1. Composition of Substances and Solutions
Chem 1A Chapter 3 Lecture Outlines
Composition of Substances and Solutions
Atomic Mass and Formula Mass;
Mole & Molar Mass;
Percent Composition of Compounds;
Determination of Empirical & Molecular Formulas;
Molarity
Other Units for Solution Concentrations

2. Atomic Masses
Absolute masses of atoms cannot be obtained – too small to measure the mass directly;
Atomic mass is expressed as relative mass – masses relative to a chosen standard or reference.
Carbon-12 was chosen as reference, and assigned an atomic mass of 12 amu exactly;
Masses of other atoms are relative to that of carbon-12 atom;
Relative atomic masses are determined using mass spectrophotometer;

3. A Schematic Diagram of Mass Spectrophotometer

4. Isotope Mass of CO2

5. Formula Mass and the Mole Concept
Atomic mass:
The average atomic mass of an element is the weighted average atomic mass of all stable isotopes, taking into consideration the abundance of those isotopes of that element.
Formula Mass:
The mass of a molecule or compound is the sum of atomic masses of all the atoms represented in the substance’s chemical formula.

6. Mass Spectrum of Chlorine: 1) Indicate 2 isotopes of chlorine, with relative masses of 35 amu and 37 amu; 2) Relative abundance: Cl-35 ~75%) & Cl-37 ~25%

7. Mass Spectrum of Mercury (Shows a total of 7 isotopes with relative masses and abundances)

8. Calculating Average Atomic Masses from the Isotope Masses and Abundances
Example-1:
Chlorine has two stable naturally occurring isotopes: chlorine-35 (with mass = amu; abundance = 75.76%) and chlorine-37 (with mass = amu; abundance = 24.24%). Calculate the average atomic mass of chlorine?
Atomic mass of chlorine
= ( u x ) + ( u x )
= u (as given in the periodic table)

9. Calculating Average Atomic Masses from the Isotope Masses and Abundances
Example-2:
Copper is composed of two naturally occurring isotopes: copper-63 (with mass = amu; abundance = 69.09%) and copper-65 (with mass = amu; abundance = 30.91%;).
Average atomic mass of copper:
= (62.93 amu x ) + (64.93 amu x )
= amu

10. Exercise-1: Calculating Average Atomic Mass
Magnesium has three stable isotopes with the following masses and natural abundances: magnesium-24 (mass = amu; abundance = 79.00%; magnesium-25 (mass = amu; abundance = 10.00%, and magnesium-26 (mass = amu; abundance = 11.00%). Calculate the average atomic mass of magnesium.
(Answer: amu)

11. Exercise-2: Calculating Relative Abundance of Isotopes
Boron two stable isotopes: boron-10 (mass = amu) and boron-11 (mass = amu). If the average atomic mass of boron is amu, calculate the relative abundance of each isotope.
(Answer: boron-10 = 20.0%; boron-11 = 80.0%)

12. Atomic Mass & Molar Mass
Examples:
Element Atomic mass Molar mass
Carbon u g/mol
Oxygen u g/mol
Aluminum u g/mol
Silicon u g/mol
Gold u g/mol

13. Chem 1A Chapter 3 Lecture Outlines

14. Molecular Mass, Formula Mass & Molar Mass
Molecular mass = mass of one molecule (in u);
Molar mass = mass of one mole of element or compound (in g/mol) = sum of atomic masses;
Examples:
Molecular Mass Molar Mass (g/mol)
N u
H2O u
C8H u
C12H22H u

15. Formula Mass for Covalent Molecules
For covalent molecules the molecular formula represents the number and types of atoms composing a single molecule of that substance.
The formula mass is also referred to as molecular mass.
Consider the compound caffeine: C8H10N4O2
What is the molar mass of the compound?
What is the percentage composition (by mass) of each element in caffeine?

16. Formula Mass for Ionic Compounds
For ionic compounds, the chemical formula represents the types of cations and anions and the ratio in which they combine to achieve electrically neutral matter.
The chemical formula does NOT represent the composition of a discrete molecule.
Consider the compound, sodium chloride (NaCl)

17. The Mole
Reporting the number of atoms, molecules, or ions in a sample is not practical because atoms are so small.
Instead chemists use the unit called the mole to represent the quantity of substances.
The mole is defined as the amount of a substance containing the same number of discrete entities (such as atoms, molecules, or ions) as the number of atoms in a sample of pure carbon-12 weighing exactly 12 g.

