Composition of Substances Chapter# 3 Composition of Substances

Stoichiometry Topics 3-1 Formula Mass and the Mole Concept 3-2 Determining Empirical and Molecular Formulas 3-3 Solution Concentration

The Chemical Package About Packages The baker uses a package called the dozen. All dozen packages contain 12 objects. The stationary store uses a package called a ream, which contains 500 sheets of paper. So what is the chemistry package?

6-2 The Chemical Package About Packages The baker uses a package called the dozen. All dozen packages contain 12 objects. The stationary store uses a package called a ream, which contains 500 sheets of paper. So what is the chemistry package? Well, it is called the mole (Latin for heap). Each of the above packages contain a number of objects that are convenient to work with, for that particular discipline.

6-3 The Mole A mole contains 6.022X1023 particles, which is the number of carbon-12 atoms that will give a mass of 12.00 grams, which is a convenient number of atoms to work with in the chemistry laboratory. The atomic weights listed on the periodic chart are the weights of a mole of atoms. For example a mole of hydrogen atoms weighs 1.00797 g and a mole of carbon atoms weighs 12.01 g.

6-4 Moles of Objects Suppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows?

6-4 Moles of Objects Suppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows? Would cover the entire 50 states 60 miles deep

6-4 Moles of Objects Suppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows? Would cover the entire 50 states 60 miles deep How about a mole of computer paper instead of a ream of computer paper, how far would that stretch?

6-4 Moles of Objects Suppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows? Would cover the entire 50 states 60 miles deep How about a mole of computer paper instead of a ream of computer paper, how far would that stretch? Way past the planet Pluto!

“the mass of one mole of that substance.” Formula Weight We can define the molar mass or molecular weight of a substance can as: “the mass of one mole of that substance.” We give molar mass the symbol M and it has units g/mol- For an atom the molar mass is equal to the atomic weight we find on the period table.

Its All About Moles For a molecule the molecular weight is equal to the sum of the atomic weights of the atoms that make it up. The mathematical relationship between mass and moles is: m = nM This can be summarized by the following diagram: moles

Formula Weight Calculation To calculate the molar mass of a compound we sum together the atomic weights of the atoms that make up the formula of the compound. This is called the formula weight (MW, M). Formula weights are the sum of atomic weights of atoms Making up the formula. The following outlines how to find the formula weight of water symbol weight number H 1.01 X 2 = 2.02 O 16.0 X 1 = 16.0 18.0 g/mole

Percent Composition Find the formula weight and the percent composition of glucose (C6H12O6) symbol weight number C 12.0 x 6 = 72.0 H 1.01 x 12 = 12.12 16.0 x 6 = 96.0 O 180.1 g/mole 72.0 %C = X = 40.0 %C 180.1 12.12 %H = X = 6.73 %H 180.1 96.0 %O = X = 53.3 %O 180.1

Mole Concepts The key equation to remember for mole calculations is: m = nM Where: M = molecular weight (gmol-1) n = number of moles (mol) m = mass of sample (g)

Mole Concepts For a mole of a molecule the number of moles of each atom is determined by how many of that atom are in each molecule. e.g. One mole of H2O contains: One mole of oxygen atoms Two moles of hydrogen atoms In 5 moles of H2SO4 how many moles of oxygen atoms is there? 20 moles of O atoms.

Mole Conversions In 50.0g of H2SO4 how many moles of oxygen atoms are there? 50.0g of H2SO4

Mole Conversions In 50.0g of H2SO4 how many moles of oxygen atoms are there? 50.0g of H2SO4 mole H2SO4 = 98.0g of H2SO4

Mole Conversions In 50.0g of H2SO4 how many moles of oxygen atoms are there? 50.0g of H2SO4 mole H2SO4 4mole O = 2.04 mole O 98.0g of H2SO4 mole H2SO4

Mole Conversions In 5 moles of H2SO4 how many atoms of oxygen are present?

Mole Conversions In 5 moles of H2SO4 how many atoms of oxygen are present? 5 moles H2SO4 =

Mole Conversions In 5 moles of H2SO4 how many atoms of oxygen are present? 5 moles H2SO4 4 mole O mole H2SO4

Mole Conversions In 5 moles of H2SO4 how many atoms of oxygen are present? 5 moles H2SO4 4 mole O 6.02 x 1023 atoms O = mole H2SO4 mole O 1.20 x 1025 atoms

Empirical Formulas Empirical formula is the smallest whole number ratio between atoms and can be calculated from the percent composition. Molecular formulas happen to be the exact number of atoms making up a molecule, and may or may no be the simplest whole number ratio. Molecular formulas are whole number multiples of the empirical formula.

Empirical Formula Steps Assume 100 g of compound. Convert percent to a mass number. Convert the mass to moles. Divide each mole number by the smallest mole number. Rounding: If the decimal is ≤ 0.1, then drop the decimals If the decimal is ≥0.9, then round up. All other decimal need to be multiplied by a whole number until roundable.

