Topic 1 Quantitative Chemistry ANSWERS
1.1
Moles, Avagadro's Constant, Formulas and Equations HL/SL
1.
2.
3.
4.
The number of moles in 500 g of water is approximately:
A.
28
B.
9000
C.
1x1025
D.
3x1026
What is the empirical formula of a compound containing 85.7 % by mass of carbon and 14.3 % by mass
of hydrogen?
A.
CH
B.
CH2
C.
CH4
D.
C2H5
An oxide of metal M contains 40 % by mass of oxygen. The metal has a relative atomic mass of 24. What is
the empirical formula of the oxide?
A.
M2O
B.
M2O
C.
MO2
D.
MO
How many molecules are present in a drop of water of mass 9.00xl02g?
A.
3.01x1021
B.
3.01 x1022
C.
9.75x1023
D.
1.20x1026
-25. 8.0 g of a pure compound contains 3.2 g of sulfur and 4.8 g of oxygen. What is its empirical formula?
A.
SO
B.
SO2
C
SO3
D.
S2O3
6. How many carbon atoms are present in 0.10 mol of ethanoic acid, CH3COOH ?
A.
6.0xl022
B.
1.2xl023
C.
6.0xl023
D.
1.2xl024
7. Which of the following contains the greatest number of molecules?
A.
lg of CH3Cl
B.
lg of CH2Cl2
C.
lg of CHCl3
D.
lg of CCl4
8. Which of the following compounds has/have the empirical formula CH2O ?
I.
CH3COOH
II.
C6H12O
III.
C11H22O11
A.
II only
B.
III only
C.
I and III only
D.
II and III only
9. What amount (in moles) is present in 2.0 g of sodium hydroxide, NaOH?
A. 0.050
B.
0.10
C.
20
D.
80.
10. How many oxygen atoms are present in 0.0500 mol carbon dioxide?
A.
3.01 x1022
B.
6.02 x1022
C.
6.02 x1023
D.
1.20 x1024
11. Indigo is a blue dye which contains only carbon, nitrogen, hydrogen and oxygen.
(a)
2.036 g of indigo was completely oxidised to produce 5.470 g of carbon dioxide and 0.697 g
of water. Calculate:
(i)
Moles CO2 = 5.470 / 44
Mass of C = nMr = x 12 = 1.4916g
% = 1.4916 / 2.036 = 73.26%
the percentage by mass of carbon in indigo;
(ii) the percentage by mass of hydrogen in indigo.
(b)
Moles of H2O = 0.697 / 18
Mass of H = x 2 = 0.07744g
% = 3.804%
[2]
If the percentage by mass of nitrogen in the indigo sample is 10.75 %, determine the
empirical formula of indigo.
(c)
[2]
C
:
73.26 /12
6.105
8
C8H5NO
N
:
10.75/14
0.768
1
H
:
3.804/1
3.804
5
If the molar mass is approximately 260 gmol -1, determine the molecular formula of indigo.
C8H5NO = 131
C16H10N2O2 = 262
(d) Nitrogen also forms an oxide on reaction with oxygen. This oxide contains 25.9 % of nitrogen
and 74.1 % of oxygen by mass. Calculate the empirical formula of this second oxide.
25.9 / 14
1.85
1
74.1/16
4.63
2.5
N2O5
[3]
O
12.19/16
0.762
1
[3]
[1]
12. An oxide of copper was reduced in a stream of hydrogen as shown below.
excess hydrogen
burning
hydrogen
gas
HEAT
oxide of copper
in a dish
After heating, the stream of hydrogen gas was maintained until the apparatus had cooled.
The following results were obtained.
Mass of empty dish = 13.80 g
Mass of dish and contents before heating = 21.75 g
Mass of dish and contents after hearing and leaving to cool = 20.15 g
(a) Explain why the stream of hydrogen gas was maintained until the apparatus cooled.
[1]
Prevents re oxidation of copper
(b)
Calculate the empirical formula of the oxide of copper using the data above, assuming
complete reduction of the oxide.
1.60/16
0.1
6.35/63.55
0.1
[3]
CuO
(c) Write an equation for the reaction that occurred.
[I]
CuO + H2 = Cu + H2O
(d) State two changes that would be observed inside the tube as it was heated.
Black copper oxide turns into red/pink/brown copper
Water/condensation produced
[2]
Topic 1 Quantitative Chemistry
1.2
Mass Relationships in Chemical Reactions
1.
SL/HL
In the decomposition of KClO3, 6.30 mol of oxygen was produced:
2KClO3->2KCl + 3O2
How many moles of KCl would be produced?
