Chapter 10 The Mole Molar Mass Mole Relationships in Chemical Equations Mass Calculations for Reactions Molar Volume Percentage Composition Empirical and Molecular Formulas

THE MOLE CONCEPT A sensitive balance can weight to the nearest 0.0001 g, but a typical atom has a mass of only 0.000 000 000 000 000 000 000 01 g Is it possible to keep track of atoms by counting them? Yes, but not directly. We count them in groups. That is we count atoms in the same way we count eggs by the dozen (12), pencils by the gross (144), pop by the case (24), and sheets of paper by the ream (500).  The relative mass of an atom is compared to a reference atom using an instrument called a mass spectrometer. Carbon-12 has been chosen as a reference and it is assigned a mass of exactly 12 amu. The masses of other atoms are compared relative to carbon-12.  

Experiments have shown that 12. 01 g of carbon contains 6 Experiments have shown that 12.01 g of carbon contains 6.02 x 1023 atoms of carbon. In fact, the gram atomic mass of each element contains 6.02x1023 atoms of that element. AVOGADRO’S Number: Number of atoms in 12.01 grams of carbon 6.02 x1023 atoms

MOLE: Amount of a substance containing 6.02x1023 representative particles  Why do we need this tool (the mole)? Chemists and many others need to know relative amounts of substances.

A Moles of Particles Contains 6.02 x 1023 particles 1 mole C = 6.02 x 1023 C atoms 1 mole H2O = 6.02 x 1023 H2O molecules 1 mole NaCl = 6.02 x 1023 Na+ ions and 6.02 x 1023 Cl– ions

Examples of Moles Moles of elements 1 mole Mg = 6.02 x 1023 Mg atoms 1 mole Au = 6.02 x 1023 Au atoms Moles of compounds 1 mole NH3 = 6.02 x 1023 NH3 molecules 1 mole C9H8O4 = 6.02 x 1023 aspirin molecules

Avogadro’s Number 6.02 x 1023 particles 1 mole or 1 mole

MOLAR MASS: Mass in grams of one mole of any substance, element, or compound Has the same numerical value as formula mass. Molar mass of any element = the gram atomic mass (gam) of that element Example: 1 mole H = 1.01g H 1 mole Ca = 40.1g Ca 1 mole O = 16.0g O 1 mole C = 12.0g C

Molar Mass Number of grams in 1 mole Equal to the numerical value of the atomic mass 1 mole of C atoms = 12.0 g 1 mole of Mg atoms = 24.3 g 1 mole of Cu atoms = 63.5 g

Learning Check Give the molar mass to 0.1 g A. 1 mole of Br atoms = ________ B. 1 mole of Sn atoms = ________

Solution Give the molar mass to 0.1 g A. 1 mole of Br atoms = 79.9 g/mole B. 1 mole of Sn atoms = 118.7 g/mole

Problems What is the molar mass for the following elements? Na Cl Fe Hg

FORMULA MASS: (Molar Mass) The sum of the atomic masses, for each element, found in the compound Exp: Calculate the formula mass of H2O 2 hydrogen atoms @ 1.008 amu 1 oxygen atom @ 15.999 amu 2.016+15.9999= 18.015 amu One molecule of water has a mass of 18.015 amu

2 atoms H x atomic mass 1.0g = 2.0 g/mole Example: H2S 2 atoms H x atomic mass 1.0g = 2.0 g/mole 1 atom S x atomic mass 32.1g = 32.1 g/mole (Molar mass) = 34.1 g/mole   2(1.0) + 32.1 = 34.1 g/mole 1 mole = 6.02 x 1023 (how many) = molar mass (grams)

Molar Mass of Compounds Mass in grams of 1 mole equal numerically to the sum of the atomic masses 1 mole of CaCl2 = 111.1 g/mole 1 mole Ca x 40.1 g/mole = 40.1 g/mole + 2 moles Cl x 35.5 g/mole = 71.0 g/mole 1 mole of N2O4 = 74.0 g/mole 2 moles N x 14.0 g/mole = 28.0 g/mole + 4 moles O x 16.0 g/mole = 64.0 g/mole

