JK-TK/Newdata huggingface dataset:
Mass of boat before heating(M2) = 19.1 Mass of empty boat(M1) = - 15.6g Mass of copper(II)Oxide 3.5 g 2. Calculate the mass of (i) Oxygen. Mass of boat before heating(M2) = 19.1 Mass of boat after heating (M3) = - 18.4g Mass of oxygen = 0.7 g (ii)Copper Mass of copper(II)Oxide = 3.5 g Mass of oxygen = 0.7 g Mass of oxygen = 2.8 g 3. Calculate the number of moles of: (i) Copper used (Cu = 63.5) number of moles of copper = mass used => 2.8 = 0.0441moles Molar mass 63.5 (ii) Oxygen used (O = 16.0) number of moles of oxygen = mass used => 0.7 = 0.0441moles Molar mass 16.0 4. Determine the mole ratio of the reactants Moles of copper = 0.0441moles = 1 => Mole ratio Cu: O = 1:1 Moles of oxygen 0.0441moles 1 5.What is the empirical, formula of copper oxide formed. CuO (copper(II)oxide 6. State and explain the observations made during the experiment. Observation Colour change from black to brown Explanation Copper(II)oxide powder is black. On heating it is reduced by a suitable reducing agent to brown copper metal. 7. Explain why magnesium oxide would be unsuitable in a similar experiment as the one above. Magnesium is high in the reactivity series. None of the above reducing agents is strong enough to reduce the oxide to the metal. 10 10 8.
Magnesium is high in the reactivity series. None of the above reducing agents is strong enough to reduce the oxide to the metal. 10 10 8. Write the equation for the reaction that would take place when the reducing agent is: (i) Hydrogen CuO(s) + H2(g) -> Cu(s) + H2O(l) (Black) (brown) (colourless liquid form on cooler parts ) (ii)Carbon(II)oxide CuO(s) + CO (g) -> Cu(s) + CO2(g) (Black) (brown) (colourless gas, form white ppt with lime water ) (iii)Ammonia 3CuO(s) + 2NH3(g) -> 3Cu(s) + N2 (g) + 3H2O(l) (Black) (brown) (colourless liquid form on cooler parts ) 9. Explain why the following is necessary during the above experiment; (i)A stream of dry hydrogen gas should be passed before heating copper (II) Oxide. Air combine with hydrogen in presence of heat causing an explosion (ii)A stream of dry hydrogen gas should be passed after heating copper (II) Oxide has been stopped. Hot metallic copper can be re-oxidized back to copper(II)oxide (iii) A stream of excess carbon (II)oxide gas should be ignited to burn Carbon (II)oxide is highly poisonous/toxic. On ignition it burns to form less toxic carbon (IV)oxide gas. 10. State two sources of error in this experiment. (i)All copper(II)oxide may not be reduced to copper. (ii)Some copper(II)oxide may be blown out the boat by the reducing agent. 4.Theoreticaly the empirical formula of a compound can be determined as in the following examples. (a)A oxide of copper contain 80% by mass of copper. Determine its empirical formula.
4.Theoreticaly the empirical formula of a compound can be determined as in the following examples. (a)A oxide of copper contain 80% by mass of copper. Determine its empirical formula. (Cu = 63.5, 16.0)
11 11 % of Oxygen = 100% - % of Copper => 100- 80 = 20% of Oxygen Element Copper Oxygen Symbol Cu O Moles present = % composition Molar mass 80 63.5 20 16 Divide by the smallest value 1.25 1.25 1.25 1.25 Mole ratios 1 1 Empirical formula is CuO (b)1.60g of an oxide of Magnesium contain 0.84g by mass of Magnesium. Determine its empirical formula(Mg = 24.0, 16.0) Mass of Oxygen = 1.60 – 0.84 => 0.56 g of Oxygen Element Magnesium Oxygen Symbol Mg O Moles present = % composition Molar mass 0.84 24 0.56 16 Divide by the smallest value 0.35 0.35 0.35 0.35 Mole ratios 1 1 Empirical formula is MgO (c)An oxide of Silicon contain 47% by mass of Silicon. What is its empirical formula(Si = 28.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2
12 12 (d)A compound contain 70% by mass of Iron and 30% Oxygen.