18. The Mole
A sample of carbon-12 isotope that weighs exactly 12 g contains x 1023 atoms.
Avogadro’s Number = x 1023
A mole is a quantity that contains the Avogadro’s Number of items (atoms, molecules, ions, etc.);

19. Avogadro’s Number and the Mole
The term “mole” is analogous to the word dozen;
Dozen – A collection of 12 items.
Mole – A collection of x 1023 items.
The relationship that 1 mole = x items is an extremely important and useful conversion factor in chemistry.

20. 1 Mole of anything is 6.022 x 1023 items
1 mole of silicon = x 1023 Si atoms
1 mole of O2 molecules = x 1023 O2 molecules
1 mole of H2O = x 1023 H2O molecules
1 mole of Na+ ions = x 1023 Na+ ions
1 mole of electrons = x 1023 electrons
1 mole of pennies = x 1023 pennies

21. The Mole and Avogadro’s Number allows us to count atoms and molecules by mass
The atomic mass of an element in the periodic table:
is equivalent to the average mass of one atom of that element in atomic mass units (amu).
but also equivalent to the mass of one mole of that element in grams.
In the lab we rarely work in amu because we deal with large collections of atoms and molecules.
It is much more practical to work in grams.

22. One mole of any element contains 6.022 x 1023 atoms of that element.
The mass of one mole of different elements are not the same because the masses of the individual atoms are different.
Each of the above samples contains 1.00 mole of atoms (or 6.02 × 1023 atoms) of the element.
(credit: modification of work by Mark Ott)

23. Just how big is x 1023?
In order to obtain a mole of sand grains (6.022 x 1023 grains of sand), it would be necessary to dig the entire surface of the Sahara desert (which has an area slightly less than that of the United States) to a depth of 6 feet.
If you had a mole of dollars (6.022 x 1023 dollars), and if you spend this money at the rate of one billion (1 x 109) dollars per second, it would take you over 19 million years to spend all of the money.

24. The number of molecules in a single droplet of water is roughly 200 billion times greater than the number of people on earth.
(credit: “tanakawho”/Wikimedia commons)

25. Molar Mass (MM)
The molar mass (MM), of an element or a compound is the mass in grams of 1 mole of that substance.
Examples:
Molar Mass (g/mol) Al
H2O
C12H22H

26. Calculating Molar Mass
Consider the compound caffeine, C8H10N4O2. What is the molar mass of the compound?
= (8 x 12.01) + (10 x 1.008) + (4 x 14.01) + (2 x 16.00)
= g/mol
What is molar mass of ammonium phosphate, (NH4)3PO4?
= (3 x 14.01) + (12 x 1.008) + (1 x 30.97) + (4 x 16.00)
= g/mol

27. Measuring Amounts
The simplest way to measure the amount of an element or a compound is by weight.
In chemistry we typically express the amount of a substance in grams or moles.

28. Calculation of Moles from Mass

29. Calculating mole of Caffeine from sample mass

30. Calculating mole of Ammonium Phosphate from sample mass

31. Calculating numbers of atoms or molecules from Mass

32. Calculating number of silicon atoms

33. Calculating number of water molecules

34. Percent Composition of a Compound
What is the composition by mass percent of caffeine, C8H10N4O2, molar mass = g/mol;
Mass percent of Carbon = (96.08 g/194.2 g) x 100% = 49.47%
Mass percent of Hydrogen = (10.08 g/194.2 g) x 100% = 5.19%
Mass percent of Nitrogen = (56.04 g/194.2 g) x 100% = 28.86%
Mass percent of Oxygen = (32.00 g/194.2 g) x 100% = %
Total percent (by mass) = 100%

35. Percent Composition of a Compound
What is the composition (by mass percent) of ammonium phosphate, (NH4)3PO4; molar mass = g/mol;
Mass percent of N = (42.03 g/149.1 g) x 100% = 28.19%
Mass percent of H = (12.10 g/149.1 g) x 100% = 8.12%
Mass percent of P = (30.97 g/149.1 g) x 100% = 20.77%
Mass percent of O = (64.00 g/149.1 g) x 100% = %
Total percent (by mass) = 100%

36. Determination of Empirical and Molecular Formulas
Empirical Formula
A formula that shows the simples whole number ratio of moles of atoms.
Examples: MgO, Cu2S, CH2O, Al2C3O9. etc.
Molecular Formula
A formula that shows the actual number of atoms of each type of element in a molecule.
Examples: C4H10, C6H6, C6H12O6, N2H4, P4O10, etc.

