1 / 74

Mass - Mole Relationships of a Compound

Mass - Mole Relationships of a Compound. For an Element. For a Compound. Mass (g) of Element. Mass (g) of compound. Moles of Element. Amount (mol) of compound. Amount (mol) of compound. Molecules (or formula units of compound). Atoms of Element.

wan
Download Presentation

Mass - Mole Relationships of a Compound

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Mass - Mole Relationships of a Compound For an Element For a Compound Mass (g) of Element Mass (g) of compound Moles of Element Amount (mol) of compound Amount (mol) of compound Molecules (or formula units of compound) Atoms of Element

  2. Calculating the Number of Moles and Atoms in a Given Mass of Element Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element 3680oC. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal? Plan: Convert mass into moles by dividing the mass by the atomic weight of the metal, then calculate the number of atoms by multiplying by Avogadro’s number! Solution: Converting from mass of W to moles: Moles of W = 35.0 mg W x = 0.00019032 mol 1.90 x 10 - 4 mol NO. of W atoms = 1.90 x 10 - 4 mol W x = = 1.15 x 1020 atoms of Tungsten 1 mol W 183.9 g W 6.022 x 1023 atoms 1 mole of W

  3. Calculating the Moles and Number of Formula Units in a Given Mass of Cpd. Problem: Sodium Phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample? Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients. Solution: The formula is Na3PO4. Calculating the molar mass: M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen = = 3 x 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol = 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/mol Converting mass to moles: Moles Na3PO4 = 38.6 g Na3PO4 x (1 mol Na3PO4) 163.94 g Na3PO4 = 0.235 mol Na3PO4 Formula units = 0.23545 mol Na3PO4 x 6.022 x 1023 formula units 1 mol Na3PO4 = 1.42 x 1023 formula units

  4. The average daily requirement of the essential amino acid leucine, C6H14O2N, is 2.2 g for an adult. What is the average daily requirement of leucine in moles? First, find the molar mass of leucine: 6 C 6(12.01) = 72.06 2 O 2(16.00) = 32.00 1 N 1(14.01) = 14.01 14 H 14(1.008) = 14.112 2 decimal places 132.18 g/mol 132.182

  5. Next, find the number of moles in 2.2 g:

  6. Flow Chart of Mass Percentage Calculation Moles of X in one mole of Compound M (g / mol) of X Mass (g) of X in one mole of compound Divide by mass (g) of one mole of compound Mass fraction of X Multiply by 100 Mass % of X

  7. Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I Problem: Sucrose (C12H22O11) is common table sugar. (a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in 24.35 g of sucrose? (a) Determining the mass percent of each element: mass of C = 12 x 12.01 g C/mol = 144.12 g C/mol mass of H = 22 x 1.008 g H/mol = 22.176 g H/mol mass of O = 11 x 16.00 g O/mol = 176.00 g O/mol 342.296 g/mol Finding the mass fraction of C in Sucrose& % C : Total mass of C 144.12 g C mass of 1 mole of sucrose 342.30 g Cpd Mass Fraction of C = = = 0.421046 To find mass % of C = 0.421046 x 100% = 42.105%

  8. Calculating Mass Percents and Masses of Elements in a Sample of Compound - II (a) continued Mass % of H = x 100% = x 100% = 6.479% H Mass % of O = x 100% = x 100% = 51.417% O (b) Determining the mass of carbon: Mass (g) of C = mass of sucrose X( mass fraction of C in sucrose) Mass (g) of C = 24.35 g sucrose X = 10.25 g C mol H x M of H 22 x 1.008 g H mass of 1 mol sucrose 342.30 g mol O x M of O 11 x 16.00 g O mass of 1 mol sucrose 342.30 g 0.421046 g C 1 g sucrose

  9. Chemical Formulas Empirical Formula- Shows the relative number of atoms of each element in the compound. It is the simplest formula, and is derived from masses of the elements. Molecular Formula - Shows the actual number of atoms of each element in the molecule of the compound. Structural Formula - Shows the actual number of atoms, and the bonds between them ; that is, the arrangement of atoms in the molecule.

  10. Empirical and Molecular Formulas Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis! The smallest set of whole numbers of atoms. Molecular Formula - The formula of the compound as it exists, it may be a multiple of the Empirical formula.

