Download Chem!stry - Chemist.sg

Name: …………………… (
Chem!stry
Class: ………………
Date: …… / …… / ……
Mole Concept Assignment One – Answers
Use decimal, not fractions. Final answers to 3 significant figures. Include units where necessary.
Question 1:
There are different nitrogen oxides with different formulae and properties.
Two oxides of nitrogen have the same empirical formulae. The table shows some data from experiments to
analyse the mass of nitrogen and oxygen in samples of both oxides:
a)
Mass of Oxygen / g
Oxide 1
0.35
0.78
Oxide 2
0.68
1.63
Use the data to work out the empirical formulae of both oxides and show that they are the same:
Steps to Take:
Oxide 1
Oxide 2
Nitrogen
Oxygen
Nitrogen
Oxygen
mass of element
0.35 g
0.78 g
0.68 g
1.63 g
 mass by Ar
0.35  14.0
= 0.0250
0.78  16.0
= 0.0488
0.68  14.0
= 0.0486
1.63  16.0
= 0.102
 by smallest ans.
= 0.0250  0.0250
= 0.0488  0.0250
= 0.0486  0.0486
= 0.102  0.0486
ratio of elements
= 1.00
= 1.95
= 1.00
2.10
empirical formula
b)
Mass of Nitrogen / g
NO2
NO2
Further experiments showed that Oxide 1 has a relative molecular mass of 46, Oxide 2 has a
relative molecular mass of 92. What are the molecular formulae of the two oxides?
For Oxide 1:
Relative molecular mass of NO2 = 14.0 + (2  16.0) = 46.0
46.0  46.0 = 1
1  NO2 = NO2
For Oxide 2:
Relative molecular mass of NO2 = 14.0 + (2  16.0) = 46.0
92.0  46.0 = 2
2  NO2 = N2O4
1
)
Question 2:
Many cars are fitted with air-bags which inflate in an accident. Air-bags contain the solid sodium azide, NaN3,
which decomposes rapidly to form sodium and nitrogen. The nitrogen that is formed fills the air-bag.
a)
Construct the equation, including state symbols, for the decomposition of sodium azide:
2NaN3(s)  2Na(s) + 3N2(g)
b)
In a crash, the air-bag fills with 72.0 dm3 of nitrogen at room temperature and pressure. What mass
of sodium azide is needed to provide the nitrogen?
moles of N2(g) = volume of gas in dm3  24.0
 moles of N2(g) = 72.0  24.0 = 3.00 mol
from the balanced chemical equation, 3 mol of N2(g) are produced by 2 mol of NaN3(s)
 3.00 mol of N2(g) are produced by 2/3  3.00 = 2.00 mol of NaN3(s)
relative molecular mass of NaN3(s) = 23.0 + (3  14.0) = 65.0
mass of NaN3(s) = moles  relative molecular mass
mass of NaN3(s) = 2.00  65.0 = 130 g
c)
Sodium azide, NaN3(s), reacts with dilute hydrochloric acid to give sodium chloride and a compound,
A, which contains 2.33% hydrogen and 97.7% nitrogen by mass.
i)
What is the empirical formula for compound A?
Steps to Take:
Compound A
Hydrogen
Nitrogen
mass of element
2.33%
97.7%
 % by Ar
2.33  1.00
= 2.33
97.7  14.0
= 6.98
 by smallest ans.
= 2.33  2.33
= 6.98  2.33
ratio of elements
= 1.00
= 3.00
empirical formula
ii)
HN3
Construct the equation for the reaction between sodium azide and dilute hydrochloric acid:
NaN3(s) + HCl(aq)  NaCl(aq) + HN3(aq)
2
Question 3:
Iron carbonyl is a volatile liquid with the formula Fe(CO) n. In an experiment, 4.9 g of iron carbonyl vapour
sample was found to have a volume of 600 cm 3 at room temperature and pressure.
a)
Calculate the number of moles of iron carbonyl present in the sample. Hence, find the value of n:
moles of Fe(CO)n = volume of gas in cm3  24000 (note: units of volume are in cm3, not dm3)
moles of Fe(CO)n = 600  24000
= 0.0250 mol
relative molecular mass of Fe(CO)n = mass in grams  moles
relative molecular mass of Fe(CO)n = 4.9  0.0250 = 196
n  (CO) = 196 – Fe
n  (CO) = 196 – 56.0
n  (CO) = 140
n = 140  (CO)
n = 140  (12.0 + 16.0)
n = 140  28.0
n=5
formula = Fe(CO)5
b)
A single tablet of a multivitamin supplement contains 14 mg of Fe in the form of iron carbonyl. What
is the mass of iron carbonyl present?
mass of Fe in grams = 14 mg  1000 = 0.0140 g
moles of Fe = mass in grams  relative atomic mass
= 0.0140  56.0
= 0.00025 mol
relative molecular mass of [Fe(CO)5] = 56.0 + (5  12.0) + (5  16.0)
relative molecular mass of [Fe(CO)5] = 196
mass of Fe(CO)5 = moles  relative molecular mass
= 0.00025  196
= 0.0490 g
3
Question 4:
10.0 cm3 of oxygen gas were mixed with 25.0 cm 3 of chlorine. A reaction took place to give a gaseous oxide
of chlorine (ClxOy). The products occupied 25.0 cm 3 and 5.0 cm3 of this were found to be unreacted chlorine.
a)
What volumes of oxygen, chlorine and the oxide of chlorine were involved in this reaction?
