lecture 4. stoichiometry (chemical formulas) (1)
DESCRIPTION
CHEM 16TRANSCRIPT
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Stoichiometry of Chemical
Formulas
GENERAL CHEMISTRY
LECTURE 4
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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
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How can we possibly count quantities that are so
small?
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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
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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
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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
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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
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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
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How do we calculate the molar mass of a compound?
Exercise 1
1.The molar mass of propane, C3H8, is:
Determining the Molar Mass
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Calcium Nitrate Ca(NO3)2
2. Calculate the formula weight of calcium nitrate:
Exercise 2 Determining the Molar Mass
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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
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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
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Summary of the mass-mole-number relationships for elements.
Interconverting Moles, Mass, and Number of
Chemical Entities
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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
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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
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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
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7. How many grams of propane contain 1.6510 x 1024 of C3H8 molecules?
SOLUTION:
Calculating the Mass and the Number of Atoms
in a Given Number of Moles of a Compound Exercise 7
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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)
Calculating the Mass and the Number of Atoms
in a Given Number of Moles of a Compound Exercise 8
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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:
Calculating the Mass and the Number of Atoms
in a Given Number of Moles of a Compound Exercise 9
_1 mol Li2CO3 _ 73.8 g Li2CO3
x
=6.49 x 1023 O atoms
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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
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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
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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?
Calculating Mass Percents and Masses of Elements in
a Sample of a Compound Exercise 10
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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
Calculating Mass Percents and Masses of Elements in
a Sample of a Compound Exercise 11
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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
Calculating Mass Percents and Masses of Elements in
a Sample of a Compound Exercise 11
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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
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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
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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.
Determining the Empirical Formula from Masses
of Elements
Exercise 13
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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.
Determining the Empirical Formula from Masses
of Elements
Exercise 13
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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.
Determining the Empirical Formula from Masses
of Elements
Exercise 13
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Determining the Empirical Formula from Masses
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
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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
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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
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Combustion apparatus for determining formulas of organic compounds.
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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
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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
Determining the Molecular Formula from
Combustion Analysis
Exercise 16
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Determining the Molecular Formula from
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?
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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
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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?
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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)