Chapter 10 Chemical Quantities

Section 10.1 The Mole: A Measurement of Matter Objectives: Describe methods of measuring the amount of something. Define Avogadro’s number as it relates to a mole of a substance. Distinguish between the atomic mass of an element and its molar mass. Describe how the mass of a mole of a compound is calculated.

How many? or How much? How much sand is in the bucket? Is it practical to count each grain of sand? How else could we measure or quantify the sand? Mass or volume

How many? or How much? Just as it is impractical to count the number of grains of sand – it is impractical to measure a sample of a chemical substance by counting the number of atoms… … but it is done by using a practical method that relates a unit to a particular quantity For example 1 dozen = 12 When we count atoms our quantity is much larger Silver Gold

Measuring Matter How do we measure matter? We can measure Mass Volume Counting pieces Measure mass in grams. Measure volume in liters. Count pieces in moles.

What is a Mole? The animal In chemistry 1 mole is 6.022 x 1023

What is a mole? A mole is a unit that is the quantity 6.022 x 1023 just as a dozen is the quantity 12, a pair is 2 and a gross is 144 A dozen roses or a dozen eggs is always 12 roses or 12 eggs A pair of shoes is always 2 shoes A gross of pencils is always 144 pencils

What is a mole? A mole is the same concept… 1 mole of eggs is 6.022 x 1023 eggs 1 mole of roses is 6.022 x 1023 roses 1 mole of carbon atom is 6.022 x 1023 carbon atoms

Representative Particles A mole is most commonly used to measure the amount of representative particles in a chemical substance It is the SI unit for measuring the amount of a substance 6.022 x 1023 is called Avogadro's number 1 mole of a substance contains Avogadro's number of representative particles (or 6.022 x 1023 representative particles)

Representative Particles Representative particles refer to the species present in a substance Most elements = atoms (diatomics = molecules) Molecular compounds = molecules H2O, SO2 Ionic compounds = formula units or ions NaCl = 1 formula unit or NaCl = 2 ions Na+ and Cl-

How many representative particles are in the following samples of matter? *1 mole of any substance contains Avogadro’s number of representative particles* 1 mole of potassium = ________________ 1 mole of Ca = ________________ 2 moles of Au = ________________ 1 mole of H2O = ________________ 1 mole of Cl2 = ________________ 1 mole of KCl = ________________ or ________________ 6.022 x 1023 atoms of K 6.022 x 1023 atoms of Ca 1.204 x 1024 atoms of Au 6.022 x 1023 molecules of H20 6.022 x 1023 molecules of Cl 6.022 x 1023 formula units of KCl 1.204 x 1024 ions

Conversion Factors This brings us back to the use of conversion factors (remember chapter 3!) A ratio of equivalent measurements used to convert a quantity from one unit to another 2 possible conversion factors for the mole 1 mole 6.022 x 1023 representative particles OR 6.022 x 1023 representative particles 1 mole

How many molecules are in 2.12 mole of carbon dioxide? Conversion Factors Using the conversion factors we can determine the number of atoms that are in a mole of a compound or how many moles are in a sample of a compound How many molecules are in 2.12 mole of carbon dioxide? Step 1: Write the quantity (including unit) that is known from the problem. Step 2: Write the unit that we are trying to convert to. Step 3: Determine conversion factor (or factors) needed to solve the problem. Step 4: Solve the problem by crossing out units and then complete the math. 2.12 mole CO2 6.022 x 10 23 molecules of CO2 = _________ molecules of CO2 1.28 x 1024 1 mole CO2

(Follow the 4 steps from the previous slide) Practice Problem #1 Magnesium is a light metal that is used in the manufacture of aircrafts, automobile wheels, and tools. How many moles of magnesium is 1.25 x 1023 atoms of magnesium? (Follow the 4 steps from the previous slide) 1 mole of Mg 1.25 x 1023 atoms Mg .208 =__________ moles of Mg 6.022 x 1023 atoms Mg

Do Now What is the total number of oxygen atoms in Fe2(SO4)3? 4 12 3 17 How many atoms total in Fe2(SO4)3?