  11. Steps to Determine Empirical Formulas Mass (g) of Element M (g/mol ) Moles of Element use no. of moles as subscripts Preliminary Formula change to integer subscripts Empirical Formula

  12. Some Examples of Compounds with the Same Elemental Ratio’s Empirical Formula Molecular Formula CH2(unsaturated Hydrocarbons) C2H4 , C3H6 , C4H8 OH or HO H2O2 S S8 P P4 Cl Cl2 CH2O (carbohydrates) C6H12O6

  13. Percentage Composition • The mass percentage of each element in the compound • The composition is determined by experiment, often by combustion. When a compound is burned, its component elements form oxides—for example, CO2 and H2O. The CO2 and H2O are captured and weighed to determine the amount of C and H in the original compound.

  14. Benzene is a liquid compound composed of carbon and hydrogen; it is used in the preparation of polystyrene plastic. A sample of benzene weighing 342 mg is burned in oxygen and forms 1158 mg of carbon dioxide. What is the percentage composition of benzene?

  15. Strategy 1. Use the mass of CO2 to find the mass of carbon from the benzene. 2. Use the mass of benzene and the mass of carbon to find the mass of hydrogen. 3. Use these two masses to find the percent composition.

  16. First, find the mass of C in 1156 mg of CO2: = 315.5 mg C

  17. Next, find the mass of H in the benzene sample: 342 mg benzene -315.5 mg C 26.5 mg H (the decimal is not significant) • Now, we can find the percentage composition:

  18. Empirical Formula (Simplest Formula) • The formula of a substance written with the smallest integer subscripts • For example: The empirical formula for N2O4 is NO2. The empirical formula for H2O2 is HO

  19. Determining the Empirical Formula Beginning with percent composition: • Assume exactly 100 g so percentages convert directly to grams. • Convert grams to moles for each element. • Manipulate the resulting mole ratios to obtain whole numbers.

  20. Manipulating the ratios: • Divide each mole amount by the smallest mole amount. • If the result is not a whole number: • Multiply each mole amount by a factor. For example: If the decimal portion is 0.5, multiply by 2. If the decimal portion is 0.33 or 0.67, multiply by 3. If the decimal portion is 0.25 or 0.75, multiply by 4.

  21. Benzene is composed of 92.3% carbon and 7.7% hydrogen. What is the empirical formula of benzene? Empirical formula: CH

  22. Molecular Formula • A formula for a molecule in which the subscripts are whole-number multiples of the subscripts in the empirical formula

  23. To determine the molecular formula: • Compute the empirical formula weight. • Find the ration of the molecular weight to the empirical formula weight. • Multiply each subscript of the empirical formula by n.

  24. Benzene has the empirical formula CH. Its molecular weight is 78.1 amu. What is its molecular formula? Molecular formula C6H6

  25. Determining Empirical Formulas from Masses of Elements - I Problem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the empirical formula and name of the compound? Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound. Solution: Finding the moles of the elements: Moles of Na = 5.678 g Na x = Moles of Cr = 6.420 g Cr x = Moles of O = 7.902 g O x = 1 mol Na 22.99 g Na 1 mol Cr 52.00 g Cr 1 mol O 16.00 g O

  26. Determining Empirical Formulas from Masses of Elements - I Problem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the empirical formula and name of the compound? Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound. Solution: Finding the moles of the elements: Moles of Na = 5.678 g Na x = 0.2469 mol Na Moles of Cr = 6.420 g Cr x = 0.12347 mol Cr Moles of O = 7.902 g O x = 0.4939 mol O 1 mol Na 22.99 g Na 1 mol Cr 52.00 g Cr 1 mol O 16.00 g O

  27. Determining Empirical Formulas from Masses of Elements - II Constructing the preliminary formula: Na0.2469 Cr0.1235 O0.4939 Converting to integer subscripts (dividing all by smallest subscript): Na1.99 Cr1.00 O4.02 Rounding off to whole numbers: Na2CrO4 Sodium Chromate

  28. Determining the Molecular Formula from Elemental Composition and Molar Mass - I Problem: The sugar burned for energy in cells of the body is Glucose (M = 180.16 g/mol), elemental analysis shows that it contains 40.00 mass % C, 6.719 mass % H, and 53.27 mass % O. (a) Determine the empirical formula of glucose. (b) Determine the molecular formula. Plan: We are only given mass %, and no weight of the compound so we will assume 100g of the compound, and % becomes grams, and we can do as done previously with masses of the elements. Solution: Mass Carbon = 40.00% x 100g/100% = 40.00 g C Mass Hydrogen = 6.719% x 100g/100% = 6.719g H Mass Oxygen = 53.27% x 100g/100% = 53.27 g O 99.989 g Cpd

