Chemical Composition Mrs. Chang Chapter 7 6/14/20161

Recall: Mole is… 2 A counting number (like a dozen) 6.02 x is called Avogadro’s number. 1 mol = 6.02  items ( representative particles) A large amount!!!!

Representative particles The smallest pieces of a substance. The smallest pieces of a substance. molecular compounds: molecule. molecular compounds: molecule. ionic compounds: formula unit. ionic compounds: formula unit. elements: atom. elements: atom. 3

Questions How many atoms are in 1 formula unit of the following cmpds? How many atoms are in 1 formula unit of the following cmpds? CaCO 3 CaCO 3 5 atoms 5 atoms Al 2 (SO 4 ) 3 Al 2 (SO 4 ) 3 17 atoms 17 atoms How many oxygen atoms are in the above cmpds? How many oxygen atoms are in the above cmpds? 3 O atoms 3 O atoms 12 O atoms 12 O atoms 4

A Moles of Particles Contains 6.02 x particles 1 mole C = 6.02 x C atoms 1 mole H 2 O = 6.02 x H 2 O molecules 1 mole NaCl = 6.02 x NaCl formula units 5

6 Sample Problem A (pg 228) Find the number of molecules in 2.5 mol of sulfur dioxide? Find the number of molecules in 2.5 mol of sulfur dioxide? (what would you add if I asked for atoms?) (what would you add if I asked for atoms?) 2.50 mol 6.02  molecules 1 mol 1.5   =atoms C =atoms C 1 molecule SO 2 3 atoms = 4.5  atoms = 4.5  atoms

7 Sample Problem B (pg 229) A sample contains 3.01  molecules of sulfur dioxide. Determine the number of moles A sample contains 3.01  molecules of sulfur dioxide. Determine the number of moles 3.01  molecules 1 mol 6.02  molecules =.500 mol SO 2

Look at the 4 practice problems on pg 228. Tell the person whether you would multiply or divide Avogardro’s #. Explain why. Look at the 4 practice problems on pg 228. Tell the person whether you would multiply or divide Avogardro’s #. Explain why. Do the same for the first 4 practice problems on pg 229. Do the same for the first 4 practice problems on pg 229. So, when do you multiply Avogadro’s #? So, when do you multiply Avogadro’s #? When do you divide Avogadro’s #? When do you divide Avogadro’s #? 6/14/20168

9 Molar Mass - Is the Mass (in Grams) of One Mole of a Substance (often called molecular weight) Molar mass or ___ is the generic term for the mass of one mole of any substance (in grams) Molar mass or ___ is the generic term for the mass of one mole of any substance (in grams) The same as: The same as: 1) gram molecular mass or ___ (molecules) 2) gram formula mass or ___ (ionic compounds) 3) gram atomic mass or ___ (elements) MM gmm gfm gam

Molar Mass of Elements = atomic mass (simply look at the periodic table) carbon carbon aluminum aluminum zinc zinc g/mol g/mol g/mol

What about compounds? Mass in grams of 1 mole equal the sum of the atomic masses Mass in grams of 1 mole equal the sum of the atomic masses What is the mass of one mole (molar mass) of CH 4 ? What is the mass of one mole (molar mass) of CH 4 ? 1 C x g/mol = g/mol 4 H x 1.01 g/mol = 4.04g/mol 1 mole CH 4 = 12.01g/mol g/mol =16.05g/mol 11

Molar Mass Example Problem Find the molar mass of sodium bicarbonate 12 NaHCO 3 NaHCO g/mol g/mol g/mol + 3(16.00 g/mol) = g/mol 22.99g/mol g/mol g/mol + 3(16.00 g/mol) = g/mol

Example Problem Prozac, C 17 H 18 F 3 NO, 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 13 Solution: 17C (12.0) 18H (1.0) 3F (19.0) 1N (14.0) + 1 O (16.0) 309 g/mole

Mass to Mole Conversion How many moles is 5.69 g of NaOH? (Find MM of NaOH) How many moles is 5.69 g of NaOH? (Find MM of NaOH) Na = g/mol Na = g/mol O = g/mol O = g/mol H = 1.01 g/ mol H = 1.01 g/ mol g/mol g/mol 14 1 mol NaOH g NaOH 5.69 g NaOH = mol NaOH

Mole to Mass Conversion How many grams are in 9.45 mol of dinitrogen trioxide? How many grams are in 9.45 mol of dinitrogen trioxide? N 2 O 3 = 2(14.01) + 3(16.00) = g/mol N 2 O 3 = 2(14.01) + 3(16.00) = g/mol mol N 2 O g N 2 O 3 1 mol N 2 O 3 = =718 g N 2 O 3 N 2 O 3

