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Chapter 3 Stoichiometry.  Mass 1. Atomic mass A. Atomic masses are representing the mass of an atom of one element compared to the mass of another. We.

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Presentation on theme: "Chapter 3 Stoichiometry.  Mass 1. Atomic mass A. Atomic masses are representing the mass of an atom of one element compared to the mass of another. We."— Presentation transcript:

1 Chapter 3 Stoichiometry

2  Mass 1. Atomic mass A. Atomic masses are representing the mass of an atom of one element compared to the mass of another. We use carbon-12 as the standard and assign it a mass of exactly 12 amu. B. Exercise: 1. Using the atomic masses in the periodic table, find an element that is four times heavier than carbon.

3 Solution: Since carbon has as an atomic mass of 12 amu, the element must have an atomic mass of about 48 amu, Ti atomic mass is amu. 2. Germanium (Ge) is about six times heavier than carbon. What is its approximate atomic mass? _______________

4 3. If helium (He) was the standard and assigned an atomic mass of 10 amu, what would the atomic mass of carbon be? ______________ C. Experimental determination of relative atomic mass 1. Instrument used: _________________ 2. Describe briefly how the instrument determines the relative atomic mass.

5 D. Atomic mass and isotopic abundance 1. Mathematical relationship: (abundance)(amu) + (abundance)(amu) = average atomic mass

6 2. Exercise: a. Antimony occurs in nature as a mixture of two isotopes: The isotope 121 Sb has a 51 mass of amu and an abundance of 57.25%. Antimony has an average atomic mass of amu. What is the abundance of the other isotope? What is the atomic mass?

7 b. Practice problems: Silicon consists of three stable isotopes whose atomic masses and abundances are as follows: amu 92.2 % amu 4.7 % amu 3.1 % Calculate the mass of Si.

8 2. Mass of a single atom a. Avogadro’s number What does it represent? ____________ ________________________________ b. Sample problems: 1. Calculate the mass of a carbon atom. Solution: Use the conversion factor: 6.02 x C atoms in g C.

9 Can be written as 6.02 x atoms g or g C 6.02 x10 23 C atoms Depending on which unit we want to cancel. In this exercise we want to eliminate the unit C atom and obtain the unit grams of C. Thus, we have 1 C atoms x g C = 2.00 x g 6.02 x C atoms

10 2. Practice Problem: How many atoms are there in g of C? 3. Formula masses a. The sum of the atomic masses of all atoms represented in the formula of any compound or polyatomic ion

11 b. Sample problem: 1. What is the formula mass of chloric acid? Solution: See packet 2. Practice problem: What is the formula mass for sodium nitrite?

12 4. Relationship of Avogadro’s number to formula masses. A formula mass in grams has 6.02 x formula units.  Mole 1. The amount of a substance that contains the same number of particles as there are atoms in exactly 12 g of carbon-12.

13 2. Molar masses a. The mass in grams of one mole of a pure substance b. Sample problem: See packet c. Practice problems: 1. Calculate the molar mass of : (NH 4 ) 2 SO 4 N = 2(14.0) H = 8 (1.0) S = 32.1 O= 4(16.0) = g/mole

14 2. Calculate the molar mass of: C 4 H 7 NO 4 C = 4(12.0) = 48.0 H= 7(1.0) = 7.0 N = 14.0 = 14.0 O = 4(16.0) = g/mole

15 3. Mole-gram conversion exercises: a. How many moles are there in 1.00 g of Al 2 O 3 ? See packet b.Practice problems: 1. Find the number of grams in 3874 mole of NH 4 Cl mole NH 4 Cl 53.0 g NH 4 Cl = 1 mole NH 4 Cl 205 g NH 4 Cl

16 2. Find the number of moles in g of Ba(OH) 2.

17 Mass Relations in Chemical Formulas 1. What a chemical reaction can tell you: a. The atom ratios of the different elements in the compound Ex. NH 4 Cl: 1 atom N 4 atoms H 1 atom Cl b. The mole ratios of the different elements in the compound Ex. NH 4 Cl: 1 mole N 4 mole H 1 mole Cl

18 c. The mass ratios of the different elements in the compound Ex. NH 4 Cl: g N, 4(1.01) g H, g Cl 2. Percent composition from formulas: a. Formula: %element = total mass element x100 total mass compound b. See packet

19 c. Practice problems: 1. Calculate the mass percent of each element in cholesterol. Its simplest formula is C 27 H 46 O. C = 27(12.0) = g C H = 46(1.0) = 46.0 g H O = 1(16.0) = 16.0 gO g

20 % mass of C = g x 100 = g % mass of H = 46.0 g x 100 = g % mass of O = 16.0 g x 100 = g

21 2. Teflon has the simplest formula CF. How many grams of fluorine are required to make g Teflon?

