Lecture 4. Stoichiometry (Chemical Formulas) (1)

Stoichiometry of Chemical

Formulas

GENERAL CHEMISTRY

LECTURE 4

4.5 Determining the Formula of an Unknown Compound

4.1 The Mole and the Avogadros number

4.2 Determining the Molar Mass of Compounds

4.3 Interconverting Moles, Mass and other chemical entities

4.4 Determining the Mass Percent from Chemical Formula

Lesson for Today

How can we possibly count quantities that are so

small?

Mole (mol) - the amount of a substance that contains

the same number of entities as there are atoms in

exactly 12 g of carbon-12.

This amount is 6.022 x 1023. The number is called

Avogadros number and is represented by N or NA.

One mole (1 mol) contains 6.022 x 1023

entities (to four significant figures)

The Mole Concept

Example: H has an atomic weight of 1.00794 g

1.00794 g of H atoms = 6.022 x 1023 H atoms Mg has an atomic weight of 24.3050 g

24.3050 g of Mg atoms = 6.022 x 1023 Mg atoms

The Atomic Mass Scale

1 amu = 1 g / mole

The atomic weight is the weighted average of the masses of its stable isotopes

24Mg: 78.99% 25Mg: 10.00% 26Mg: 11.01%

The Atomic Weight Scale

Mass in grams numerically equal to the atomic weight of the element in grams.

For elements atomic mass

For compounds sum of the molar masses of the atoms of the elements in the formula

Unit: g/mol

The Molar Mass

Water 18.02 g

CaCO3 100.09 g

Oxygen 32.00 g

Copper 63.55 g

One mole of some familiar substances.

The Molar Mass

How do we calculate the molar mass of a compound?

Exercise 1

1.The molar mass of propane, C3H8, is:

Determining the Molar Mass

Calcium Nitrate Ca(NO3)2

2. Calculate the formula weight of calcium nitrate:

Exercise 2 Determining the Molar Mass

3. Calculate the MW of the following:

a. Sodium hydroxide

b. Glucose, C6H12O6 c. Barium phosphate

d. Silver Nitrate

e. Aluminum chloride

NaOH: 39.997 or 40 g/mol

180.16 g/mol

Ba3(PO4)2 : 601.93 g/mol

AgNO3 : 169.87 g/mol

AlCl3 : 133.34 g/mol

Exercise 3 Determining the Molar Mass

Mass (g) = no. of moles x no. of grams

1 mol

No. of moles = mass (g) x no. of grams

1 mol

No. of entities = no. of moles x 6.022 x 1023 entities

1 mol

No. of moles = no. of entities x 6.022 x 1023 entities

1 mol

g

mol

Interconverting Moles, Mass, and Number of

Chemical Entities

Atom/

molecules

Summary of the mass-mole-number relationships for elements.

Interconverting Moles, Mass, and Number of

Chemical Entities

4. Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342 mol of Ag?

SOLUTION:

a) 0.0342 mol Ag x mol Ag

107.9 g Ag 3.69 g Ag =

Calculating the Mass and the Number of Atoms in

a Given Number of Moles of an Element

Exercise 4

5. Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8 g of Fe?

SOLUTION:

95.8 g Fe x 55.85 g Fe 1 mol Fe 6.022 x1023 atoms Fe

mol Fe = 1.04 x1024 atoms Fe x

Calculating the Mass and the Number of Atoms in

a Given Number of Moles of an Element Exercise 5

Calculating the Mass and the Number of Atoms

in a Given Number of Moles of a Compound

6. Calculate the number of C3H8 molecules in 74.6 g of propane.

SOLUTION:

Exercise 6

7. How many grams of propane contain 1.6510 x 1024 of C3H8 molecules?

SOLUTION:


in a Given Number of Moles of a Compound Exercise 7

8. How many (a) moles, (b) molecules, and (c) oxygen atoms are contained in 60.0 g of ozone, O3? The layer of ozone in the stratosphere is very beneficial to life on earth.

SOLUTION:

(a)

(b)

(c)



9. Calculate the number of O atoms in 26.5 g of Li2CO3

26.5 g Li2CO3 __3 O mole__ 1 mole Li2CO3

x 6.02 x 1023 O molecules 1 O mole

x

SOLUTION:



_1 mol Li2CO3 _ 73.8 g Li2CO3

x

=6.49 x 1023 O atoms

Mass % of element X =

moles of X in formula x molar mass of X (g/mol)

mass (g) of 1 mol of compound x 100

Mass Percent from the Chemical Formula

Calculating Mass Percents and Masses of Elements in

a Sample of a Compound

10. In mammals, lactose (milk sugar) is metabolized to glucose (C6H12O6), the key nutrient for generating chemical potential energy.