(Cu = 63.5, 16.0)
11 11 % of Oxygen = 100% - % of Copper => 100- 80 = 20% of Oxygen Element Copper Oxygen Symbol Cu O Moles present = % composition Molar mass 80 63.5 20 16 Divide by the smallest value 1.25 1.25 1.25 1.25 Mole ratios 1 1 Empirical formula is CuO (b)1.60g of an oxide of Magnesium contain 0.84g by mass of Magnesium. Determine its empirical formula(Mg = 24.0, 16.0) Mass of Oxygen = 1.60 – 0.84 => 0.56 g of Oxygen Element Magnesium Oxygen Symbol Mg O Moles present = % composition Molar mass 0.84 24 0.56 16 Divide by the smallest value 0.35 0.35 0.35 0.35 Mole ratios 1 1 Empirical formula is MgO (c)An oxide of Silicon contain 47% by mass of Silicon. What is its empirical formula(Si = 28.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2
12 12 (d)A compound contain 70% by mass of Iron and 30% Oxygen. What is its empirical formula(Fe = 56.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2 2.During heating of a hydrated copper (II)sulphate(VI) crystals, the following readings were obtained: Mass of evaporating dish =300.0g Mass of evaporating dish + hydrated salt = 305.0g Mass of evaporating dish + anhydrous salt = 303.2g Calculate the number of water of crystallization molecules in hydrated copper (II)sulphate(VI) (Cu =64.5, S = 32.0,O=16.0, H = 1.0) Working Mass of Hydrated salt = 305.0g -300.0g = 5.0g Mass of anhydrous salt = 303.2 g -300.0g = 3.2 g Mass of water in hydrated salt = 5.0g -3.2 g = 1.8g Molar mass of water(H2O) = 18.0g Molar mass of anhydrous copper (II)sulphate(VI) (CuSO4) = 160.5g Element/compound anhydrous copper (II) sulphate(VI) Oxygen Symbol Si O Moles present = composition by mass Molar mass 3,2 160.5 1.8 18 Divide by the smallest value 0.0199 0.0199 0.1 18 Mole ratios 1 5 The empirical formula of hydrated salt = CuSO4.5H2O
13 13 Hydrated salt has five/5 molecules of water of crystallizations 4.
Determine its empirical formula(Mg = 24.0, 16.0) Mass of Oxygen = 1.60 – 0.84 => 0.56 g of Oxygen Element Magnesium Oxygen Symbol Mg O Moles present = % composition Molar mass 0.84 24 0.56 16 Divide by the smallest value 0.35 0.35 0.35 0.35 Mole ratios 1 1 Empirical formula is MgO (c)An oxide of Silicon contain 47% by mass of Silicon. What is its empirical formula(Si = 28.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2
12 12 (d)A compound contain 70% by mass of Iron and 30% Oxygen. What is its empirical formula(Fe = 56.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2 2.During heating of a hydrated copper (II)sulphate(VI) crystals, the following readings were obtained: Mass of evaporating dish =300.0g Mass of evaporating dish + hydrated salt = 305.0g Mass of evaporating dish + anhydrous salt = 303.2g Calculate the number of water of crystallization molecules in hydrated copper (II)sulphate(VI) (Cu =64.5, S = 32.0,O=16.0, H = 1.0) Working Mass of Hydrated salt = 305.0g -300.0g = 5.0g Mass of anhydrous salt = 303.2 g -300.0g = 3.2 g Mass of water in hydrated salt = 5.0g -3.2 g = 1.8g Molar mass of water(H2O) = 18.0g Molar mass of anhydrous copper (II)sulphate(VI) (CuSO4) = 160.5g Element/compound anhydrous copper (II) sulphate(VI) Oxygen Symbol Si O Moles present = composition by mass Molar mass 3,2 160.5 1.8 18 Divide by the smallest value 0.0199 0.0199 0.1 18 Mole ratios 1 5 The empirical formula of hydrated salt = CuSO4.5H2O
13 13 Hydrated salt has five/5 molecules of water of crystallizations 4. The molecular formula is the actual number of each kind of atoms present in a molecule of a compound.