A.
4.20
B.
6.30
C.
12.6
D.
18.9
2. Formation of polyethene from calcium carbide, CaC2, can take place as follows:
CaC2 + 2 H2O -> Ca(OH)2 + C2H2
C2H2+H2->C2H4
nC2H4->H-CH2-CH2-)nWhat mass of polyethene is obtained from 64 kg of CaC2 ?
A.
7kg
B.
14 kg
C
21 kg
D.
28 kg
3. Ammonia is manufactured by the synthesis of nitrogen and hydrogen as follows:
N2(g) + 3H2(g)-+2NH3(g)
56.0 g of N2 produces 34.0 g of NH3.
What is the percentage yield of ammonia?
A.
50
B.
68
C.
74
D.
100
'
4. What amount of H2(g) is produced when 12 g of magnesium reacts completely with dilute HCl(aq)?
Mg(s) + 2HCl(aq) -> MgCl2 (aq) + H2 (g)
A.
¼ mol
B.
½ mol
C.
1 mol
D.
2 mol
5.
What amount (in moles) of FeS2 (s) are required to produce 64 g of SO2 (g) according to the following
equation?
4FeS2(s) + 1102(g) -> 2Fe203(s) + 8SO2(g)
A.
0.40
B.
0.50
C.
1.0
D.
2.0
6.
PbS(s) + O2(g) ->PbO(s) + SO2(g)
The reaction of lead (II) sulfide with oxygen at high temperatures is represented by the unbalanced
equation above. What is the sum of the coefficients in the balanced equation?
7.
A.
4
B.
5
C.
8
D.
9.
8.0 g of a pure compound contains 3.2 g of sulfur and 4.8 g of oxygen. What is its empirical formula?
A.
SO
B.
SO2
C.
SO3
D.
S2O3.
8.
Consider the following equation.
2C4H10(g)+13O2(g) -> 8CO2(g)+10H2O(l)
How many moles of CO2(g) are produced by the complete combustion of 58 g of butane, C4H10 (g) ?
A.
4
B.
8
C.
12
D.
16
9.
Copper can react with nitric acid as follows.
3Cu + _ HNO3 ->_ Cu(NO3)2 + _ H2O + _ NO What is the coefficient for HNO3
when the equation is balanced?
A.
4
B.
6
C
8
D.
10
10.
Consider the equation below.
Fe(s)+S(s) ->FeS(s)
If 10.0 g of iron is heated with 10.0g of sulfur to form iron (II) sulfide, what is the theoretical yield of FeS
in grams?
A.
10.0 +10.0
87.91x10.0
B
55.85
C.
87.91x10.0
32.06
D
55.85x10.0
32.06
11.
6.0 moles of Fe2O3 (s) reacts with 9.0 moles of carbon in a blast furnace according to the equation below.
Fe2O3(s) + 3C(s) -+2Fe(s) + 3CO(g)
What is the limiting reagent and hence the theoretical yield of iron?
Limiting reagent
Theoretical yield of iron
A
Fe203
6.0 mol
B
Fe203
12.0 mol
C
carbon
9.0 mol
D
carbon
6.0 mol
12. Calcium carbonate decomposes on heating as shown below.
CaCO
CaO + CO2

When 50 g of calcium carbonate are decomposed, 7 g of calcium oxide are formed.
What is the percentage yield of calcium oxide?
A.
7%
B.
25 %
C.
50 %
D
75%
13. Chloroethene, C2H3Cl, reacts with oxygen according to the equation below:
2C2H3Cl + 5O2 -> 4CO2 + 2H2O + 2HCl
How many moles of CO2 are produced when 3.0 mol of C2H3Cl and 3.0 mol of O2 are reacted?
A.
2.4
B
3.0
C
4.0
D.
6.0
14. When the equation C4H10 + O2 -» CO2 + H2O is balanced correctly, what is the coefficient for O2 ?
A.
9
B.
13
C.
18
D.
24
15. A student was asked to make some copper(II) sulfate-5-water (CuSO4.5H2O) by reacting
copper(II) oxide (CuO) with sulfuric acid.
(a)
Calculate the molar mass of copper(II) sulfate-5-water.
[1]
CuSO4.5H2O = 249.71
(b)
Calculate the amount (in mol) of copper(II) sulfate-5-water in a 10.0 g sample.
[1]
10 / 249.71 = 0.0400
(c)
Calculate the mass of copper(II) oxide needed to make this 10.0 g sample.
moles of CuO = n Mr = 0.04 x 79.55 = 3.18g
16. The reaction below represents the reduction of iron ore to produce iron.
2Fe2O3 + 3C
4Fe + 3CO2
A mixture of 30 kg of Fe2O3 and 5.0 kg of C was heated until no further reaction occurred.
