Measuring matter 11.1

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Transcription:

The Mole Ch 11

Measuring matter 11.1

Review 11.1 Vocabulary o molecule: two or more atoms that covalently bond together to form a unit New mole Avogadro s number Main Idea - Chemists use the mole to count atoms, molecules, ions, and formula units. NOTE YOU WILL NEED A SCIENTIFIC CALCULATOR FOR THIS CHAPTER!

How do we measure items? You can measure mass, or volume, or you can count pieces. We measure mass in grams. We measure volume in liters. We count pieces in MOLES.

What is the mole? We re not talking about this kind of mole!

Counting Particles Chemists need a convenient method for accurately counting the number of atoms, molecules, or formula units of a substance. The mole is the SI base unit used to measure the amount of a substance.

Moles (is abbreviated: mol) It is an amount, defined as the number of carbon atoms in exactly 12 grams of carbon- 12. 1 mole = 6.022 x 10 23 of the representative particles. Treat it like a very large dozen! 6.022 x 10 23 number. is called Avogadro s

Similar Words for an amount Pair: 1 pair of shoelaces = 2 shoelaces Dozen: 1 dozen oranges = 12 oranges Case: 1 case of Dr. Pepper = 24 cans Dr. Pepper Gross: 1 gross of pencils = 144 pencils Ream: 1 ream of paper = 500 sheets of paper

What are Representative Particles? The smallest pieces of a substance: 1) For a molecular compound: it is the molecule. 2) For an ionic compound: it is the formula unit (made of ions). 3) For an element: it is the atom. Remember the 7 diatomic elements? (made of molecules)

Types of questions How many oxygen atoms in the following? CaCO 3 3 atoms of oxygen Al 2 (SO 4 ) 3 12 (3 x 4) atoms of oxygen How many ions in the following? CaCl 2 3 total ions (1 Ca 2+ ion and 2 Cl 1- ions) NaOH Al 2 (SO 4 ) 3 2 total ions (1 Na 1+ ion and 1 OH 1- ion) 5 total ions (2 Al 3+ + 3 SO 4 2- ions)

Converting Between Moles and Particles Conversion factors must be used. Moles to particles Example: Number of molecules in 3.50 mol of sucrose

Converting Between Moles and Particles (cont.) Particles to moles o Use the inverse of Avogadro s number as the conversion factor.

Practice problems (round to 3 sig. figs.) How many molecules of CO 2 are in 4.56 moles of CO 2? 2.75 x 10 24 molecules How many moles of water is 5.87 x 10 22 molecules? 0.0975 mol (or 9.75 x 10-2 ) How many atoms of carbon are in 1.23 moles of C 6 H 12 O 6? 4.44 x 10 24 atoms C How many moles is 7.78 x 10 24 formula units of MgCl 2? 12.9 moles

10.1 Check What does the mole measure? A. mass of a substance B. amount of a substance C. volume of a gas D. density of a gas

10.1 Check What is the conversion factor for determining the number of moles of a substance from a known number of particles? A. B. C. 1 particle 6.02 10 23 D. 1 mol 6.02 10 23 particles

Mass and the mole 11.2

Review o New o 11.2 Vocabulary conversion factor: a ratio of equivalent values used to express the same quantity in different units molar mass Main Idea - A mole always contains the same number of particles; however, moles of different substances have different masses.

The Mass of a Mole 1 mol of copper and 1 mol of carbon have different masses. One copper atom has a different mass than 1 carbon atom.

Measuring Moles Remember relative atomic mass? - The amu was one twelfth the mass of a carbon-12 atom. Since the mole is the number of atoms in 12 grams of carbon-12, the decimal number on the periodic table is also the mass of 1 mole of those atoms in grams.

The Mass of a Mole (cont.) Molar mass (MM) is the mass in grams of one mole of any pure substance. Also called Formula Weight (FW)

Molar Mass Equals the mass of 1 mole of an element in grams (from periodic table) 12.011 grams of C has the same number of pieces as 1.008 grams of H 55.85 grams of iron. We can write this as: 12.011 g C = 1 mole C We can count things by weighing them.

Moles to mass Using Molar Mass

Using Molar Mass (cont.) Convert mass to moles with the inverse molar mass conversion factor.

