Actual and theoretical yields
When a chemical reaction occurs to get certain amount of chemical substances, the chemical equation of the reaction determines theoretically the amount of the products. But practically, the amount of the products will be less than the theoretical amount because:-
1- The products may be volatile and parts of them spread in the air
2- Parts of the products may stick to the glass containers walls
3- The reactants may be impure
4- side reactions may occur
Practical yield: The amount of substances we get practically from the reaction
Theoretic yield: The amount of substances we expect to get form the reaction
The percentage yield (the percentage of practical yield)
100 X Practical yield
Theoretical yield
Example:-
39.4 gm of solid barium sulphate BaSO4 precipitated when 40 gm of barium chloride solution BaCl2 reacted with potassium sulphate. Calculate the percentage yield of barium sulphate
Solution
2BaCl2 + 2KSO4 2BaSO4 + 2KCl
Mole of BaCl2 = 137 + 35.5 + 35.5 = 208 gm
Mole of BaSO4 = 137 + 32 + 16+16+16+16 = 233 gm
The no. of BaCl2 moles = mass of substance / molar mass = 40/208 = 0.19 mol
2 moles of BaCl2 2 moles of BaSO4
0.19 mole of BaCl2 0.19 mol of BaSO4
The mass of BaSO4 = 0.19 x molar mass = 0.19 x 233 = 44.8 gm
The practical yield = 39.4 gm
The theoretical yield = 44.8 gm
The percentage yield = the practical yield / the theoretical yield x 100
= 39.4/44.8 x 100 = 87.95%
1- Chemical equation: The representation of chemical reaction using
chemical symbols,. formulas of reactants and products, and the conditions of it
2- Avogadro's number: The no. of ions, molecules and atoms in one mole of matter (6.02x1023 matter unit/mole)
3- Mole: The mass of atoms, molecules, or formula units of matter in grams which contain Avogadro's number of them
4- Avogadro's law: Equal volumes of gases in the same conditions of pressure and temperature have the same no. of molecules
5- Gay-Lussac's law: The volumes of reactant and product gases have certain volumes expressed in whole number
6- Empirical formula: The formula that describe the simplest ratio between the atoms of the elements forming the compound molecules
7- Molecular formula: The symbolic formula of the molecule of a compound which describes the kind and the actual no. of the atoms forming that molecule
8- Practical yield: The amount of substances we get practically from the reaction
9- Theoretical yield: The amount of substances we expect to get form the reaction
1- The volume of 26 gm Acetylene gas (C2H2) is equal to the volume of 2g of hydrogen gas in (STP) conditions
Because the mole of Acetylene molecule equals 26 gm, and the mole of hydrogen molecule equals 2g. By applying Avogadro's law, we'll find that the volumes of both gases are equal in (STP) conditions (because they contain the same no. of moles)
2- The molar mass of phosphorus differs according to its physical state
Because in gaseous state, phosphorus molecule consists of 4 atoms, while in solid state, It consists of 1 atom. So, the molar mass of gaseous phosphorus is different from that of solid phosphorus
3- Litre of oxygen gas has the same no. of molecules in a litre of chlorine gas in STP conditions
Because according to Avogadro's law, equal volumes of gases in STP conditions have the same no. of molecules
4- the no. of molecules in 9 gm of water H2O is equal to that in 39 gm of Aromatic Benzene (C2H2)
Because the mass of one mole of water = 9 gm, whereas the mass of one mole of Aromatic Benzene = 39 gm, so they have the same no. of molecules (Avogadro's number) because they have the same no. of moles
5- Chemical equations should be balanced
In order to get the required amounts of products
6- Gas should be in STP conditions in order to calculate its volume using its molar mass
Because in STP conditions, one mole of any gas occupies volume of 22.4 litres
7- The molar mass of sulphur in solid state is different from that in gas state
Because a molecule of gaseous sulphur contains 8 atoms of sulphur, while that of solid sulphur contains only 1 atom. So, They have different molar masses
8- The actual (practical) yield is always less than the theoretical yield
Because the reactants may be impure, side reactions may occur, the products may be volatile and spread in the air, or they can stick to the glass containers walls
1- The reaction of sodium chloride with silver nitrates forming a white ppt. of silver chloride and sodium nitrates
NaCl (aq) + AgNO3 NaNO3 (aq) + AgCl
2- The reaction of Nitric acid with potassium hydroxide solution forming potassium nitrate solution and water
HNO3 (aq) + KOH(aq) KNO2(aq) + H2O (l)
3- The reaction of potassium chromate with silver nitrate solution forming insoluble silver chromate ( red ppt.) and potassium nitrite
K2Cr2O7(aq) + 2AgNO3(aq) --------- 2KNO3(aq) + Ag2Cr2O7
4- The reaction of sulphuric acid with sodium hydroxide forming sodium sulphate and water (Neutralization reaction) ,
2NaOH(aq) + H2SO4(aq)------ Na2SO4(aq) + 2H2O (l)
1- Choose the correct answer
1- When 50 gm of CaCO3 decomposes thermally, ……..gm of CaO is formed (Ca = 40, C=12, O=16)
A- 28 B- 16 C- 76 D- 35
2- The volume of hydrogen required to form 11.2 L of water is……….