Calculate the maximum mass of iron that can be obtained from these masses of reactants.
[5]
Moles of Fe2O3 = 30 000 / 159.7 = 188 moles
Moles of C = 5000 / 12 = 416 moles
3 moles of C react with 2 moles of Fe2O3…therefore Fe2O3 is limiting
2 moles of Fe2O3 react with 4 moles of Fe
Moles of Fe = 2 x 188
Mass of fe = x 55.85 = 21Kg
[I]
17. An element X reacts with oxygen to form the oxide X2O3.
(a) Write a balanced equation for the reaction.
2X + 1 ½ O2 = X2O3
[1]
(b) If 2.199 g of the oxide was obtained from 1.239 g of X, calculate the relative atomic mass of
X and identify the element.
[5]
Mass of O = 2.199 – 1,239 = 0.96g
Moles of O = 0.96 / 16 = 0.06
From formula .. 3 moles of O react with 2 moles of X
Moles of X = 2/3 x 0.06 = 0.04 moles
Mr = m / n = 1.239 / 0.04 = 30.975
Phosphorus
18.
100 cm3 of ethene, C2H4, is burned in 400 cm3 of oxygen, producing carbon dioxide and some liquid
water. Some oxygen remains unreacted.
(a) Write the equation for the complete combustion of ethene.
C2H4 +
3O2 = 2 CO2
+
[2]
2 H2O
(b) Calculate the volume of carbon dioxide produced and the volume of oxygen remaining.
[2]
1 mole of C2H4 = 2 moles of CO2
100 cm3
200 cm3
1 mole C2H4 = 3 moles O2
100cm3
300cm3
Therefore 100cm3 O2 left unreacted
19.
(a) Write an equation for the formation of zinc iodide from zinc and iodine.
Zn + I2 =
[I]
ZnI2
(b) 100.0 g of zinc is allowed to react with 100.0 g of iodine producing zinc iodide. Calculate
the amount (in moles) of zinc and iodine, and hence determine which reactant is in excess.
[3]
Moles Zn = 100 / 65.37 = 1.53 moles
Moles I2 = 100 / 253.8 = 0.3940 moles
Zn in excess
(c) Calculate the mass of zinc iodide that will be produced.
m = n Mr = 0.3940 x (65.37 + 253.8) = 125.8g
[I]
Topic 1 Quantitative Chemistry
1.3
Gaseous Volume Relationships
SL/HL
1. When the pressure is increased at constant temperature, the particles in a gas will
A.
become smaller.
B.
become larger.
C.
move faster.
D
Closer together
2. For which set of conditions does a fixed mass of an ideal gas have the greatest volume?
A.
B.
C.
D.
Temperature
Pressure
low
low
low
high
high
high
high
low
.
3. Which expression represents the density of a gas sample of relative molar mass, Mr, at
temperature, T, and pressure, P?
A.
B.
C.
D.
Ma
4.
What will happen to the volume of a fixed mass of gas when its pressure and temperature
(in Kelvin) are both doubled?
A. It will not change.
B. It will increase.
C. It will decrease.
D. The change cannot be predicted.
5. A cylinder of gas is at a pressure of 40 kPa. The volume and temperature (in K) are both doubled.
What is the pressure of the gas after these changes?
A. 10 kPa
B. 20 kPa
C. 40 kPa
D. 80 kPa
6. A fixed mass of an ideal gas has a volume of 800 cm3 under certain conditions. The pressure (in
kPa) and temperature (in K) are both doubled. What is the volume of the gas after these changes with
other conditions remaining the same?
A. 200 cm3
B. 800 cm3
C. 1600 cm3
D. 3200 cm3
7. When a small quantity of a strongly smelling gas such as ammonia is released into the air, it can be
detected several metres away in a short time.
(a) Use the kinetic molecular theory to explain why this happens.
Particles are in rapid, random, continuous motion
[2]
Causing mixing/diffusion/spreading out
(b) State and explain how the time taken to detect the gas changes when the temperature is
increased. [2]
Less time/quicker
Higher temp = grater ke /speed = diffuse/mix faster
8.
