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# The vertices of a tetrahedron correspond to four ISBN: 9780321696724 27

## Solution for problem 91AE Chapter 9

Chemistry: The Central Science | 12th Edition

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Problem 91AE

Problem 91AE

The vertices of a tetrahedron correspond to four alternating corners of a cube. By using analytical geometry, demonstrate that the angle made by connecting two of the vertices to a point at the center of the cube is 109.5°, the characteristic angle for tetrahedral molecules.

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Chemistry 116 Test 3 Study guide (Bangbo Yan) Lecture 10  Molecular Mass • Sum of atomic masses of all atoms in molecule of substance • Unit: measured in amu EXAMPLE: Molecular mass of H O is218.02 amu H 2 2 x 1.008= 2.016 amu O= 1 x16.00= 16.00 amu 18.02 amu (add 2.016 and 16.00; round to correct amt. of sigfigs)  Formula Mass • Sum of atomic masses of all atoms in formula unit of the compound, whether molecular or not • Unit: measure in amu EXAMPLE: Formula mass of CaCl 2 is 110.98 Ca= 1 x 40.08= 70.40 amu Cl 2 2 x 35.45= 40.08 amu 110.98 amu (add and round)  The mole Concept • Mole: quantity of a given amount of substance that contains as many molecules or formula units as the number of atoms in exactly 12g of carbon-12 ♦ Symbol: mol ( no unit for mole; similar to when we say 1 pair of shoes or 12 dozen eggs, they are not measured in units) • Avogadro’s number: number of atoms in exactly 12g of carbon-12. Avogadro’s number is 6.022 x 103 • Molar mass: mass of one mole of substance. The molar mass in grams per mol is numerical equal to the formula mass in amu ♦ Unit: grams (g) EXAMPLE: 1 mol O= 16.00 g O= 16.00 amu (same but different unit) ♦ How to calculate the mass of one atom • Take the mass of the atom and divide it by the Avogadro’s number (6.022 x10 ) 23 EXAMPLE: Mass in grams of one oxygen atom is 2.657 x 10 g -23 23 1 mol O= 6.022x 10 O atoms Mass of O= 16.00g 23 -23 16.00g O/6.022 x 10 O atoms= 2.657 x 10 g ♦ How to calculate the mass of one molecule • Take the mass of the entire the molecule (each atom in the molecule) and++ divide it by Avogadro’s number. -22 EXAMPLE: Mass in grams of a hydrogen bromide molecule is 1.343 x 10 g 23 1 mol of HBr= 6.022x 10 Mass of HBr= 80.91g 23 -22 80.91g HBr/ 6.022x 10 HBr atoms= 1.343x 10 g ♦ How to convert moles of substances to grams • Take the amount of moles given and multiply it by the molar mass of the substance EXAMPLE: Convert 0.721 mol of C H O (g6u12se6 to grams Molar mass of C H6O12 160.16g 0.721 mol x 180.16g=129.90 Lecture 11  Mole concept • Molar mass ♦ To find molar mass, convert grams of substances to moles EXAMPLE: 646 grams of PbCrO 4is how many moles Divide amount of grams of PbCrO (644g) by the mass of PbCrO (323g) 4 646g/323g=2 mol  Mass percentage from the formula • Mass percentage ♦ To find mass percentage, divide mass of the whole compound by mass of A in compound. EXAMPLE: What is the mass percentage of Na in NaCl (Hint: A is Na) Divide mass of Na (22.99g) by mass of NaCl (58.44). Then multiply by 100 ***If you were trying to find the mass percentage of Na2then you would multiply the mass of Na by 2 then divide by the mass of the whole compound*** 22.99g/58.44g x 100=39.34 is the mass % of A • Percentage composition from the formula EXAMPLE: What is the mass percentage of each