q01

Finding Composition of Minerals:

Finding Composition of Minerals The mineral hydromagnesite is a hydrated carbonate of magnesium, with the formula Mg x ( CO 3 ) y (OH) z . n H 2 O and a molar mass of 466. When 2 grams of a pure sample of hydromagnesite is heated until its mass is constant, 0.756 grams of carbon dioxide was given off, together with steam. The remaining white solid from the above decomposition was completely dissolved in 100 cm 3 of solution of 1.0 mol dm -3 hydrochloric acid. The resultant solution was transferred into a volumetric flask and diluted to 500 cm 3 with deionised water. A 50 cm 3 aliquot of this diluted solution required 57.0 cm 3 of a 0.10 mol dm -3 solution of sodium hydroxide for complete neutralization. Calculate the value of x and y. Deduce the values of z and n.

Finding Composition of Minerals:

Finding Composition of Minerals The mineral hydromagnesite is a hydrated carbonate of magnesium, with the formula Mg x ( CO 3 ) y (OH) z . n H 2 O and a molar mass of 466. When 2 grams of a pure sample of hydromagnesite is heated until its mass is constant, 0.756 grams of carbon dioxide was given off, together with steam. The remaining white solid from the above decomposition was completely dissolved in 100 cm 3 of solution of 1.0 mol dm -3 hydrochloric acid. The resultant solution was transferred into a volumetric flask and diluted to 500 cm 3 with deionised water. A 50 cm 3 aliquot of this diluted solution required 57.0 cm 3 of a 0.10 mol dm -3 solution of sodium hydroxide for complete neutralization. Calculate the value of x and y. Deduce the values of z and n. There are 3 reactions in this questions

Slide 3:

Reactions involved in this question are : Reaction 1 : Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O Reaction 2 : MgO + 2 HCl → MgCl 2 + H 2 O Reaction 3 : NaOH + HCl → NaCl + H 2 O The mineral hydromagnesite is a hydrated carbonate of magnesium, with the formula Mg x ( CO 3 ) y (OH) z . n H 2 O and a molar mass of 466. When 2 grams of a pure sample of hydromagnesite is heated until its mass is constant, 0.756 grams of carbon dioxide was given off, together with steam. The remaining white solid from the above decomposition was completely dissolved in 100 cm 3 of solution of 1.0 mol dm -3 hydrochloric acid. The resultant solution was transferred into a volumetric flask and diluted to 500 cm 3 with deionised water. A 50 cm 3 aliquot of this diluted solution required 57.0 cm 3 of a 0.10 mol dm -3 solution of sodium hydroxide for complete neutralization. Calculate the value of x and y. Deduce the values of z and n.

Slide 4:

Reactions involved in this question are : Reaction 1 : Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O Reaction 2 : MgO + 2 HCl → MgCl 2 + H 2 O Reaction 3 : NaOH + HCl → NaCl + H 2 O The mineral hydromagnesite is a hydrated carbonate of magnesium, with the formula Mg x ( CO 3 ) y (OH) z . n H 2 O and a molar mass of 466. When 2 grams of a pure sample of hydromagnesite is heated until its mass is constant, 0.756 grams of carbon dioxide was given off, together with steam. The remaining white solid from the above decomposition was completely dissolved in 100 cm 3 of solution of 1.0 mol dm -3 hydrochloric acid. The resultant solution was transferred into a volumetric flask and diluted to 500 cm 3 with deionised water. A 50 cm 3 aliquot of this diluted solution required 57.0 cm 3 of a 0.10 mol dm -3 solution of sodium hydroxide for complete neutralization. Calculate the value of x and y. Deduce the values of z and n.

Slide 5:

Reactions involved in this question are : Reaction 1 : Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O Reaction 2 : MgO + 2 HCl → MgCl 2 + H 2 O Reaction 3 : NaOH + HCl → NaCl + H 2 O The mineral hydromagnesite is a hydrated carbonate of magnesium, with the formula Mg x ( CO 3 ) y (OH) z . n H 2 O and a molar mass of 466. When 2 grams of a pure sample of hydromagnesite is heated until its mass is constant, 0.756 grams of carbon dioxide was given off, together with steam. The remaining white solid from the above decomposition was completely dissolved in 100 cm 3 of solution of 1.0 mol dm -3 hydrochloric acid. The resultant solution was transferred into a volumetric flask and diluted to 500 cm 3 with deionised water. A 50 cm 3 aliquot of this diluted solution required 57.0 cm 3 of a 0.10 mol dm -3 solution of sodium hydroxide for complete neutralization. Calculate the value of x and y. Deduce the values of z and n.

