1 / 22

Chem 1215 Workshop Series Mastering the Basics of Chemistry

Chem 1215 Workshop Series Mastering the Basics of Chemistry. Stoichiometry. Dr. Courtney Dodd. Stoichiometry. The relationships among the quantities of reactants and products involved in chemical reactions. 3CaCl 2 (aq) + 2H 3 PO 4 (aq)  6HCl(aq) + Ca 3 (PO 4 ) 2 (s)

constance-morse
Download Presentation

Chem 1215 Workshop Series Mastering the Basics of Chemistry

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. Chem 1215 Workshop Series Mastering the Basics of Chemistry Stoichiometry Dr. Courtney Dodd

  2. Stoichiometry The relationships among the quantities of reactants and products involved in chemical reactions. 3CaCl2(aq) + 2H3PO4(aq) 6HCl(aq) + Ca3(PO4)2(s) 3 mol 2 mol 6 mol 1 mol 3 moles of CaCl2 react with 2 moles H3PO4 of to produce 6 moles of HCl and 1 mole of Ca3(PO4)2.

  3. Chemical Equations Deal in Moles… but we cannot count that high! 3CaCl2(aq) + 2H3PO4(aq) 6HCl(aq) + Ca3(PO4)2(s) moles (reactants)  moles (products) In the laboratory, we cannot count moles. In the laboratory, we calculate moles after measuring: • mass (for pure solids, liquids, or gases), • volume and concentration (for solutions), or • volume, pressure, and temperature (for gases).

  4. How to Get from the Measurement to Moles… If you measure the mass divide by the molar mass volume and molarity multiply the molarity by the volume in liters moles  moles (reactants) (products) chemical equation pressure, volume, and temperature of a gas PV RT

  5. …and Back Again multiply by the molar mass mass divide by the volume in liters moles moles (reactants) (products) molarity chemical equation pressure, volume, or temperature of a gas use PV = nRT

  6. Possible Stoichiometry Problems Since we can start with: moles mass molarity P, V, and T and we can end with: moles mass molarity P, V, and T, there are 16 possible types of stoichiometry problems. Here are eight of them: moles  moles mass  moles moles  mass mass  mass moles  molarity mass  molarity moles  P,V,T mass  P,V,T

  7. Possible Stoichiometry Problems moles  moles mass  moles moles  mass mass  mass moles  molarity mass  molarity moles  P,V,T mass  P,V,T The good news is that ONE method serves to solve ANY type of stoichiometry problem. We will start with massmass problems, so we must be able to calculate a molar mass.

  8. Calculating Molar Masses • The formula for calcium phosphate is Ca3(PO4)2. • The formula mass for Ca3(PO4)2 is calculated by adding the masses of every atom in Ca3(PO4)2: atomic mass from periodic table Write the masses so that they have four significant figures. 1 formula unit of Ca3(PO4)2 has: 3 atoms Ca, which have a mass of 3 x 40.08 amu = 120.24 amu 2 atoms P, which have a mass of 2 x 30.97 amu = 61.94 amu 8 atoms O, which have a mass of 8 x 16.00 amu = 128.00 amu formula mass of Ca3(PO4)2 310.18 amu

  9. Calculating Molar Masses The molar mass for Ca3(PO4)2 is the mass of 1 mole of Ca3(PO4)2. The molar mass has the same value as the formula mass, but the units are grams. The formula mass of Ca3(PO4)2 is 310.18 amu. The molar mass of Ca3(PO4)2 is 310.18 g.

  10. Converting from Mass to Moles Once we know the molar mass of a compound, we can calculate the number of moles present in ANY MASS of that compound.Simply treat the molar mass as a conversion factor. For example,1 mol Ca3(PO4)2 = 310.18 g. How many moles of Ca3(PO4)2are in 29.16 g of the compound? 29.16 g x 1 mol Ca3(PO4)2 = 0.09401 mol Ca3(PO4)2 310.18 g

  11. How to Solve any Stoichiometry Problem in Five Steps Example: What mass of oxygen is consumed in the combustion of 100.0 g of propane, C3H8? (This is a mass  mass stoichiometry problem.) 1. Write and balance the equation for the reaction. C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l) (In the combustion of a hydrocarbon, the other reactant is always oxygen, and the products are always carbon dioxide and water.)

