chapter 3 topics

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© 2006 Brooks/Cole - Thomson

Kull Spring 2007 Chem 105 Lsn 5

Chapter 3Topics

• Announcements– All homework quizzes completed on line

– Extension for chapter 2 until Jan 28, 8 pm to allow for minor computer issues to be resolved

– Monday is a review day

• Topics– Using percent composition

» Emperical and molecular formulas

– Calculating a formula

– Formula of a compound from combining masses

– Formula from mass data

– Hydrated compounds

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Using percent compositionEmperical and molecular

formulasA pure compound always consists of the A pure compound always consists of the

same elements combined in the same same elements combined in the same proportions by weight.proportions by weight.

Therefore, we can express molecular Therefore, we can express molecular composition as composition as PERCENT BY PERCENT BY WEIGHTWEIGHT

Ethanol, CEthanol, C22HH66OO

52.13% C52.13% C13.15% H 13.15% H 34.72% O34.72% O

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Percent CompositionPercent CompositionPercent CompositionPercent CompositionConsider NOConsider NO22, Molar mass = ?, Molar mass = ?

What is the weight percent of N and of What is the weight percent of N and of O?O?

Wt. % O 2 (16 .0 g O per mole )46 .0 g

x 100 % 69 .6%Wt. % O 2 (16 .0 g O per mole )46 .0 g

x 100 % 69 .6%

Wt. % N = 14.0 g N

46.0 g NO2 • 100% = 30.4 %Wt. % N =

14.0 g N46.0 g NO2

• 100% = 30.4 %

What are the weight percentages of What are the weight percentages of N and O in NO?N and O in NO?

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Calculating a formulaCalculating a formula

In In chemical analysischemical analysis we determine we determine the % by weight of each element in a given the % by weight of each element in a given amount of pure compound and derive the amount of pure compound and derive the

EMPIRICALEMPIRICAL or or SIMPLESTSIMPLEST formula.formula.

PROBLEMPROBLEM: A compound of B : A compound of B and H is 81.10% B. What is its and H is 81.10% B. What is its empirical formula?empirical formula?

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• Because it contains only B and H, it Because it contains only B and H, it must contain 18.90% H.must contain 18.90% H.

• In 100.0 g of the compound there are In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H.81.10 g of B and 18.90 g of H.

• Calculate the number of moles of each Calculate the number of moles of each constitutent.constitutent.

A compound of B and H is 81.10% B. A compound of B and H is 81.10% B. What is its empirical formula?What is its empirical formula?A compound of B and H is 81.10% B. A compound of B and H is 81.10% B. What is its empirical formula?What is its empirical formula?

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Calculate the number of moles of each Calculate the number of moles of each element in 100.0 g of sample.element in 100.0 g of sample.

81.10 g B • 1 mol

10.81 g = 7.502 mol B81.10 g B •

1 mol10.81 g

= 7.502 mol B

18.90 g H • 1 mol

1.008 g = 18.75 mol H18.90 g H •

1 mol1.008 g

= 18.75 mol H


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Now, recognize that Now, recognize that atoms combine in atoms combine in the ratio of small whole numbers.the ratio of small whole numbers.

1 atom B + 3 atoms H --> 1 molecule BH1 atom B + 3 atoms H --> 1 molecule BH33

oror

1 mol B atoms + 3 mol H atoms ---> 1 mol B atoms + 3 mol H atoms ---> 1 mol BH1 mol BH33 molecules molecules

Find the ratio of moles of elements Find the ratio of moles of elements in the compound.in the compound.


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But we need a whole number ratio. But we need a whole number ratio.

2.5 mol H/1.0 mol B = 5 mol H to 2 mol B2.5 mol H/1.0 mol B = 5 mol H to 2 mol B

EMPIRICAL FORMULA = BEMPIRICAL FORMULA = B22HH55

Take the ratio of moles of B and H. Take the ratio of moles of B and H. AlwaysAlwaysdivide by the smaller number.divide by the smaller number.

18.75 mol H7.502 mol B

= 2.499 mol H1.000 mol B

= 2.5 mol H1.0 mol B

18.75 mol H7.502 mol B

= 2.499 mol H1.000 mol B

= 2.5 mol H1.0 mol B


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A compound of B and H is 81.10% B. A compound of B and H is 81.10% B. Its Its empirical formulaempirical formula is B is B22HH55. .

What is its What is its molecular molecular formulaformula??Is the molecular formula BIs the molecular formula B22HH55, B, B44HH1010, ,

BB66HH1515, B, B88HH2020, etc.? , etc.?

