CH1010 Lecture 4

CH1010 Lecture 4 - Molecularity Molecularity CH1010 Lecture...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Molecularity Molecularity CH1010 Lecture 4 James P. Dittami Molecular and Formula Mass Molecular and Formula Mass Use the formula for a compound and the atomic masses from the periodic table to calculate overall mass For Covalent Compounds – Molecular Mass For Ionic Compounds – Formula Mass Commonly referred to as Molecular Weight Molecular Mass Molecular Mass Molecular Mass = (1 x atomic mass N) + (3 x atomic mass F) = 14.01 amu + (3 x 19.00 amu) = 71.01 amu NF3 – Nitrogen trifluoride Formula Mass Formula Mass NaF – Sodium Fluoride Formula Mass = (1 x atomic mass Na) + (1 x atomic mass F) = 22.99 amu + 19.00 amu = 41.99 amu Formula Mass Formula Mass Formula Mass = (1 x atomic mass Ba) + (2 x atomic mass N) + (6 x atomic mass O) = 137.3 amu + (2 x 14.01 amu) + (6 x 16.00 amu) = 261.3 amu Ba(NO3)2 – Barium Nitrate Counting by Weighing Counting by Weighing Problem­ we want a way to count large number of small things­ pills, paper clips, atoms. Solution­ if they are all of similar weight and we know an average weight then Number of items = Total Mass Average Mass Counting Atoms Counting Atoms Should be same as counting by weighing Caveat – Mass of all atoms in an element is not the same due to isotopes Carbon – Consists of 12 C – 98.89 % 12 amu and 13C – 1.11 % 14 C present is negligible Average Atomic Mass C = 12.01 amu Average Atomic Mass Average Atomic Mass Calculate the average mass for any element Avg. Mass = (Abundance A)(Isotope Mass A) + (Abundance B)(Isotope Mass B) + (Abundance C) (Isotope Mass C) ….etc Average Atomic Mass Average Atomic Mass Calculate the average mass for Neon which is found in nature as 90.92 % 20Ne – mass 20 amu 0.257 % 21Ne – mass 21 amu 8.82 % 22Ne – mass 22 amu Avg. Mass = (0.9092)(20) + (0.00257)(21) + (0.0882)(22) Avg. Mass = 20.18 amu The Mole The Mole Mole – Defined as the number of carbon atoms in exactly 12 grams of pure 12C Experimentally determined as 6.02214 x 1023 atoms We will use 6.022 x 1023 We call this quantity Avogadro’s number Recall his assumption !! Moles Moles By definition 1 mole of anything contains 6.022 x 1023 units of that substance. Because of the way we define a mole we can count atoms by weighing. Mole – Defined as the number of carbon atoms in exactly 12 grams of pure 12C where an atom of 12C is defined to have a mass of 12 atomic mass units. Counting atoms Counting atoms By analogy a 12.01 gram sample of naturally occurring carbon will contain 6.022 x 1023 atoms each with an average weight of 12.01 amu Careful – if you could pick a single atom out what would it’s mass be? Ans. Either that of 12C or 13C but not 12.01 Moles – expanded definition Moles – expanded definition If we take a sample of a natural element whose Mass is equal to the element’s Atomic Mass expressed in grams, the sample will contain 1 mole of atoms. Molar Mass Molar Mass If we take a sample of a compound whose Mass is equal to it’s molecular Mass expressed in grams, the sample will contain 1 mole of molecules. One mole of CH = 16.04 g as follows 4 1 mole C =12.01 g 4 moles H = 4 x 1.008 g Mass of 1 mole CH4 = 16.04 g Oxygen 32.00 g One mole of common substances. CaCO3 100.09 g Water 18.02 g Copper 63.55 g Table 3.1 Term Isotopic mass Atomic mass (also called atomic weight) Summary of Mass Terminology Definition Mass of an isotope of an element Average of the masses of the naturally occurring isotopes of an element weighted according to their abundance Sum of the atomic masses of the atoms (or ions) in a molecule (or formula unit) Unit amu amu Molecular (or formula) mass (also called molecular weight) amu Molar mass (M) Mass of 1 mole of chemical entities (also called (atoms, ions, molecules, formula units) gram-molecular weight) g/mol Interconverting Moles, Mass, and Number of Chemical Entities no. of grams 1 mol 1 mol no. of grams No. of entities = no. of moles x 6.022x1023 entities 