37. Determination of Empirical Formula #1

38. Determination of Empirical Formula #2
Example-2: A compound made up of carbon, hydrogen, and oxygen has the following composition (by mass percent): 68.1% C, 13.7% H, and 18.2% O. Determine its empirical formula.
Solution: Treat mass percent like mass and calculate mole of each element in 100-g sample of the compound.
Mole of C = 68.1 g x (1 mol C/12.01 g) = 5.67 mol;
Mole of H = 13.7 g x (1 mol H/1.008 g) = 13.6 mol;
Mole of O = 18.2 g x (1 mol O/16.00 g) = 1.14 mol;
Divide throughout by the mole of O (the smallest mole value):
5.67 mol C/1.14 = ~ 5;
13.6 mol H/1.14 = 11.9 ~ 12;
1.14 mol O/1.14 = 1;
Empirical formula = C5H12O

39. Determination of Empirical Formula #3
Example-3: A 2.32-g sample of a compound composed of carbon, hydrogen, and oxygen is completely combusted to yield 5.28 g of CO2 gas and 2.16 g of water. Calculate the composition (in mass percent) of the compound and determine its empirical formula.
Solution:
Determine the mass of C, H, and O, respectively, in the sample:
Mass of C = 5.28 g CO2 x (12.01 g C/44.01 g CO2) = 1.44 g;
Mass of H = 2.16 g H2O x (2 x g/18.02 g H2O) = g;
Mass of O = 2.32 g sample – 1.44 g C – 0.24 g H = 0.64 g;
Calculate the mass percent of each element:
Mass % of C = (1.44 g C/2.32 g sample) x 100% = 62.1% ;
Mass % of H = (0.242 g H/2.32 g sample) x 100% = 10.4% ;
Mass % of O = 100 – 62.1% C – 10.4% H = 27.5%
Composition (by mass %): 62.1% C; 10.4% H, and 27.5% O;
(continue next slide for empirical formula determination)

40. Determination of Empirical Formula #3
(from previous slide)
Calculate mole of each element from mass percent:
Mole of C = 62.1 g C x (1 mol/12.01 g) = 5.17 mol
Mole of H = 10.4 g H x (1 mol/1.008 g) = 10.3 mol
Mole of O = 27.5 g O x (1 mol/16.00 g) = 1.72 mol
Dividing throughout by smallest mole of O yield a simple ratio:
5.17 mol C/1.72 = 3 mol C;
10.3 mol H/1.72 = 6 mol H;
1.72 mol O/1.72 = 1 mol O;
Simple mole ratio of the elements: 3 mol C : 6 mol H : 1 mol O
Empirical formula: C3H6O

41. Molecular Formula

42. Exercise #1: Determination of empirical formula of a Copper Sulfide
A 5.00-gram copper sample is completely reacted with excess sulfur, which yields 6.26 g of product composed of only copper and sulfur. (a) Determine the empirical formula of the compound and give its correct chemical name. (b) Write the chemical equation for the reaction.
Answer: (a) Cu2S = Copper(I) sulfide;
(b) 2Cu(s) + S(s)  Cu2S(s)

43. Exercise #2: Determination of empirical and molecular formula of phosphorus oxide
A 2.50-gram sample of phosphorus is completely burned in air, which yields 5.72 g of product composed of only phosphorus and oxygen. In a separate analysis, the compound was found to have molar mass of 284 g/mol.
(a) Determine the empirical formula and molecular formula of the compound.
(b) What is the name of the compound?
(c) Write an equation for the combustion of phosphorus.
Answer: (a) Empirical formula = P2O5; molecular formula = P4O10; (b) Tetraphosphorus decoxide; (c) 4P(s) + 5 O2(g)  P4O10(s)

44. Solution Concentrations
The concentration of a solution may be expressed in:
Molarity, mass percent, volume percent, mass-volume percent, ppm & ppb.