Empirical Formula Example A compound is composed of 75.0% C and 25.0% H. Find its empirical formula. Step #1 Assume 100 g of compound 75.0 g C 25.0 g H

Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #2 Convert grams to moles. 75.0 g C Mole C = 6.225 mole C 12.01 g 25.0 g H Mole H = 24.802 mole H 1.008 g H

Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #3 Divide each mole number by the smallest. 75.0 g C Mole C = 6.225 mole C 12.01 g 25.0 g H Mole H = 24.802 mole H 1.008 g H 6.225 24.802 = 1.00 = 3.98 6.225 6.225

Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≤ 0.1, drop decimals 75.0 g C Mole C = 6.225 mole C 12.01 g 25.0 g H Mole H = 24.802 mole H 1.008 g H 6.225 24.802 = 1.00 = 3.98 6.225 6.225

Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≤ 0.1, drop decimals 75.0 g C Mole C = 6.225 mole C 12.01 g 25.0 g H Mole H = 24.802 mole H 1.008 g H 6.225 24.802 = 1 C = 3.98 6.225 6.225

Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≥ 0.9, round up 75.0 g C Mole C = 6.225 mole C 12.01 g 25.0 g H Mole H = 24.802 mole H 1.008 g H 6.225 24.802 = 1 C = 3.98 6.225 6.225

Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≥ 0.9, round up 75.0 g C Mole C = 6.225 mole C 12.01 g 25.0 g H Mole H = 24.802 mole H 1.008 g H 6.225 24.802 = 1 C = 3.98 6.225 6.225

Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≥ 0.9, round up 75.0 g C Mole C = 6.225 mole C 12.01 g 25.0 g H Mole H = 24.802 mole H 1.008 g H 6.225 24.802 = 1 C = 4 H 6.225 6.225

Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≥ 0.9, round up 75.0 g C Mole C = 6.225 mole C 12.01 g 25.0 g H Mole H = 24.802 mole H 1.008 g H 6.225 24.802 = 1 C = 4 H 6.225 6.225 Empirical Formula = CH4

Molecular Formulas Empirical formula, is the smallest ratio between atoms in a molecular or formula unit. Molecular formula, is the exact number of atoms in a molecule; a whole number multiple of an empirical formula

Possible Molecular Formulas Assume an empirical formula of C3H5O Empirical formula Integer Molecular Formula C3H5O 1 2 3 4 5 C3H5O C3H5O C3H5O C3H5O C3H5O

6-9 Possible Molecular Formulas Assume an empirical formula of C3H5O Empirical formula Integer Molecular Formula C3H5O 1 2 3 4 5 C3H5O C3H5O C6H10O2 C3H5O C3H5O C3H5O

Possible Molecular Formulas Assume an empirical formula of C3H5O Empirical formula Integer Molecular Formula C3H5O 1 2 3 4 5 C3H5O C3H5O C6H10O2 C9H15O3 C3H5O C3H5O C3H5O

Possible Molecular Formulas Assume an empirical formula of C3H5O Empirical formula Integer Molecular Formula C3H5O 1 2 3 4 5 C3H5O C3H5O C6H10O2 C9H15O3 C3H5O C3H5O C12H20O4 C3H5O C15H25O5

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #1 Assume 100g of compound 83.6 g C 16.3 g H

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #2 Convert grams to moles 83.6 g C mole 12.01 g C 16.3 g H mole 1.008 g H

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #2 Convert grams to moles 83.6 g C mole = 6.961 mole 12.01 g C 16.3 g H mole = 16.17 mole 1.008 g H

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #3 Divide each mole number by the smallest. 83.6 g C mole = 6.961 mole 12.01 g C 16.3 g H mole = 16.17 mole 1.008 g H

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #3 Divide each mole number by the smallest. 83.6 g C mole 6.961 = 6.961 mole = 1.00 12.01 g C 6.961 16.3 g H mole = 16.17 mole = 2.32 1.008 g H

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #4 Round if---Not Roundable 83.6 g C mole 6.961 = 6.961 mole = 1.00 12.01 g C 6.961 16.3 g H mole = 16.17 mole = 2.32 1.008 g H Step #4, Multiply by an integer until roundable 1.00 X 3 = 3 2.32 X 3 = 7 Empirical formula C3H7

Molecular Formula Integer Divide empirical weight into molecular weight 3x12 + 7x1 =43 2 43 86 Now multiply the empirical formula by 2

Molecular Formula Integer Divide empirical weight into molecular weight 3x12 + 7x1 =43 2 43 86 Now multiply the empirical formula by 2 Molecular Formula is C6 H14

Solutions Solutions are homogeneous mixtures of two or more substances. The solvent is the substance in greatest quantity. Solutes are the other ingredients in the mixture. Solutions can exist in all states of matter.

Solution Examples Margarine Tap Water Steel 18 Carat Gold Air Sterling Silver

Solution Examples Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. What is the solvent in 18 ct gold ?