  29. Determining the Molecular Formula from Elemental Composition and Molar Mass - II Converting from Grams of Elements to moles: Moles of C = Mass of C x = 3.3306 moles C Moles of H = Mass of H x = 6.6657 moles H Moles of O = Mass of O x = 3.3294 moles O Constructing the preliminary formula C 3.33 H 6.67 O 3.33 Converting to integer subscripts, divide all subscripts by the smallest: C 3.33/3.33 H 6.667 / 3.33 O3.33 / 3.33 = CH2O 1 mole C 12.01 g C 1 mol H 1.008 g H 1 mol O 16.00 g O

  30. Determining the Molecular Formula from Elemental Composition and Molar Mass - III (b) Determining the Molecular Formula: The formula weight of the empirical formula is: 1 x C + 2 x H + 1 x O = 1 x 12.01 + 2 x 1.008 + 1 x 16.00 = 30.03 M of Glucose empirical formula mass Whole-number multiple = = = = 6.00 = 6 180.16 30.03 Therefore the Molecular Formula is: C 1 x 6 H 2 x 6 O 1 x 6 =C6H12O6

  31. Adrenaline Is a Very Important Compound in the Body - I • Analysis gives : • C = 56.8 % • H = 6.50 % • O = 28.4 % • N = 8.28 % • Calculate the Empirical Formula

  32. Adrenaline - II • Assume 100g! • C = 56.8 g C/(12.01 g C/ mol C) = 4.73 mol C • H = 6.50 g H/( 1.008 g H / mol H) = 6.45 mol H • O = 28.4 g O/(16.00 g O/ mol O) = 1.78 mol O • N = 8.28 g N/(14.01 g N/ mol N) = 0.591 mol N • Divide by 0.591 = • C = 8.00 mol C = 8.0 mol C or • H = 10.9 mol H = 11.0 mol H • O = 3.01 mol O = 3.0 mol O C8H11O3N • N = 1.00 mol N = 1.0 mol N

  33. Combustion Train for the Determination of the Chemical Composition of Organic Compounds. m 2 m 2 CnHm + (n+ ) O2 = n CO(g) + H2O(g) Fig. 3.4

  34. Ascorbic Acid ( Vitamin C ) - I Contains C , H , and O • Upon combustion in excess oxygen, a 6.49 mg sample yielded 9.74 mg CO2 and 2.64 mg H2O • Calculate it’s Empirical formula! • C: 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2) = 2.65 x 10-3 g C • H: 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O) = 2.92 x 10-4 g H • Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg = 3.54 mg O

  35. Vitamin C Combustion - II • C = 2.65 x 10-3 g C / ( 12.01 g C / mol C ) = = 2.21 x 10-4 mol C • H = 0.295 x 10-3 g H / ( 1.008 g H / mol H ) = = 2.92 x 10-4 mol H • O = 3.54 x 10-3 g O / ( 16.00 g O / mol O ) = = 2.21 x 10-4 mol O • Divide each by 2.21 x 10-4 • C = 1.00 Multiply each by 3 = 3.00 = 3.0 • H = 1.32 = 3.96 = 4.0 • O = 1.00= 3.00 = 3.0 C3H4O3

  36. Determining a Chemical Formula from Combustion Analysis - I Problem: Erthrose (M = 120 g/mol) is an important chemical compound as a starting material in chemical synthesis, and contains Carbon Hydrogen, and Oxygen. Combustion analysis of a 700.0 mg sample yielded 1.027 g CO2 and 0.4194 g H2O. Plan: We find the masses of Hydrogen and Carbon using the mass fractions of H in H2O, and C in CO2. The mass of Carbon and Hydrogen are subtracted from the sample mass to get the mass of Oxygen. We then calculate moles, and construct the empirical formula, and from the given molar mass we can calculate the molecular formula.