element in CH 2O (formaldehyde). Keep 3 sigfigs. C: 12g/30.0 x 100 = 40% H 2:2 x 1.01 (2.02) /30.0 x 100 = 6.73% O: 16/ 30.0 x 100 = 53.3 % • Empirical formula (simplest formula) ♦ Empirical formula: The formula of a substance written with the smallest integer subscripts. EXAMPLE: What is the empirical formula of hydrogen peroxide H 2O 2molecular formula) HO (empirical formula= What is the empirical formula of benzene C 6 (6olecular formula) CH (empirical formula)  Elemental analysis • Used when grams of each element is unknown/ not given. EXAMPLE: combust 11.5 g ethanol containing only C, H, and O Collect 22.0g CO 2nd 13.5 g H O.2What is the empirical formula CO 2 22.0 g CO 2 1 mol CO 2 12.01 g C 1 mol C 44.01 g CO 2 1 mol C 12.01 g C =.500 mol C .500 mol C x 12 g C = 6.00 g C H 2 13.5 g H2O 1 mol H2O 1.01 g H 2 mol H 18.02 g H2O 1 mol H 2 1.01 g H =1.50 mol H 1.50 mol H x 1.01 g H = 1.51 g H O: g of O = g of sample (11.5) – g of C (6.00) + g of H (1.51) = 4.00 g O 4.00g O 1 mol O 16 g O = .25 mol O C 0.5 1.5 0.25 Divide by the smallest subscript (0.25) ANSWER: C H O 2 6 Lecture 12  Empirical formula from percentage composition • An analysis of sodium dichromate give the following mass percentages: 17.5% Na, 39.7% Cr, and 42.8 O. What is the empirical formula of this compound ♦ In 100 g… 1. • Na: 100g x (17.5/100)= 17.5 g Na • Cr: 100g x (39.7/100)=39.7g Cr • O:100g x (42.8/100)=42.8g O **The above was done to change the percentages to be expressed as a decimal.** 2. • 17.5 Na x (1 mol/23.0g Na)=0.761 mol Na • 39.7g Cr x (1 mol/52.0g Cr)= 0.763 mol Cr • 42.8 g O x (1 mol/16g O)= 2.68 mol O **The above is used to convert grams to moles** 3. • 2.68 mol O/ 0.761 mol Na=3.52 • 0.763 mol Cr/ 0.761 mol Na=1 • 0.761 mol Na/ 0.761 mol Na=1 **Divide each by the smallest amount of moles to find out how many moles are in each. ** 4. • NaCrO 3.52*3.52 is not a whole number so multiply each by the simplest whole number which in this case is 2** • ANSWER: Na Cr O 2 2 7  Molecular formula from empirical formula • Molecular mass= n x empirical formula mass • n = molecular mass/empirical formula mass ♦ Acetic acid is said to be 39.9% C, 6.7% H and 53.45% O. What is its empirical formula .What is its molecular fomula with the molecular mass of acetic acid being 60.0 amu. ♦ Do steps 1-3 for C, H, & O ♦ You should get the following answers… 1. • C: 39.9 g • H: 6.7 g • O: 53.45 g 2. • C: 3.32 mol • H: 6.63 mol • O: 3.34 mol 3. • C: 1.00 mol • H: 2.00 mol • O: 1.00 mol ♦ Based on the answers above the empirical formula is CH O. 2 ♦ Multiply by the smallest whole number to find the molecular formula which is C 2 4 .2 • What is the formula mass of a 0.50 mol sample of a molecular compound weighing 9.40g ♦ Us algebra to create new equation.: empirical formula mass= molecular mass/ moles • Empirical formula mass= 9.40g/0.50 mol ◊ =18.8g  Stoichiometry :quantitative relations • Stoichiometry: the quantitative study of reactants and products in a chemical reaction.  Molar interpretation of a chemical equation • In terms of ♦ Number of moles ♦ Number of molecules ♦ Mass of substance • Mole method ♦ Use moles to calculate the amount of product in a reaction ♦ How many moles of NH 3can be produced from 2.7 mol of H 2 N 2(g) + 3 H2(g) NH (3) • Balance the equation N 2g) + 3 H (2) 2NH (3) ANSWER: 2 moles • How many gram of HCL react with 5.00 g of manganese dioxide, accoriding to the following equation ♦ 4HCl (aq) +MnO 2 2H 2 (l) + MnCl (aq) + Cl 2 5.00 g MnO 2 4 mol HCl 36. 