Atomic Mass l Atoms are so small, it is difficult to discuss how much they weigh in grams. l Use atomic mass units. l an atomic mass unit (amu) is one twelth the mass of a carbon-12 atom. l This gives us a basis for comparison. l The decimal numbers on the table are atomic masses in amu.

They are not whole numbers l Because they are based on averages of atoms and of isotopes. l can figure out the average atomic mass from the mass of the isotopes and their relative abundance. l add up the percent as decimals times the masses of the isotopes.

Examples l There are two isotopes of carbon 12 C with a mass of amu(98.892%), and 13 C with a mass of amu (1.108%). l There are two isotopes of nitrogen, one with an atomic mass of amu and one with a mass of amu. What is the percent abundance of each?

The Mole l The mole is a number. l A very large number, but still, just a number. l x of anything is a mole l A large dozen. l The number of atoms in exactly 12 grams of carbon-12.

The Mole l Makes the numbers on the table the mass of the average atom.

More Stoichiometry

Molar mass l Mass of 1 mole of a substance. l Often called molecular weight. l To determine the molar mass of an element, look on the table. l To determine the molar mass of a compound, add up the molar masses of the elements that make it up.

Find the molar mass of l CH 4 l Mg 3 P 2 l Ca(NO 3 ) 3 l Al 2 (Cr 2 O 7 ) 3 l CaSO 4 · 2H 2 O

Percent Composition l Percent of each element a compound is composed of. l Find the mass of each element, divide by the total mass, multiply by a 100. l Easiest if you use a mole of the compound. l Find the percent composition of CH 4 l Al 2 (Cr 2 O 7 ) 3 l CaSO 4 · 2H 2 O

Working backwards l From percent composition, you can determine the empirical formula. l Empirical Formula the lowest ratio of atoms in a molecule. l Based on mole ratios. l A sample is 59.53% C, 5.38%H, 10.68%N, and 24.40%O what is its empirical formula.

Pure O 2 in CO 2 is absorbed H 2 O is absorbed Sample is burned completely to form CO 2 and H 2 O

l A gram sample of a compound (vitamin C) composed of only C, H, and O is burned completely with excess O g of CO 2 and g of H 2 O are produced. What is the empirical formula?

More Stoichiometry

Empirical To Molecular Formulas l Empirical is lowest ratio. l Molecular is actual molecule. l Need Molar mass. l Ratio of empirical to molar mass will tell you the molecular formula. l Must be a whole number because...

l Empirical Formula- lowest whole number ratio of elements in a compound l Molecular formula- # of elements in a compound as it appears in nature –May or may not be the same as empirical l HO is empirical formula l H 2 O 2 is molecular l CO 2 empirical & molecular are the same

l To find empirical formula: –1. Change % to g –2. Change g to mol –3. Find ratio of moles (divide by lowest # moles) –4. Ratios become subscripts in formula »Tricks- if molar ratio is: –.5 X by 2 –.3 or.6 X 3 –.25 or.75 X 4 –.2 or lower, round down –.8 or above round up

l Determine the empirical formula of a compound that contains35.98%Al & 64.02%S. l

l What is the empirical formula of a compound that is 25.9 % N & 74.1%O?

l What is the empirical formula of a compound that contains 10.89%Mg, 31.77%Cl & 57.34%O?

l If they give you molecular mass: –1. Get mass of empirical formula (what you just found) –2. Divide molecular mass/empirical –3. Multiply that # by subscripts

l Calculate molecular formula of compound whose molar mass is 60.0g & the empirical formula is CH 4 N.

l A compound is composed of 40.68% C, 5.08% H & 54.24%O. It has a molecular mass of 118.1g/mol. Find the empirical & molecular formula.

Example l A compound is made of only sulfur and oxygen. It is 69.6% S by mass. Its molar mass is 184 g/mol. What is its formula?

Chemical Equations l Are sentences. l Describe what happens in a chemical reaction. Reactants  Products l Equations should be balanced. l Have the same number of each kind of atoms on both sides because...