  37. Determining a Chemical Formula from Combustion Analysis - II Calculating the mass fractions of the elements: Mass fraction of C in CO2 = = = = 0.2729 g C / 1 g CO2 Mass fraction of H in H2O = = = = 0.1119 g H / 1 g H2O Calculating masses of C and H Mass of Element = mass of compound x mass fraction of element mol C xM of C mass of 1 mol CO2 1 mol C x 12.01 g C/ 1 mol C 44.01 g CO2 mol H xM of H mass of 1 mol H2O 2 mol H x 1.008 g H / 1 mol H 18.02 g H2O

  38. Determining a Chemical Formula from Combustion Analysis - III 0.2729 g C 1 g CO2 Mass (g) of C = 1.027 g CO2 x = 0.2803 g C Mass (g) of H = 0.4194 g H2O x = 0.04693 g H Calculating the mass of O: Mass (g) of O = Sample mass -( mass of C + mass of H ) = 0.700 g - 0.2803 g C - 0.04693 g H = 0.37277 g O Calculating moles of each element: C = 0.2803 g C / 12.01 g C/ mol C = 0.02334 mol C H = 0.04693 g H / 1.008 g H / mol H = 0.04656 mol H O = 0.37277 g O / 16.00 g O / mol O = 0.02330 mol O C0.02334H0.04656O0.02330 = CH2O formula weight = 30 g / formula 120 g /mol / 30 g / formula = 4 formula units / cpd = C4H8O4 0.1119 g H 1 g H2O

  39. Some Compounds with Empirical Formula CH2O (Composition by Mass 40.0% C, 6.71% H, 53.3%O) Molecular M Formula (g/mol) Name Use or Function CH2O 30.03 Formaldehyde Disinfectant; Biological preservative C2H4O2 60.05 Acetic acid Acetate polymers; vinegar ( 5% solution) C3H6O3 90.08 Lactic acid Causes milk to sour; forms in muscle during exercise C4H8O4 120.10 Erythrose Forms during sugar metabolism C5H10O5 150.13 Ribose Component of many nucleic acids and vitamin B2 C6H12O6 180.16 Glucose Major nutrient for energy in cells

  40. Two Compounds with Molecular Formula C2H6O Property Ethanol Dimethyl Ether M (g/mol) 46.07 46.07 Color Colorless Colorless Melting point - 117oC - 138.5oC Boiling point 78.5oC - 25oC Density (at 20oC) 0.789 g/mL 0.00195 g/mL Use Intoxicant in In refrigeration alcoholic beverages H H H H H C C O H H C O C H H H H H Table 3.4

  41. Molecular Formula Atoms Molecules Avogadro’s Number 6.022 x 1023 Moles Moles

  42. Chemical Equations Qualitative Information: Reactants Products States of Matter: (s) solid (l) liquid (g) gaseous (aq) aqueous 2 H2 (g) + O2 (g) 2 H2O (g)

  43. Chemical Equation Calculation - I Atoms (Molecules) Avogadro’s Number 6.02 x 1023 Molecules Reactants Products

  44. Chemical Equation Calculation - II Mass Atoms (Molecules) Molecular Weight Avogadro’s Number g/mol 6.02 x 1023 Molecules Reactants Products Moles

  45. Stoichiometry • The calculation of the quantities of reactants and products involved in a chemical reaction • Interpreting a Chemical Equation • The coefficients of the balanced chemical equation may be interpreted in terms of either (1) numbers of molecules (or ions or formula units) or (2) numbers of moles, depending on your needs.

  46. Information Contained in a Balanced Equation Viewed in Reactants Products terms of: 2 C2H6 (g) + 7 O2 (g) = 4 CO2 (g) + 6 H2O(g) + Energy Molecules 2 molecules of C2H6 + 7 molecules of O2 = 4 molecules of CO2 + 6 molecules of H2O Amount (mol) 2 mol C2H6 + 7 mol O2 = 4 mol CO2 + 6 mol H2O Mass (amu) 60.14 amu C2H6 + 224.00 amu O2 = 176.04 amu CO2 + 108.10 amu H2O Mass (g) 60.14 g C2H6 + 224.00 g O2 = 176.04 g CO2 + 108.10 g H2O Total Mass (g) 284.14g = 284.14g

  47. To find the amount of B (one reactant or product) given the amount of A (another reactant or product): 1. Convert grams of A to moles of A  Using the molar mass of A 2. Convert moles of A to moles of B  Using the coefficients of the balanced chemical equation 3. Convert moles of B to grams of B  Using the molar mass of B

  48. Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation: • C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) • How many grams of CO2 are produced when 20.0 g of propane is burned?

  49. Molar masses C3H8: 3(12.01) + 8(1.008) = 44.094 g CO2: 1(12.01) + 2(16.00) = 44.01 g 59.9 g CO2 (3 significant figures)

  50. Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation: • C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) • How many grams of O2 are required to burn 20.0 g of propane?

More Related