(a) What is the mass percent of each element in glucose?

(b) How many grams of carbon are in 16.55 g of glucose?

SOLUTION: 6 mol C x 12.01 g C

mol C = 72.06 g C

12 mol H x 1.008 g H mol H

= 12.096 g H

6 mol O x 16.00 g O

mol O = 96.00 g O

M = 180.16 g/mol

Exercise 10

mass percent of C = 72.06 g C

180.16 g glucose = 0.4000 x 100 = 40.00 mass % C

mass percent of H = 12.096 g H

180.16 g glucose = 0.06714 x 100 = 6.714 mass % H

mass percent of O = 96.00 g O

180.16 g glucose = 0.5329 x 100 = 53.29 mass % O

16.55 g C6H12O6 x 0.4000 = 6.620 g C

(b) How many grams of carbon are in 16.55 g of glucose?


a Sample of a Compound Exercise 10

11. Which of the following compounds has the highest percentage of N (AW 14.00)?

a. Ca(NO3)2 (MW 164.03)

b. N2O5 (MW 100.02)

c. (NH4)2SO4 (MW 132.16)

d. C3H7NH2 (MW 59.4)

a. Ca(NO3)2

2 mol N (14.00 g/mol)= (28.00 g N/164.03 g) x 100

= 17.07% N



b. N2O5 (MW 100.02)

2 mol N (14.00 g/mol)= (28.00 g N/100.02 g) x 100

= 27.99 % N

c. (NH4)2SO4 (MW 132.16)

2 mol N (14.00 g/mol)= (28.00 g N/132.16 g) x 100

= 21.19 % N

d. C3H7NH2 (MW 59.4)

1 mol N (14.00 g/mol)= (14.00 g N/59.4 g) x 100

= 23.57 % N



The simplest formula for a compound that agrees with the

elemental analysis and gives rise to the smallest set of whole

numbers of atoms.

The formula of the compound as it exists; it may be a

multiple of the empirical formula.

Empirical and Chemical Formula

Empirical Formula

Molecular Formula

12. Elemental analysis of a sample of an ionic compound showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What is the empirical formula and name of the compound?

Determining the Empirical Formula from Masses

of Elements

SOLUTION:

2.82 g Na x mol Na

22.99 g Na = 0.123 mol Na

4.35 g Cl x mol Cl

35.45 g Cl = 0.123 mol Cl

7.83 g O x mol O

16.00 g O = 0.489 mol O

Na1.00 Cl1.00 O3.98 NaClO4

NaClO4 is sodium perchlorate.

Exercise 12

13. During physical activity, lactic acid (M = 90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental analysis shows that this compound contains 40.0 mass % C, 6.71 mass % H, and 53.3 mass % O.

(a) Determine the empirical formula of lactic acid.

(b) Determine the molecular formula.

PLAN:

a) Assume 100 g of lactic acid and find the mass of each element. Convert mass of each to moles, get a ratio and convert to integer subscripts.

b) Divide molar mass by empirical mass to get the multiplier then write the molecular formula accordingly.


of Elements

Exercise 13

SOLUTION: In 100.0 g of lactic acid, there are: 40.0 g C; 6.71 g H; 53.3 g O

40.0 g C x

6.71 g H x

53.3 g O x

mol C

12.01 g C

mol H

1.008 g H

mol O

16.00 g O

= 3.33 mol C

= 6.66 mol H

= 3.33 mol O

We convert the grams to moles and get a ratio for the empirical formula.


of Elements

Exercise 13

SOLUTION:

3.33 mol C 6.66 mol H 3.33 mol O

C3.33 H6.66 O3.33

3.33 3.33 3.33 CH2O empirical formula

mass of CH2O

molar mass of lactate 90.08 g

30.03 g 3

C3H6O3 is the

molecular formula

We now divide by the smallest number and get a ratio for the empirical formula.


of Elements

Exercise 13


of Elements

Exercise 14

14. Determine the empirical and molecular formula of a compound that contains 71.65 % Cl, 24.27 % C and 4.07 % H. The molecular weight of the compound is known to be 98.96 g/mol.

Answer: Emprical formula: ClCH2 Molecular formula: Cl2C2H4

Determining the Molecular Formula from Mass Percentage

15. Cortisol (MW=362.46 g/mol), one of the major steroid hormones, is used in the treatment of rheumatoid athritis. Cortisol is 69.6% C, 8.34% H, 22.1 % O by mass. What is the molecular formula?

