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:
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

9. Chem 1A Chapter 3 Lecture Outlines
Calculating Average Atomic Masses from the Isotope Masses and Abundances
Exercise-#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
= ( amu x ) + ( amu x )
= u (as given in the periodic table)

10. Calculating Average Atomic Mass
Chem 1A Chapter 3 Lecture Outlines
Calculating Average Atomic Mass
Exercise-#2:
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. Calculating Relative Abundance of Isotopes
Exercise-#3:
Boron has 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. Calculating Number of Atoms from Mass
Example-2:
The atomic mass of carbon is amu. (a) How many carbon atoms are present in a 5.00-g carbon? (b) How many carbon atoms are present in a g carbon? (1 amu = x g)
Solution:
First, we find what is amu in gram: amu x 𝑥 10 −24 𝑔 1 𝑎𝑚𝑢
= x g
Then, divide the mass of sample by the mass of a C-atom in gram:
(a) Number of C-atoms = 5.00 g x 1 𝐶−𝑎𝑡𝑜𝑚 𝑥 10 −23 𝑔 = 2.51 x 1023 atoms;
(b) # of C-atoms = g x 1 𝐶−𝑎𝑡𝑜𝑚 𝑥 10 −23 𝑔 = x 1023 atoms

13. Calculating Number of Atoms from Mass
Exercise-#4:
The atomic mass of silicon is amu. (a) How many silicon atoms are present in a 5.00-g silicon? (b) How many silicon atoms are present in g of silicon?
(1 amu = x 10–24 g)

14. Atomic Mass & Molar Mass
Chem 1A Chapter 3 Lecture Outlines
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

15. Chem 1A Chapter 3 Lecture Outlines

16. 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

17. 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?

18. 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)

19. 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.

20. 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.);

21. 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.

22. 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

23. 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.

24. 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)

25. 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.

26. 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)

27. 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

28. Calculating Molar Mass
Chem 1A Chapter 3 Lecture Outlines
Calculating Molar Mass
Example-3:
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

29. 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.

30. Calculation of Moles from Mass
The quantity in moles of a substance can be calculated from the sample mass in grams and the molar mass of the substance as follows:
Mole = 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐 𝑖𝑛 𝑔𝑟𝑎𝑚𝑠 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒

31. Calculating Moles from Sample Mass
Chem 1A Chapter 3 Lecture Outlines
Calculating Moles from Sample Mass
Example-4:
How many moles of caffeine, C8H10N4O2, are present in 5.00 g of this compound?
Solution:
Molar mass of caffeine =
(8x12.01)+(10x1.008)+(4x14.01)+(2x16.00) = g/mol;
Mole of caffeine in a 5.00-g sample = 5.00 g x 1 𝑚𝑜𝑙 𝑔 = mol;

32. Calculating mole of Ammonium Phosphate from sample mass
Example-5:
How many moles of ammonium phosphate, (NH4)3PO4, are present in 5.00 x 102 g of this compound?
Solution:
Molar mass of ammonium phosphate =
(3x14.01) + (12x1.008) + (1x30.97) + (4x16.00) = g/mol;
Moles of (NH4)3PO4 in a 5.00 x 102-g sample
= 5.00 x 102 g x 1 𝑚𝑜𝑙 𝑔 = 3.35 moles

33. Calculating Moles and Number of Atoms
Chem 1A Chapter 3 Lecture Outlines
Calculating Moles and Number of Atoms
Exercise-#5:
Calculate the molar mass of sucrose, C12H22O11.
How many moles of sucrose are present in 5.00 g of sucrose?
How many sucrose molecules are present in 5.00 g of sucrose?
How many atoms of carbon, hydrogen, and oxygen, respectively, are present in a 5.00-g sample?
(Answer: (a) g/mol; (b) mol; (c) 8.80 x 1021 molecules;
(d) 1.06 x 1023 C-atoms; 1.94 x 1023 H-atoms; 9.68 x 1022 O-atoms)