Solution Examples Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. What is the solvent in 18 ct gold ? Gold

Solution Examples Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. What is the solvent in 18 ct gold ? Gold What are the solutes?

Solution Examples Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. What is the solvent in 18 ct gold ? Gold What are the solutes? Silver and Copper

Solution Properties Some general properties of solutions include: Solutions may be formed between solids, liquids or gases. They are homogenous in composition They do not settle under gravity They do not scatter light (Called the Tyndall Effect) Solute particles are too small to scatter light and therefore light will go right through a solution like is shown on the next slide.

Tyndall Effect Laser light reflected by a colloid. In a solution you would not see any red light.

Solutions

Aqueous Solutions Water is the dissolving medium 104.5o  

Some Properties of Water Water is “bent” or V-shaped. Water is a molecular compound. Water is a polar molecule. Hydration occurs when ionic compounds dissolve in water.

SOLUTION CONCENTRATION The ratio of the amount of solute to amount of solution, or solvent is defined by the concentration. solute solute Concentration = = solution solvent There are various combinations of units that are used in these rations. Ratio X 102 X 103 X 106 X 109 g solute % (w/w) ppt (w/w) ppm (w/w) ppb (w/w) = g solution g solute = % (w/v) ppt (w/v) ppm (w/v) ppb (w/v) mL solution mL solute = % (v/v) ppt (v/v) ppm (v/v) ppb (v/v) mL solution

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 25.2 g NaCl

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 25.2 g NaCl 33.6g H2O + 25.2 g NaCl

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl 33.6g H2O + 25.2 g NaCl

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O + 25.2 g NaCl

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O + 25.2 g NaCl 44.6 g NaCl 100 g solution

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O + 25.2 g NaCl 44.6 g NaCl 333 g solution 100 g solution

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O + 25.2 g NaCl 44.6 g NaCl 333 g solution 100 g solution

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O + 25.2 g NaCl 44.6 g NaCl 333 g solution = 149 g NaCl 100 g solution

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O + 25.2 g NaCl 44.6 g NaCl 333 g solution = 149 g NaCl 100 g solution Mass of water?

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O + 25.2 g NaCl 44.6 g NaCl 333 g solution = 149 g NaCl 100 g solution Mass of water? 333 g solution – 149 g NaCl = 184 g H2O

SAMPLE SOLUTION PROBLEMS 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 100 25.2 g NaCl = 43.0 % NaCl 33.6g H2O + 25.2 g NaCl 44.6 g NaCl 333 g solution = 149 g NaCl 100 g solution Mass of water? 333 g solution – 149 g NaCl = 184 g H2O

SAMPLE SOLUTION PROBLEMS How many grams of NaCl are required to dissolve in 88.2 g of water to make a 29.2% (w/w) solution. A sugar solution is 35.2%(w/v) find the mass of sugar contained in a 432 mL sample of this sugar solution.

SOLUTION CONCENTRATION The solution concentration can also be defined using moles. The most common example is molarity (M). The molarity of a solution is defined as: “The number of moles of solute in 1 L of solution” and is given the formula: Moles solute Molarity (M) = Liters solution

MOLARITY SAMPLE PROBLEMS A student dissolves 25.8 g of NaCl in a 250 mL volumetric flask. Calculate the molarity of this solution. (picture of volumetric flask is on the next slide) Find the mass of HCl required to form 2.00 L of a 0.500 M solution of HCl. A student evaporates the water form a 333 mL sample of a 0.136 M solution of NaCl. What mass of salt remains?

SOLUTION PREPARATION In the lab we would use a piece of glassware called a volumetric flask to prepare this solution.

SOLUTION PREPARATION

VOLUMETRIC FLASK

SOLUTION DILUTION Often we will want to make a dilute solution from a more concentrated one. To determine how to do this we use the formula : C1V1 = C2V2 Where: C1 = concentration of more concentrated solution V1 = volume required of more concentrated solution C2 = concentration of more dilute solution V2 = volume of more dilute solution We can use any units in this equation but they must be the same on both sides.

DILUTION PROBLEM How would one prepare 50.0 mL of a 3.00 M solution of NaOH using a 7.10 M stock solution? C1V1 = C2V2 (7.10 M)V1 = (3.00 M) (50.0 mL) This means that you add 21.1 mL of the concentrated stock solution to a 50.0 mL volumetric flask and add water until the bottom of the meniscus touches the line on the volumetric flask.

DILUTION PROBLEM How would one prepare 50.0 mL of a 3.00 M solution of NaOH using a 7.10 M stock solution? C1V1 = C2V2 (7.10 M)V1 = (3.00 M) (50.0 mL) (3.00 M) (50.0 mL) (7.10 M)V1 = (7.10 M) (7.10 M) V1 = 21.1 mL This means that you add 21.1 mL of the concentrated stock solution to a 50.0 mL volumetric flask and add water until the bottom of the meniscus touches the line on the volumetric flask.

THE END