What is its empirical formula(Si = 28.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2
12 12 (d)A compound contain 70% by mass of Iron and 30% Oxygen. What is its empirical formula(Fe = 56.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2 2.During heating of a hydrated copper (II)sulphate(VI) crystals, the following readings were obtained: Mass of evaporating dish =300.0g Mass of evaporating dish + hydrated salt = 305.0g Mass of evaporating dish + anhydrous salt = 303.2g Calculate the number of water of crystallization molecules in hydrated copper (II)sulphate(VI) (Cu =64.5, S = 32.0,O=16.0, H = 1.0) Working Mass of Hydrated salt = 305.0g -300.0g = 5.0g Mass of anhydrous salt = 303.2 g -300.0g = 3.2 g Mass of water in hydrated salt = 5.0g -3.2 g = 1.8g Molar mass of water(H2O) = 18.0g Molar mass of anhydrous copper (II)sulphate(VI) (CuSO4) = 160.5g Element/compound anhydrous copper (II) sulphate(VI) Oxygen Symbol Si O Moles present = composition by mass Molar mass 3,2 160.5 1.8 18 Divide by the smallest value 0.0199 0.0199 0.1 18 Mole ratios 1 5 The empirical formula of hydrated salt = CuSO4.5H2O
13 13 Hydrated salt has five/5 molecules of water of crystallizations 4. The molecular formula is the actual number of each kind of atoms present in a molecule of a compound. The empirical formula of an ionic compound is the same as the chemical formula but for simple molecular structured compounds, the empirical formula may not be the same as the chemical formula.
What is its empirical formula(Fe = 56.0, 16.0) Mass of Oxygen = 100 – 47 => 53% of Oxygen Element Silicon Oxygen Symbol Si O Moles present = % composition Molar mass 47 28 53 16 Divide by the smallest value 1.68 1.68 3.31 1.68 Mole ratios 1 1.94 = 2 Empirical formula is SiO2 2.During heating of a hydrated copper (II)sulphate(VI) crystals, the following readings were obtained: Mass of evaporating dish =300.0g Mass of evaporating dish + hydrated salt = 305.0g Mass of evaporating dish + anhydrous salt = 303.2g Calculate the number of water of crystallization molecules in hydrated copper (II)sulphate(VI) (Cu =64.5, S = 32.0,O=16.0, H = 1.0) Working Mass of Hydrated salt = 305.0g -300.0g = 5.0g Mass of anhydrous salt = 303.2 g -300.0g = 3.2 g Mass of water in hydrated salt = 5.0g -3.2 g = 1.8g Molar mass of water(H2O) = 18.0g Molar mass of anhydrous copper (II)sulphate(VI) (CuSO4) = 160.5g Element/compound anhydrous copper (II) sulphate(VI) Oxygen Symbol Si O Moles present = composition by mass Molar mass 3,2 160.5 1.8 18 Divide by the smallest value 0.0199 0.0199 0.1 18 Mole ratios 1 5 The empirical formula of hydrated salt = CuSO4.5H2O
13 13 Hydrated salt has five/5 molecules of water of crystallizations 4. The molecular formula is the actual number of each kind of atoms present in a molecule of a compound. The empirical formula of an ionic compound is the same as the chemical formula but for simple molecular structured compounds, the empirical formula may not be the same as the chemical formula. The molecular formula is a multiple of empirical formula .It is determined from the relationship: (i) n = Relative formular mass Relative empirical formula where n is a whole number.