Practice Problem #2 How many atoms are in 1.5 mole of SO2? (follow the same steps but note step #3) 1.5 mole SO2 6.022 x 1023 molecules SO2 3 atoms SO2 = atoms of SO2 2.7 x 1024 1 mole SO2 1 molecule SO2 What kind of compound is SO2? What kind of representative particles are molecular compounds composed of? Therefore our first conversion factor is… Our second conversion factor will convert molecules to atoms How many atoms are in 1 molecule of SO2? Molecular Molecules (not atoms) 3 atoms in molecule

Measuring a Mole …but we can not count this many atoms We know 1 mole of a substance is 6.022 x 1023 particles …but we can not count this many atoms How then do we measure moles of a chemical substance? There must be another relationship Molar Mass

Molar Mass Molar mass is defined as the mass in grams of 1 mole of a particular substance In other words, molar mass is the mass of 6.022 x 1023 particles of that substance Molar mass can be found on the periodic table for all elements Remember the average atomic mass measured in amu…molar mass is the same number except measure in grams The molar mass of hydrogen is 1.008 g The molar mass of carbon is 12.011g

We can now count atoms by weighing them! Molar Mass of Elements What is the molar mass of oxygen? 15.998g What does this mean? 1 mole of oxygen has a mass of 15.998g It also means… A sample of oxygen with a mass of 15.998g contains 6.022 x 1023 atoms We can now count atoms by weighing them!

Molar Mass of Elements Why do 1 mole of 2 different elements, such as carbon and oxygen, have different molar masses if they contain the same number of atoms? An atom of oxygen contains 8 protons and 8 neutrons in the nucleus. An atom of carbon contains 6 protons and 6 neutrons in the nucleus, if 1 atom of O has more mass than 1 atom of C, it holds true that 1 mole of O should have more mass than 1 mole of C

Molar Mass of Compounds Molar mass of compounds is found by adding the individual atomic masses of all the elements in the compound (ionic or molecular) To find the mass of 1 mole of a compound Determine the kinds and numbers of elements in the compound Find out the mass of each element in the compound Multiply the masses of the elements by the number of atoms Add all elements together

Calculating Molar Mass of Compounds Calculate the molar mass of magnesium carbonate, MgCO3. 24.305 g Mg 12.011 g C + 3 (15.999) g O + = 84.32 g 1 mole of MgCO3 has a mass of 84.32 g. This is the mass of 6.022 x 1023 molecules of MgCO3

Calculating Molar Mass of Compounds Calculate the molar mass of H2O. 2 (1.008) g H 15.999 g O + = 18.02 g H2O

Do Now What is the molar mass (aka gram formula mass) of Ca3(PO4)2?

Section 10.2 - Mole-Mass and Mole-Volume Relationships Objectives: Describe how to convert the mass of a substance to the number of moles of a substance, and moles to mass Identify the volume of a quantity of gas at STP

Molar Mass is… The number of grams of one mole of atoms, ions, or molecules. We can make conversion factors from these to change grams of a compound to moles of a compound or moles a compound to grams of a compound Use the molar mass of an element or compound to convert between the mass of a substance and the moles of a substance

Conversion Factors Involving Molar Mass There are two possible conversion factors relating molar mass and number of moles 1 mole molar mass OR 1 mole molar mass What is the mass of 3.25 mole of potassium? Follow the steps for conversion problems 39.098 g of K 3.25 mole K 127.1 = ___________ g of K 1 mole K

Practice Problem #1 How many moles is 5.69 g of NaOH? .142 mole NaOH 5.69 g NaOH 39.997 g NaOH

Practice Problem #2 When iron is exposed to air it corrodes to form red-brown rust. Rust is iron(III) oxide. How many moles of iron(III) oxide are contained in 92.2 grams of iron(III) oxide? 1 mole .577 92.2 g Fe2O3 = moles of Fe2O3 159.691 g Fe2O3

Mole-Volume Relationship Many of the chemicals we deal with are gases. They are difficult to weigh (or mass). Need to know how many moles of gas we have. Two things effect the volume of a gas Temperature and pressure We know this from chapter 13 & 14

Standard Temperature and Pressure (STP) We need to compare moles of a gas at the same temperature and pressure 0ºC (32ºF) and 1 atm pressure Abbreviated STP Table A