Examples How much would 2.34 moles of carbon weigh? 28.1 grams C How many moles of magnesium is 24.31 g of Mg? 1 mol Mg How many atoms of lithium is 1.00 g of Li? 8.72 x 10 22 atoms Li How much would 3.45 x 10 22 atoms of U weigh? 13.6 grams U

Using Molar Mass (cont.) This figure shows the steps to complete conversions between mass and atoms.

11.2 Check The mass in grams of 1 mol of any pure substance is: A. molar mass B. Avogadro s number C. atomic mass D. 1 g/mol

11.2 Check Molar mass is used to convert what? A. mass to moles B. moles to mass C. atomic weight D. particles

Moles of compounds 11.3

Vocabulary Review o representative particle: an atom, molecule, formula unit, or ion Main Idea -The molar mass of a compound can be calculated from its chemical formula and can be used to convert from mass to moles of that compound.

Chemical Formulas and the Mole Chemical formulas indicate the numbers and types of atoms contained in one unit of the compound. One mole of CCl 2 F 2 contains one mole of C atoms, two moles of Cl atoms, and two moles of F atoms.

The Molar Mass of Compounds The molar mass of a compound equals the molar mass of each element, multiplied by the moles of that element in the chemical formula, added together. The molar mass of a compound demonstrates the law of conservation of mass.

The Molar Mass of Compounds in 1 mole of H 2 O molecules there are two moles of H atoms and 1 mole of O atoms (think of a compound as a molar ratio) To find the mass of one mole of a compound odetermine the number of moles of the elements present o Multiply the number times their mass (from the periodic table) oadd them up for the total mass

Calculating Molar Mass Calculate the molar mass of magnesium carbonate, MgCO 3. 24.3050 g + 12.0107 g + 3 x (15.9994 g) = 84.3139 Thus, 84.33139 grams is the formula mass for MgCO 3.

Examples Calculate the molar mass of the following and tell what type it is: Na 2 S N 2 O 4 C Ca(NO 3 ) 2 C 6 H 12 O 6 (NH 4 ) 3 PO 4 = 78.045 g/mol = 92.011 g/mol = 12.011 g/mol = 164.088 g/mol = 180.156 g/mol = 149.087 g/mol

Moles to Mass Conversion for Compounds For elements, the conversion factor is the molar mass of the elements. The procedure is the same for compounds, except that you must first calculate the molar mass of the compound.

Mass to Moles Conversion for Compounds The conversion factor is the inverse of the molar mass of the compound.

For example How many moles is 5.69 g of NaOH? 37

For example How many moles is 5.69 g of NaOH? 5. 69 g 38

For example How many moles is 5.69 g of NaOH? 5. 69 g mole g We need to change 5.69 grams NaOH to moles 39

For example How many moles is 5.69 g of NaOH? 5. 69 g mole g We need to change 5.69 grams NaOH to moles 1mole Na = 22.990 g 1 mol O = 15.999 g 1 mole of H = 1.008 g 40

For example 41 How many moles is 5.69 g of NaOH? 5. 69 g mole g We need to change 5.69 grams NaOH to moles 1mole Na = 22.990 g 1 mol O = 15.999 g 1 mole of H = 1.008 g 1 mole NaOH = 39.997 g

For example 1 mole NaOH = 39.997 g NaOH = 0.1422606696 mol NaOH = 0.142 mol NaOH 3 Sig Figs 42

Mass to Particles Conversion for Compounds Convert mass to moles of compound with the inverse of molar mass. Convert moles to particles with Avogadro s number. This figure summarizes the conversions between mass, moles, and particles.

11.3 Check How many moles of OH ions are in 2.50 moles of Ca(OH) 2? A.2.00 B. 2.50 C.4.00 D.5.00

11.3 Check How many particles of Mg are in 10 moles of MgBr 2? A.6.02 10 23 B. 6.02 10 24 C.1.20 10 24 D.1.20 10 25

Empirical and Molecular formulas 11.4

Vocabulary Review o percent by mass: the ratio of the mass of each element to the total mass of the compound expressed as a percent New percent composition empirical formula molecular formula Main Idea - A molecular formula of a compound is a whole-number multiple of its empirical formula.

Percent Composition The percent by mass of any element in a compound can be found by dividing the mass of the element by the mass of the compound and multiplying by 100.

