Download presentation

Presentation is loading. Please wait.

Published byRandall Langstaff Modified over 2 years ago

1
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

2
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

3
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 %

4
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

5
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.

6
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

7
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

8
–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.

9
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

10
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.

11
–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

12
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 %

Similar presentations

OK

Stoichiometry & the Mole. The Mole __________ - SI base unit used to measure the amount of a substance. A mole of anything contains __________ representative.

Stoichiometry & the Mole. The Mole __________ - SI base unit used to measure the amount of a substance. A mole of anything contains __________ representative.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on polynomials in maths the ratio Ppt on business environment nature concept and significance of dreams Ppt on cleanliness of surroundings Ppt on personality development and motivation Army ppt on react to possible ied Ppt on aerobics and fitness Ppt on chromosomes and genes model Ppt on cmos image sensor Download ppt on say no to crackers Ppt on acute coronary syndrome algorithm