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Eg. Relative formula mass of MgSO4?
Mr = 24 + 32 + 4(16) = 120
Mr(Fe2O3) = 2(56) + 3(16) = 160
Percentage of Fe in Fe2O3
Mr(Fe3O4) = 3(56) + 4(16) = 232
Percentage of Fe in Fe2O3 = x 100 %
= x 100%
Therefore, Fe3O4 has more iron composition than that of Fe2O3.
Calculating the Mass of an Element in a Compound
Use the example of Fe2O3 in the example above.
The percentage mass of iron in iron(III) oxide is 70%.
Therefore to calculate mass of iron in a 200g compound of iron(III) oxide is (0.7 x 200)g = 140g
Eg. Determine the mass of iron in 200g of Fe2O3.
Mr(Fe2O3)= 2(56) + 3(16) = 160
Mass of Fe in Fe2O3
Calculating the Mass of Water in a Compound
Compound with water mass is ‘hydrated’ and has H2O in their formula.
Eg. Calculate water mass in 12.5g hydrated copper sulfate, CuSO4.5H2O
Mass of 5H2O in CuSO4.5H2O
Eg 2: Calculate the number of molecules in 0.25 mole of CO2. Hence, how many atoms are present?
Number of particles = 0.25 mol x 6 x 1023
= 1.5 x 1023 molecules
Number of atoms = total number of atoms in CO2 x number of particles
= 3 x 1.5 x 1023
= 4.5 x 1023 atoms
n = m / Mr
m = n x Mr
m = 0.4 x 56 = 22.4 g
Eg. Argon Fluorohydride gas, HArF, first known noble gas compound, has molar mass of 60g. Find the number of moles Argon atom in 6.66g of HArF.
n (HArF) = 6.66/60
= 0.111 mol
n (Ar) = 0.111 mol x 1 Ar atom in HArF
= 0.111 mol
Step 1: find the number of moles of the 2 elements.
n(Mg) =0.96 / 24
= 0.04 mol
n(O) = 0.32 / 16
= 0.02 mol
Step 2: Divide the moles by the smallest number.
Mg = 0.04 / 0.02
O = 0.02 / 0.02
Therefore, the empirical formula is Mg2O
Take the total 100% to be 100g.
Step 1: find the number of moles of the 2 elements.
n(S) = 40 / 32
= 1.25 mol
n(O) = 60 / 16
= 3.75 mol
Step 2: Divide the moles by the smallest number.
S = 1.25 / 1.25
O = 3.75 / 1.25
Therefore, the empirical formula is SO3
Molecular formula will be CnH2n
Relative molecular mass = 12n(from carbon Ar) + 2n(2 x hydrogen Ar) = 14n
14 n = 42
n = 42 / 14 = 3
Therefore --> C3H6
Equal volume of gases at same temperature and volume contain equal number of particles or molecules.
= 0.01 mol
Example: Hydrogen -> 2g, Carbon Dioxide -> 44g
Eg. Find the volume of 7g of N2 at r.t.p.
Step 1: Find the number of moles from the mass of nitrogen
n = 7 / 28
= 0.25 mol
Step 2: Find the volume of nitrogen, now with formula of gas
0.25 mol = volume of gas / 24
Volume of gas = 0.25 mol x 24
= 6 dm3 (or 6000cm3)
Eg. What is the mass of solute in 600cm3 of 1.5 NaOH solution?
Volume of solution in dm3 = 0.60 dm3
n = 1.5 x 0.60
= 0.9 mol
Number of moles of NaOH = m / Mr
0.9 = m / 40
m = 0.9 x 40
Word Equation: Oxygen + Hydrogen --> Water
To write the chemical equation, we use symbols of atoms/molecules:
O2 + H2 --> H2O
Balance the equation!
O2 + 2H2 --> 2H2O
Eg. X + 2Z --> 2Y
1. Find the ratio first:
2. Then multiply the ratio by no. of moles of Y to find the reacting mole of X.
1/2 x 0.25 = 0.125 mole
Therefore 0.125 mole of X reacted with 0.25 mole of Y.