34. Calculating Moles and Number of Atoms
Exercise-#6:
Calculate the molar mass of aluminum sulfate, Al2(SO4)3.
How many moles of Al2(SO4)3 are present in 25.0 g of this compound?
How many Al2(SO4)3 units are present in 25.0 g of this compound?
How many aluminum, sulfur, and oxygen atoms, respectively, are present in a 25.0-g sample?
(Answer: (a) g/mol; (b) mol; (c) 4.40 x 1022 units;
(d) 8.80 x 1022 Al-atoms; 1.32 x 1023 S-atoms; 5.28 x 1023 O-atoms)

35. Calculating numbers of atoms or molecules from Mass
The number of atoms or molecules in a sample may be calculated from the sample mass in grams, the molar mass of the substance, and the Avogadro’s number. This involves converting the mass into mole and then multiplying the moles with Avogadro’s number.
Number of atoms or molecules =
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 𝑖𝑛 𝑔𝑟𝑎𝑚𝑠 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 x 𝑥 10^23 1 𝑚𝑜𝑙𝑒

36. Calculating Number of Atoms or Molecules
Example-5:
How many silicon atoms are present in a 5.00-g sample of pure silicon?
Molar mass of silicon = g/mol;
Number of Si-atoms in 5.00 g of silicon
= 5.00 g Si x 1 𝑚𝑜𝑙 𝑆𝑖 𝑔 x 𝑥 𝑎𝑡𝑜𝑚𝑠 1 𝑚𝑜𝑙𝑒
= 1.07 x 1023 Si atoms

37. Calculating Number of Atoms or Molecules
Example-6:
How many H2O molecules are present in a 1.00-g sample (~1 mL) of water?
Solution:
Molar mass of water = g/mol;
Number of H2O molecules in 1.00 g of water:
= 1.00 g x 1 𝑚𝑜𝑙 𝑔 x 𝑥 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 1 𝑚𝑜𝑙 = 3.34 x 1023 molecules

38. Percent Composition of a Compound
What is the percent composition by mass of caffeine, C8H10N4O2? (molar mass = g/mol)
Mass percent of carbon = 𝑔 𝐶 𝑔 𝑐𝑎𝑓𝑓𝑒𝑖𝑛𝑒 x 100% = 49.47%
Mass percent of hydrogen = 𝑔 𝐻 𝑔 𝑐𝑎𝑓𝑓𝑒𝑖𝑛𝑒 x 100% = 5.19%
Mass percent of nitrogen = 𝑔 𝑁 𝑔 𝑐𝑎𝑓𝑓𝑒𝑖𝑛𝑒 x 100% = 28.86%
Mass percent of oxygen = 𝑔 𝑂 𝑔 𝑐𝑎𝑓𝑓𝑒𝑖𝑛𝑒 x 100% = %
Total percent (by mass) = 100%

39. Calculation of Percent Composition
What is the percent composition by mass 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%

40. Percent Composition of a Compound
What is the percent composition by mass of ammonium phosphate, (NH4)3PO4? (molar mass = g/mol)
Mass percent of nitrogen = 𝑔 𝑁 𝑔 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 x 100% = 28.19%
Mass percent of hydrogen = 𝑔 𝐻 𝑔 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 x 100% = 8.12%
Mass percent of phosphorus = 𝑔 𝑃 𝑔 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 x 100% = 20.77%
Mass percent of oxygen = 𝑔 𝑂 𝑔 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 x 100% = 42.92%
Total percent (by mass) = 100%

41. Calculation of Percent Composition
What is the percent composition by mass 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% = 42.92%
Total percent (by mass) = 100%

42. 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.

