CHEM 111 1st Edition Lecture 3Avogadro’s number: NAAvogadro’s number is used to find the moles of an elementNA=6.0221415× 1023C❑12atom=12 uexactlyAvogadro’s number is the number of C❑12∈12 g of C❑12NA= C❑12atoms=12 gramsC❑12atom=12 uNAC❑12atoms=12 NAu12 NAu=12 gramsNAu=1 gramSo 1 amu=1 g /molWhat is the “weight” or mass of He?4.0 u which is the same as 4.0 g/molMolar mass=mass of 1 mole of “the formula”Elements-----Atomic Weight AWMolecules-----Molecular Weight MWIonic and Network Substances-----Formula WeightFinding % Composition from a formula1. Find the molar mass using periodic table2. Calculate the % weight for each elementExample: Calculate MW of C2H6O(2x12.011u)(6x1.0079u)+(1x15.9999u)46.068 u%C=(2x12.011u/46.068)x100%=52.145%%H=(6x1.0079u/46.068)x100%=13.127%%O=(1x15.999u/46.068)x100%=34.729%Calculate Empirical Formula given percent compositionGiven: N=30.45%O=69.55%1. Assume weight of 100gN=60.45gO=69.55g2. Convert from Grams to Moles30.45gN x (1mol/14.007g)=2.17 mol69.55gO x (1 mol/15.999g)=4.35 molThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.N2.17O4.353. Find the smallest ratio divide by smallest numberN: 2.17/2.17=1O: 4.35/2.17=2Anwer : N O2A balanced reactionReactants----ProductEvery atom on the left side (reactant) MUST end up on the right side (product) balanceC2H6+O2−−−→ C O2+H2OC2H6+72O2−−−→2 C O2+3 H2O2C2H6+7 O2−−−→ 4 C O2+6 H2OThe coefficients in the balanced equation may be either # of molecules or # of moles. For the Latter, fractions are acceptable.You burn 8 lbs of butane (C4H10) . How many pounds of CO2 are produced.1. Write a balanced equationC4H10+O2−→ C O2+H2OC4H10+6.5 O2−→ 4 C O2+5 H2O2. Convert pounds to moles8lbs x (453.9 g/lb) x (mol/58.123g)=62.47 moles butane3. Determine the # of moles CO2 producedC4H10+6.5 O2−−→ 4 C O2+5 H2O62.47 moles C4H10x(44.01 gmol)=249.9 moles C O24. Convert back to pounds62.47 moles C4H10x(44.01 gmol)x(lb453.9 g)=24.23 lbof C