46 g HCl 86.94 g 1 mol HCl MnO 2 = 8.39 g HCL Lecture 12  Limiting reactant; Theoretical and Percentage yields • Limiting reactant: the reactant that us entirely consumed when a reaction goes to completion EXMPLE: 2H 2+ O 2 2H O2 O 2s the limiting reactant ♦ Think of the process of making a sandwich. If you have 4 pieces of bread (2H 2) but only 2 pieces of cheese (O 2, will you have extra bread left over or extra cheese left over The answer is extra Bread (H ).2Which is why H is 2n the reactant and the product side while O 2s only on the reactant side because it was entirely consumed when the reaction went to completion. ♦ Once the reactant has been completely consumed the reaction stops. ♦ Any problem giving the starting amount for more than one reactant is the limiting reactant problem ♦ The moles of product are always determined by the staring moles of limiting reactant EXAMPLE: Zn(s) + 2HCl (aq) ZnCl (aq) + H (g) 2 2 If 0.30 mol of Zn is added to hydrochloric acid containg 0.52 mol of HCl, how many moles of H 2are produced (Both reactants are given so it’s a limited reactant problem). Find out which reactant produces les product. 0.30 mol Zn 1 mol H 2 1 mol Zn = 0.30 mol H 2 0.52 mol HCl 1 mol H 2 2 mol HCl = 0.26 mol H 2 **HCL is the limiting reactant** EXAMPLE (limiting reactant involving masses): 2CH CHO (l) +3O (g) 2 2HC H2O 3 2 (l) In a laboratory test of this reaction, 20.0g CH C3O and 10.0 g O were p2t into a reduction vessel. (A)How many grams of acetic acid (HC H O ) c2n 3e 2roduced by this reaction from these amounts of reactants (B) How many grams of this excess reactant remain after the reaction is complete **This is a limiting reactant problem because both reactants are given** 20.0g CH CH3 1 mol CH C3O 2 mol HC H 2 3 2 44.1g CH C3O 2 mol CH CH3CHO3 =0.454 mol HC H O2 3 2 **CH 3CHO is the limiting reactant** 10.0g O 2 1 mol O 2 2 mol HC H 2 3 2 32g O 2 1 mol O 2 =0.625 mol HC H O2 3 2 0.454 mol HC H 2 3 2 1 mol HC H 2 3 2 2 mol HC H 2 3 2 = 27.3 g HC H2O 3 2 0.454 mol HC H 2 3 2 6.01g HC H2O 3 2 1 mol HC H2O 3 2 = 7.26g O 2  Percentage yield • Percentage yield: the actual yield expressed as a percentage of the theoretical yield. • (Actual yield/theoretical yield) x 100 ♦ Theoretical yield: maximum amount of product that can be obtained by a reaction from given amount of reactants. Based on limiting reactant. ♦ Actual yield: amount of product actually obtained from reaction. Normally less than theoretical yield. EXAMPLE: CH 3OH (l) + CO (g) HC 2 O3(l2 In an experiment, 15.0 g of methanol and 10.0 g of carbon monoxide were placed in a reaction vessel. (A) What is the theoretical yield of the acetic acid (B) If the actual yield is 19.1 g, what is the percentage yield 15.0 g MeOH 1 mol MeOH 1 mol HC H 2 3 2 32.0 g MeOH 1 mol MeOH = 0.469 mol HC H2O 3 2 10.0 g CO 1 mol CO 1 mol HC H 2 3 2 28.0 g CO 1 mol CO = 0.357 mol HC H2O 3 2 **CO is the limiting reactant** 0.357 mol HC H 2 3 2 601 g HC H2O 3 2 1 molHC H 2 3 2 = 21.5 g HC H2O3(t2eoretical yield) Percentage yield= (19.1g/21.5g) x 100 = 88.8% Chemistry 116 Test 3 Study guide (Bangbo Yan) Lecture 10  Molecular Mass • Sum of atomic masses of all atoms in molecule of substance • Unit: measured in amu EXAMPLE: Molecular mass of H O is218.02 amu H 2 2 x 1.008= 2.016 amu O= 1 x16.00= 16.00 amu 18.02 amu (add 2.016 and 