BB22HH66 is one example of this class of compounds. is one example of this class of compounds.

B2H6

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A compound of B and H is 81.10% B. Its empirical A compound of B and H is 81.10% B. Its empirical

formula is Bformula is B22HH55. What is its molecular formula. What is its molecular formula??A compound of B and H is 81.10% B. Its empirical A compound of B and H is 81.10% B. Its empirical

formula is Bformula is B22HH55. What is its molecular formula. What is its molecular formula??

We need to do an We need to do an EXPERIMENTEXPERIMENT to to find the MOLAR MASS.find the MOLAR MASS.

Here experiment gives Here experiment gives 53.3 g/mol53.3 g/molCompare with the mass of BCompare with the mass of B22HH55

= = 26.66 g/unit26.66 g/unit

Find the ratio of these masses.Find the ratio of these masses.

53.3 g/mol26.66 g/unit of B2H5

= 2 units of B2H5

1 mol

53.3 g/mol26.66 g/unit of B2H5

= 2 units of B2H5

1 mol

Molecular formula = BMolecular formula = B44HH1010

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Formula of a compound from combining masses

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DETERMINE THE FORMULA OF DETERMINE THE FORMULA OF A COMPOUND OF Sn AND IA COMPOUND OF Sn AND I

DETERMINE THE FORMULA OF DETERMINE THE FORMULA OF A COMPOUND OF Sn AND IA COMPOUND OF Sn AND I

Sn(s) + some ISn(s) + some I22(s) ---> (s) ---> SnISnIxx

44++, 2, 2++ 1 1--

Formula of a compound from combining masses

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Data to Determine the Data to Determine the formula of a Sn—I Compoundformula of a Sn—I Compound

Data to Determine the Data to Determine the formula of a Sn—I Compoundformula of a Sn—I Compound

• Reaction of Sn and IReaction of Sn and I22 is done using excess Sn. is done using excess Sn.

• Mass of Sn in the beginning = 1.056 gMass of Sn in the beginning = 1.056 g

• Mass of Sn remaining (recovered) = 0.601 gMass of Sn remaining (recovered) = 0.601 g

• Mass of iodine (IMass of iodine (I22) used = 1.947 g) used = 1.947 g

(See p. 125)(See p. 125)

Convert these masses to molesConvert these masses to moles

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• Reaction of Sn and IReaction of Sn and I22 is done using excess Sn. is done using excess Sn.

Mass of iodine (IMass of iodine (I22) used = 1.947 g) used = 1.947 g

Mass of Sn initially = 1.056 gMass of Sn initially = 1.056 g

Mass of Sn recovered = 0.601 gMass of Sn recovered = 0.601 g

Mass of Sn used = 0.455 gMass of Sn used = 0.455 g

Find the mass of Sn that combined with 1.947 g IFind the mass of Sn that combined with 1.947 g I22..

Find moles of Sn used:Find moles of Sn used:

0.455 g Sn • 1 mol

118.7 g = 3.83 x 10-3 mol Sn0.455 g Sn •

1 mol118.7 g

= 3.83 x 10-3 mol Sn

Tin and Iodine Tin and Iodine CompoundCompound

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Now find the number of moles of INow find the number of moles of I22 that that combined with 3.83 x 10combined with 3.83 x 10-3-3 mol Sn. Mass mol Sn. Mass of Iof I22 used was 1.947 g. used was 1.947 g.

1.947 g I2 • 1 mol

253.81 g = 7.671 x 10-3 mol I21.947 g I2 •

1 mol253.81 g

= 7.671 x 10-3 mol I2

How many mol of How many mol of iodine atomsiodine atoms? ?

= 1.534 x 10-2 mol I atoms= 1.534 x 10-2 mol I atoms

7.671 x 10-3 mol I2 2 mol I atoms

1 mol I2

7.671 x 10-3 mol I2 2 mol I atoms

1 mol I2

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Now find the ratio of number of moles of moles Now find the ratio of number of moles of moles of I and Sn that combined.of I and Sn that combined.

1.534 x 10-2 mol I

3.83 x 10-3 mol Sn =

4.01 mol I1.00 mol Sn

1.534 x 10-2 mol I

3.83 x 10-3 mol Sn =

4.01 mol I1.00 mol Sn

Empirical formula is Empirical formula is

SnISnI44

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Formula from mass data

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Hydrated compounds

chapter 3 topics

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