43. Determination of Empirical Formula
Example-7:
A 5.00 sample of an oxide of phosphorus is found to contain 2.18 g of phosphorus and 2.82 g of oxygen. Calculate the empirical formula of the compound.
Solution:
Mol of P = 2.18 g x 1 𝑚𝑜𝑙 𝑃 𝑔 = mol;
Mol of O = 2.82 g x 1 𝑚𝑜𝑙 𝑂 𝑔 = mol;
Mole ratio: 𝑚𝑜𝑙 𝑂 𝑚𝑜𝑙 𝑃 = 𝑚𝑜𝑙 𝑂 𝑚𝑜𝑙 𝑃 = 2.5/1 = 5/2
Empirical formula = P2O5

44. Determination of Empirical Formula
Chem 1A Chapter 3 Lecture Outlines
Determination of Empirical Formula
Example-8:
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

45. Determination of Empirical Formula
Chem 1A Chapter 3 Lecture Outlines
Determination of Empirical Formula
Example-9:
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. (a) Calculate the composition of the compound (in mass percent); (b) 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)

46. Determination of Empirical Formula
Example-9 (continued):
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

47. Molecular Formula
Molecular formula of a compound is determined from the empirical formula and the molar mass of the compound; the latter is determined independently.
If empirical formula = CxHyOz
Molecular formula = (CxHyOz)n
Where n = 𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝐸𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠 ; (n is an integer)

48. Determination of Molecular Formula
Example-10:
A compound with empirical formula C3H6O has molecular mass of 116 amu. What is the molecular formula?
Solution:
Empirical formula mass = (3 x 12.01) + (6 x 1.008) amu
= 58.1 amu
n = 116 𝑎𝑚𝑢 58.1 𝑎𝑚𝑢 = 2; Molecular formula = C6H12O2

49. Determination of empirical and molecular formula of phosphorus oxide
Chem 1A Chapter 3 Lecture Outlines
Determination of empirical and molecular formula of phosphorus oxide
Exercise-#7:
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)

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

51. Solution Concentrations
Mass-volume percent = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑔𝑟𝑎𝑚𝑠 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑚𝐿 x 100%
Parts per million (ppm) = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 x 106
Parts per billion (ppb) = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 x 109

52. Molar Concentration Example-11:
9.00 g of sodium chloride, NaCl, is dissolved in enough water to make mL of salt solution. Calculate the molarity of NaCl.
Mole of NaCl = 9.00 g NaCl x 1 𝑚𝑜𝑙𝑒 𝑁𝑎𝐶𝑙 𝑔 = mol;
Molarity of NaCl = 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙 𝐿 = M

53. Calculation of Solute Mass in Solution
Chem 1A Chapter 3 Lecture Outlines
Calculation of Solute Mass in Solution
Example-12:
How many grams of NaOH are required to prepare mL of 2.0 M NaOH solution?
(Molar mass of NaOH = 40.0 g/mol) ?
Answer:
Mole of NaOH needed = L x 2.0 𝑚𝑜𝑙 1 𝐿 = 1.5 mol;
Mass of NaOH needed = 1 5 mol x 𝑔 1 𝑚𝑜𝑙 = 60. g

54. 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”.)

55. Preparing Solution from Solid
Example-13:
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.

56. Calculating Molarity Exercise-#8:
10.55 g of silver nitrate, AgNO3 (molar mass = g/mol), is dissolved in enough water to make a mL solution. What is the molar concentration of AgNO3?
(Answer: M)

57. Calculating Mass of Solute
Chem 1A Chapter 3 Lecture Outlines
Calculating Mass of Solute
Exercise-#9:
How many grams of glucose, C6H12O6, are needed to prepare mL of M glucose solution?
Briefly explain the method you would use to prepare this solution. (molar mass of glucose = g/mol)
(Answer: g)

58. 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.)