Slide 6:

Reactions involved in this question are : Reaction 1 : Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O Reaction 2 : MgO + 2 HCl → MgCl 2 + H 2 O Reaction 3 : NaOH + HCl → NaCl + H 2 O The mineral hydromagnesite is a hydrated carbonate of magnesium, with the formula Mg x ( CO 3 ) y (OH) z . n H 2 O and a molar mass of 466. When 2 grams of a pure sample of hydromagnesite is heated until its mass is constant, 0.756 grams of carbon dioxide was given off, together with steam. The remaining white solid from the above decomposition was completely dissolved in 100 cm 3 of solution of 1.0 mol dm -3 hydrochloric acid. The resultant solution was transferred into a volumetric flask and diluted to 500 cm 3 with deionised water. A 50 cm 3 aliquot of this diluted solution required 57.0 cm 3 of a 0.10 mol dm -3 solution of sodium hydroxide for complete neutralization. Calculate the value of x and y. Deduce the values of z and n.

Slide 7:

Reactions involved in this question are : Reaction 1 : Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O Reaction 2 : MgO + 2 HCl → MgCl 2 + H 2 O Reaction 3 : NaOH + HCl → NaCl + H 2 O The mineral hydromagnesite is a hydrated carbonate of magnesium, with the formula Mg x ( CO 3 ) y (OH) z . n H 2 O and a molar mass of 466. When 2 grams of a pure sample of hydromagnesite is heated until its mass is constant, 0.756 grams of carbon dioxide was given off, together with steam. The remaining white solid from the above decomposition was completely dissolved in 100 cm 3 of solution of 1.0 mol dm -3 hydrochloric acid. The resultant solution was transferred into a volumetric flask and diluted to 500 cm 3 with deionised water. A 50 cm 3 aliquot of this diluted solution required 57.0 cm 3 of a 0.10 mol dm -3 solution of sodium hydroxide for complete neutralization. Calculate the value of x and y. Deduce the values of z and n.

Slide 8:

Reactions involved in this question are : Reaction 1 : Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O Reaction 2 : MgO + 2 HCl → MgCl 2 + H 2 O Reaction 3 : NaOH + HCl → NaCl + H 2 O

Slide 9:

Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 Mol ratio 1 x y Mol Mr 466 - 44 Mass 2 gram - 0.756 gram Reaction 1 : Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O

Slide 10:

Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 Mol ratio 1 x y Mol Mr 466 - 44 Mass 2 gram - 0.756 gram Reaction 1 : Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O

Slide 11:

Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 Mol ratio 1 x y Mol Mr 466 - 44 Mass 2 gram - 0.756 gram Reaction 1 : Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O from question

Slide 12:

Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 Mol ratio 1 x y Mol Mr 466 - 44 Mass 2 gram - 0.756 gram Reaction 1 : Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O mol = mass  Mr

Slide 13:

Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 Mol ratio 1 x y Mol Mr 466 - 44 Mass 2 gram - 0.756 gram Reaction 1 : Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O follow mol ratio

Slide 14:

Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 Mol ratio 1 x y Mol Mr 466 - 44 Mass 2 gram - 0.756 gram Reaction 1 : Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O from question

Slide 15:

Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 Mol ratio 1 x y Mol Mr 466 - 44 Mass 2 gram - 0.756 gram Reaction 1 : Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 Mol ratio 1 x y Mol Mr 466 - 44 Mass 2 gram - 0.756 gram mol = mass  Mr

Slide 16:

Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 Mol ratio 1 x y Mol Mr 466 - 44 Mass 2 gram - 0.756 gram Reaction 1 : Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 Mol ratio 1 x y Mol Mr 466 - 44 Mass 2 gram - 0.756 gram

Slide 17:

Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 Mol ratio 1 x y Mol Mr 466 - 44 Mass 2 gram - 0.756 gram Reaction 1 : Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 Mol ratio 1 x y Mol Mr 466 - 44 Mass 2 gram - 0.756 gram

Slide 18:

MgO HCl Mol ratio 1 2 Mol Reaction 2 : MgO + 2 HCl → MgCl 2 + H 2 O

Slide 19:

MgO HCl Mol ratio 1 2 Mol Reaction 2 : MgO + 2 HCl → MgCl 2 + H 2 O

Slide 20:

HCl available HCl in reaction 2 HCl after reaction 2 HCl in 50 cm 3 aliquot Mol 0.1 0.1 - 0.01 - Concentration 1.0 M - - - Volume 0.10 dm 3 - - - Dilution process : from question

Slide 21:

HCl available HCl in reaction 2 HCl after reaction 2 HCl in 50 cm 3 aliquot Mol 0.1 0.1 - 0.01 - Concentration 1.0 M - - - Volume 0.10 dm 3 - - - Dilution process : mol = mass  Mr

