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CH 3: Stoichiometry Moles. Mole Defined Mole = number equal to the # of carbon atoms in exactly 12 grams of pure C-12. –Mole = 6.022 x 10 23 –Called Avogadro’s.

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Presentation on theme: "CH 3: Stoichiometry Moles. Mole Defined Mole = number equal to the # of carbon atoms in exactly 12 grams of pure C-12. –Mole = 6.022 x 10 23 –Called Avogadro’s."— Presentation transcript:

1 CH 3: Stoichiometry Moles

2 Mole Defined Mole = number equal to the # of carbon atoms in exactly 12 grams of pure C-12. –Mole = x –Called Avogadro’s number One mole of protons = 1g One mole of neutrons = 1g One mole of an element = atomic mass in grams

3 Molar Mass and related terms Molar mass – mass in grams of one mole of a substance – molecular weight, mass of 1 molecule in amu –Formula weight, mass of 1 formula unit for an ionic compound in amu Calculate by summing the masses of the component atoms, units: –molar mass – grams or grams/mole –molecular wt & formula wt - amu

4 Moles, mass and # particles Molar mass links the mass of a substance to the number of particles present 1 mole Mg = 24.31g = x Mg atoms

5 Molar Mass Related Calculations: 1.Molar mass of a substance. 2.Moles present in a given mass of substance. 3.Mass of a given number of moles of a substance. 4.Number of particles in a given mass or a given number of moles of a substance.

6 1.What is the molar mass of glucose. The formula for glucose is C 6 H 12 O 6. 2.What is the mass of moles of glucose?

7 3.How many glucose molecules are present in moles of the compound? 4.How many moles of glucose are present in 3.5 x grams of the substance?

8 Mass Percent Compounds are typically described by either their chemical formula or their percent by mass of the component elements. Mass % X = n (molar mass X) x 100% molar mass of compound

9 Mass Percent Calculations Calculate the mass percent of each element in C 6 H 12 O 6 –Assume one mole of the substance.

10 Empirical and Molecular Formulas Molecular formula – ratio of atoms in a molecule Empirical formula – simplest ratio of elements in a molecule Glucose: –Molecular formula: C 6 H 12 O 6 –Empirical Formula: Molecular formula = n (empirical formula)

11 Mass %  Empirical Formula Given mass percent data: 1.“Calculate” mass in grams of each element in 100 g of compound. 2.Convert each mass into moles of the element. 3.Divide each molar answer by the smallest of the values. 4.If the numbers obtained in step 3 are not whole numbers, multiply all by an integer so the results are whole numbers 5.Whole numbers obtained in step ¾ are the subscripts in the empirical formula.

12 Empirical Formula  Molecular Formula Molar mass needed Calculate mass of one mole of the empirical formula. Molar Mass = integer (n) empirical formula weight (n)(empirical formula) = molecular formula

13 Caffeine Molar mass = 194 g/mol % C 5.19 % H % N % O What is the empirical and molecular formula of caffeine?

14 Acetaminophen (Tylenol) Molar mass = 151 g/mol % C 6.00 % H 9.27 % N % O What is the empirical and molecular formula of acetaminophen?

15 CH 3: #86 Urea: Molar Mass 60 g/mol (I looked this up – not in question) g N g H g C g O What is the empirical and molecular formula of urea?

16 3.7/3.8 Chemical Reactions Writing chemical reactions –Write the correct chemical formula for each reactant and product. –Use a subscript after each formula to indicate the state of each substance (s) – solid ( l ) – liquid (g) – gas (aq) – aqueous (dissolved in water)

17 Chemical equations must obey the law of conservation of matter – balancing does this. Balance the equation by adding coefficients in front of reactants and products as needed –DO NOT CHANGE THE FORMULAS FOR THE COMPOUNDS

18 One of my favorite chemical reactions: Mg (s) + HCl (aq)  MgCl 2 (aq) + H 2 (g) As written the reaction does not obey the law of conservation of matter – reaction needs to be balanced

19 Mg (s) + 2 HCl (aq)  MgCl 2 (aq) + H 2 (g) Now the reaction is balanced!

20 Balance the Reactions N 2 (g) + H 2 (g)  NH 3 (g) Na (s) + Cl 2 (g)  NaCl (s) Pb(NO 3 ) 2(aq) + NaCl (aq)  PbCl 2(s) + NaNO 3(aq)

21 Meaning of the Balanced Reaction Atomic level Molar level Stoichiometry - relationship between moles, the balanced reaction and mass –Page 10? outlines the needed steps

22 2 Na + Cl 2  2 NaCl 2 mol Na + 1 mol Cl 2  2 mol NaCl

23 Cookie example! 2 eggs + 1 bag chips  50 cookies How many cookies could you make if you had only one egg? How many cookies could you make if you had 2 bags of chips?

24 Molar Ratios Balanced reactions lead to molar ratios N H 2  2 NH 3

25 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) 1.If 3.5 grams of N 2 to react, how many moles and how many grams of NH 3 would be made?

26 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) 1.How many grams of H 2 are needed to make 37.8 grams of NH 3 ?

27 2 LiOH + CO 2  Li 2 CO 3 + H 2 O 1.How many grams of LiOH are needed to react 750. grams of CO 2 ? 2.How many grams of Li 2 CO 3 will be made if 750. grams of CO 2 react?

28 Limiting Reagent Reactants are not always combined in stoichiometric quantities. Often one reactant is used up before others. This reactant is said to be the limiting reagent (or reactant.) this reactant limits how much product can be made.

29 Limiting Reagent Calculations When you are given masses for each reactant you must determine which reactant is limiting. 1.Convert grams of each reactant to moles. 2.Use the molar ratios to determine which reactant is limiting. 3.Base the amount of product that can be made on the limiting reagent.

30 Another Approach… 1.Convert grams of each reactant to moles. 2.Calculate the moles (or grams) of a product that could be made from each reactant. 3.Whichever reactant results in the smaller quantity of product is the limiting reagent. 4.Base the amount of product made on the limiting reagent.

31 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) 12.5 grams of nitrogen is combined with 2.30 grams of hydrogen. Which reactant is in excess and which is limiting? How many grams of ammonia will be made?

32 Percent Yield % Yield = actual yield x 100 % theoretical yield


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