# 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)

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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 6.022x10 23 atoms/n = 2.3x10 23 atoms

Percent composition –Ex. –What is the percent comp. of H 2 O? –%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 –S 2 O

Molecular Formula Ex. –S 2 O has a molar mass of 160 g/n, what’s the molecular formula –160 g/n / 80 g/n = 2 so S 2 O becomes –S 4 O 2 –If given percentages assume its out of 100 g and change the sign from % to g.

Chemical Equations – Reactants  Products Balancing –Ex.s –H 2 + O 2  H 2 O –2H 2 + O 2  2H 2 O –C 2 H 6 + O 2  CO 2 + H 2 O –C 2 H 6 + 7/2O 2  2CO 2 + 3H 2 O –2C 2 H 6 + 7O 2  4CO 2 + 6H 2 O

Stoichiometry Mole ratio –2H 2 + O 2  2H 2 O –2n H 2 / 1n O 2 –2n H 2 / 2n H 2 O –1n O 2 / 2n H 2 O

–How much NaCl is formed from 23 g of Na and excess Cl 2 ? –2Na + Cl 2  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 CH 4 ? CH 4 + O 2 → CO 2 + H 2 O CH 4 + 2O 2 → CO 2 + 2H 2 O 15g CH 4 x 1n / 16.05 g x 2n O 2 / 1n CH 4 1.9 n O 2 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 H 2 and 10. g of O 2 ? –2H 2 + O 2  2H 2 O –10g H 2 x 1n/2g x 1n O 2 / 2n H 2 x 32 g/n = 80 g O 2 –O 2 is the limiting reactant need 80g have 10g! –10.g O 2 x 1n/32g x 2n H 2 O/1n O 2 x 18g/n= 11g H 2 O

If we want to know the remaining reactant we just start with the limiting reactant and calculate the amount of reactant used. 10.g O 2 x 1n/32g x 2n H 2 /1n O 2 x 2g/n= 1.25g H 2 10 g – 1.25 g = 8.75 g of H 2 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 %

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