16.00; round to correct amt. of sigfigs)  Formula Mass • Sum of atomic masses of all atoms in formula unit of the compound, whether molecular or not • Unit: measure in amu EXAMPLE: Formula mass of CaCl 2 is 110.98 Ca= 1 x 40.08= 70.40 amu Cl 2 2 x 35.45= 40.08 amu 110.98 amu (add and round)  The mole Concept • Mole: quantity of a given amount of substance that contains as many molecules or formula units as the number of atoms in exactly 12g of carbon-12 ♦ Symbol: mol ( no unit for mole; similar to when we say 1 pair of shoes or 12 dozen eggs, they are not measured in units) • Avogadro’s number: number of atoms in exactly 12g of carbon-12. Avogadro’s number is 6.022 x 103 • Molar mass: mass of one mole of substance. The molar mass in grams per mol is numerical equal to the formula mass in amu ♦ Unit: grams (g) EXAMPLE: 1 mol O= 16.00 g O= 16.00 amu (same but different unit) ♦ How to calculate the mass of one atom • Take the mass of the atom and divide it by the Avogadro’s number (6.022 x10 ) 23 EXAMPLE: Mass in grams of one oxygen atom is 2.657 x 10 g -23 23 1 mol O= 6.022x 10 O atoms Mass of O= 16.00g 23 -23 16.00g O/6.022 x 10 O atoms= 2.657 x 10 g ♦ How to calculate the mass of one molecule • Take the mass of the entire the molecule (each atom in the molecule) and++ divide it by Avogadro’s number. -22 EXAMPLE: Mass in grams of a hydrogen bromide molecule is 1.343 x 10 g 23 1 mol of HBr= 6.022x 10 Mass of HBr= 80.91g 23 -22 80.91g HBr/ 6.022x 10 HBr atoms= 1.343x 10 g ♦ How to convert moles of substances to grams • Take the amount of moles given and multiply it by the molar mass of the substance EXAMPLE: Convert 0.721 mol of C H O (g6u12se6 to grams Molar mass of C H6O12 160.16g 0.721 mol x 180.16g=129.90 Lecture 11  Mole concept • Molar mass ♦ To find molar mass, convert grams of substances to moles EXAMPLE: 646 grams of PbCrO 4is how many moles Divide amount of grams of PbCrO (644g) by the mass of PbCrO (323g) 4 646g/323g=2 mol  Mass percentage from the formula • Mass percentage ♦ To find mass percentage, divide mass of the whole compound by mass of A in compound. EXAMPLE: What is the mass percentage of Na in NaCl (Hint: A is Na) Divide mass of Na (22.99g) by mass of NaCl (58.44). Then multiply by 100 ***If you were trying to find the mass percentage of Na2then you would multiply the mass of Na by 2 then divide by the mass of the whole compound*** 22.99g/58.44g x 100=39.34 is the mass % of A • Percentage composition from the formula EXAMPLE: What is the mass percentage of each element in CH 2O (formaldehyde). Keep 3 sigfigs. C: 12g/30.0 x 100 = 40% H 2:2 x 1.01 (2.02) /30.0 x 100 = 6.73% O: 16/ 30.0 x 100 = 53.3 % • Empirical formula (simplest formula) ♦ Empirical formula: The formula of a substance written with the smallest integer subscripts. EXAMPLE: What is the empirical formula of hydrogen peroxide H 2O 2molecular formula) HO (empirical formula= What is the empirical formula of benzene C 6 (6olecular formula) CH (empirical formula)  Elemental analysis • Used when grams of each element is unknown/ not given. EXAMPLE: combust 11.5 g ethanol containing only C, H, and O Collect 22.0g CO 2nd 13.5 g H O.2What is the empirical formula CO 2 22.0 g CO 2 1 mol CO 2 12.01 g C 1 mol C 44.01 g CO 2 1 mol C 12.01 g C =.500 mol C .500 mol C x 12 g C = 6.00 g C H 2 13.5 g H2O 1 mol H2O 1.01 g H 2 mol H 18.02 g H2O 1 mol H 2 1.01 g H =1.50 mol H 1.50 mol H x 1.01 g H = 1.51 