3. To find the reacting mass of X, e.g. Y is given as 35g, we just multiply the mole by the mass of Y as they are always in ratio:
0.125 x 35 = 4.375 g
Reacting Masses and Volumes
First, find the ratio of moles and multiply the mole of the gas volume you want to find with the volume of gas at room temperature (24dm3)
MgCl2 is formed by reacting Mg and HCl according to equation:
Mg (s) + 2HCl (aq) --> MgCl2 (s) + H2 (g)
Find the amount of hydrogen gas, in cm3, formed when 14.6g of HCl is reacted.
m(HCl) = 14.6/36.5
Multiply ratio by mole of HCl = 1/2 x 0.4 = 0.2 mol
Multiply mole by molar volume of gas at r.t.p. = 0.2 x 24 dm3= 4.8 dm3
1dm3 = 1000cm3
Therefore, 4.8dm3 x 1000 = 4800 cm3
4800cm3 of gas is formed
- 6.02 x 1023 molecules
- 2 moles of O
- 1 mole of C
- 44.0095 grams
- 22.4 L at r.t.p.
- 31.9988 grams of O
- 12.0107 grams of C
2. 20 cm3 of carbon monoxide are reacted with 10 cm3 of oxygen. The equation for the reaction is shown:
2CO + O2 --> 2CO2
Which volume of carbon dioxide will be produced?
a. 10 cm3
b. 20 cm3
c. 30 cm3
d. 40 cm3
3. What has a mass equal to that of one mole of water?
a. 24 dm3 of water
b. one mole of steam
c. one molecule of water
d. two moles of hydrogen molecules and one mole of hydrogen molecules
4. An 8 g sample of oxygen contains the same number of atoms as 16 g of element X. What is the relative atomic mass, Ar of X?
5. Which quantity is the same for one mole of ethanol and one mole of ethane?
b. number of atoms
c. number of molecules
d. volume at r.t.p.
6. Which ion is present in the highest concentration in a 2 mol/dm3 aqueous solution of sodium sulphate?
a. hydrogen ions
b, hydroxide ions
c. sodium ion
d. sulphate ion
7. Magnesium reacts with hydrochloric acid. Which solution would give the fastest initial rate of reaction?
a. 40 g of HCl in 1000 cm3 of water
b. 20 g of HCl in 1000 cm3 of water
c. 10 g of HCl in 100 cm3 of water
d. 4 g of HCl in 50 cm3 of water
8. One mole of hydrated copper(II) sulphate, CuSO4.5H2O is dissolved in water. How many moles of ions does the solution contain?
9. What is the ratio of the volume of 2 g of hydrogen to the volume of 16 g of methane, both volumes at r.t.p.?
a. 1 to 1
b. 1 to 2
c. 1 to 8
d. 2 to 1
10. Which of the following contains the same number of molecules as 9 g of water?
a. 2 g of hydrogen gas
b. 14 g of nitrogen gas
c. 32 g of oxygen gas
d. 44 g of carbon dioxide gas
11. Calcium reacts with water as shown.
Ca (s) + 2H2O (l) --> Ca(OH)2 (aq) + H2 (g)
What is the total mass of the solution that remains when 40g of calcium reacts with 100 g of water?
a. 58 g
b. 74 g
c. 138 g
d. 140 g
12. 20 cm3 of oxygen are reacted with 20 cm3 of carbon monoxide. What are the volumes of the gases remaining, at the original temperature and pressure?
13. What is the mass of oxygen contained in 72 g of pure water?
a. 16 g
b. 32 g
c. 64 g
d. 70 g
14. A volume of ethane, C2H6, at r.t.p. has a mass of 20 g. What is the mass of an equal volume of propene, C3H6 at r.t.p.?
a. 20 g
b. 21 g
c. 28 g
d. 42 g
15. Sodium reacts with water according to the equation below.
2Na + 2H2O --> 2NaOH + H2
Which volume of hydrogen is produced at r.t.p. when 0.2 mol of sodium reacts?
a. 1.2 dm3
b. 2.4 dm3
c. 4.8 dm3
d. 9.6 dm3
16. What is the mass of aluminium in 204g of aluminium oxide, Al2O3?
17. The equation for the burning of hydrogen is: 2H2 (g) + O2 (g) ---> 2H2O (g)
One mole of hydrogen gas is made to react with one mole of oxygen gas. What will be present after the reaction?
a. 1 mol of steam only
b. 1 mol of steam + 0.5 mol of oxygen gas
c. 1 mol of steam + 1 mol of hydrogen gas
d. 2 mol of steam + 0.5 mol of oxygen gas
18. The compound SO2Cl2 reacts with water according to the equation:
SO2Cl2 + 2H2O ---> H2SO4 + 2HCl
How many moles of sodium hydroxide will neutralise the solution produced by one mole of SO2Cl2 and excess water?
