1 / 14

Stoichiometry Notes

Stoichiometry Notes. Equations to MEMORIZE. 1 mol = (molar mass) g Get it from the atomic mass. Add it up if you have a molecule. (O=15.999, so O 2 =31.998) 1 mol = 6.022 x 10 23 atoms or molecules 1 mol = 22.4 L (of a gas). Avagadro’s Number. Congruent Triangles.

artan
Download Presentation

Stoichiometry Notes

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Stoichiometry Notes

  2. Equations to MEMORIZE • 1 mol = (molar mass) g • Get it from the atomic mass. Add it up if you have a molecule. (O=15.999, so O2=31.998) • 1 mol = 6.022 x 1023 atoms or molecules • 1 mol = 22.4 L (of a gas) Avagadro’s Number

  3. Congruent Triangles What is the mass in grams of 3.5 moles of water ? Molar mass = mass Of 2 H’s and 1 O 3.5 mol H2O • 3.5 mol H2O • 3.5 mol H2O • __X g H2O__ • __X g H2O__ • ______2___ • = 1 mol H2O • 18.015 g H2O 1 mol H2O • 1 x X = 3.5 x 18.015 • X=63.05 g H2O 18.015 g H2O X g H2O Question Amount • = • Question Amount • Equation Amount • Equation Amount

  4. Cross Multiplication Review A x Y = Z x B A • B • =  • Z • Y 3 • 5 3 x Y = 4 x 5 • =  • 4 • Y 3 x Y = 4 x 5 3 3 3 x Y = 4 x 5 3 3 Y = 4 x 5 3 Y = 6.67

  5. Basic conversions Question Amount • = • Question Amount • Equation Amount • Equation Amount • [Question amount/Equation amount = Question amount/Equation amount] • Mole  Grams • 2.5 moles of O2 = how many grams? • 1 mole O2 = 31.998 g O2 (the weight of 2 Oxygen) • 2.5 mol O2= __X g O2__ • 8 g of O2 = how many moles • __8 g O2= X mol O2 X = 80.00 g O2 • 1 mol O2 31.998 g O2 X = .25 mol O2 • 31.998 g O2 1 mol O2

  6. Basic conversions Question Amount • = • Question Amount • Equation Amount • Equation Amount • [Question amount/Equation amount = Question amount/Equation amount] • Moles  Atoms (or molecules) • 7.3 moles of H2O = how many molecules? • 1 mole H2O = 6.022 x 1023 molecules H2O • (This is Avagadro’s Number) • 7.3 mol H2O = ____X Molecules H2O____ • 3.92 x 1023 molecules of H2O = how many Moles • 3.92x1023 Molecules H2O = X mol H2O X = 4.40 x 1024 molecules H2O • 1 mol H2O 6.022x10 23Molecules H2O X = .65 mol H2O • 6.022x10 23 Molecules H2O 1 mol H2O

  7. 2 step conversion • MoleculesGrams (Have to convert to moles first) • 3 g of C4H10 = how many molecules? • __3 g C4H10 __= X mol C4H10 • X = .05 mol C4H10 • .05 mol C4H10 = X molecules C4H10 ____ • 3.058 x 1024 molecules of C4H10 = how many g? • 3.058x1024 molecules C4H10= X mol C4H10 6.022x1023 molecules C4H10 1 mol C4H10 • X = 5.08 mol C4H10 • 5.08 mol C4H10 = X g C4H10 1 mol C4H10 58.124 g C4H10 • 58.124 g C4H10 1 mol C4H10 X = 3.01 x 1022 molecules C4H10 • 1 mol 6.022x1023 molecules C4H10 X = 295.27 g C4H10

  8. Stoichiometry Moles to moles Question Amount • = • Question Amount • Equation Amount • Equation Amount • 2 C4H10 +  13 O2 8 CO2 +  10 H2O • How many moles H2O will 3.5 moles of C4H10 make? • 3.5 mol C4H10 = X mol H2O X = 17.5 mol H2O • 2 mol C4H10 10 mol H2O

  9. Moles to grams • 2 C4H10 +  13 O2 8 CO2 +  10 H2O • How many moles of CO2 will 2.4 grams of C4H10 make? • Same as before but have to substitute in the mass of 2 moles • 2.4 g C4H10 = X mol CO22 mol C4H10 8 mol CO2 • 2 x (4(12.011) + 10(1.008)) = 116.248 {2 moles x 4 Carbons + 10Hydrogens} • 2.4 g C4H10 = X mol CO2116.248 g C4H10 8 mol CO2 X = .17 mol CO2

  10. Moles to grams • 2 C4H10 +  13 O2 8 CO2 +  10 H2O • How many grams of CO2 will 2.4 moles of C4H10 make? • 2.4 mol C4H10 = _X g CO2 _ 2 mol C4H108 mol CO2 • 8 x (1(12.011) + 2(15.999)) = 352.072 • 2.4 mol C4H10 = X g CO2 2 mol C4H10 352.072 g CO2 X = 422.49 g CO2

  11. Grams to grams • 2 C4H10 +  13 O2 8 CO2 +  10 H2O • 17 g of Oxygen will produce how many grams of Carbon Dioxide? • Have to substitute on both ends • _17 g O2_ = X g CO213 mol O28 mol CO2 • 13 x 2(15.999) = 415.974 • 8 x (1(12.011) + 2(15.999)) = 352.072 • 17 g O2 = X g CO2 415.974 g O2 352.072 g CO2 X = 14.39 g CO2

  12. Volume to moles • 2 C4H10 +  13 O2 8 CO2 +  10 H2O • How many liters of CO2 gas are produced by 5 mol of C4H10? • Good luck measuring the weight of a gas! Gasses are usually dealt with in volume. • 22.4 L of any gas (at standard pressure) = 1 mol of that substance. • 5 mol C4H10 = _X L CO2_ 2 mol C4H10 8 mol CO2 • 8 x 22.4 = 179.2 • 5 mol C4H10 = X L CO2 2 mol C4H10 = 179.2 L CO2 • (Moles to volume would be the same but the left side would have the substitution) X = 448 L CO2

  13. Volume to grams • 2 C4H10 +  13 O2 8 CO2 +  10 H2O • How many Liters of O2 do you need to produce 105 g of H2O? • Same as before but you will have two substitutions. • 105 g H2O = X L O210 mol H2O = 13 Mol O2 • 10 x (2(1.008) + 1(15.999) ) = 180.15 • 13 x 22.4 = 291.2 • 105 g H2O = X L O2180.15 g H2O 291.2 L O2 X = 169.73 L O2

  14. Limiting Reagents =what you run out of first • 2 C4H10 +  13 O2 8 CO2 +  10 H2O • If you react 80 g of C4H10with 15 g of O2, which is the limiting reagent? • Question/Equation. Then compare the numbers • 80 g C4H10= 80 g C4H10= .69 2 mol C4H10(2 x (4(12.011) + 10(1.008))) g C4H10 • 15 g O2 = 15 g O2= .04 13 mol O2(13 x 2(15.999)) g O2 • Lowest answer is the Limiting Reagent! • If you want to know how much product you will make, you need to use the numbers of the limiting reagent. You will end up with extra of the other one that does not react.

More Related