE CAMPUS
3. One of the steps in the commercial process for converting ammonia to nitric acid is the
conversion of NH3 to NO.
In a certain experiment 1.50 g of NH3 reacts with 2.75 g of O2.
(a) Which is the limiting reactant?
(b) How many grams of NO is form?
4 NH 3 (g) + 5 O 2 (g)  4 NO(g) + 6 H 2 O(g)
(a)
Find the no. of mol of NH3 and O2 then determine the amount of NH3 required to
react with O2
n = mass/ MM
1.50 g NH3 ×
2.75 g O2 ×
1 mol NH3
= 0.08808 mol NH3 = 0.0881 mol NH3
17.03 g NH3
1 mol O2
= 0.08594 mol O2 = 0.0859 mol O2
32.00 g O2
0.08594 mol O2 ×
4 mol NH3
= 0.06875 mol NH3 = 0.0688 mol NH3 required
5 mol O2
More than 0.0688 mol NH 3 is available, so O 2 is the limiting reactant.
(b)
no. of moles of NO = no. of moles of O2
4
5
X 30.01 g mol-1 = 2.063 g NO
Mass of NO = 0.08594 mol O2 X
4. When benzene (C6H6) reacts with bromine (Br2), bromobenzene (C6H5Br) is obtained.
C6H6 + Br2 → C6H5Br + HBr
(a) What is the theoretical yield of bromobenzene in this reaction when 30.0 g of
benzene reacts with 65.0 g of bromine?
(b) If the actual yield of bromobenzene was 56.7 g, what was the percentage yield?
Find the no. of moles of C6H6 and Br2. Then determine the limiting reactant and the
maximum amount of product it could produce.
Page 3 of 4
SCHOOL OF CHEMISTRY & PHYSICS
UNIVERSITY OF KWAZULU-NATAL, WESTVILLE CAMPUS
(a)
n = m/M
30.0 g C6H6
= 0.3841 mol C6H6 = 0.384 mol C6H6
78.11 g mol-1
65.0 g Br2
= 0.4068 mol Br2 = 0.407 mol Br2
159.8 g mol-1
Since C6H6 and Br2 react in a 1:1 mole ratio, C6H6 is the limiting reactant and
determines the theoretical yield.
m=nxM
0.3841 mol C6H6 ×
(b)
1 mol C6H5Br
× 157.0 g mol-1 = 60.30 g C6H5Br = 60.3 g C6H5Br
1 mol C6H6
% Yield = actual yield/ theoretical yield x 100
% yield =
56.7 g C6H5Br actual
× 100 = 94.0%
60.3 g C6H5Br theoretical
Page 4 of 4