Assuming ideal behaviour, calculate the volume occupied by 2.00g of carbon monoxide at 20
degree C under a pressure of 6250 N m-2
PV = nRT
V = mRT / MrP
= 2 x 8.314 x 293 / 28 x 6250
= 0.0278 m3
9. A 0.230 g sample of A, when vaporized, had a volume of 0.0785 dm3 and a pressure of 102 kPa. at
95 C ° Determine the relative molecular mass of A.
Mr = mRT /PV
= 0.230 x 8.314 x 368 / 102 x 10 3 x 0.0785 x 10-3
= 87.89
10. A 1.00 g sample of the hydrocarbon at a temperature of 273 K and a pressure of 1.01 x105 Pa (1.00
atm has a volume of 0.399 dm3
(i) Calculate the molar mass of the hydrocarbon.
n = PV / RT
1.01 x 105 x 0.399 x 10-3 / 8.314 x 273
= 0.178 moles
Mr = m / n = 1 / 0.178 = 56.3
11. This question deals with gases and liquids.
(a)
The mass of a gas sample is measured under certain conditions. List the variables that must
be measured and show how these can be used to determine the molar mass of the gas.
Volume
(b)
temp
pressure
[4]
Mr = mRT /PV
As a volatile liquid in an isolated container evaporates, its temperature drops. Account for
this observation in terms of the behaviour of the molecules.
[2]
Since particles of gas with a high KE escape
The average KE of particles decrease, hence temp decreases
(c) In hydrogen gas what happens to the average speed of the molecules if the temperature is
increased?
[1]
increases
.
(d)
Explain, in terms of molecules, what happens to the pressure of a sample of hydrogen gas if its volume
is halved and the temperature kept constant.
[2]
Pressure doubles
Since particles collide with inside container with twice the frequency (since ½ the size)
Topic 1 Quantitative Chemistry
1.4
Solutions
SL/HL
Zn(s) + Cu2+(aq) -> Zn2+ (aq) + Cu(s)
1.
Powdered zinc reacts with Cu2+ ions according to the equation above. What will be the result of adding
3.25 g of Zn to 100 cm3 of 0.25 moldm -3 CuS04 solution?
A.
All the Cu + ions react and some solid zinc remains.
B.
All the Cu2+ ions react and no solid zinc remains.
C.
All the solid zinc reacts and Cu2+ ions remain.
D.
Neither solid zinc nor Cu2+ ions remain.
2. Consider the following reaction:
CaCl2(aq) + 2AgNO3(aq) -> 2AgCl(s) + Ca(NO3)2(aq)
2.0 dm3 of 0.50 moldm -3 CaCl2(aq) is mixed with 1.0 dm3of 2.0 mol dm -3 AgNO3(aq). What are the
concentrations of Ca2+(aq) and NO3-(aq) after mixing?
[Ca2+]/moldm3
3.
[NO3-]/moldm3
A.
0.66
0.33
B.
0.33
0.66
C.
1.0
2.0
D.
3.0
1.5
Which sample contains the greatest number of ions?
A.
25 cm3 of 0.40 moldm -3 NaCl
B.
50 cm3 of 0.20 moldm -3 MgCl2
C.
100 cm3 of 0.10 moldm -3 KNO3
D.
200 cm3 of 0.05 moldm -3 CuSO4
4. 10.0 cm3 of 0.200 moldm -3 H3PO4(aq) is converted into Na2HPO4(aq). What volume (in cm3) of 0.200
moldm -3 NaOH(aq) is required?
A. 10.0
B. 13.3
C. 20.0
D. 30.0
5. 25.0 cm3 of 2.00 mol dm -3 HNO3(aq) reacts completely with 20.0 cm3 of Ba(OH)2(aq). What is the
concentration of barium hydroxide solution?
A. 0.800 mol dm-3
B. 1.25 mol dm--3
C. 2.00 mol dm-3
D. 2.50 mol dm-3
6. Which solution contains the greatest amount (in mol) of solute?
A.
10.0 cm3 of 0.500 mol dm-3 NaCl
B.
20.0 cm3 of 0.400 mol dm-3 NaCl
C.
30.0 cm3 of 0.300 mol dm-3 NaCl
D.
40.0 cm3 of 0.200 mol dm-3 NaCl
16.20X10-3 dm3 of 0.1020 moldm-3 aqueous AgNO3 is added to 14.80x10-3 dm3 of 0.1250
moldm-3 aqueous NaCl. Calculate the maximum mass (g) of AgCl which could be obtained from
this reaction. (Relative Atomic Masses are Ag = 107.87, Cl = 35.45.)
7.