22 3. Empirical formula from chemical analysis: a. Steps in determining the simplest formula. 1. Determine the mass of each element. 2. Convert each mass to moles. 3. Pick out the smallest number of moles. 4. Divide the number of moles of each element by the smallest number of moles. 5. The answers are usually integers and become subscripts.

23 b. Sample problem: See packet c. Practice problems: 1. A compound is made up of g of sulfur, g of sodium, and g oxygen. What is the empirical formula? 2. A compound is made up of C, Cl and O atoms. It has 12.13% of C and 70.91% of Cl by mass. What is the empirical formula?

24 4. Empirical formula from experimental analysis. See packet b. Practice problems: An unknown compound contains only carbon, hydrogen, and oxygen. If g of the compound is burned with an excess of oxygen, g CO 2 and g H 2 O are produced. What is the empirical formula of the compound?

25 2. A compound is made up of C, Cl, and O atoms. It has 12.13% of C and 70.91% Cl by mass. What is the empirical formula?

26 4. Empirical formula from experimental analysis The result of an experimental analysis to determine the simplest formula of a compound can be given in terms of masses of compounds. You have to determine the mass of the element desired from the given mass of compound obtained in the analysis. All the other steps to empirical formulas are the same.

27 a. Sample problem: See packet b. Practice problem: An unknown compound contains only carbon, hydrogen, and oxygen. If g of the compound is burned with an excess of oxygen, g CO 2 and H 2 O are produced. What is the empirical formula of the compound?

28 5.Molecular Formula and empirical formula To determine the molecular formula from the simplest formula, one has to know the molar mass of the compound. Since the molecular formula is a whole number multiple of the simplest formula, it follows that Multiple = molar mass empirical mass

29 Here the empirical formula mass is the sum of the atomic masses of the elements that make up the empirical formula. The subscripts in the simplest formula are multiplied by the multiple to get the subscripts for the molecular formula. Ex. See packet

30 Practice problem: Determine the molecular formula of ethylene glycol. It has an atom ratio of 1 C : 1 O : 3 H. Its molar mass is g.

31  Reactions 1. Writing and balancing equations A. Steps in writing a balanced equation 1. Determine the correct formulas for all the reactants and products. Include states of matter. 2. Balance the elements one at a time by using coefficients. 3. Make sure all the coefficients are in the lowest possible ratio.

32 B. Thing to remember: 1. Use only coefficients, do not change the subscripts 2. Leave H and O go to last 3. Remember the diatomic molecules ( H O F Br I N Cl) 4. All metals are solids at room temperature except Hg (l). 5. All ionic compounds (+ -) are solids.

33 C. Sample Problem D. Practice problems: 1. Write a balanced equation for the reaction of chromium with sulfur. Cr (s) + S (s)  Cr 2 S 3 2. Write a balanced equation for the reaction of bismuth with oxygen

34 2. Mass Relationship with equations a. Mole-mole relationship 1. The coefficients indicate how many moles of one reactant are required to combine with another reactant. They also show how many moles of product are obtained. 2. These coefficients can be used as conversion factors for the particular equation you are dealing with.

35 3. Ex. See packet 3.Gram-mole relationships Ex. See packet 4.Gram-gram packet Ex. See packet

36 5. Practice problems:  Limiting reactant and theoretical yield 1. Definitions: a. limiting reactant: the reactant that is used up first in the equation b. Theoretical yield: the amount of product that would be formed if the reaction went to completion

37 2. Steps to be followed in determining the limiting reactant of a reaction. a. Convert the amounts to moles or g b. Compare what you calculated to what was given c. If the calculated > given = limiting If the calculated < given = excess

38 3. Sample problems: See packets 4. Practice problems: g of NH 3 reacts with g of H 2 S a. Write and balance the equation. b. What is the limiting reactant?

39 c. What is the theoretical yield of (NH 4 ) 2 S? d. What is the reactant in excess and how much is left over?

40 Percent Yield 1. % yield = actual mass of product x 100 expected mass of product 2. Actual yield: the quantity of a product that is obtained from a chemical reaction 3. See packet

41 4. Practice problems: When g of carbon reacts with hydrogen, 1.67 g of ethane gas (C 2 H 6 ) are produced. a. Write a balanced equation for the reaction. b. What is the theoretical yield?

42 c. What is the percent yield?


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