g H O: g of O = g of sample (11.5) – g of C (6.00) + g of H (1.51) = 4.00 g O 4.00g O 1 mol O 16 g O = .25 mol O C 0.5 1.5 0.25 Divide by the smallest subscript (0.25) ANSWER: C H O 2 6 Lecture 12  Empirical formula from percentage composition • An analysis of sodium dichromate give the following mass percentages: 17.5% Na, 39.7% Cr, and 42.8 O. What is the empirical formula of this compound ♦ In 100 g… 1. • Na: 100g x (17.5/100)= 17.5 g Na • Cr: 100g x (39.7/100)=39.7g Cr • O:100g x (42.8/100)=42.8g O **The above was done to change the percentages to be expressed as a decimal.** 2. • 17.5 Na x (1 mol/23.0g Na)=0.761 mol Na • 39.7g Cr x (1 mol/52.0g Cr)= 0.763 mol Cr • 42.8 g O x (1 mol/16g O)= 2.68 mol O **The above is used to convert grams to moles** 3. • 2.68 mol O/ 0.761 mol Na=3.52 • 0.763 mol Cr/ 0.761 mol Na=1 • 0.761 mol Na/ 0.761 mol Na=1 **Divide each by the smallest amount of moles to find out how many moles are in each. ** 4. • NaCrO 3.52*3.52 is not a whole number so multiply each by the simplest whole number which in this case is 2** • ANSWER: Na Cr O 2 2 7  Molecular formula from empirical formula • Molecular mass= n x empirical formula mass • n = molecular mass/empirical formula mass ♦ Acetic acid is said to be 39.9% C, 6.7% H and 53.45% O. What is its empirical formula .What is its molecular fomula with the molecular mass of acetic acid being 60.0 amu. ♦ Do steps 1-3 for C, H, & O ♦ You should get the following answers… 1. • C: 39.9 g • H: 6.7 g • O: 53.45 g 2. • C: 3.32 mol • H: 6.63 mol • O: 3.34 mol 3. • C: 1.00 mol • H: 2.00 mol • O: 1.00 mol ♦ Based on the answers above the empirical formula is CH O. 2 ♦ Multiply by the smallest whole number to find the molecular formula which is C 2 4 .2 • What is the formula mass of a 0.50 mol sample of a molecular compound weighing 9.40g ♦ Us algebra to create new equation.: empirical formula mass= molecular mass/ moles • Empirical formula mass= 9.40g/0.50 mol ◊ =18.8g  Stoichiometry :quantitative relations • Stoichiometry: the quantitative study of reactants and products in a chemical reaction.  Molar interpretation of a chemical equation • In terms of ♦ Number of moles ♦ Number of molecules ♦ Mass of substance • Mole method ♦ Use moles to calculate the amount of product in a reaction ♦ How many moles of NH 3can be produced from 2.7 mol of H 2 N 2(g) + 3 H2(g) NH (3) • Balance the equation N 2g) + 3 H (2) 2NH (3) ANSWER: 2 moles • How many gram of HCL react with 5.00 g of manganese dioxide, accoriding to the following equation ♦ 4HCl (aq) +MnO 2 2H 2 (l) + MnCl (aq) + Cl 2 5.00 g MnO 2 4 mol HCl 36. 46 g HCl 86.94 g 1 mol HCl MnO 2 = 8.39 g HCL Lecture 12  Limiting reactant; Theoretical and Percentage yields • Limiting reactant: the reactant that us entirely consumed when a reaction goes to completion EXMPLE: 2H 2+ O 2 2H O2 O 2s the limiting reactant ♦ Think of the process of making a sandwich. If you have 4 pieces of bread (2H 2) but only 2 pieces of cheese (O 2, will you have extra bread left over or extra cheese left over The answer is extra Bread (H ).2Which is why H is 2n the reactant and the product side while O 2s only on the reactant side because it was entirely consumed when the reaction went to completion. ♦ Once the reactant has been completely consumed the reaction stops. ♦ Any problem giving the starting amount for more than one reactant is the limiting reactant problem ♦ The moles of product are always determined by the staring moles of limiting reactant EXAMPLE: Zn(s) + 2HCl (aq) ZnCl (aq) + H (g) 2 2 If 0.30 mol of Zn is added to hydrochloric acid containg 0.52 mol of HCl, how many moles of H 2are produced (Both reactants are given so it’s a limited reactant problem). Find out which reactant produces les product. 