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Chapter 4

Chapter 4. Reaction Stoichiometry. Multiplying the chemical formulas in a balanced chemical equation reflect the fact that atoms are neither created nor destroyed in a chemical reaction. 2H 2(g) +O 2(g) ---  2H 2 O

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Chapter 4

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  1. Chapter 4 Reaction Stoichiometry

  2. Multiplying the chemical formulas in a balanced chemical equation reflect the fact that atoms are neither created nor destroyed in a chemical reaction. • 2H2(g) +O2(g)--- 2H2 O • These coefficients can be interpreted in terms of mole. In the above equation we can say that 1 mole of oxygen reacts with two moles of hydrogen to form 2 moles of water.

  3. Mole to mole calculations • This procedure is based on two principles: • 1. the number of atoms of each element is conserved in a chemical reaction. • 2. the coefficients in a given chemical equation give the relative numbers of moles of each reactant and product. These numbers of moles are used to construct conversion factors.

  4. Class Practice • Find the number of moles of N2 needed to produce 5.0 mol NH3 by reaction with H2. • Step 1. Write the balanced chemical equation: N2(g) +3H2(g)---- 2NH3(g) • Step 2. It follows from the equation that 1 mol of N2 approx. equals to 2mol of NH3 so the mol ratio is • Substance required/ substance given =1mol N2/2mol NH3.

  5. Step 3. We now apply this conversion factor to the information given and obtain the information required: • Moles of N2 =(5.0 mol NH3)x1mol N2/2mol NH3 =2.5 mol of nitrogen.

  6. How many moles of NH3 molecules can be produced from 2.0 mol H2 in the same reaction as in the previous reaction if all the hydrogen reacts. • ANS 1.3mol ammonia

  7. Mass to mass calculations • By using molar masses we can also find the mass of one substance that a given mass of another substance can produce. We first convert the given mass to the corresponding number of moles by using the molar mass of that substance. We then convert the number of moles of the required substance to mass by using its molar mass.

  8. STEPS • 1. Convert the given mass of one substance (in grams) to number of moles by using its molar mass. • Number of moles of substance A=mass of A in grams/molar mass of A (grams per mole) • 2. Write the balanced equation for the reaction. • 3. Use the mol ratio to convert from the given number of moles of one substance to the number of moles of the other substance. • Substance required/substance given

  9. 4. Convert from number of moles of the second substance to mass (in grams) by using the molar mass of the substance. • Mass of B required (grams) =number of moles of B x molar mass of B (grams per mole) • mB = nB x MB

  10. Class Practice • Calculate the mass of oxygen needed to react with 0.450 g of hydrogen gas in the reaction 2H2(g) + O2 (g)-- 2H2O(l).

  11. Reaction yield • Example: • A major source of atmospheric carbon dioxide is the burning of gasoline in automobiles, so we need to know how much CO2 is likely to be produced from a liter of gasoline. • 2C8 H18(l) + 25O2(g)---16CO2(g) + 18H2O

  12. Theoretical yield is the maximum mass of product that can be obtained from a given mass of reactant. The theoretical yield of carbon dioxide is based on the assumptions that this is the only reaction taking place and that every molecule of octane is converted into carbon dioxide and water.

  13. Class Practice • Calculate the theoretical yield of carbon dioxide when 702 g of octane(C8H18) is burned. • Steps: • 1. Convert mass to moles of C8H18 molecules then moles of C8H18 to moles of carbon dioxide, then moles of carbon dioxide to mass of carbon dioxide.

  14. Home work • Page 166 • # 4.2,4.3,4.8,4.10,

  15. Percentage yield • The percentage yield is the percentage of the theoretical yield actually achieved. • When 24 g of potassium nitrate was heated with lead, 13.8 g of potassium nitrite was formed in the reaction Pb(s) + KNO3 (s)--- PbO(s) +KNO2(s). Calculate the percentage yield of potassium nitrite. Ans 68.3%

  16. Limiting reactant • Limiting Reactant - The reactant in a chemical reaction that limits the amount of product that can be formed.  The reaction will stop when all of the limiting reactant is consumed. • Excess Reactant - The reactant in a chemical reaction that remains when a reaction stops when the limiting reactant is completely consumed.  The excess reactant remains because there is nothing with which it can react.

  17. 8car bodies + 48 tires -- 8cars with tires + 16 tires in excess. • No matter how many tires there are, if there are only 8 car bodies, then only 8 cars can be made.  • Likewise with chemistry, if there is only a certain amount of one reactant available for a reaction, the reaction must stop when that reactant is consumed whether or not the other reactant has been used up.