Another Example How many atoms of oxygen are there in 1.23 moles of CO 2 ? How many atoms of oxygen are there in 1.23 moles of CO 2 ? mol CO 2 1 mol CO x molecules CO 2 1 molecule CO 2 2 atoms O = 1.48 x atoms O

Molar Conversions 17 molar mass (g/mol) MASS IN GRAMS MOLES NUMBER OF PARTICLES 6.02  (particles/mol) ÷ X÷ X

More Molar Conversion Examples Find the mass of 2.1  molecules of NaHCO 3. Find the mass of 2.1  molecules of NaHCO  molecules 1 mol 1 mol 6.02  molecules = 290 g NaHCO g g 1 mol 1 mol

Atomic Mass average atomic mass. average atomic mass. Based on abundance of each element in nature. Based on abundance of each element in nature. Don’t use grams because the numbers would be too small. Don’t use grams because the numbers would be too small. Each isotope has its own atomic mass, thus we determine the average from percent abundance. Each isotope has its own atomic mass, thus we determine the average from percent abundance. Is not a whole number because it is an average. Is not a whole number because it is an average. are the decimal numbers on the periodic table. are the decimal numbers on the periodic table.

Calculating averages Multiply the individual atomic mass of each isotope by its abundance (expressed as a decimal), then add the results. Multiply the individual atomic mass of each isotope by its abundance (expressed as a decimal), then add the results. n If not told otherwise, the mass of the isotope is the mass number in amu

Example Problem 1 Chlorine is a mixture of chlorine-35 with a mass of amu and chlorine-37 with a mass of amu. If the natural abundance of chlorine-35 is 75.77% and the natural abundance of chlorine-37 is 24.23%, calculate the atomic mass of chlorine. Chlorine is a mixture of chlorine-35 with a mass of amu and chlorine-37 with a mass of amu. If the natural abundance of chlorine-35 is 75.77% and the natural abundance of chlorine-37 is 24.23%, calculate the atomic mass of chlorine.

Example Problem 1 Solution: For chlorine-35: amu x = amu For chlorine-37: amu x = amu _______________ Add them up = amu This is the “average” atomic mass

Example Problem Calculate the atomic mass of copper if copper has two naturally occurring isotopes: copper-63 (62.93 amu)has an abundance of 69.2% and copper-65 (64.93 amu) has an abundance of 30.8% Calculate the atomic mass of copper if copper has two naturally occurring isotopes: copper-63 (62.93 amu)has an abundance of 69.2% and copper-65 (64.93 amu) has an abundance of 30.8% Answer: amu

(Use as S&A Q’s) 1.Magnesium has three isotopes % magnesium-24 with a mass of amu, 10.00% magnesium- 25 with a mass of amu, and the rest magnesium-26 with a mass of amu. What is the atomic mass of magnesium? ….answer is amu

2. Identify the unknown element by finding the average atomic mass of element X and comparing to atomic mass given on the Periodic Table. element X-10: amu, % element X-11: amu, % 6/14/201625

Calculating Percent Composition of a Compound Calculating Percent Composition of a Compound 6/14/ Like all percent problems: Like all percent problems: Part Part whole whole Find the mass of each component, Find the mass of each component, then divide by the total mass (assume one mole). then divide by the total mass (assume one mole). x 100

Percent Composition What is the % composition of water. What is the % composition of water. Find MM: 2(1.01) = g/mol Find MM: 2(1.01) = g/mol Double check: %’s should add up to 100. Double check: %’s should add up to /14/  100 = %H =%H =%H =%H = 2.02 g H g H 2 O  100 = %O =%O =%O =%O = g O g H 2 O 11.2 % H 88.8 % O

Practice Problem What is the percent carbon in C 5 H 8 NO 4 (MSG monosodium glutamate), a compound used to flavor foods and tenderize meats? 1. Find mass of C 1. 5 x 12.0 g = 60.0 g C 2. Find mass of MSG 1. 5(12.0) + 8(1.0) (16.0)=146.0 g 3. Mass of element/ mass of cmpd x g C/146.0 g x100 = 41.1% C 6/14/201628

Percent Composition Percent Composition Find the percentage composition of a sample that is 28 g Fe and 8.0 g O. Find the percentage composition of a sample that is 28 g Fe and 8.0 g O. Determine total mass: Determine total mass: 6/14/ %Fe = 28 g 36 g  100 = 78% Fe %O = 8.0 g 36 g  100 = 22% O 28 g g = 36 g