Slide 22:

HCl available HCl in reaction 2 HCl after reaction 2 HCl in 50 cm 3 aliquot Mol 0.1 0.1 - 0.01 - Concentration 1.0 M - - - Volume 0.10 dm 3 - - - Dilution process : from previous calculation

Slide 23:

HCl available HCl in reaction 2 HCl after reaction 2 HCl in 50 cm 3 aliquot Mol 0.1 0.1 - 0.01 - Concentration 1.0 M - - - Volume 0.10 dm 3 - - - Dilution process : The mol of HCl in 50cm3 aliquot is 1/10 of mol of HCl in 500 cm3 solution.

Slide 24:

NaOH HCl in reaction 3 Mol 0.0057 0.0057 = 0.01 - Volume 0.057 dm 3 - Concentration 0.1 mol dm -3 - Reaction 3 : 0.0057 = 0.01 - x = 5 from previous calculation from question

Slide 25:

NaOH HCl in reaction 3 Mol 0.0057 0.0057 = 0.01 - Volume 0.057 dm 3 - Concentration 0.1 mol dm -3 - Reaction 3 : 0.0057 = 0.01 - x = 5 mol = mass  Mr

Slide 26:

NaOH HCl in reaction 3 Mol 0.0057 0.0057 = 0.01 - Volume 0.057 dm 3 - Concentration 0.1 mol dm -3 - Reaction 3 : 0.0057 = 0.01 - x = 5

Slide 27:

NaOH HCl in reaction 3 Mol 0.0057 0.0057 = 0.01 - Volume 0.057 dm 3 - Concentration 0.1 mol dm -3 - Reaction 3 : 0.0057 = 0.01 - x = 5

Slide 28:

NaOH HCl in reaction 3 Mol 0.0057 0.0057 = 0.01 - Volume 0.057 dm 3 - Concentration 0.1 mol dm -3 - Reaction 3 : 0.0057 = 0.01 - x = 5

Slide 29:

If 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O decomposed according to reaction 1 : Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 H 2 O Mol ratio 1 5 4 Mol 1 5 4 Mr 466 40 44 18 Mass 466 gram 200 gram 176 gram 18 18 = 466-200-176 = 90 18 n + 9z = 90 18 n = 90 – 9z Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O

Slide 30:

If 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O decomposed according to reaction 1 : Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 H 2 O Mol ratio 1 5 4 Mol 1 5 4 Mr 466 40 44 18 Mass 466 gram 200 gram 176 gram 18 18 = 466-200-176 = 90 18 n + 9z = 90 18 n = 90 – 9z Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O

Slide 31:

If 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O decomposed according to reaction 1 : Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 H 2 O Mol ratio 1 5 4 Mol 1 5 4 Mr 466 40 44 18 Mass 466 gram 200 gram 176 gram 18 18 = 466-200-176 = 90 18 n + 9z = 90 18 n = 90 – 9z Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O

Slide 32:

If 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O decomposed according to reaction 1 : Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 H 2 O Mol ratio 1 5 4 Mol 1 5 4 Mr 466 40 44 18 Mass 466 gram 200 gram 176 gram 18 18 = 466-200-176 = 90 18 n + 9z = 90 18 n = 90 – 9z Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O

Slide 33:

If 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O decomposed according to reaction 1 : Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 H 2 O Mol ratio 1 5 4 Mol 1 5 4 Mr 466 40 44 18 Mass 466 gram 200 gram 176 gram 18 18 = 466-200-176 = 90 18 n + 9z = 90 18 n = 90 – 9z Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O

Slide 34:

If 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O decomposed according to reaction 1 : Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 H 2 O Mol ratio 1 5 4 Mol 1 5 4 Mr 466 40 44 18 Mass 466 gram 200 gram 176 gram 18 18 = 466-200-176 = 90 18 n + 9z = 90 18 n = 90 – 9z Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O

Slide 35:

If 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O decomposed according to reaction 1 : Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 H 2 O Mol ratio 1 5 4 Mol 1 5 4 Mr 466 40 44 18 Mass 466 gram 200 gram 176 gram 18 = 466-200-176 = 90 18 n + 9z = 90 18 n = 90 – 9z Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O 18

Slide 36:

If 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O decomposed according to reaction 1 : Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 H 2 O Mol ratio 1 5 4 Mol 1 5 4 Mr 466 40 44 18 Mass 466 gram 200 gram 176 gram 18 = 466-200-176 = 90 18 n + 9z = 90 18 n = 90 – 9z Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O 18

Slide 37:

If 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O decomposed according to reaction 1 : Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 H 2 O Mol ratio 1 5 4 Mol 1 5 4 Mr 466 40 44 18 Mass 466 gram 200 gram 176 gram 18 = 466-200-176 = 90 18 n + 9z = 90 18 n = 90 – 9z Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O 18

Slide 38:

If 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O decomposed according to reaction 1 : Mg x (CO 3 ) y (OH) z . n H 2 O MgO CO 2 H 2 O Mol ratio 1 5 4 Mol 1 5 4 Mr 466 40 44 18 Mass 466 gram 200 gram 176 gram 18 = 466-200-176 = 90 18 n + 9z = 90 18 n = 90 – 9z Mg x ( CO 3 ) y (OH) z . n H 2 O → x MgO + y CO 2 + (n+ z/2) H 2 O 18

Slide 39:

Mg CO 3 OH H 2 O Mol ratio 5 4 z n Mol 5 4 z n Mr 24 60 17 18 Mass 120 240 17 z 18 n 17 z + 18 n = 466 -120 – 240 = 106 17 z + 90 – 9z = 106 8 z = 16 z = 2 n = 4 Composition of mass of elements in 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O :

Slide 40:

Mg CO 3 OH H 2 O Mol ratio 5 4 z n Mol 5 4 z n Mr 24 60 17 18 Mass 120 240 17 z 18 n 17 z + 18 n = 466 -120 – 240 = 106 17 z + 90 – 9z = 106 8 z = 16 z = 2 n = 4 Composition of mass of elements in 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O :

Slide 41:

Mg CO 3 OH H 2 O Mol ratio 5 4 z n Mol 5 4 z n Mr 24 60 17 18 Mass 120 240 17 z 18 n 17 z + 18 n = 466 -120 – 240 = 106 17 z + 90 – 9z = 106 8 z = 16 z = 2 n = 4 Composition of mass of elements in 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O :

Slide 42:

Mg CO 3 OH H 2 O Mol ratio 5 4 z n Mol 5 4 z n Mr 24 60 17 18 Mass 120 240 17 z 18 n 17 z + 18 n = 466 -120 – 240 = 106 17 z + 90 – 9z = 106 8 z = 16 z = 2 n = 4 Composition of mass of elements in 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O :

Slide 43:

Mg CO 3 OH H 2 O Mol ratio 5 4 z n Mol 5 4 z n Mr 24 60 17 18 Mass 120 240 17 z 18 n 17 z + 18 n = 466 -120 – 240 = 106 17 z + 90 – 9z = 106 8 z = 16 z = 2 n = 4 Composition of mass of elements in 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O :

Slide 44:

Mg CO 3 OH H 2 O Mol ratio 5 4 z n Mol 5 4 z n Mr 24 60 17 18 Mass 120 240 17 z 18 n 17 z + 18 n = 466 -120 – 240 = 106 17 z + 90 – 9z = 106 8 z = 16 z = 2 n = 4 Composition of mass of elements in 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O :

Slide 45:

Mg CO 3 OH H 2 O Mol ratio 5 4 z n Mol 5 4 z n Mr 24 60 17 18 Mass 120 240 17 z 18 n 17 z + 18 n = 466 -120 – 240 = 106 17 z + 90 – 9z = 106 8 z = 16 z = 2 n = 4 Composition of mass of elements in 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O : 18 n = 90 – 9z from previous calculation

Slide 46:

Mg CO 3 OH H 2 O Mol ratio 5 4 z n Mol 5 4 z n Mr 24 60 17 18 Mass 120 240 17 z 18 n 17 z + 18 n = 466 -120 – 240 = 106 17 z + 90 – 9z = 106 8 z = 16 z = 2 n = 4 Composition of mass of elements in 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O : 18 n = 90 – 9z

Slide 47:

Mg CO 3 OH H 2 O Mol ratio 5 4 z n Mol 5 4 z n Mr 24 60 17 18 Mass 120 240 17 z 18 n 17 z + 18 n = 466 -120 – 240 = 106 17 z + 90 – 9z = 106 8 z = 16 z = 2 n = 4 Composition of mass of elements in 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O : 18 n = 90 – 9z

Slide 48:

Mg CO 3 OH H 2 O Mol ratio 5 4 z n Mol 5 4 z n Mr 24 60 17 18 Mass 120 240 17 z 18 n 17 z + 18 n = 466 -120 – 240 = 106 17 z + 90 – 9z = 106 8 z = 16 z = 2 n = 4 Composition of mass of elements in 466 grams of Mg x ( CO 3 ) y (OH) z . n H 2 O : 18 n = 90 – 9z