  18. Example Limiting Reactant Calculation: • A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen.  Which is the limiting reactant and how much excess reactant remains after the reaction has stopped? • First, we need to create a balanced equation for the reaction: • 4 NH 3(g) + 5 O2(g) ---4 NO (g) + 6 H 2 O (g) • Next we can use stoichiometry to calculate how much product is produced by each reactant. . • 2.0 g of NH3 x 1mol of NH3 / 17.0 g of NH3 x 4mol NO/4mol NH3 x 30.0g of NO/1 mol NO =3.53 g • 4.0 g of O2x 1 mol O2 /32.0 g of O2 x 4 mol NO/5 mol O2 x 30.0 g NO/1mol NO=3.00 g of NO

  19. The reactant that produces the lesser amount of product in this case the oxygen. • Next, to find the amount of excess reactant, we must calculate how much of the non-limiting reactant (oxygen) actually did react with the limiting reactant (ammonia).

  20. 4.0g of O2 x 1 mol of O2 /32 g of O2x 1mol NH3 /5 mol O2 x 17 g NH3/1mol NH3=1.70 g of NH3   1.70 g is the amount of ammonia that reacted, not what is left over.  • To find the amount of excess reactant remaining, subtract the amount that reacted from the amount in the original sample. • 2.00g of NH3 – 1.70 g (reacted) =0.30 g remaining.

  21. Class Practice • In the synthesis of ammonia which is the limiting reactant when 100 kg of hydrogen reacts with 800 kg of nitrogen?

  22. Empirical formula • Empirical Formula - A formula that gives the simplest whole-number ratio of atoms in a compound. • Steps for Determining an Empirical Formula • 1. Start with the number of grams of each element, given in the problem.  • 2. Convert the mass of each element to moles using the molar mass from the periodic table.  • Divide each mole value by the smallest number of moles calculated.  • Round to the nearest whole number.  This is the mole ratio of the elements and is represented by subscripts in the empirical formula. 

  23. If the number is too far to round (x.1 ~ x.9), then multiply each solution by the same  factor to get the lowest whole number multiple.  • e.g.  If one solution is 1.5, then multiply each solution in the problem by 2 to get 3.  • e.g.  If one solution is 1.25, then multiply each solution in the problem by 4 to get 5.  • Once the empirical formula is found, the molecular formula for a compound can be determined if the molar mass of the compound is known.  • Simply calculate the mass of the empirical formula and divide the molar mass of the compound by the mass of the empirical formula to find the ratio between the molecular formula and the empirical formula.  Multiply all the atoms (subscripts) by this ratio to find the molecular formula. 

  24. A compound was analyzed and found to contain 13.5 g Ca, 10.8 g O, and 0.675 g H.  What is the empirical formula of the compound? • Start with the number of grams of each element, given in the problem. • Convert the mass of each element to moles using the molar mass from the periodic table. • Divide each mole value by the smallest number of moles calculated.  Round to the nearest whole number. • In this we divide by the moles of Ca obtained.

  25. This is the mole ratio of the elements and is represented by subscripts in the empirical formula. • CaO2H2 = Ca(OH)2

  26. Homework • Page 168 • # 4.34, 4.36

  27. Solutions • Molarity : Molarity is the number of moles of solute dissolved in one liter of solution. The units, therefore are moles per liter, specifically it's moles of solute per liter of solution. • molarity = amount of solute (moles) volume of solution (liter)

  28. Class Practice • Calculate the molarity of glucose in a solution made by dissolving 1.368 g of glucose (C6 H12 O6) in enough water to make 50.00 ml of solution.

  29. Dilution • The most important thing to remember concerning dilution is that you are only adding solvent. You are not adding solute when you dilute. • Therefore: • moles of solute before dilution = moles of solute after dilution • or Number of millimoles of solute before = millimoles of solute after • or • Number of grams of solute before dilution = number of grams of solute after dilution • Since the definition for Molarity is: • Molarity = moles of solute / volume of solution in liters

  30. Solving for moles of solute gives: • moles of solute = M x V of solution in liters • or • millimoles of solute = M x V of solution in ml • If Moles of solute before dilution = moles of solute after dilution • then M x V in liters before dilution = M x V in liters after dilution • or • M1V1 = M2V2 • where M1 = Molarity before dilution • V1 is volume of solution before dilution • M2 = Molarity of solution after dilution • V2 = Volume of solution after dilution • We can use Molarity or mass % as the concentration term so you could have the following alternative where mass % is used: • mass % x grams solution before dilution = mass % x grams of solution after dilution

  31. Homework • Page 169 • # 4.40,

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