Using Percent as a Conversion Factor How many grams of copper are in a 38.0-gram sample of Cu 2 S? How many grams of copper are in a 38.0-gram sample of Cu 2 S? Multiply mass of cmpd with % element Multiply mass of cmpd with % element 6/14/ (38.0 g Cu 2 S)x( ) = 30.3 g Cu Cu 2 S is 79.8% Cu (Sample 7-10 p 227)

EF vs MF Empirical Formula (EF) Empirical Formula (EF) Def’n: lowest whole # ratio Def’n: lowest whole # ratio H 2 O H 2 O HO HO CH CH CH 2 O CH 2 O Molecular Formula (MF) Def’n: actual ratio H 2 O H 2 O 2 C 6 H 6 C 6 H 12 O 6 6/14/ Water peroxide benzene sugar

B. Empirical Formula Smallest whole number ratio of atoms in a compound Smallest whole number ratio of atoms in a compound 6/14/ C2H6C2H6C2H6C2H6 CH 3 reduce subscripts

Calculating Empirical Formula It is not just the ratio of atoms, it is also the ratio of moles of atoms. It is not just the ratio of atoms, it is also the ratio of moles of atoms. In 1 mole of CO 2 there is 1 mole of carbon and 2 moles of oxygen. In 1 mole of CO 2 there is 1 mole of carbon and 2 moles of oxygen. 6/14/201633

Steps for Calculating the Empirical Formula We get a ratio from the percent composition. We get a ratio from the percent composition. Assume you have a 100 g sample. Assume you have a 100 g sample. 1. % = g 2. Convert grams to moles (÷ atomic mass) 3. Find lowest whole number ratio. (÷ by smallest # of moles) (if this step gives a decimal, multiply by 2, 3, or 4 to get whole #’s) 4. Use ratio to write the EF 6/14/201634

Example Calculate the empirical formula of a compound composed of % C, % H, and %N. Calculate the empirical formula of a compound composed of % C, % H, and %N. Assume 100 g so Assume 100 g so g C x 1mol C = mole C g C g C x 1mol C = mole C g C g H x 1mol H = mole H 1.01 g H g H x 1mol H = mole H 1.01 g H g N x 1mol N = mole N g N g N x 1mol N = mole N g N 6/14/201635

Example The ratio is mol C = 1 mol C mol The ratio is mol C = 1 mol C mol The ratio is mol H = 5 mol H mol The ratio is mol H = 5 mol H mol The ratio is mol N = 1 mol N mol The ratio is mol N = 1 mol N mol = CH 5 N = CH 5 N 6/14/201636

Empirical Formula Find the empirical formula for a sample of 25.9% N and 74.1% O. Find the empirical formula for a sample of 25.9% N and 74.1% O. 6/14/ g 1 mol g = 1.85 mol N 74.1 g 1 mol g = 4.63 mol O 1.85 mol = 1 N = 2.5 O

Empirical Formula N 1 O 2.5 6/14/ Need to make the subscripts whole numbers  multiply by 2 N2O5N2O5N2O5N2O5

A compound is 43.64g P and 56.36g O. What is the empirical formula? A compound is 43.64g P and 56.36g O. What is the empirical formula? 6/14/ g 1 mol g = 1.4 mol P g 1 mol g = 3.5 mol O 1.4 mol = 1 P = 2.5 O x 2 = 2 P = 5 O P2O5P2O5P2O5P2O5

Molecular Formula “True Formula” - the actual number of atoms in a compound “True Formula” - the actual number of atoms in a compound 6/14/ CH 3 C2H6C2H6C2H6C2H6 empiricalformula molecularformula ?

Steps to calculating Molecular Formula 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empirical mass. 4. Multiply each subscript by the answer from step 3. 6/14/201641

Molecular Formula The empirical formula for ethylene is CH 2. Find the molecular formula if the molecular mass is 28.1 g/mol? The empirical formula for ethylene is CH 2. Find the molecular formula if the molecular mass is 28.1 g/mol? 6/14/ g/mol g/mol = 2.00 empirical mass = g/mol (CH 2 ) 2  C 2 H 4

Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula? Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula? EF: C 4 H 5 N 2 O EF: C 4 H 5 N 2 O 6/14/201643

Example A compound is known to be composed of % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) to be g. What is its molecular formula? A compound is known to be composed of % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) to be g. What is its molecular formula? EF: CClH 2 EF: CClH 2 MF: C 2 Cl 2 H 4 MF: C 2 Cl 2 H 4 6/14/201644