Learning Check A. 1 mole of K2O = ______g B. 1 mole of antacid Al(OH)3 = ______g

Solution A. 1 mole of K2O = 94.2 g/mole 2 atoms K x 39.1 g/mole + 1 atom O x 16.0 g/mole B. 1 mole of antacid Al(OH)3 = 78.0 g/mole 1 atom Al x 27.0 g/mole + 3 atoms O x 16.0 g/mole + 3 atoms x 1.0 g/mole

Learning Check Prozac, C17H18F3NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. It has a molar mass of 1) 40.0 g/mole 2) 262 g/mole 3) 309 g/mole

Solution Prozac, C17H18F3NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. It has a molar mass of 3) 309 g/mole 17 atoms C (12.0) + 18 atoms H (1.0) + 3 atoms F (19.0) + 1 atom N (14.0) + 1atom O (16.0)

Problems State how many moles and calculate the molar mass of each substance. 2 Cu 5.00 NaCl 2.5 H2SO4

Molar Mass Factors Methane CH4 known as natural gas is used in gas cook tops and gas heaters. Express the molar mass of methane in the form of conversion factors. Molar mass of CH4 = 16.0 g 16.0 g CH4 and 1 mole CH4 1 mole CH4 16.0 g CH4

Learning Check Acetic acid CH3COOH is the acid in vinegar . It has a molar mass of 60.0 g/mole. 1 mole of acetic acid = ____________ 1 mole acetic acid or g acetic acid g acetic acid 1 mole acetic acid

Solution Acetic acid CH3COOH is the acid in vinegar . It has a molar mass of 60.0 g/mole. 1 mole of acetic acid = 60.0 g acetic acid 1 mole acetic acid or 60.0 g acetic acid 60.0 g acetic acid 1 mole acetic acid

Problems The formula of sodium hydrogen carbonate (baking soda) is NaHCO3. What is its formula mass? Answer the following: Write the formula for Magnesium Acetate. Determine its formula mass. Determine the molar mass for the following: NaCl CaCO3 H2SO4

Calculations with Molar Mass Grams Moles 1 mole = molar mass

Moles and Grams 3.00 moles Al ? g Al Aluminum is often used for the structure of light-weight bicycle frames. How many grams of Al are in 3.00 moles of Al? 3.00 moles Al ? g Al

Known: Given: 3.00 Al Molar mass of Al 1 mole Al = 27.0 g Al Conversion factors for Al 27.0g Al or 1 mol Al 1 mol Al 27.0 g Al

Unknown: ? g Al M(mass) = n(moles) x molar mass or n(moles) = M(mass) / molar mass Setup: M(mass) = n(moles) x molar mass

Solution 1. Molar mass of Al 1 mole Al = 27.0 g Al 2. Conversion factors for Al 27.0g Al or 1 mol Al 1 mol Al 27.0 g Al 3. Setup 3.00 moles Al x 27.0 g Al 1 mole Al Answer = 81.0 g Al

Learning Check The artificial sweetener aspartame (Nutri-Sweet) formula C14H18N2O5 is used to sweeten diet foods, coffee and soft drinks. How many moles of aspartame are present in 225 g of aspartame?

Solution 1. Molar mass of Aspartame C14H18N2O5 (14 x 12.0) + (18 x 1.01) + (2 x 14.0) + (5 x 16.0) = 294 g/mole 2. Setup 225 g aspartame x 1 mole aspartame 294 g aspartame = 0.765 mole aspartame

Calculations with Molar Volume Liter Moles 1 mole = 22.4L

RELATE MOLES TO PARTICLES AND PARTICLES TO MOLES Factor Label Method: unit needed = given x conversion factor Determine the given and unit needed. Find relationship between given unit and needed unit that is your conversion factor. Substitute into factor label equation. Unit that are identical cancel out.