69.6 g C x 1 mol C

12.01 g C = 5.795 mol C

8.34 g H x 1 mol H

1.008 g H = 8.275 mol H

1. Assuming 100 g of sample

1 mol O

16.00 g O

22.1 g O x =1.381 mol O

C4.20

H5.99

O1.00

~ 4.00 ?

~ 6.00 ok!

C4.2H6O 362.46 g/mol

72.49 g = 5.0 C21H30O5

Exercise 15

2. Using mass percentage

362. 46 g/mol x 0.696 = 252. 27 g C x 1 mol C

12.01 g C = 21 mol C

15. Cortisol (MW=362.46 g/mol), one of the major steroid hormones, is used in the treatment of rheumatoid athritis. Cortisol is 69.6% C, 8.34% H, 22.1 % O by mass. What is the molecular formula?

362. 46 g/mol x 0.0834 = 30.229 g H x 1 mol H

1.008 g H = 30 mol H

362. 46 g/mol x 0.221 = 80.104 g H x 1 mol O

16.0 g O = 5 mol O

Molecular formula: C21H30O5

Determining the Molecular Formula from Mass Percentage Exercise 15

Combustion apparatus for determining formulas of organic compounds.

PLAN:

16. Vitamin C (M = 176.12 g/mol) is a compound of C,H, and O found in many natural sources, especially citrus fruits. When a 1.000-g sample of vitamin C is placed in a combustion chamber and burned, the following data are obtained:

mass of CO2 absorber after combustion = 85.35 g

mass of CO2 absorber before combustion = 83.85 g

mass of H2O absorber after combustion = 37.96 g

mass of H2O absorber before combustion = 37.55 g

What is the molecular formula of vitamin C?

The difference in absorber mass before and after combustion is the mass of oxidation product of the element. Find the mass of each element from its combustion product, convert each to moles and determine the formula. Using the molar mass and empirical mass, determine the molecular formula.

Determining the Molecular Formula from

Combustion Analysis

Exercise 16

SOLUTION:

CO2 85.35 g - 83.85 g = 1.50 g H2O 37.96 g - 37.55 g = 0.41 g

There are 12.01 g C per mol CO2. 1.50 g CO2 x 12.01 g C

44.01 g CO2 = 0.409 g C

0.41 g H2O x 2.016 g H

18.02 g H2O

= 0.046 g H There are 2.016 g H per mol H2O.

O must be the difference: 1.000 g - (0.409 + 0.046) = 0.545 g O

0.409 g C

12.01 g C

0.046 g H

1.008 g H

0.545 g O

16.00 g O

= 0.0341 mol C = 0.0456 mol H = 0.0341 mol O

C1.00H1.3O1.00 C3H4O3 176.12 g/mol

88.06 g = 2.000 = 2 C6H8O6


Combustion Analysis

Exercise 16


Combustion Analysis

Exercise 17

16. Menthol (MW =156.3 g/mol), a strong smelling substance using in cough drops , is a compound of carbon, hydrogen and oxygen. When 0.1595 g of menthol was subjected to combustion analysis, it produced 0.449 g CO2 and 0.184 g H2O. What is menthols molecular formula?

Quiz 4 June 26, 2012; Total points: 13/13

1. Calculate each of the following quantities: 2 pts each

a. Mass in grams of 6.44 x10-2 mol of MnSO4 b. Number of moles in 15.8 kg Fe(ClO4)3 c. Number of N atoms in 92.6 mg NH4NO2

2. Ethyl acetate, the compound responsible for the characteristic

smell of plastic balloon, has an empirical formula of C2H4O. If the

molar mass is about 88 g/mole, give the molecular formula of ethyl

acetate. (AW C = 12.01, H = 1.01, O = 16.00) 2 pts

3. Determine the molecular formula of a compound that contains

26.7% P, 12.1% N and 61.2 % Cl and has a molar mass of 580

g/mol. 5 pts

Additional Exercises: Stoichiometry of Chemical Formulas

1. An analytical balance can detect a mass of 0.1 mg. What is the total number of ions present in this minimally detectable quantity of MgCl2?

Additional Exercises: Stoichiometry of Chemical Formulas

2. The volatile liquid ethyl mercaptan, C2H6S, is one of the most odoriferous substances known. It is used in natural gas to make gas leaks detectable. How many C2H6S molecules are contained in a 1.0 L sample? (d = 0.84 g/mL)

Lecture 4. Stoichiometry (Chemical Formulas) (1)

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