Avogadro’s Hypothesis Avogadro hypothesized that equals volumes of gases at the same temperature and pressure contained equal numbers of particles The particles are not the same size but are so far apart that size is irrelevant At STP 1 mole of any gas occupies 22.4 L Called the molar volume We now have 2 more conversion factors 22.4 L of any gas at STP 1 mole OR 22.4 L of any gas at STP 1 mole

AVOGADRO’S HYPOTHESIS REMEMBER @ STP: 1 mol = 22.4L BUT What if a gas is not at STP?????? THEN “Equal Volumes of any gas at the SAME temperature and pressure contain an equal # of particles”

AVOGADRO’S HYPOTHESIS – AN EXAMPLE The data table below gives the temperature and pressure for four different gas samples, each in a 2-liter container. Which two gas samples contain the same total number of particles? CH4 and Ne

Practice Problem #1 How many moles is 5.67 L of O2 at STP? 1 mole .253 1 mole 5.67 L O2 = moles of O2 22.4 L

Practice Problem #2 What is the volume of 8.8 g of CH4 gas at STP? 12.27 8.8 g CH4 1 mole 22.4 L = _______ L CH4 16.063 g CH4 1 mole

Density of a Gas Density = Mass / Volume For a gas the units will be g / L We can determine the density of any gas at STP if we know the gases formula. If you assume you have 1 mole, then the mass is the molar mass and at STP the volume is 22.4 L. OR The density of a gas at STP and molar volume at STP can be used to calculate the molar mass of a gas

Practice Problem #2 The density of a gaseous compound containing carbon and oxygen is found to be 1.964 g/L at STP. What is the molar mass of the compound? X = 1.964 g/L 22.4 L X = 43.994 g

Summary of Conversion Factors These four items are all equal: a) 1 mole b) molar mass (in grams/mole) c) 6.022 x 1023 representative particles d) 22.4 L of gas at STP

Summary of Conversion Factors We have now examined a mole in terms of particles, mass and volume of gases at STP We can now convert between any of these quantities. The mole is the center of the chemical calculations and to convert from one unit to another we must use the mole as the intermediate step The form of the conversion factor depends on what we know and what we want to calculate

The Mole Road Map for Conversion Factors

Section 10.3 – Percent Composition and Chemical Formulas Objectives: Describe how to calculate the percent by mass of an element in a compound. Interpret an empirical formula. Distinguish between empirical and molecular formulas.

Percent Composition Percent Composition Relative amounts of the elements in a compound Percent by mass of each element in the compound Percent composition of a compound consists of a percent value for each different element in the compound Percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound multiplied by 100%

Calculating Percent Composition of a Compound Like all percent problems: Part Whole Find the mass of each of the components (the elements) Divide by the total mass of the compound Multiply by 100 % x 100 % Mass of element % mass of element = x 100 % Mass of compound

Calculating Percent Composition from Mass Data When a 13.60g sample of a compound containing only magnesium and oxygen is decomposed, 5.40g of oxygen is obtained. What is the percent composition of this compound? Sample of Mg and O = 13.60 g O = 5.40 g therefore Mg = 8.20 g O - 5.40 g Mg – 8.20 39.7 % = 60.3% = 13.60 g 13.60 g This 13.60 g sample of magnesium and oxygen is 39.7% oxygen and 60.3% magnesium

Finding Percent Composition from the Formula If we know the formula, assume you have 1 mole. Then you know the mass of the pieces and the whole (these values come from the periodic table) Mass of element in 1 mole of compound = X 100% % mass Molar mass of compound

Finding Percent Composition from the Formula – an example Calculate the percent composition of chlorine in mercury (II) chloride? Step 1: Calculate the molar mass of HgCl2 1 x (200.59 g for Hg) + 2 x (35.45 g for Cl) = 271.49 g of HgCl2 Step 2: Calculate % of molar mass for chlorine atoms in HgCl2 Mass due to Cl2 70.90 g Cl2 = X 100% = 26.12% Cl2 Molar mass 271.49 g of HgCl2