A- 22.4 L B- 11.2 L C- 68.2 L D- 44.8 L
3- One atomic unit equals …………. Gm
A- 1.66x10-24 B- 2.73x10-23 C- 1.75x10-15 D- 3.65x10-13
4- The unit used in IS for measuring the quantity of matter is…..
A- Mole B- Joule C- Calenda D- Kelvin
5- The mass of 44.8L of ammonia gas (NH3) in STP conditions is ………gm.
(N=14, H=1)
A- 0.5 B-2 C- 17 D- 34
6- If an amount of sodium has 3.01x1023 atoms, so its mass is ……..gm
A- 11.5 B- 0.5 C- 23 D-46
7- The chemical equation should be balanced according to ..............
A- Avogadro's law B- Gay-Lussac's law C- law of mass conservation
D- Law of energy conservation
8- 0.5 mole of carbon dioxide gas (CO2) weighs….. gm (C=12, O=16)
A- 22 B- 44 C-66 D-88
9- When 64 gm of oxygen reacts with hydrogen, ….. litres of water vapour (H2O) are formed
A- 11.2 B- 22.4 C- 44.8 D- 89.6
10- The no. of moles in 36g of water equals…………. (H=1,O=16)
A- 1 B-2 C- 3 D- 4
11- The no. of molecules in 128g of sulphur dioxide (SO2) equals…… (S=32, O=16)
A- 2 B- 6.02x1023 C- 3.01x1023 D- 12.04x1023
12- The no. of sodium ions resulted from the dissolution of 40g of sodium hydroxide (NaOH) equals…..(Na=23,O=16,H=1)
A- 2 B- 6.02x1023 C- 3.01x1023 D- 12.04x1023
13- The volume of 4g of hydrogen in STP conditions equals …
A- 11.2 B- 22.4 C- 44.8 D- 89.6
14- The volumes of reactant gases are inversely proportional to those of products according to……….
A- Avogadro's law B- Gay-Lussac's law C- law of mass conservation
D- Law of energy conservation
2- Solve the following problems
1- Find the no. of sodium ions resulted from the dissolution of 117g of sodium chloride (NaCl) in water (Na=23,Cl=35.5)
2- 26.5g of sodium carbonate (Na2CO3) reacted with an abundant amount of hydrochloric acid in STP conditions (Na=23,C=12, O=16) find:-
a- The no. of water molecules
b- The volume of carbon dioxide gas
3- Calculate the no. of moles in 144gm. of carbon (C=12)
4- Calculate the mass of 2.4 moles of calcium carbonate (CaCO3) (Ca=40, C=12, O=16)
5- Calculate the volume of 56g of nitrogen gas in STP conditions (N=14)
6- 23g of sodium (Na) reacted with water(H2O) forming sodium hydroxide (NaOH) and hydrogen gas (Na=23, O=16, H=1), Find:-
a- The no. of sodium ions resulted from the reaction
b- The volume of evolving hydrogen gas
7- Calculate the molar mass of gaseous phosphorus in STP conditions, and the no. of atoms in one mole of it.
3- Balance the following equations
1- N2(g) + H2(g) Δ NH3(g)
2- Cu(NO3)2(s) Δ CuO(s) + NO2(g) + O2(g)
3- 3 Al(s) + O2(g) Δ Al2O3(s)
The Answers
1- Choose
1- 28gm
The molar mass of calcium carbonate = 40 + 12 + 16+16+16 = 100 gm
The molar mass of calcium oxide (CaO) = 40+16 = 56
The no. of moles = the mass of the sample/ the molar mass = 50/100 = 0.5 mol
1 mole of calcium carbonate 1 mole of calcium oxide CaO
0.5 mole of calcium carbonate 0.5 mole of calcium oxide CaO
The mass of calcium oxide = 0.5 x molar mass = 0.5 x 56 = 28g
2- 11.2 L
2H2 + O2 2H2O
2 moles of hydrogen 2 moles of water
The volume of hydrogen gas = the volume of water
The volume of water = 11.2 L
The volume of hydrogen gas = 11.2 L
3- 1.66x10-24
4- Mole
5- 34g
The molar mass of ammonia gas = 14+1+1+1= 17gm
The no. of moles = the volume of gas in STP / 22.4 = 44.8/22.4 = 2 moles
The mass of 44.8 L of ammonia gas = no. moles x molar mass = 2 x 17 = 34g
6- 11.5gm
the molar mass of sodium = 23 gm
The no. of moles = no. of atoms / Avogadro's number = 0.5 mol.