The molecular formula is the actual number of each kind of atoms present in a molecule of a compound. The empirical formula of an ionic compound is the same as the chemical formula but for simple molecular structured compounds, the empirical formula may not be the same as the chemical formula. The molecular formula is a multiple of empirical formula .It is determined from the relationship: (i) n = Relative formular mass Relative empirical formula where n is a whole number. (ii) Relative empirical formula x n = Relative formular mass where n is a whole number. Practice sample examples 1. A hydrocarbon was found to contain 92.3% carbon and the remaining Hydrogen. If the molecular mass of the compound is 78, determine the molecular formula(C=12.0, H =1.0) Mass of Hydrogen = 100 – 92.3 => 7.7% of Oxygen Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 92.3 12 7.7 1 Divide by the smallest value 7.7 7.7 7.7 7.7 Mole ratios 1 1 Empirical formula is CH The molecular formular is thus determined : n = Relative formular mass = 78 = 6
14 14 Relative empirical formula 13 The molecular formula is (C H ) x 6 = C6H6 2. A compound of carbon, hydrogen and oxygen contain 54.55% carbon, 9.09% and remaining 36.36% oxygen. If its relative molecular mass is 88, determine its molecular formula(C=12.0, H =1.0, O= 16.0) Element Carbon Hydrogen Oxygen Symbol C H O Moles present = % composition Molar mass 54.55 12 9.09 1 36.36 16 Divide by the smallest value 4.5458 2.2725 9.09 2.2725 2.2725 2.2725 Mole ratios 2 4 1 Empirical formula is C2H4O The molecular formula is thus determined : n = Relative formular mass = 88 = 2 Relative empirical formula 44 The molecular formula is (C2H4O ) x 2 = C4H8O2.
If the molecular mass of the compound is 78, determine the molecular formula(C=12.0, H =1.0) Mass of Hydrogen = 100 – 92.3 => 7.7% of Oxygen Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 92.3 12 7.7 1 Divide by the smallest value 7.7 7.7 7.7 7.7 Mole ratios 1 1 Empirical formula is CH The molecular formular is thus determined : n = Relative formular mass = 78 = 6
14 14 Relative empirical formula 13 The molecular formula is (C H ) x 6 = C6H6 2. A compound of carbon, hydrogen and oxygen contain 54.55% carbon, 9.09% and remaining 36.36% oxygen. If its relative molecular mass is 88, determine its molecular formula(C=12.0, H =1.0, O= 16.0) Element Carbon Hydrogen Oxygen Symbol C H O Moles present = % composition Molar mass 54.55 12 9.09 1 36.36 16 Divide by the smallest value 4.5458 2.2725 9.09 2.2725 2.2725 2.2725 Mole ratios 2 4 1 Empirical formula is C2H4O The molecular formula is thus determined : n = Relative formular mass = 88 = 2 Relative empirical formula 44 The molecular formula is (C2H4O ) x 2 = C4H8O2. 4.A hydrocarbon burns completely in excess air to form 5.28 g of carbon (IV) oxide and 2,16g of water. If the molecular mass of the hydrocarbon is 84, draw and name its molecular structure. Since a hydrocarbon is a compound containing Carbon and Hydrogen only.
4.A hydrocarbon burns completely in excess air to form 5.28 g of carbon (IV) oxide and 2,16g of water. If the molecular mass of the hydrocarbon is 84, draw and name its molecular structure. Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 => Molar mass of CO2 12 x 5.28 = 1.44g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 2.16 = 0.24g√ 18 Element Carbon Hydrogen
15 15 Symbol C H Moles present = mass Molar mass 1.44g 12 0.24g√ 1 Divide by the smallest value 0.12 0.12 0.24 0.12 Mole ratios 1 2√ Empirical formula is CH2√ The molecular formular is thus determined : n = Relative formular mass = 84 = 6√ Relative empirical formula 14 The molecular formula is (CH2 ) x 6 = C6H12. √ molecular name Hexene√/Hex-1-ene (or any position isomer of Hexene) Molecular structure H H H H H H H C C C C C C H√ H H H H 5. Compound A contain 5.2% by mass of Nitrogen .The other elements present are Carbon, hydrogen and Oxygen. On combustion of 0.085g of A in excess Oxygen,0.224g of carbon(IV)oxide and 0.0372g of water was formed.