0.30 mol Zn 1 mol H 2 1 mol Zn = 0.30 mol H 2 0.52 mol HCl 1 mol H 2 2 mol HCl = 0.26 mol H 2 **HCL is the limiting reactant** EXAMPLE (limiting reactant involving masses): 2CH CHO (l) +3O (g) 2 2HC H2O 3 2 (l) In a laboratory test of this reaction, 20.0g CH C3O and 10.0 g O were p2t into a reduction vessel. (A)How many grams of acetic acid (HC H O ) c2n 3e 2roduced by this reaction from these amounts of reactants (B) How many grams of this excess reactant remain after the reaction is complete **This is a limiting reactant problem because both reactants are given** 20.0g CH CH3 1 mol CH C3O 2 mol HC H 2 3 2 44.1g CH C3O 2 mol CH CH3CHO3 =0.454 mol HC H O2 3 2 **CH 3CHO is the limiting reactant** 10.0g O 2 1 mol O 2 2 mol HC H 2 3 2 32g O 2 1 mol O 2 =0.625 mol HC H O2 3 2 0.454 mol HC H 2 3 2 1 mol HC H 2 3 2 2 mol HC H 2 3 2 = 27.3 g HC H2O 3 2 0.454 mol HC H 2 3 2 6.01g HC H2O 3 2 1 mol HC H2O 3 2 = 7.26g O 2  Percentage yield • Percentage yield: the actual yield expressed as a percentage of the theoretical yield. • (Actual yield/theoretical yield) x 100 ♦ Theoretical yield: maximum amount of product that can be obtained by a reaction from given amount of reactants. Based on limiting reactant. ♦ Actual yield: amount of product actually obtained from reaction. Normally less than theoretical yield. EXAMPLE: CH 3OH (l) + CO (g) HC 2 O3(l2 In an experiment, 15.0 g of methanol and 10.0 g of carbon monoxide were placed in a reaction vessel. (A) What is the theoretical yield of the acetic acid (B) If the actual yield is 19.1 g, what is the percentage yield 15.0 g MeOH 1 mol MeOH 1 mol HC H 2 3 2 32.0 g MeOH 1 mol MeOH = 0.469 mol HC H2O 3 2 10.0 g CO 1 mol CO 1 mol HC H 2 3 2 28.0 g CO 1 mol CO = 0.357 mol HC H2O 3 2 **CO is the limiting reactant** 0.357 mol HC H 2 3 2 601 g HC H2O 3 2 1 molHC H 2 3 2 = 21.5 g HC H2O3(t2eoretical yield) Percentage yield= (19.1g/21.5g) x 100 = 88.8%

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##### ISBN: 9780321696724

The full step-by-step solution to problem: 91AE from chapter: 9 was answered by , our top Chemistry solution expert on 04/03/17, 07:58AM. This full solution covers the following key subjects: vertices, Cube, angle, demonstrate, characteristic. This expansive textbook survival guide covers 49 chapters, and 5471 solutions. The answer to “The vertices of a tetrahedron correspond to four alternating corners of a cube. By using analytical geometry, demonstrate that the angle made by connecting two of the vertices to a point at the center of the cube is 109.5°, the characteristic angle for tetrahedral molecules.” is broken down into a number of easy to follow steps, and 45 words. This textbook survival guide was created for the textbook: Chemistry: The Central Science, edition: 12. Since the solution to 91AE from 9 chapter was answered, more than 356 students have viewed the full step-by-step answer. Chemistry: The Central Science was written by and is associated to the ISBN: 9780321696724.

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The vertices of a tetrahedron correspond to four