The mass of sodium = 0.5 x 23 = 11.5 gm
7- law of mass conservation
8- 22 g
The molar mass of carbon dioxide gas = 16 + 16+12 = 44g
The mass of carbon dioxide = 0.5 x 44 = 22g
9- 89.6 L
2H2 + O2 2H2O
Molar mass of oxygen O2 = 16+16 = 32 gm
The no. of moles = 64 / 32 = 2 moles
1 mole of oxygen 2 moles of water
2 moles of oxygen 4 moles of water
The volume of water vapour in STP = 22.4 x the no. of moles = 22.4 x 4 = 89.6L
10- 2 moles
11- 12.04x1023 molecules
The molar mass of sulphur oxide = 32 +16+16 = 64 g
The no. of moles = 128/64 = 2 moles
The no. of molecules = no. of moles x Avogadro's no. = 12.04x1023 molecules
12- 6.02x1023 ions
One mole of sodium hydroxide = 23 + 16+1 = 40g
One mole of NaOH one mole of sodium ions
The no. of ions = 6.02x1023 ions
13- 44.8 L
The molar mass of hydrogen gas = 2g
The no. of moles = 4/2 = 2 moles
The volume of gas in STP = 2 x 22.4 = 44.8 L
14- Gay-Lussac's law
Solve the following problems
1- The molar mass of sodium chloride = 23+35.5 = 58.5
The no. of moles in 117 gm of NaCl = 117/58.5 = 2 moles
One mole of NaCl One mole of sodium positive ions
Two moles of NaCl Two moles of sodium positive ions
The no. of sodium ions = 2 x 6.02x1023 = 12.04x1023 ions
2- Na2CO3 +2 HCl NaCl +H2O+ CO2
a- The no. of water molecules
The molar mass of sodium carbonate = 23+23+16+16+16+12 = 106g
The no. of moles in 26.5 g = 26.5 / 106 = 0.25 mole
One mole of Na2CO3 one mole of water
0.25 mole of Na2CO3 0.25 mole of water
The no. of water molecules = 0.25 x 6.02x1023 = 1.505x1023 molecules
b- The volume of CO2 gas in STP conditions
One mole of Na2CO3 one mole of carbon dioxide
0.25 mole of Na2CO3 0.25 mole of carbon dioxide
The volume of the gas = 0.25 x 22.4 = 5.6 Litres
3- The molar mass of carbon element = 12+12 = 24
(N.B: the molar masses of diatomic nonmetal elements are calculated this way)
The no. of moles in 144 gm of carbon = 144/24 = 6 moles
4- The molar mass of CaCO3 = 40+12+16+16+16= 100g
The mass of 2.4 moles = 100 x 2.4 = 240 g
5- The molar mass of nitrogen gas = 14+14=28 gm
The no. of moles in 56 g = 56/28 = 2 moles
The volume of nitrogen gas in STP conditions = 22.4 x 2 = 44.8 L
6- 2Na + 2H2O 2NaOH +2 H2O
a- the no. of sodium ions
The molar mass of sodium = 23 g
One mole of sodium one mole of sodium hydroxide solution one mole of sodium ions
The no. of ions = 6.02x1023 ions
b- The volume of hydrogen gas in STP conditions
One mole of sodium one mole of hydrogen gas
The volume of hydrogen gas in STP conditions = 22.4L
7- Gaseous phosphorus molecule consists of 4 atoms
The molar mass of gaseous phosphorus = 4 x 31 = 124 gm
The no. of atoms in one mole = Avogadro's no. x the no. of atoms per molecule
= 6.02x1023 x 4 = 24.08x1023 atoms
Balance the following equations
1- N2(g) + H2(g) Δ NH3(g)
The no. of hydrogen atoms on the right side of the equation is 3, while that on the left side of it is 2. To balance the no. of hydrogen atoms on both sides, we increase it on both of them to 6
N2(g) +3 H2(g) Δ 2NH3(g)
2- Cu(NO3)2(s) Δ CuO(s) + NO2(g) + O2(g)
There are 2 nitrogen atoms on the left side of the equation, while there is only 1 nitrogen atom on the right side of it. To balance the no. of nitrogen atoms, we increase the no. of nitrogen atoms on the right side to 2
Cu(NO3)2(s) Δ CuO(s) + 2NO2(g) + O2(g)
3- Al + 3 O2 Al2
we find that there are 3 oxygen atoms on the right side of the equation, while there are only 2 on the left side of it. To balance the no. of oxygen atoms on both
sides, we should increase the no. of oxygen atoms on both of them to 6 (6 is the least common multiple of 2 and 3)
Al + 3 O2 2 Al2O3
There are four aluminium atoms on the right side of the equation, while there's a single atom on the left size. To balance the no. of aluminium atoms on both sides, we increase the no. of atoms in left size to 4 aluminium atoms
4Al + 3 O2 2 Al2O3
1- Choose
1- The empirical formula of C4H8O2 is………….