1 mol 1 mol 6.022x1023 entities Mass (g) = no. of moles x g No. of moles = mass (g) x M No. of moles = no. of entities x Figure 3.3 MASS(g) of element Summary of the mass-molenumber relationships for elements. M (g/mol) AMOUNT(mol) of element Avogadro’s number (atoms/mol) ATOMS of element Sample Problem 3.1 Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element PROBLEM: (a) Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342mol of Ag? (b) Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8g of Fe? PLAN: (a) To convert mol of Ag to g we have to use the #g Ag/mol Ag, the molar mass M . amount(mol) of Ag multiply by M of Ag (107.9g/mol) mass(g) of Ag mass(g) of Fe divide by M of Fe (55.85g/mol) amount(mol) of Fe multiply by 6.022x1023 atoms/mol atoms of Fe SOLUTION: 0.0342mol Ag x 107.9 g Ag = 3.69g Ag mol Ag PLAN: (b) To convert g of Fe to atoms we first have to find the #mols of Fe and then convert mols to atoms. SOLUTION: 95.8g Fe x mol Fe = 1.72mol Fe 55.85g Fe 6.022x1023atoms Fe = 1.04x1024 atoms 1.72mol Fe x Fe mol Fe Figure 3.3 MASS(g) of compound M (g/mol) AMOUNT(mol) of compound Summary of the mass-molenumber relationships for compounds. chemical formula AMOUNT(mol) of elements in compound Avogadro’s number (molecules/mol) MOLECULES (or formula units) of compound Sample Problem 3.2 Calculating the Moles and Number of Formula Units in a Given Mass of a Compound PROBLEM: Ammonium carbonate is white solid that decomposes with warming. Among its many uses, it is a component of baking powder, fire extinguishers, and smelling salts. How many formula units are in 41.6 g of ammonium carbonate? compound, we find its M by adding the masses of the elements. Convert the given mass, 41.6 g to mols using M and then the mols to formula units with Avogadro’s number. SOLUTION: PLAN: After writing the formula for the mass(g) of (NH4)2CO3 divide by M amount(mol) of (NH4)2CO3 multiply by 6.022x1023 formula units/mol number of (NH4)2CO3 formula units The formula is (NH4)2CO3. M = (2 x 14.01 g/mol N)+(8 x 1.008 g/mol H) +(12.01 g/mol C)+(3 x 16.00 g/mol O) = 96.09 g/mol x 41.6 g (NH4)2CO3 x 96.09 g (NH4)2CO3 mol (NH4)2CO3 6.022x1023 formula units (NH4)2CO3 mol (NH4)2CO3 = 2.61x1023 formula units (NH4)2CO3 Mass percent from the chemical formula Mass % of element X = atoms of X in formula x atomic mass of X (amu) molecular (or formula) mass of compound(amu) x 100 Mass % of element X = moles of X in formula x molar mass of X (g/mol) Gram in one mole of compound (g) x 100 Sample Problem 3.3 Calculating the Mass Percents and Masses of Elements in a Sample of Compound PROBLEM: Glucose (C6H12O6) is the most important nutrient in the living cell for generating chemical potential energy. (a) What is the mass percent of each element in glucose? (b) How many grams of carbon are in 16.55g of glucose? PLAN: We have to find the total mass of glucose and the masses of the constituent elements in order to relate them. amount(mol) of element X in 1mol compound multiply by M (g/mol) of X mass(g) of X in 1mol of compound divide by mass(g) of 1mol of compound mass fraction of X multiply by 100 mass % X in compound SOLUTION: (a) Per mole glucose there are 6 moles of C 12 moles H 6 moles O Sample Problem 3.3 continued 12.01 g C mol C 16.00 g O mol O Calculating the Mass Percents and Masses of Elements in a Sample of Compound 6 mol C x = 72.06 g C 12 mol H x 1.008 g H mol H = 12.096 g H 6 mol O x = 96.00 g O M = 180.16 g/mol (b) mass percent of C = 72.06 g C 180.16 g glucose 12.096 g H 180.16 g glucose 96.00 g O 180.16 g glucose = 0.3999 x 100 = 39.99 mass %C mass percent of H = = 0.06714 x 100 = 6.714 mass %H mass percent of O = = 0.5329 x 100 = 53.29 mass %O Mass of C in sample Mass of C in sample Mass (g) C = mass sample X Mass % C = 16.55 g glucose x 0.3999 g C/g glucose =6.618 g C ...
View Full Document

This note was uploaded on 02/14/2011 for the course CH 1010 taught by Professor Kumar during the Spring '06 term at WPI.

Ask a homework question - tutors are online