59. Preparing Solution from Stock
Example-14:
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.

60. Calculating Volume of Stock Solution
Exercise-#10:
35.0 mL of 6.0 M NaOH solution is added into a mL volumetric flask, and enough deionized water is then added to make the final solution to mL. Calculate the molar concentration of NaOH in the dilute solution.
(Answer: 0.42 M)

61. Calculating Volume of Stock Solution
Exercise-#11:
How many milliliters (mL) of concentrated sulfuric acid, H2SO4, are needed to prepare 5.0 x 102 mL of 3.0 M H2SO4 solution? (Molarity of concentrated H2SO4 is 18 M)
Explain the safe method you would use to prepare this solution.
(Answer: (a) 83 mL)

62. Other Units for Solution Concentration

63. Calculating of Mass Percent
Example-15:
A sugar solution contains 25.0 g sugar dissolved in g of water. What is the mass percent of sugar in the solution?
Answer:
Percent sugar = 𝑔 𝑔 x 100% = 20.0% (by mass)

64. Calculating of Mass Percent
Chem 1A Chapter 3 Lecture Outlines
Calculating of Mass Percent
Example-16:
Seawater contains 3.5% NaCl, by mass. How many grams of salt are present in 5.00 gallons of seawater (sw)?
Solution: (1 gallon = 3785 mL; assume density = 1.00 g/mL)
Mass of seawater = 5.00 gall x 𝑚𝐿 1 𝑔𝑎𝑙𝑙𝑜𝑛 x 𝑔 𝑚𝐿 = 1.89 x 104 g
Mass of salt = 1.89 x 104 g sw x 3.5 𝑔 𝑁𝑎𝐶𝑙 100 𝑔 𝑠𝑤 = 660 g NaCl

65. Calculation of Mass Percent
Exercise-#12:
20.00 g of sugar and 5.00 g of salt (NaCl) are dissolved in g of water. Calculate the mass percent of sugar and salt, respectively, in the solution.
(Answer: 16.00% sugar; 4.00% salt)

66. Calculating Volume Percent
Example-17:
150 mL of methanol, 120 mL of acetone and 230 mL of water are mixed to form a homogeneous 5.0 x 102 mL solution. Calculate the percentage (by volume) of each component in the mixture.
Solution:
Percent methanol = 150 𝑚𝐿 5.0 𝑥 𝑚𝐿 x 100 = 30.%
Percent acetone = 120 𝑚𝐿 5.0 𝑥 𝑚𝐿 x 100 = 24%
Percent water = 230 𝑚𝐿 5.0 𝑥 𝑚𝐿 x 100 = 46%

67. Calculation Mass Percent
Chem 1A Chapter 3 Lecture Outlines
Calculation Mass Percent
Exercise-#13:
Gasohol is a mixture containing 10.% ethanol (C2H5OH) and 90.% gasoline (by volume). The density of ethanol and gasoline are 0.79 g/mL and 0.70 g/mL, respectively. What are the composition of gasohol by mass percent?
(Answer: 11% ethanol; 89% gasoline)

68. Calculation using Mass Percent
Exercise-#14:
Gasohol contains 10.% ethanol, C2H5OH, and 90.% gasoline (by volume). The density of ethanol is 0.79 g/mL.
How many grams of ethanol are present in 1.00 gallon of gasohol? (1 gallon = L)
What is the molarity of ethanol in gasohol?
(Answer: (a) 3.0 x 102 g; (b) 1.7 M)

69. Calculating Mass-Volume Percent
Example-18:
9.0 g of NaCl is dissolved in 473 mL of solution. What is the percent (w/v) of NaCl in the solution?
28 g of glucose is dissolved in 475 mL of solution. What is the percent (w/v) of glucose in the solution?
Solutions:
Percent of NaCl = 9.0 𝑔 473 𝑚𝐿 x 100 = 1.9% (w/v)
Percent of glucose = 28 𝑔 475 𝑚𝐿 x 100 = 5.9% (w/v)

70. Chem 1A Chapter 3 Lecture Outlines
ppm and ppb

71. 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)

72. 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)

73. 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)

74. Chemical Equation #4

75. Determination of empirical formula of a Copper Sulfide
Exercise-#15:
Write a balanced equation of each reaction described below:
Copper reacts with sulfur to form copper(I) sulfide;
Sodium metal reacts with water to form aqueous sodium hydroxide solution.
Butane gas (C4H10(g)) burns and reacts with atmospheric oxygen gas to produce carbon dioxide and water vapor.