volume of O2 involved = 10.0 cm 3
volume of Cl2 involved = 25.0 cm 3 (original volume) – 5.0 cm3 (unreacted) = 20.0 cm 3
volume of ClxOy = 25.0 cm3 (volume of products) – 5.0 cm3 (unreacted chlorine) = 20.0 cm 3
Note: O2 is the limiting reagent as there is an excess of chlorine (some chlorine remains unreacted).
b)
Calculate the formula of the oxide of chlorine:
moles of chemical = volume in cm 3  24000
moles of ClxOy = 20.0  24000 = 0.000833 mol
moles of O2 = 10.0  24000 = 0.000417 mol
moles of Cl2 = 20.0  24000 = 0.000833 mol
 1 mol of ClxOy contains 0.000833  0.000833 = 1.00 mol of Cl2 = 2.00 mol of Cl atoms
 1 mol of ClxOy contains 0.000417  0.000833 = 0.500 mol O2 = 1.00 mol of O atoms
 formula for the compound = Cl2O
Note: An alternative method is to substitute numbers into the following chemical equation:
x/ Cl
2 2
+ y/2O2  ClxOy
Question 5:
50 cm3 of hydrogen peroxide, which has a concentration of 2.0 mol/dm3, is decomposed. The equation for
the reaction is:
2H2O2(aq)  2H2O(l) + O2(g)
a)
How many moles of hydrogen peroxide are used in the reaction?
moles of chemical in solution = concentration  volume  103
= 2.0  50  103
= 0.100 mol
b)
How many moles of oxygen are produced in the reaction?
from the balanced chemical equation, 2 mol of H2O2 produce 1 mol of O2
 0.100 mol of O2 produces 1/2  0.100 = 0.0500 mol of O2
c)
Calculate the volume of oxygen that is produced in the reaction at r.t.p.:
volume of gas = moles  24.0
= 0.0500  24.0
= 1.20 dm3
4
Question 6:
Calcium carbonate decomposes on heating according to the following balanced chemical equation:
CaCO3(s)  CaO(s) + CO2(g)
25.0 g of calcium carbonate was heated until there was no change in its mass. If 9.50 g of calcium oxide was
obtained, find the percentage yield of calcium oxide:
first, calculate the theoretical yield of calcium oxide:
relative molecular mass of CaCO3 = 40.0 + 12.0 + (3  16.0) = 100
moles = mass in grams  relative molecular mass
 moles of CaCO3 = 25.0  100 = 0.250 mol
from the balanced chemical equation, 1 mol of CaCO3 produces 1 mol of CaO
 0.250 mol of CaCO3 produces 1/1  0.250 = 0.250 mol of CaO
relative molecular mass of CaO = 40.0 + 16.0 = 56.0
mass in grams of CaO = moles  relative molecular mass
= 0.250  56.0
= 14.0 g
the theoretical yield of CaO = 14.0 g
% yield = (experimental yield  theoretical yield)  100
= (9.50  14.0)  100
= 67.9 %
Question 7:
Ammonia is used to produce nitric acid. The first step in the manufacturing process is the reaction between
ammonia and oxygen to produce gaseous nitrogen monoxide and steam.
a)
Write the balanced chemical equation, including state symbols, for the reaction:
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
b)
0.5 mol of ammonia gas was allowed to react with 0.5 mol of oxygen.
i)
Which is the limiting reagent in the reaction?
from the balanced chemical equation, 4 mol of NH3 react with 5 mol of O2
 0.5 mol of NH3 react with 5/4  0.5 = 0.625 mol of O2
0.5 mol of NH3 require 0.625 mol of O2 to react with, but only 0.5 mol of O2 are present
 O2 is the limiting reagent
ii)
Calculate the theoretical yield, in grams, of gaseous nitrogen oxide:
from the balanced chemical equation, 5 mol of O2 produce 4 mol of NO
 0.5 mol of O2 produce 4/5  0.5 = 0.400 mol of NO
relative molecular mass of NO = 14.0 + 16.0 = 30.0
mass in grams = moles  relative molecular mass
= 0.400  30.0
= 12.0 g
5
Question 8:
In the manufacture of the fertilizer ammonium sulphate, 20 dm 3 of 1 mol/dm3 sulphuric acid and 1440 dm 3 of
ammonia were reacted according to the equation given below:
H2SO4(aq) + 2NH3(g)  (NH4)2SO4(aq)
a)
What is the limiting reagent in the mixture?
moles of chemical in solution = concentration  volume  103
 moles of sulphuric acid = 1  20000  103
= 20.0 mol
moles of gas = volume of gas in dm3  24.0
 moles of NH3 = 1440  24.0
= 60.0 mol
From the balanced chemical equation, 1 mol of H2SO4 reacts with 2 mol of NH3
 20 mol of H2SO4 will react with 2/1  20 = 40 mol of NH3
there are 60 mol of NH3 present  NH3 is in excess and H2SO4 is the limiting reagent
b)
What is the maximum mass of ammonium sulphate that can be obtained from the reaction?
from the balanced chemical equation, 1 mol of H2SO4 produces 1 mol of (NH4)2SO4
 20.0 mol of H2SO4 will produce 1/1  20.0 = 20.0 mol of (NH4)2SO4
relative molecular mass of (NH4)2SO4 = (2  14.0) + (8  1.0) + 32.0 + (4  16.0) = 132
mass in grams = moles  relative molecular mass = 20.0  132 = 2640 g or 2.64 kg
6