Chapter 3

• Atomic masses

• Average atomic mass– Ex. What is the avg. atomic mass of a sample

that is 69.09% 62.93 amu and 30.91% 64.93 amu?

– 0.6909(62.93amu) + 0.3091(64.93amu) = 63.55 amu

• Mole

• Molar mass

– What is the mass of 3 moles of KOH?

• 3 n KOH x 56 g/n = 168 g = 200 g s.f.

– How many atoms are in 15 g of K?

• 15 g x 1n/39.1g x 6.022x1023atoms/n = 2.3x1023

atoms

• Percent composition– Ex.– What is the percent comp. of H2O?

– %H = 2.02 g / 18.02 g x 100 = 11.2 %

– %O = 16.0 g / 18.02 g x 100 = 88.8 %

• Determining Formulas• Empirical Formula• Steps

– Mass– Moles– Divide

– Ex. 8 g of O and 32 g of S– 8 g O x 1n/16g = 0.5 n O / 0.5 n = 1 O– 32 g S x 1n/32g = 1.0 n S / 0.5 n = 2 S– S2O

• Molecular Formula

• Ex.– S2O has a molar mass of 160 g/n, what’s the

molecular formula

– 160 g/n / 80 g/n = 2 so S2O becomes– S4O2

– If given percentages assume its out of 100 g and change the sign from % to g.

• Chemical Equations– Reactants Products

• Balancing– Ex.s– H2 + O2 H2O– 2H2 + O2 2H2O

– C2H6 + O2 CO2 + H2O– C2H6 + 7/2O2 2CO2 + 3H2O– 2C2H6 + 7O2 4CO2 + 6H2O

• Stoichiometry

• Mole ratio– 2H2 + O2 2H2O

– 2n H2 / 1n O2 – 2n H2 / 2n H2O– 1n O2 / 2n H2O

– How much NaCl is formed from 23 g of Na and excess Cl2?

– 2Na + Cl2 2NaCl

– 23g Na x 1n/23g x 2n NaCl / 2n Na = 1n NaCl

– If we wanted to go to mass we would have multiplied by the molar mass of NaCl.

• How much oxygen is needed to burn 15 g of CH4?

• CH4 + O2 → CO2 + H2O

• CH4 + 2O2 → CO2 + 2H2O

• 15g CH4 x 1n / 16.05 g x 2n O2 / 1n CH4

• 1.9 n O2 if we want mass then use molar mass

• Limiting Reactant

• Cheeseburgers– 24 buns, 20 patties, 48 slices of cheese

• With reactions we have to prove which reactant runs out by doing a stoichiometry problem.

– How much water is produced from 10. g of H2 and 10. g of O2?

– 2H2 + O2 2H2O

– 10g H2 x 1n/2g x 1n O2/ 2n H2 x 32 g/n = 80 g O2

– O2 is the limiting reactant need 80g have 10g!

– 10.g O2 x 1n/32g x 2n H2O/1n O2 x 18g/n= 11g H2O

• If we want to know the remaining reactant we just start with the limiting reactant and calculate the amount of reactant used.

• 10.g O2 x 1n/32g x 2n H2/1n O2 x 2g/n= 1.25g H2 10 g – 1.25 g = 8.75 g of H2 left over

• Percent Yield– actual / theoretical x 100– If a lab using the previous example is conducted and

only 5.0 g of water is produced what’s the percent yield?

– 5.0 g / 11 g x 100 = 46 %