45. Solution Concentrations

46. Molar Concentration Example-1:
4.0 g of sodium hydroxide, NaOH, is dissolved in enough water to make 100. mL of solution. Calculate the molarity of NaOH.
Mole of NaOH = 4.0 g NaOH x (1 mole/40.0 g)
= 0.10 mole
Molarity of NaOH = 0.10 mol/0.100 L = 1.0 M

47. Molar Concentration

48. Calculation of Solute Mass in Solution
Example-2:
How many grams of NaOH are present in 35.0 mL of 6.0 M NaOH solution?
Mole of NaOH = (6.0 mol/L) x 35.0 mL x (1 L/1000 mL)
= 0.21 mol
Mass of NaOH = 0.21 mol x (40.0 g/mol)
= 8.4 g NaOH

49. Preparing of Solutions from Pure Solids
From the volume (in liters) and the molarity of solution, calculate the mole and mass of solute needed;
Weigh the mass of pure solute accurately;
Transfer solute into a volumetric flask of appropriate size;
Add deionized water to the volumetric flask, well below the narrow neck, and shake well (or use a magnetic stirrer) to dissolve the solute;
When solid is completely dissolved, add more deionized water to fill the flask to the mark and mix the solution well.
(Note: If the solution is very warm due to exothermic reaction, let it cools down to room temperature before adding more water to the “mark”.)

50. Preparing Solution from Solid
Example-1:
Explain how you would prepare a mL of M NaCl solution.
Calculate the mole and mass of NaCl needed:
Mole of NaCl = L x (0.154 mol/L) = mol
Mass of NaCl = mol x (58.44 g/mol) = 4.50 g
Preparing the solution:
Weigh accurately 4.50 g of NaCl and transfer solid into 500-mL volumetric flask. Fill the flask half way with distilled water, shake well until all solid has dissolved. Add deionized water to fill the flask to the 500-mL level, place a stopper on the flask and mix the solution well by inverting and shaking the flask several times.

51. Preparing Solution from Stock
Calculate volume of stock solution needed using the formula: MiVi = MfVf (i = initial; f = final)
Measure accurately the needed volume of stock solution and transfer to a volumetric flask of appropriate size;
Dilute stock solution with distilled water to the required volume (or to the “mark” on volumetric flask)
Mix solution well.
(Note: for diluting concentrated acid, place some deionized water in the flask (to about a quarter full), add the required volume of concentrated acid, and then add more deionized water to the required volume.)

52. Preparing Solution from Stock
Example-2:
Explain how you would prepare 1.0 L of 3.0 M H2SO4 solution from concentrated H2SO4, which is 18 M.
Calculate volume of concentrated H2SO4 needed:
Vol. of conc. H2SO4 = (1.0 L x 3.0 M/18 M) = 0.17 L = 170 mL
Preparing the solution:
Place about 200 mL of deionized water in the 1-liter volumetric flask.
Measure accurately 170 mL of concentrated H2SO4 and transfer carefully to the volumetric flask containing some deionized water.
Allow solution to cool down to room temperature; then fill the flask to the 1-liter mark with more distilled water;
Mix solution well by placing a stopper on the flask and inverting the flask back and forth several times.

53. Other Units for Solution Concentration

54. Percent by Mass

55. Calculation of Mass from Percent

56. Percent by Volume

57. Mass-Volume Percentage

58. ppm and ppb

59. Chemical Equation #1 4Fe(s) + 3 O2(g)  2Fe2O3(s)
Description of reaction:
Iron reacts with oxygen gas and forms solid iron(III) oxide:
Identity: reactants = iron (Fe) and oxygen gas (O2); product = iron(III) oxide
Chemical equation: Fe(s) + O2(g)  Fe2O3(s)
Balanced equation:
4Fe(s) + 3 O2(g)  2Fe2O3(s)

60. Chemical Equation #2 Description of reaction:
Phosphorus reacts with oxygen gas to form solid tetraphosphorus decoxide.
Equation: P(s) + O2(g)  P4O10(s)
Balanced eqn.: 4P(s) + 5 O2(g)  P4O10(s)

61. Chemical Equation #3 Description of reaction:
Propane gas (C3H8) burns in air (reacts with oxygen gas) to form carbon dioxide gas and water vapor;
Identity: reactants = C3H8(g) and O2(g);
products = CO2(g) and H2O(g);
Equation: C3H8(g) + O2(g)  CO2(g) + H2O(g);
Balanced equation:
C3H8(g) + 5 O2(g)  3CO2(g) + 4H2O(g)

62. Chemical Equation #4

63. Balancing Chemical Equations
Rules for balancing equations:
Use smallest integer coefficients in front of each reactants and products as necessary; coefficient “1” need not be indicated;
Never change the formula of any substance in the equation.
Helpful steps in balancing equations:
Begin with the compound that contains the most atoms or types of atoms.
Balance elements that appear only once on each side of the arrow.
Balance last elements that appear more than once on either side.
Balance free elements last.
Finally, check that the smallest whole number coefficients are used.