[4]
AgNO3 + NaCl = AgCl + NaNO3
Moles of AgNO3 = CV = 0.1020 x 16.2 x 10 -3 = 1.6524 x 10-3…limits
Moles of NaCl = CV = 0.125 x 14.8 x 10-3 = 1.85 x 10-3….excess
Moles of AgCl = 1.6524 x 10-3
Mass of AgCl = n x Mr = x 143.32
= 0.2368g
8. Vinegar is a solution of ethanoic acid. A 10.0 cm3 portion of a certain brand of vinegar needed 55.0
cm3 of 0.200 mol dm-3 sodium hydroxide solution to neutralise the ethanoic acid in it. [3]
Ethanoic + Sodium
_____ ► Sodium
acid
hydroxide
ethanoate
+ Water
CH3CO2H(aq) + NaOH(aq) --------► CH3CO2Na(aq) + H2O(1)
a) Calculate the concentration of ethanoic acid in the vinegar in -mol dm-3.
Moles of NaOH = CV = 0.2 x 0.055 = 0.011 moles
Moles of acid = 0.011
Conc of acid = n / V = 0.011 / 0.01 = 1.1 mol dm-3
9
We placed 20.0 cm3 of a solution of barium hydroxide, Ba(OH)2, of unknown
concentration in a conical flask and titrated with a solution of hydrochloric acid, HC1,
which has a concentration of 0.0600, mol dm-3. The volume of acid required is 25.0 cm3.
Calculate the concentration of the barium hydroxide solution.
[3]
Ba(OH)2 (aq) + 2HC1 (aq) -» BaCl2 (aq) + 2H2O (1)
Moles of acid = CV = 0.06 x 0.025 = 1.5 x 10 -3
Moles of alkali = ½ x 1.5 x 10-3 = 7.5 x 10-4
Conc of alkali = n / V = 7.5 x 10-4 / 0.02 = 0.0375 mol dm-3
10. A household cleaner contains aqueous ammonia. A 2.447 g sample of the cleaner is diluted with
water to 20.00 cm3. This solution requires 28.51 cm3 of 0.4040 mol dm-3 sulfuric acid to reach
the equivalence point.
(i)
Write a balanced chemical equation for the reaction of sulfuric acid with ammonia to
form ammonium sulfate.
[1]
(ii) Calculate the amount (moles) of sulfuric acid required for this reaction, and the amount
(moles), mass and percentage by mass of ammonia present in the household cleaner.
[4]
H2SO4 + 2NH4OH = (NH4)2 SO4 + 2H2O
Moles of acid = CV = 0.404 x 0.02851 = 0.01152
Moles of NH3 = 2 x 0.01152 = 0.02304
Mass of NH3 = n x Mr = 0.02304 x 17 = 0.3916g
% mass = 0.3916 / 2.447 = 16.00 %
HL ONLY
11.
A student wished to determine the percentage of calcium carbonate in a small sea shell found
on a local beach. The clean, dry shell that weighed 1.306g was placed in a small beaker and 10
cm3 of 5.00 mol dm-3 hydrochloric acid was added. After the shell had completely dissolved, the
resulting solution was carefully transferred to a volumetric flask and the volume was made up to
25.0 cm3 with distilled water. A 10.0 cm3 sample of this solution required 11.2 cm of 1.00 mol
dm-3 sodium hydroxide solution for complete neutralisation.
(a) Calculate the number of moles of NaOH present in 11.2 cm3 of 1.00 mol dm-3 NaOH
solution.
Moles NaOH = CV
= 1.00 x 0.0112 = 0.0112
.....................................................................................................................[2]
(b) How many moles of acid remained in the beaker after the reaction with the shell?
Moles of acid neutralised = 0.0112 in 10cm3 sample
Moles of acid in 25cm3 beaker = 0.0112 x 25/10 = 0.028 moles
......................................................................................................................[2]
(c) How many moles of acid reacted with the shell?
Original moles of acid = CV = 5.00 x 0.01 = 0.05 moles
Difference = 0.05 – 0.028 = 0.022 moles
....................................................................................................................... [2]
(d) Write the equation for the reaction of calcium carbonate and hydrochloric acid.
CaCO3 + 2 HCl = CaCl2 + H2O .......................................................
+ CO2
...............................................................
[1]
(e) What mass of calcium carbonate was present in the shell?
2 moles of acid react with 1 mole CaCO3
Moles of CaCO3 = ½ x 0.022 = 0.011 moles
Mass of caCO3 = n x Mr = 0.011 x 100 = 1.10g
....................................................................................................................... [2]
(f) What was the percentage of calcium carbonate in the shell?
1.10 / 1.306 x 100 = 84.2 %
....................................................................................................................... [1]
(g) What assumptions have you made about the composition of the shell in arriving at
your answer?
Acid only reacts with CaCO3 in the shell
[1]