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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 use carbon-12 as the standard and assign it a mass of exactly 12 amu . B. Exercise:

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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.

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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 47.90 amu.

2. Germanium (Ge) is about six times heavier than carbon. What is its approximate atomic mass? _______________

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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.

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D. Atomic mass and isotopic abundance

1. Mathematical relationship:

(abundance)(amu) + (abundance)(amu) = average atomic mass

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2. Exercise:

a. Antimony occurs in nature as a mixture of two isotopes: The isotope 121Sb has a

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mass of 120.9 amu and an abundance of 57.25%. Antimony has an average atomic mass of 121.75 amu. What is the abundance of the other isotope? What is the atomic mass?

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b. Practice problems: Silicon consists of three stable isotopes whose atomic masses and abundances are as follows:

27.98 amu 92.2 %

28.98 amu 4.7 %

29.97 amu 3.1 %

Calculate the mass of Si.

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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 1023 C atoms in 12.01 g C.

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Can be written as 6.02 x 1023atoms

12.01 g

or 12.01 g C

6.02 x1023 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 12.01 g C = 2.00 x 10-23g

6.02 x 1023 C atoms

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2. Practice Problem: How many atoms are there in 2.483 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

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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?

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4. Relationship of Avogadro’s number to formula masses.

A formula mass in grams has 6.02 x 1023 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.

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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 : (NH4)2SO4

N = 2(14.0) H = 8 (1.0) S = 32.1 O= 4(16.0)

28.0 +8.0 + 32.1 + 64.0 = 132.1 g/mole

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2. Calculate the molar mass of: C4H7NO4

C = 4(12.0) = 48.0

H= 7(1.0) = 7.0

N = 14.0 = 14.0

O = 4(16.0) = 64.0

133.07 g/mole

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3. Mole-gram conversion exercises:

a. How many moles are there in 1.00 g of Al2O3? See packet

b.Practice problems:

1. Find the number of grams in 3874 mole of NH4Cl.

3.874 mole NH4Cl

53.0 g NH4Cl

=

205 g NH4Cl

1 mole NH4Cl

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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. NH4Cl: 1 atom N 4 atoms H

1 atom Cl

b. The mole ratios of the different elements in the compound

Ex. NH4Cl: 1 mole N 4 mole H 1 mole Cl

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c. The mass ratios of the different elements in the compound

Ex. NH4Cl: 14.01 g N, 4(1.01) g H, 35.45 gCl

2. Percent composition from formulas:

a. Formula: %element = total mass element x100

total mass compound

b. See packet

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c. Practice problems:

1. Calculate the mass percent of each element in cholesterol. Its simplest formula is C27H46O.

C = 27(12.0) = 324.0 g C

H = 46(1.0) = 46.0 g H

O = 1(16.0) = 16.0 gO

386.0 g

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% mass of C = 324.0 gx 100 = 83.9

386.0 g

% mass of H = 46.0 gx 100 = 11.9

386.0 g

% mass of O = 16.0 gx 100 = 4.15

386.0 g

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2. Teflon has the simplest formula CF. How many grams of fluorine are required to make 100.0 g Teflon?

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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.

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b. Sample problem: See packet

c. Practice problems:

1. A compound is made up of 11.80 g of sulfur, 16.92 g of sodium, and 23.55 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?

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4. Empirical formula from experimental analysis.

See packet

b. Practice problems: An unknown compound contains only carbon, hydrogen, and oxygen. If 1.500 g of the compound is burned with an excess of oxygen, 1.433 g CO2 and 0.582 g H2O are produced. What is the empirical formula of the compound?

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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?

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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.

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a. Sample problem: See packet

b. Practice problem: An unknown compound contains only carbon, hydrogen, and oxygen. If 1.500 g of the compound is burned with an excess of oxygen, 1.433 g CO2 and 0.582 H2O are produced. What is the empirical formula of the compound?

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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

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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

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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

62.06 g.

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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.

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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.

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C. Sample Problem

D. Practice problems:

1. Write a balanced equation for the reaction of chromium with sulfur.

Cr (s) + S (s)  Cr2S3

2. Write a balanced equation for the reaction of bismuth with oxygen

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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.

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3. Ex. See packet

  • Gram-mole relationships

Ex. See packet

  • Gram-gram packet

Ex. See packet

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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

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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

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3. Sample problems: See packets

4. Practice problems: 3.837 g of NH3 reacts with 4.129 g of H2S

a. Write and balance the equation.

b. What is the limiting reactant?

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c. What is the theoretical yield of (NH4)2S?

d. What is the reactant in excess and how much is left over?

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Percent Yield

1. % yield = actual mass of productx 100

expected mass of product

2. Actual yield: the quantity of a product that is obtained from a chemical reaction

3. See packet

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4. Practice problems:

When 1.417 g of carbon reacts with hydrogen, 1.67 g of ethane gas (C2H6) are produced.

a. Write a balanced equation for the reaction.

b. What is the theoretical yield?

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