Percent Composition The percent by mass of each element in a compound is the percent composition of a compound. Percent composition of a compound can also be determined from its chemical formula.

Calculating Percent Composition of a Compound 50 Like all percent problems: part x 100 % = percent whole 1) Find the mass of each of the components (the elements), 2) Next, divide by the total mass of the compound; then x 100%

Example Calculate the percent composition of a compound that is made of 29.0 grams of Ag with 4.30 grams of S (Assume you have one mol of substance) 51 29.0 g Ag 33.3 g total 4.30 g S 33.3 g total X 100 = 87.1 % Ag X 100 = 12.9 % S Total = 100 %

Examples Calculate the percent composition of C 2 H 4? 85.7% C, 14.3 % H How about Aluminum carbonate? 23.1% Al, 15.4% C, and 61.5 % O 52

Empirical Formula The empirical formula for a compound is the smallest whole-number mole ratio of the elements. You can calculate the empirical formula from percent by mass by assuming you have 100.00 g of the compound. Then, convert the mass of each element to moles. o SO 3 has a percent composition of 40.05% S and 59.95% O. o To find mole ratio assume 100 g of Sulfur Trioxide; this means 40.05 g S and 59.95 g O. Multiply each by molar mass to determine moles. o o 1.249 mol S and 3.747 mol O. Turn this into a whole number ratio by dividing by the smallest. S becomes 1 and O becomes 3 [If you still don t get whole number, multiply by smallest number that will produce a whole number. If you had 1.5 mol C, 3 mol H, 1 mol O, multiply all by 2 to get 3 mol C, 6 mol H, 3 mol O = C 3 H 6 O 2 The empirical formula may or may not be the same as the molecular formula. Molecular formula of hydrogen peroxide = H 2 O 2 Empirical formula of hydrogen peroxide = HO

Molecular Formula The molecular formula specifies the actual number of atoms of each element in one molecule or formula unit of the substance. Molecular formula is always a whole-number multiple of the empirical formula. To distinguish the molecular formula from empirical formula, chemists must experimentally determine the molar mass of the compound. Then, this relationship is used: o o Molecular formula = (empirical formula)n n is the factor by which molecular formula is obtained. Unknown substance X has formula weight of 26.04 g/mol and its empirical formula is CH. Determine the molecular formula. o (26.04g/mol)/ (13.02 g/mol) = 2 o ANSWER C 2 H 2

Empirical vs. Molecular Formula What is the empirical formula for the compound C 6 H 12 O 6? Which is the empirical formula for hydrogen peroxide? A.CHO B. C 2 H 3 O 2 C.CH 2 O D.CH 3 O A.H 2 O 2 B. H 2 O C.HO D.none of the above

Formulas of hydrates 11.5

Vocabulary Review crystal lattice: a three-dimensional geometric arrangement of particles New hydrate Main Idea -Hydrates are solid ionic compounds in which water molecules are trapped.

Naming Hydrates A hydrate is a compound that has a specific number of water molecules bound to its atoms. The number of water molecules associated with each formula unit of the compound is written following a dot. Sodium carbonate decahydrate = Na 2 CO 3 10H 2 O

Naming Hydrates

Analyzing Hydrates When heated, water molecules are released from a hydrate leaving an anhydrous compound. To determine the formula of a hydrate, find the number of moles of water associated with 1 mole of hydrate.

Analyzing Hydrates (cont.) 1. Weigh hydrate. 2. Heat to drive off the water. 3. Weigh the anhydrous compound. 4. Subtract and convert the difference to moles. 5. The ratio of moles of water to moles of anhydrous compound is the coefficient for water in the hydrate.

Use of Hydrates Anhydrous forms of hydrates are often used to absorb water, particularly during shipment of electronic and optical equipment. In chemistry labs, anhydrous forms of hydrates are used to remove moisture from the air and keep other substances dry.

11.5 Check Heating a hydrate causes what to happen? A.Water is driven from the hydrate. B. The hydrate melts. C.The hydrate conducts electricity. D.There is no change in the hydrate.

11.5 Check A hydrate that has been heated and the water driven off is called: A.dehydrated compound B. antihydrated compound C.anhydrous compound D.hydrous compound

11.5 Check Two substances have the same percent by mass composition, but very different properties. They must have the same. A.density B. empirical formula C.molecular formula D.molar mass