√ molecular name Hexene√/Hex-1-ene (or any position isomer of Hexene) Molecular structure H H H H H H H C C C C C C H√ H H H H 5. Compound A contain 5.2% by mass of Nitrogen .The other elements present are Carbon, hydrogen and Oxygen. On combustion of 0.085g of A in excess Oxygen,0.224g of carbon(IV)oxide and 0.0372g of water was formed. Determine the empirical formula of A (N=14.0, O=16.0 , C=12.0 , H=1.0) Mass of N in A = 5.2% x 0.085 = 0.00442 g Mass of C in A = 12 x 0.224 = 0.0611g 44 Mass of H in A = 2 x 0.0372 = 0.0041g 18 Mass of O in A = 0.085g – 0.004442g = 0.0806g (Mass of C,H,O) => 0.0611g + 0.0041g = 0.0652g (Mass of C,H) 0.0806g (Mass of C,H,O)- 0.0652g (Mass of C,H) = 0.0154 g Element Nitrogen Carbon Hydrogen Oxygen Symbol N C H O Moles present = mass Molar mass 0.00442 g 14 0.0611g 12 0.0041g 1 0.0154 g 16 Divide by the smallest value 0.00032 0.00509 0.0041g 0.00096
16 16 0.00032 0.00032 0.00032 0.00032 Mole ratios 1 16 13 3 Empirical formula = C16H13NO3 (d)Molar gas volume The volume occupied by one mole of all gases at the same temperature and pressure is a constant.It is: (i) 24dm3/24litres/24000cm3 at room temperature(25oC/298K)and pressure(r.t.p). i.e.
22.4dm3 -> 6.0 x1023 2.24dm3 -> 2.24 x 6.0 x1023 22.4 =6.0 x1022 molecules = (CO2) = 3 x 6.0 x1022. = 1.8 x1023 atoms 2. 0.135 g of a gaseous hydrocarbon X on complete combustion produces 0.41g of carbon(IV)oxide and 0.209g of water.0.29g of X occupy 120cm3 at room temperature and 1 atmosphere pressure .Name X and draw its molecular structure.(C=12.0,O= 16.O,H=1.0,1 mole of gas occupies 24dm3 at r.t.p) Molar mass CO2= 44 gmole-1√ Molar mass H2O = 18 gmole-1√ Molar mass X = 0.29 x (24 x 1000)cm3 = 58 gmole-1√ 120cm3 Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 => Molar mass of CO2 12 x 0.41 = 0.1118g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 0.209 = 0.0232g√ 18 Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 0.g118 12 0.0232g√ 1 Divide by the smallest value 0.0093 0.0093 0.0232 0.0093√ Mole ratios 1 x2 2.5x2
18 18 2 5√ Empirical formula is C2H5√ The molecular formular is thus determined : n = Relative formular mass = 58 = 2√ Relative empirical formula 29 The molecular formula is (C2H5 ) x 2 = C4H10.√ Molecule name Butane Molecula structure H H H H H C C C C H√ H H H H (e)Gravimetric analysis Gravimetric analysis is the relationship between reacting masses and the volumes and /or masses of products.
= 1.8 x1023 atoms 2. 0.135 g of a gaseous hydrocarbon X on complete combustion produces 0.41g of carbon(IV)oxide and 0.209g of water.0.29g of X occupy 120cm3 at room temperature and 1 atmosphere pressure .Name X and draw its molecular structure.(C=12.0,O= 16.O,H=1.0,1 mole of gas occupies 24dm3 at r.t.p) Molar mass CO2= 44 gmole-1√ Molar mass H2O = 18 gmole-1√ Molar mass X = 0.29 x (24 x 1000)cm3 = 58 gmole-1√ 120cm3 Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 => Molar mass of CO2 12 x 0.41 = 0.1118g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 0.209 = 0.0232g√ 18 Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 0.g118 12 0.0232g√ 1 Divide by the smallest value 0.0093 0.0093 0.0232 0.0093√ Mole ratios 1 x2 2.5x2
18 18 2 5√ Empirical formula is C2H5√ The molecular formular is thus determined : n = Relative formular mass = 58 = 2√ Relative empirical formula 29 The molecular formula is (C2H5 ) x 2 = C4H10.√ Molecule name Butane Molecula structure H H H H H C C C C H√ H H H H (e)Gravimetric analysis Gravimetric analysis is the relationship between reacting masses and the volumes and /or masses of products. All reactants are in mole ratios to their products in accordance to their stoichiometric equation.