A- C2H4O B- C4H2O C- CH4O2 D- C2H8O2
2- The no. of empirical formulas in C2H2O4 is …..
A-1 B-2 C- 3 D-4
3- If the empirical formula of a compound is CH2 and its molar mass if 56g, its molecular formula is…..
A- C2H4 B- C4H8 C- C3H6 D- C5H10
4- If the molecular formula of Vitamin C is C6H8O6, its empirical formula is….
A- C3H4O3 B- C3H4O6 C- C3H8O3 D- C3H6O
5- The empirical formula CH2O describes……
A- CH3 COOH B- C6H12O6 C- HCHO D- All the previous answers
6- The hydrocarbon compound formed from the reaction of 0.1 mol. of carbon atoms with 0.4 mol. of hydrogen atoms is……
A- CH4 B- C2H4 C- C4H8 D- C3H8
Write the scientific term
1- A method to describe chemical formula, the quantities of reactants
and products, and the conditions for chemical reaction
2- The mass of atoms or molecules in grams
3- A constant no. of the ions, molecules or atoms in one mole of matter
4- A formula describes the actual no. of atoms in molecules
5- The amount of matter we get practically from the reaction
6- The sum of the atoms masses forming the molecule
7- The volumes of the reactant and products gases have certain ratios
8- The equals volume of gases in the same conditions of temperature and pressure have the same no. of molecules
9- A formula which describe the simplest ratios between the atoms forming molecules
10- The amount of reactants we expect to get from the reaction
Solve the following problems
1- Find the molecular formula of a compound containing 85.7% carbon and 14.3% hydrogen whose molar mass is 42g
2- 130g of silver chloride (AgCl) precipitated when a mole of sodium chloride (NaCl) reacted with silver nitrates (AgNO3), calculate the percentage yield (percentage of actual yield) (Ag=108, N=14, Cl= 35.5, Na= 23,O=16)
3- Calculate the weight percent of iron in FeCO3 (Fe=56, C=12, O=16)
4- Calculate the weight percent of the elements forming Glucose sugar C6H12O6
(C=12, H=1, O=16)
The Answers
1- Choose the correct answer
1- C2H4O
2- 2
3- C4H8
The molar mass of CH2 the empirical formula = 12+1+1 = 14g
The no. of units = molar mass of the compound / the molar mass of the empirical formula = 56/14 = 4 units
The molecular formula = 4 x CH2 = C4H8
4- C3H4O3
5- All the previous answers
6- CH4
Write the scientific term
1- Chemical reaction
2- Mole
3- Avogadro's number
4- Molecular formula
5- Practical yield
6- Molecular mass
7- Gay-Lussac's law
8- Avogadro's law
9- Empirical formula
10- Theoretical formula
Solve the following problems
1-
Hydrogen : Carbon
14.3 85.7
12 1
14.3 7.1416
2 1
The empirical formula: CH2
The molar mass of the compound = 12 + 1+1+ = 14 gm
The no. of units = 42/14 = 3 units
The molecular formula = CH2 x 3 = C3H6
2- NaCl (aq) + AgNO3 (aq) NaNO3 + AgCl(s)
Mole of sodium chloride = 23+35.5 = 58.5
Mole of silver chloride = 108+58.5 = 143.5
The practical yield = 140g
The theoretical yield = 143.5g
The percentage yield = 140/143.5 x 100 = 97.56%
3- The molar mass of FeCO3 = 56+12+16+16+16 = 116g
The mass of iron forming one mole of FeCO3 = 56g
The weight percent = 56/116 x 100 = 48.26%
4- The molar mass of glucose = 12x6 + 12x1 +16x6 = 72+12+96 = 180g
The mass of carbon = 12x6 = 72g
The mass of hydrogen = 12x1 = 12g
The mass of oxygen = 16x6= 96g
The weight percent of carbon = 72/180 x 100 = 40%
The weight percent of hydrogen = 12/180 x 100 = 6.7%
The weight percent of oxygen = 96/180 x 100 = 53.3%