How many H2 molecules are there in 0.025 moles of H2? Known: Given = 0.025 moles of H2 1 mole H2 = 6.022x1023 molecules H2 Unknown: unit needed = H2 molecules Setup: molecules of H2 =

Learning Check 1. Number of atoms in 0.500 mole of Al 1) 500 Al atoms 2) 6.02 x 1023 Al atoms 3) 3.01 x 1023 Al atoms 2.Number of moles of S in 1.8 x 1024 S atoms 1) 1.0 mole S atoms 2) 3.0 mole S atoms 3) 1.1 x 1048 mole S atoms

Solution 1. Number of atoms in 0.500 mol of Al 3) 3.01 x 1023 Al atoms 0.500 mol Al x 6.02 x 1023 Al atoms 1 mol Al 2. Number of moles of S if a sample of S contains 4.50 x 1024 S atoms 2) 3.0 mole S atoms 4.50 x 1024 S atoms x 1 mol S 6.02 x 1023 S atoms

Percent Composition The percent composition is a list of the mass percent of each element in a compound. Percent Composition of a Compound = Mass of the element Mass of the compound x 100

Calculate the percent composition of proply chloride (C3H7Cl) First, calculate the molar mass for the compound by adding up the masses of all the elements: 3 carbons at 12.01amu each = 36.03 amu 7 hydrogen at 1.01amu each= 7.07 amu 1 chlorine atom at 35.45 amu Summing these values yield: 36.03+7.07+35.45= 78.55 amu = formula mass of C3H7Cl

Second, for each element found in the compound set up ratio of its mass over the total mass times 100 to convert to percent. Carbon (36.03 / 78.55) x 100 = 45.87% Carbon Hydrogen (7.07 / 78.55) x 100 = 9.00% Hydrogen Chlorine (35.45 / 78.55) x 100 = 45.13% Chlorine These values add up to a hundred percent (100%).

Learning Check What is the percent carbon in C5H8NO4 (MSG monosodium glutamate), a compound used to flavor foods and tenderize meats? 1) 8.22 %C 2) 24.3 %C 3) 41.1 %C

Solution Molar mass = 146.0 g/mole % = total g C x 100 total g compound = 60.0 g C x 100 = 41.1% C 146.0 g MSG

Empirical formula Simplest formula for a compound; it is the smallest whole-number ratio of the atoms present. Molecular formula Expression of the formula for a compound; it shows the actual number of atoms of each element present in one molecule of the compound.

Calculating Empirical Formulas Steps: Change % to g (change signs). Convert to Moles. Divide the mass by 1 mol/atomic mass of each element. Find mole ratio. Divide each by the lowest moles. If ratio is not a whole number multiple by a number that converts the fraction to a whole number.

Calculate the empirical formula for a compound composed of 26 Calculate the empirical formula for a compound composed of 26.6% potassium, 35.4% chromium, and 38.1% oxygen. Percent means a fraction of 100. Thus, if we assume 100 grams of sample we can then use the percent values as grams of each element. From grams we find moles of each element in the formula.

Find moles potassium. Next find moles of chromium. Then find moles oxygen.

Next, set up a mole ratio that relates the moles of each element to the moles of the element that is least present. Because a chemical formula must have only whole numbers we multiply by 2 to yield: K2Cr2O7 as the empirical formula.

Calculating Molecular Formulas Molecular mass must be known before a molecular formula can be determined. Molecular mass is a whole-number multiple of the empirical formula mass. (empirical formula mass)x = molecular mass

Molecular formulas Solve for x. Multiply empirical formula by whatever is x. Examples: (CH4)x (CH4 mass)x = 16 {12.0 + 4(1.0)}x = 16 (16.0)x = 16 x = 16.0 16.0 x = 1 (CH4)x (CH4)1 = CH4  

(HO)x (HO mass)x = 34.0 (1.0 + 16.0)x = 34.0 (17.0)x = 34.0 x = 34.0 17.0 x = 2 (HO)x (HO)2 = H2O2

Problems A certain hydrocarbon compound is found by analysis to have the empirical formula CH3. Its molecular weight is 30.0. What is the molecular formula? A compound is found by analysis to consist of 40.1% S and 59.9% O. Its molecular weight is 80.1. (Change % to g, just change the signs) What is its empirical formula? What is its molecular formula? What is the name of the compound?

A compound of nitrogen and sulfur is found by analysis to have the empirical formula NS2. Its molecular weight is 156.4. What is its molecular formula? What is the name of the compound?