Balancing equations CH 4 + O 2  CO 2 + H 2 O ReactantsProducts C11 O23 H42

Balancing equations CH 4 + O 2  CO H 2 O ReactantsProducts C11 O23 H424

Balancing equations CH 4 + O 2  CO H 2 O ReactantsProducts C11 O23 H424 4

Balancing equations CH 4 + 2O 2  CO H 2 O ReactantsProducts C11 O23 H

Abbreviations ( s )  ( g )  l ( aq ) l heat  l catalyst

Practice Ca(OH) 2 + H 3 PO 4  H 2 O + Ca 3 (PO 4 ) 2 Cr + S 8  Cr 2 S 3 KClO 3 ( s )  Cl 2 ( g ) + O 2 ( g ) l Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas. Fe 2 O 3 ( s ) + Al( s )  Fe( s ) + Al 2 O 3 ( s )

Meaning l A balanced equation can be used to describe a reaction in molecules and atoms. l Not grams. l Chemical reactions happen molecules at a time l or dozens of molecules at a time l or moles of molecules.

Stoichiometry l Given an amount of either starting material or product, determining the other quantities. l use conversion factors from –molar mass (g - mole) –balanced equation (mole - mole) l keep track.

Examples l One way of producing O 2 ( g ) involves the decomposition of potassium chlorate into potassium chloride and oxygen gas. A 25.5 g sample of Potassium chlorate is decomposed. How many moles of O 2 (g) are produced? l How many grams of potassium chloride? l How many grams of oxygen?

Examples l A piece of aluminum foil 5.11 in x 3.23 in x in is dissolved in excess HCl(aq). How many grams of H 2 ( g ) are produced? How many grams of each reactant are needed to produce 15 grams of iron form the following reaction? Fe 2 O 3 ( s ) + Al( s )  Fe( s ) + Al 2 O 3 ( s )

Examples K 2 PtCl 4 ( aq ) + NH 3 ( aq )  Pt(NH 3 ) 2 Cl 2 ( s )+ KCl( aq ) l what mass of Pt(NH 3 ) 2 Cl 2 can be produced from 65 g of K 2 PtCl 4 ? l How much KCl will be produced? l How much from 65 grams of NH 3 ?

Yield How much you get from an chemical reaction

Limiting Reagent l Reactant that determines the amount of product formed. l The one you run out of first. l Makes the least product. l Book shows you a ratio method. l It works. l So does mine

Limiting reagent l To determine the limiting reagent requires that you do two stoichiometry problems. l Figure out how much product each reactant makes. l The one that makes the least is the limiting reagent.

Example Ammonia is produced by the following reaction N 2 + H 2  NH 3 What mass of ammonia can be produced from a mixture of 100. g N 2 and 500. g H 2 ? l How much unreacted material remains?

Excess Reagent l The reactant you don’t run out of. l The amount of stuff you make is the yield. l The theoretical yield is the amount you would make if everything went perfect. l The actual yield is what you make in the lab.

Percent Yield l % yield = Actual x 100% Theoretical l % yield = what you got x 100% what you could have got

Examples l Aluminum burns in bromine producing aluminum bromide. In a laboratory 6.0 g of aluminum reacts with excess bromine g of aluminum bromide are produced. What are the three types of yield.

Examples Years of experience have proven that the percent yield for the following reaction is 74.3% Hg + Br 2  HgBr 2 If 10.0 g of Hg and 9.00 g of Br 2 are reacted, how much HgBr 2 will be produced? l If the reaction did go to completion, how much excess reagent would be left?

Examples Commercial brass is an alloy of Cu and Zn. It reacts with HCl by the following reaction Zn(s) + 2HCl(aq)  ZnCl 2 (aq) + H 2 (g) Cu does not react. When g of brass is reacted with excess HCl, g of ZnCl 2 are eventually isolated. What is the composition of the brass?