Practice Problem Propane (C3H8), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum. Calculate the percent composition of propane. Propane molar mass = 44.097 g Mass due to carbon (C3) = 36.033 g Mass due to hydrogen (H8) = 8.064 g 36.033g % composition of carbon 8.064 g % composition of hydrogen = 18.29% = 81.7% C 44.097 g 44.097 g

Percent Composition as a Conversion Factor We can use percent composition to calculate the number of grams of any element in a specific mass of compound Multiply the mass of the compound by a conversion factor based on the percent composition of the elements in the compound % composition of element Conversion Factor = 100

Percent Composition as a Conversion Factor – Practice Problem How much carbon and hydrogen are contained in 82.0g of propane? Use percent composition of propane from previous practice problem % composition of H = 18.29% (18.29/100) % composition of C = 81.71% (81.71/100) 82. 0g propane 18.29 14.998 + 66.994 = 14.998 g H 100 82. 0g propane 81.71 = 66.994 g C 100 81.992 = 82

Formulas Empirical Formula Molecular Formula The simplest whole number ratio of atoms in a compound Molecular Formula The true number of atoms of each element in the formula of a compound Molecular Formula = (empirical formula)n n = integer Molecular formula = C6H6 = (CH)6 Empirical formula = CH

Formulas Formulas for ionic compounds are always empirical (lowest whole number ratio because charges positive & negative charges must balance). Examples: NaCl MgCl2 Al2(SO4)3 K2CO3

Formulas Formulas for molecular compounds might be empirical (lowest whole number ratio). Molecular: H2O C6H12O6 C12H22O11 Empirical: H2O CH2O C12H22O11

Calculating Empirical Formula We can calculate a ratio of the elements from the percent composition. We must assume there are 100 g Therefore the percentage equals the number of grams. Convert grams to moles. Find lowest whole number ratio by dividing by the smallest value.

Calculating Empirical Formula – an example Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Assume 100 g (thus % = grams) so 38.67 g C x 1mole C = 3.220 mole C 12.01g C 16.22 g H x 1mole H = 16.09 mole H 1.01g H 45.11 g N x 1mole N = 3.219 mole N 14.01g N Now divide each value by the smallest value

Calculating Empirical Formula – an example The ratio is 3.220 mole C 3.219 mole N The ratio is 16.09 mole H The ratio is 3.219 mole N = C1H5N1 = CH5N = 1 mole carbon = 5 mole hydrogen = 1 mole nitrogen

Practice Problem Calculate the empirical formula of a compound that contains 67.6% Hg, 10.8% S, and 21.6% O 67.6 g Hg 1 mole = .337 mole Hg 200.59 g Hg Find smallest value 1 mole 10.8 g S = .337 mole S 32.06 g S 21.6 g O 1 mole = 1.35 mole O 15.99 g O

Practice Problem Continued Divide each mole value by smallest value .337 mole Hg = 1 mole Hg .337 mole .337 mole S Empirical Formula = HgSO4 = 1 mole S .337 mole 1.35 mole O = 4 mole O .337 mole

Converting From Empirical Formula to Molecular Formula The molecular formula of a compound is either the same as its experimentally determined empirical formula, or it is a simple whole number multiple of its empirical formula The empirical formula is the lowest ratio, the actual molecule has more mass by a whole number multiple.

Converting From Empirical Formula to Molecular Formula Once the empirical formula is determined the molecular formula can be calculated but the molar mass of the compound must be known Divide the actual molar mass of the compound by the empirical formula mass – you get a whole number to increase each coefficient in the empirical formula

Converting From Empirical Formula to Molecular Formula – an example A compound with empirical formula CH3 is found to have a gram molecular mass of 30 grams. Find its molecular formula? Step 1: Find the empirical formula mass of CH3 1 x (12.01 g for C) + 3 x (1.01 g for H) = 15.04 g of CH3 Step 2: Find Molecular Formula: Molecular Formula = Empirical Formula X Molar Mass Empirical Mass

Practice Problem Find the molecular formula of ethylene glycol, which is used in antifreeze. The molar mass is 62 g/mole and the empirical formula is CH3O. Empirical formula mass = 31.0 g 62 g (molar mass) = 2 31 g (empirical mass) Molecular Formula CH3O x 2 = C2H6O2