1 / 31

Stoichiometry Notes

Stoichiometry Notes. Chemistry 2013-2014. Stoichiometry is the calculation of quantities in chemical equations.

ervin
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 Chemistry 2013-2014

  2. Stoichiometry is the calculation of quantitiesin chemical equations. • Stoichiometry can be used to predict the amount of product that will form in a reaction. It can also be used to determine the amount of each reactant you need to use, if you want to get a certain amount of a product.

  3. The “heart” of stoichiometry is the mole ratio, which is the ratio of the coefficients of two things in a chemical reaction. For example, if you look at the equation in the box for the formation of water from hydrogen gas and oxygen gas, the mole ratio of hydrogen to water is 2:2 or 1:1.

  4. What is the mole ratio of H2 to O2? Or O2 to H2? What is the mole ratio of O2 to H2O? Or H2O to O2? 2:1 1:2 1:2 2:1

  5. Besides mole ratios, we will also need our old conversion factors. • 1 mole = GFM • 1 mole = 6.02 x 1023atoms (elements), molecules (molecular compounds), or formula units (ionic compounds). • 1 mole = 22.4 L at STP

  6. One of the governing laws of Chemistry is the Law of Conservation of Mass, which states that matter cannot be created or destroyed. Another way of saying this is that atoms cannot be created or destroyed (though they can be changed, combined, or split through nuclear reactions).

  7. Are these conserved in a chemical reaction? • MassYes No • Number of atoms Yes No • Moles Yes No • Molecules Yes No • Volume YesNo

  8. Before you start any stoichiometry problem, you need to write a balanced equation. Never assume an equation is balanced—always check for yourself. If the equation is not balanced, or if it’s balanced incorrectly, you will get the wrong coefficients, which will give you the wrong mole ratio, which will give you the wrong answer.

  9. For example, if you used the equation in the top-right corner of this page as-is, you would think the mole ratio of nitrogen to hydrogen is 1:1. If you used this mole ratio in a stoichiometry problem, you would get a wrong answer. Balance the equation. What is the real mole ratio of nitrogen to hydrogen? 1:3

  10. There are several types of stoichiometry problems. Let’s start off simple with mole-mole calculations. In mole-mole calculations, you need to convert from moles of one substance to moles of another. This type of problem only uses one conversion factor, the mole ratio.

  11. How many moles of aluminum are needed to form 2.3 moles of aluminum oxide? 4Al + 3O2 2Al2O3 x mol 2.3mol 2.3 mol Al2O3 4 mol Al = 2 mol Al2O3 4.6 mol Al

  12. How many moles of oxygen are required to react completely with 0.84 mol of Aluminum? 4Al + 3O2 2Al2O3 0.84 mol x mol 0.84 mol Al 3 mol O2 = 4 mol Al 0.63 mol O2

  13. Mole-mole calculations are the simplest because they only use one conversion factor. Here are other types of stoichiometry problems: • Mass-moleor mole-mass calculations are two-step problems that use GFM and mole ratio • Mass-mass calculations are three-step problems that use GFM and mole ratio • Mole-volume calculations are two-step problems that use 1 mole = 22.4 L and mole ratio • Volume-volume calculations are three-step problems that use 1 mole = 22.4 L and mole ratio

  14. Volume-massor mass-volumecalculations are three-step problems that use GFM, 1 mole = 22.4 L, and mole ratio • Sometimes problems begin, end, or begin AND end with “atoms”, “molecules”, “formula units”, or “particles”. In this case, you will need to use Avogadro’s number, 1 mole = 6.02 x 1023 atoms (elements), molecules (molecular compounds), or formula units (ionic compounds).

  15. Remember that the mole ratiois the “heart” of stoichiometry. If your given is not in moles, you will need to convert it to moles (this will be the first step of the problem). Use the mole ratio to figure out how many moles of the substance you want are present, and then change those moles or grams, liters, etc. according to the problem statement.

  16. Example 1: Calculate the number of grams of solid sodium oxide that will be produced when 115 g of solid SODIUM reacts with excess O2. 4Na + O2 2Na2O 115g x g BUT HOW????

  17. Steps to a stoichiometry problem:(mass-mass)

  18. 1. Write a balanced chemical equation.4Na + O2 2Na2O

  19. 2. Convert the given mass to moles by dividing by the gfm. 115g Na 1 mol Na 23.0g Na

  20. 3. Convert moles of the first substance to moles of the second substance by using the mole ratio (coefficients from the balanced equation). Mole ratio = moles “wanted” moles “given” 2 mol Na2O 4 mol Na

  21. 4. Convert moles of the second substance into grams by multiplying by the gfm. 62.0g Na2O 1 mol Na2O

  22. Calculate the number of grams of solid sodium oxide that will be produced when 115 g of solid sodium reacts with excess O2. 4Na + O2 2Na2O 115g x g 115g Na 1 mol Na 2 mol Na2O 62.0g Na2O 23.0g Na 4 mol Na 1 mol Na2O = 155g Na2O

  23. Example 2: How many grams of oxygen are required to “burn”13.0 g of C2H2?2C2H2(g)+5O2(g) 4CO2(g) +2H2O(g)13.0g x g 13.0g C2H2 1 mol C2H2 5 mol O2 32.0g O2 26.0 g C2H2 2 mol C2H2 1 mol O2 =40.0g

  24. EXAMPLE 3: How many grams of CO2and water are produced when 13.0 g of C2H2 reacts with an excess amount of oxygen? 2C2H2(g)+5O2(g)4CO2(g)+2H2O(g) 13.0g x g 13.0g C2H2 1 mol C2H2 4 mol CO2 44.0g CO2 26.0g C2H2 2 mol C2H2 1 mol CO2 =44.0g CO2 =9.00g H2O

  25. Example 4 How many grams of SnF2 can be made by reacting 7.42 x 1024 molecules of HF with tin? Balance the equation _____ HF + _____ Sn  _____ SnF2 + _____ H2 What conversion factors do you need to solve this problem? ________________________

  26. EXAMPLE 4: How many grams of SnF2 can be made by reacting 7.42 x 1024 molecules of HF with tin? Sn(s) + 2HF(g)  SnF2(s) + H2(g) 7.42 x 1024molec X g 7.42 x1024molec HF 1 mol HF 1 mol SnF2 156.7g SnF2 6.02 x1023molec HF 2 mol HF 1 mol SnF2 = 966g SnF2

  27. Example 5 How many liters of hydrofluoric acid, HF, are needed to produce 14.2 L of hydrogen gas at STP? Balance the equation _____ Sn+ _____ HF  _____ H2 + ________ SnF2 What conversion factors do you need to solve this problem? ________________________

  28. Example 5: How many liters of HF are needed to produce 14.2 L of hydrogen at STP ?Sn(s) + 2HF(g) H2(g) + SnF2(s) x L 14.2 L 14.2L H2 1 mol H2 2 mol HF 22.4 L HF 22.4 L H2 1 mol H2 1 mol HF = 28.4L HF

  29. Example 6 How many molecules of hydrogen gas are needed to produce 80. L of hydrofluoric acid (HF) at STP? H2+ F2 HF What conversion factors do you need to solve this problem? ________________________ Set up the problem and solve. Always show your work.

  30. Example 6: How many molecules of hydrogen are Needed to produce 80.0 L of HF (STP)? H2(g) + F2(g)  2HF(g) x molec. 80.0L 80.0L HF 1 mol HF 1 mol H2 6.02 x 1023molec 22.4L HF 2 mol HF 1 mol H2 = 1.08 x 1024 molecules H2

More Related