0.135 g of a gaseous hydrocarbon X on complete combustion produces 0.41g of carbon(IV)oxide and 0.209g of water.0.29g of X occupy 120cm3 at room temperature and 1 atmosphere pressure .Name X and draw its molecular structure.(C=12.0,O= 16.O,H=1.0,1 mole of gas occupies 24dm3 at r.t.p) Molar mass CO2= 44 gmole-1√ Molar mass H2O = 18 gmole-1√ Molar mass X = 0.29 x (24 x 1000)cm3 = 58 gmole-1√ 120cm3 Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 => Molar mass of CO2 12 x 0.41 = 0.1118g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 0.209 = 0.0232g√ 18 Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 0.g118 12 0.0232g√ 1 Divide by the smallest value 0.0093 0.0093 0.0232 0.0093√ Mole ratios 1 x2 2.5x2
18 18 2 5√ Empirical formula is C2H5√ The molecular formular is thus determined : n = Relative formular mass = 58 = 2√ Relative empirical formula 29 The molecular formula is (C2H5 ) x 2 = C4H10.√ Molecule name Butane Molecula structure H H H H H C C C C H√ H H H H (e)Gravimetric analysis Gravimetric analysis is the relationship between reacting masses and the volumes and /or masses of products. All reactants are in mole ratios to their products in accordance to their stoichiometric equation. Using the mole ration of reactants and products any volume and/or mass can be determined as in the examples: 1.
Then: Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 => Molar mass of CO2 12 x 0.41 = 0.1118g√ 44 Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O => Molar mass of H2O 2 x 0.209 = 0.0232g√ 18 Element Carbon Hydrogen Symbol C H Moles present = % composition Molar mass 0.g118 12 0.0232g√ 1 Divide by the smallest value 0.0093 0.0093 0.0232 0.0093√ Mole ratios 1 x2 2.5x2
18 18 2 5√ Empirical formula is C2H5√ The molecular formular is thus determined : n = Relative formular mass = 58 = 2√ Relative empirical formula 29 The molecular formula is (C2H5 ) x 2 = C4H10.√ Molecule name Butane Molecula structure H H H H H C C C C H√ H H H H (e)Gravimetric analysis Gravimetric analysis is the relationship between reacting masses and the volumes and /or masses of products. All reactants are in mole ratios to their products in accordance to their stoichiometric equation. Using the mole ration of reactants and products any volume and/or mass can be determined as in the examples: 1. Calculate the volume of carbon(IV)oxide at r.t.p produced when 5.0 g of calcium carbonate is strongly heated.(Ca=40.0, C= 12.0,O = 16.0,1 mole of gas =22.4 at r.t.p) Chemical equation CaCO3(s) -> CaO(s) + CO2(g) Mole ratios 1: 1: 1 Molar Mass CaCO3 =100g Method 1 100g CaCO3(s) -> 24dm3 CO2(g) at r.t.p 5.0 g CaCO3(s) -> 5.0 g x 24dm3 = 1.2dm3/1200cm3 100g Method 2
19 19 Moles of 5.0 g CaCO3(s) = 5.0 g = 0.05 moles 100 g Mole ratio 1:1 Moles of CO2(g) = 0.05moles Volume of CO2(g) = 0.05 x 24000cm3 =1200cm3 /1.2dm3 2.
All reactants are in mole ratios to their products in accordance to their stoichiometric equation. Using the mole ration of reactants and products any volume and/or mass can be determined as in the examples: 1. Calculate the volume of carbon(IV)oxide at r.t.p produced when 5.0 g of calcium carbonate is strongly heated.(Ca=40.0, C= 12.0,O = 16.0,1 mole of gas =22.4 at r.t.p) Chemical equation CaCO3(s) -> CaO(s) + CO2(g) Mole ratios 1: 1: 1 Molar Mass CaCO3 =100g Method 1 100g CaCO3(s) -> 24dm3 CO2(g) at r.t.p 5.0 g CaCO3(s) -> 5.0 g x 24dm3 = 1.2dm3/1200cm3 100g Method 2
19 19 Moles of 5.0 g CaCO3(s) = 5.0 g = 0.05 moles 100 g Mole ratio 1:1 Moles of CO2(g) = 0.05moles Volume of CO2(g) = 0.05 x 24000cm3 =1200cm3 /1.2dm3 2. 1.0g of an alloy of aluminium and copper were reacted with excess hydrochloric acid.