  12. How to Solve any Stoichiometry Problem in Five Steps Example: What mass of oxygen is consumed in the combustion of 100.0 g of propane, C3H8? 2. Write what you know and what you need to know, including units. 100.0 g ? g C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l)

  13. How to Solve any Stoichiometry Problem in Five Steps Example: What mass of oxygen is consumed in the combustion of 100.0 g of propane, C3H8? 3. Convert “what you know” to moles. 100.0 g ? g C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l) 100.0 g C3H8 x 1 molC3H8 44.10 gC3H8 mass  moles

  14. How to Solve any Stoichiometry Problem in Five Steps Example: What mass of oxygen is consumed in the combustion of 100.0 g of propane, C3H8? 4. Using the stoichiometry of the reaction, convert to moles of “what you need to know.” 100.0 g ? g C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l) 100.0 g C3H8 x 1 molC3H8 x 5 molO2 44.10 gC3H81 molC3H8 from the chemical equation mass  moles

  15. How to Solve any Stoichiometry Problem in Five Steps Example: What mass of oxygen is consumed in the combustion of 100.0 g of propane, C3H8? 5. Convert moles of “what you need to know” to the appropriate final units. 100.0 g ? g C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l) 100.0 g C3H8 x 1 molC3H8 x 5 molO2 x 32.00 g O2 = 362.8 g O2 44.10 gC3H81 molC3H81 molO2 mass  moles from the chemical equation moles  mass

  16. How to Solve any Stoichiometry Problem in Five Steps Example: What mass of oxygen is consumed in the combustion of 100.0 g of propane, C3H8? Notice that steps 3,4, and 5 are combined into one calculation. This helps avoid rounding errors. 100.0 g C3H8 x 1 molC3H8 x 5 molO2 x 32.00 g O2 = 362.8 g O2 44.10 gC3H81 molC3H81 molO2

  17. How to Solve any Stoichiometry Problem in Five Steps 1. Write and balance the equation for the reaction. 2. Write what you know and what you need to know, including units. 3. Convert “what you know” to moles. 4. Using the stoichiometry of the reaction, convert to moles of “what you need to know.” 5. Convert moles of “what you need to know” to the appropriate final units.

  18. Using Stoichiometry to Handle Problems Involving Thermochemical Equations Example: How many kJ of heat are released in the combustion of 25.0 g of methanol (CH3OH), a liquid, at 25°C? ΔH°(25°C) = -726.56 kJ / mol methanol (This isn’t exactly massmoles, but we still use most of the same steps.) 1. Write and balance the equation for the reaction. 2CH3OH (l) + 3O2 (g)  2CO2 (g) + 4H2O (l) This is another combustion reaction. Although methanol contains O as well as C and H, the products are again carbon dioxide and water.

  19. Using Stoichiometry to Handle Problems Involving Thermochemical Equations Example: How many kJ of heat are released in the combustion of 25.0 g of methanol (CH3OH), a liquid, at 25°C? ΔH°(25°C) = -726.56 kJ / mol methanol 2. Write what you know and what you need to know, including units. 25.0 g ? kJ 2CH3OH(l) + 3O2(g)  2CO2(g) + 4H2O(l) When the enthalpy change is included with the chemical equation, the result is called a thermochemical equation. Note this equation is for 2 mol CH3OH, so ΔH doubled. ΔH° = -1453.12 kJ

  20. Using Stoichiometry to Handle Problems Involving Thermochemical Equations Example: How many kJ of heat are released in the combustion of 25.0 g of methanol (CH3OH), a liquid, at 25°C? ΔH°(25°C) = -726.56 kJ / mol methanol 3. Convert “what you know” to moles. 25.0 g ? kJ 2CH3OH(l) + 3O2(g)  2CO2(g) + 4H2O(l) 25.0 g CH3OH x 1 molCH3OH 32.04 g CH3OH ΔH° = -1453.12 kJ mass  moles

  21. Using Stoichiometry to Handle Problems Involving Thermochemical Equations Example: How many kJ of heat are released in the combustion of 25.0 g of methanol (CH3OH), a liquid, at 25°C? ΔH°(25°C) = -726.56 kJ / mol methanol 4 & 5. Using the stoichiometry of the reaction, convert to “what you need to know” in the appropriate units. 25.0 g ? kJ 2CH3OH(l) + 3O2(g)  2CO2(g) + 4H2O(l) 25.0 g CH3OH x 1 molCH3OH x -1453.12 kJ = - 567 kJ 32.04 g CH3OH 2 mol CH3OH ΔH° = -1453.12 kJ mass  moles from the thermochemical equation

  22. Resources for Stoichiometry 1. Dr. Bailey’s web page http://www.occc.edu/kmbailey click on General Chemistry Tutorials (stoichiometry is #5) -or- click on the Chem 1115 web page, practice problems, Unit 3 2. Dr. Dodd’s Chem I web page http://www.occc.edu/cdodd/chem1115.htm click on the two Unit 3 worksheets 3. Don’t forget…your textbook is a valuable resource!

More Related