19. How many atoms are there in (NH4)SO4?
20. 3Cu + 8HNO3 --> XCu(NO3)2 + 2NO + YH2O
What are the appropriate values of X and Y for the balanced reaction equation above?
a. X = 3, Y = 8
b. X = 4, Y = 3
c. X = 3, Y = 4
d. X = 4, Y = 8
21. What is the percentage by mass of calcium in calcium carbonate?
22. Which of the following compounds contain the same percentage by mass of nitrogen as ammonium cyanate, NH4CNO?
23. A hydrocarbon contains 20% of carbon and 80% of hydrogen. What is its molecular formula if it has a relative molecular mass of 30?
24. 0.1 mol of metal X (Ar = 27) was burned in oxygen to give an oxide with mass of 5.1g. What is the formula of the metal oxide if it has a relative molecular mass of 30?
25. 100cm3 of gaseous hydrogen contains n molecules. How many molecules are there in 100cm3 of gaseous methane (CH4) under the same temperature and pressure?
26. Which of the following consists of the greatest number of atoms?
a. 4g of H2
b. 8g of O2
c. 71g of Cl2
d. 72dm3 of argon gas at rtp
27. Which of the following consists of the most number of molecules?
a. 32g of O2
b. 18g of water
c. 28g of N2
d. 4g of H2
28. Which of the following formulae for ionic compounds is not correct?
29. What is the maximum mass of chromium, Cr, that can be extracted from 76g of chromium(III) oxide?
30. How many oxygen atoms are present in 0.2 moles of N2O5?
a. 6.02 x 1023
b. 4.05 x 1023
c. 3.01 x 1022
d. 6.02 x 1021
5. c (Avogadro's law)
11. c (no. of moles of Ca = 40/40 = 1; no. of moles of water = 100/18. So Ca is the limiting reactant and H2O was present in excess. Mass of hydrogen gas liberated/lost = 2g. Mass of solution = 140 - 2 = 138 g)
1. When iron is heated in a steam of dry chlorine, it produces a chloride that contains 34.5% by mass of iron.
a. calculate the empirical formula of this chloride
b. the relative molecular mass of this chloride (Mr) is 325.
i. what is the molecular formula of this chloride?
ii. hence construct an equation, including state symbols, for the reaction of iron with chlorine.
1a. molar ratio of Fe : Cl = 34.5/56 : (100 - 34.5)/35.5
= 0.616 : 1.845
= 1 : 3
Hence empirical formula is FeCl3
1bi. let the molecular formula be (FeCl3)n
n x (56 + 3 x 35.5) = 325
--> n = 2
Hence molecular formula is Fe2Cl6
1bii. 2Fe (s) + 3Cl2 (g) --> Fe2Cl6 (s)
2. CFCs are compounds that contain only carbon, chlorine and fluorine. They are atmospheric pollutants and destroy ozone in the upper atmosphere.
a. 'CFC11' has the following composition by mass
Calculate the empirical formula of CFC11
b. 'CFC12' has the molecular formula CF2Cl2. It can be made by the reaction of hydrogen fluoride, HF, with tetrachloromethane, CCl4
CCl4 + 2HF --> CCl2F2 + 2HCl
What is the maximum mass of CFC12 that can be made from 10.0 g of hydrogen fluoride?
Empirical formula: CFCl3
2b. Mr of HF = 20
Mr of CF2Cl2 = 121
2 x 20 g of HF give 121 g of CF2Cl2
Therefore, 10 g of HF give (121/40) x 10 = 30.25 g of CF2Cl2
3. Potassium superoxide, KO2 is an ionic solid. It can be used in spacecraft to supply oxygen according to the following equation.
4KO2 (s) + 2H2O (l) --> 4KOH (s) + 3O2 (g)
The potassium hydroxide formed removes carbon dioxide.
a. show that 1.0 g of potassium superoxide will supply about 0.25 dm3 of oxygen at room temperature and pressure.
bi. name the compound formed when carbon dioxide reacts with solid potassium hydroxide
bii. give the equation for the formation of this compound
c. Supplies of oxygen in hospitals are stored in cylinders.
i. state one other use for oxygen
ii. describe briefly how oxygen is obtained from air
3a. Mr of KO2 = 71
4 x 71 = 284 g of KO2 give 3 x 24 = 72 dm3 of O2 at r.t.p.
hence, 1.0 g of KO2 give (72/284) x 1 = 0.25 dm3 of O2 at r.t.p
3bi. potassium carbonate
3bii. CO2 (g) + 2KOH (s) --> K2CO3 (s) + H2O (l)
3ci. together with acetylene in welding
3cii. oxygen is obtained by fractional distillation of liquid air
4. 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 fills the air-bag.
a. construct the equation, including state symbols, for the decomposition of sodium azide.
b. in a crash, an air-bag fills with 72 dm3 of nitrogen at room temperature and pressure. What mass of sodium azide is needed to provide the nitrogen?
c. Sodium azide, NaN3 reacts with dilute hydrochloric acid to give sodium chloride and a compound A.
Compound A contains 2.33% hydrogen and 97.7% nitrogen by mass.
i. what is the empirical formula of compound A?
ii. construct the equation for the reaction between sodium azide and dilute hydrochloric acid.
4a. 2NaN3 (s) --> 2Na (s) + 3N2 (g)
4b. no. of moles of nitrogen = 72/24 = 3
no. of moles of NaN3 required = 2
mass of NaN3 = 2 x (23 + 3 x 14) = 130 g
Empirical formula: N3H
4cii. NaN3 + HCl --> NaCl + N3H
5. Lead white, a white pigment used in old paintings, contains lead(II) carbonate. It darkens when exposed to air containing traces of hydrogen sulphide, due to the formation of black lead(II) sulphide, PbS. The white colour can be restored by treating the painting with aqueous hydrogen peroxide which converts lead(II) sulphide into lead(II) sulphate and water.
ai. Write the chemical equation of the reaction between lead(II) sulphide and hydrogen peoxide.
aii. Calculate the volume of 0.100 mol/dm3 hydrogen peroxide required to react with 0.25 g of lead(II) sulphide.
b. The element germanium (Ge) was once an important component of transistors. The flow chart below shows how germanium can be made from its ore germanite.
i. Determine the empirical formula of the chloride.
ii. Write down the valency of germanium in the chloride.
iii. Calculate the percentage of germanium in germanite
ai. PbS + 4H2O2 ---> PbSO4 + 4H2O
aii. no. of moles of PbS = Mass / Mr
= 0.25 / (207 + 32)
1 mole of PbS reacts with 4 moles of H2O2
0.001046 moles of PbS react with 0.001046 / 1 x 4 = 0.004184 moles of H2O2
Volume of H2O2 required = no. of moles / concentration
= 0.004184 / 0.100
= 0.0418 dm3
--> Empirical formula = GeCl4
bii. valency = 4
biii. Mass of germanium in germanium chloride = 33.9 / 100 x 0.177
= 0.0600 g
% of germanium in germanite = 0.0600 / 1.00 x 100%
= 6.00% (3 sf)
6. Calculate the number of moles of the following:
a. 62.1g of BaCl2
b. H2SO4 in 55cm3 of 0.25 mol/dm3 H2SO4 solution
c. 17dm3 of nitrogen gas N2
6a. Mr of BaCl2 = 208
no. of mol of BaCl2 = 62.1/208 = 0.2985 mol --> 0.3mol (2 dp)
6b. no. of mol of H2SO4 = concentration x volume = 0.25 x 55/1000 = 0.01375 --> 0.01 mol (2 dp)
6c. Mr of N2 gas = 28
no. of mol of N2 gas = 17/28 = 0.6071 mol --> 0.61 mol (2 dp)
a. the concentration of 32g of KOH in 500cm3 of water
b. the mass of iron (Fe) that can be extracted from 14.9g of iron(II) oxide (Fe2O3)
c. the volume of 100g of CO2 at normal atmospheric pressure and temperature
7a. Mr of KOH = 56
no. of mol of KOH = 32/56 = 0.5714mol --> 0.57mol (2 dp)
conc of KOH = no. of mol/(500/1000) = 1.14 mol/dm3
7b. Mr of Fe2O3 = 160
Mass of Fe extracted = (56 x 2)/160 x 14.9 = 10.43g
7c. Mr of CO2 = 44
no. of mol of CO2 = 100/44 = 2.272 mol --> 2.27 mol (2 dp)
vol of CO2 = 2.27mol x 24dm3 = 54.48dm3
8. A solution of 250cm3 of ethanol in water contained 22.3g of ethanol, C2H5OH. Calculate the concentration of ethanol in g/dm3 and mol/dm3.
8. Conc of ethanol in g/dm3 = 22.3 / (250/1000) = 89.2 g/dm3
Mr of ethanol = 46
conc of ethanol in mol/dm3 = 89.2/46 = 1.94 mol/dm3