Mass Mole Relationships of a Compound

1
Mass - Mole Relationships of a Compound
For an Element
For a Compound
Mass (g) of Element
Mass (g) of compound
Moles of Element
Amount (mol) of compound
Amount (mol) of compound
Molecules (or formula units of compound)
Atoms of Element
2
Calculating the Number of Moles and Atoms in a
Given Mass of Element
Problem Tungsten (W) is the element used as the
filament in light bulbs, and
has the highest melting point of any element
3680oC. How many moles of
tungsten, and atoms of the
element are contained in a 35.0 mg sample of the
metal? Plan Convert mass into moles by dividing
the mass by the atomic weight of the
metal, then calculate the number of atoms by
multiplying by Avogadros
number! Solution Converting from mass of W to
moles Moles of W 35.0
mg W x 0.00019032
mol
1.90 x
10 - 4 mol NO. of W atoms 1.90 x 10 - 4 mol
W x


1.15 x 1020 atoms of Tungsten
1 mol W 183.9 g W
6.022 x 1023 atoms 1 mole of W
3
Calculating the Moles and Number of Formula
Units in a Given Mass of Cpd.
Problem Sodium Phosphate is a component of some
detergents. How many moles and
formula units are in a 38.6 g sample? Plan We
need to determine the formula, and the molecular
mass from the atomic masses of each
element multiplied by the coefficients. Solution
The formula is Na3PO4. Calculating the molar
mass M 3x Sodium 1 x
Phosphorous 4 x Oxygen
3 x 22.99 g/mol 1 x 30.97 g/mol 4 x 16.00
g/mol 68.97 g/mol 30.97
g/mol 64.00 g/mol 163.94 g/mol
Converting mass to moles
Moles Na3PO4 38.6 g Na3PO4 x (1 mol Na3PO4)

163.94 g Na3PO4 0.235 mol
Na3PO4
Formula units 0.23545 mol Na3PO4 x 6.022 x 1023
formula units
1 mol Na3PO4
1.42 x 1023 formula units
4

• The average daily requirement of the essential
  amino acid leucine, C6H14O2N, is 2.2 g for an
  adult. What is the average daily requirement of
  leucine in moles?

First, find the molar mass of leucine 6
C 6(12.01) 72.06 2 O 2(16.00) 32.00 1
N 1(14.01) 14.01 14 H 14(1.008) 14.112
2 decimal places 132.18 g/mol
132.182
5
Next, find the number of moles in 2.2 g
6
Flow Chart of Mass Percentage Calculation
Moles of X in one mole of Compound
M (g / mol) of X
Mass (g) of X in one mole of compound
Divide by mass (g) of one mole
of compound
Mass fraction of X
Multiply by 100
Mass of X
7
Calculating Mass Percentage and Masses of
Elements in a Sample of a Compound - I
Problem Sucrose (C12H22O11) is common table
sugar. ( a) What is the mass percent of each
element in sucrose? ( b) How many grams of
carbon are in 24.35 g of sucrose?

(a) Determining the mass percent of each
element mass of C 12 x 12.01
g C/mol 144.12 g C/mol
mass of H 22 x 1.008 g H/mol
22.176 g H/mol mass of O 11 x
16.00 g O/mol 176.00 g O/mol

342.296 g/mol Finding
the mass fraction of C in Sucrose C
Total mass
of C 144.12 g C
mass of 1 mole of sucrose
342.30 g Cpd
Mass Fraction of C

0.421046 To find mass of C 0.421046 x
100 42.105
8
Calculating Mass Percents and Masses of Elements
in a Sample of Compound - II
(a) continued
Mass of H
x 100 x 100
6.479 H Mass of O
x 100
x 100
51.417 O (b) Determining the
mass of carbon Mass (g) of C mass of
sucrose X( mass fraction of C in sucrose) Mass
(g) of C 24.35 g sucrose X
10.25 g C
mol H x M of H 22 x
1.008 g H mass of 1 mol sucrose
342.30 g
mol O x M of O 11 x
16.00 g O mass of 1 mol sucrose
342.30 g
0.421046 g C 1 g sucrose
9
Chemical Formulas
Empirical Formula - Shows the relative number of
atoms of each element in the compound.
It is the simplest formula, and is
derived from masses of the elements. Molecular
Formula - Shows the actual number of atoms
of each element in the molecule of the
compound. Structural Formula - Shows the actual
number of atoms, and the bonds between
them that is, the arrangement of
atoms in the molecule.
10
Empirical and Molecular Formulas
Empirical Formula - The simplest formula for a
compound that agrees
with the elemental analysis! The
smallest set of whole numbers of
atoms. Molecular Formula - The formula of the
compound as it exists,
it may be a multiple of the Empirical
formula.
11
Steps to Determine Empirical Formulas
Mass (g) of Element
M (g/mol )
Moles of Element
use no. of moles as subscripts
Preliminary Formula
change to integer subscripts
Empirical Formula
12
Some Examples of Compounds with the Same
Elemental Ratios
Empirical Formula
Molecular Formula
CH2(unsaturated Hydrocarbons) C2H4 ,
C3H6 , C4H8 OH or HO
H2O2 S

S8 P
P4
Cl
Cl2 CH2O
(carbohydrates)
C6H12O6
13

• Percentage Composition
• The mass percentage of each element in the
  compound
• The composition is determined by experiment,
  often by combustion. When a compound is burned,
  its component elements form oxidesfor example,
  CO2 and H2O. The CO2 and H2O are captured and
  weighed to determine the amount of C and H in the
  original compound.

14

• Benzene is a liquid compound composed of carbon
  and hydrogen it is used in the preparation of
  polystyrene plastic. A sample of benzene weighing
  342 mg is burned in oxygen and forms 1158 mg of
  carbon dioxide. What is the percentage
  composition of benzene?

15
Strategy 1. Use the mass of CO2 to find the mass
of carbon from the benzene. 2. Use the mass of
benzene and the mass of carbon to find the mass
of hydrogen. 3. Use these two masses to find the
percent composition.
16
First, find the mass of C in 1156 mg of CO2
315.5 mg C
17
Next, find the mass of H in the benzene sample
342 mg benzene -315.5 mg C 26.5 mg H (the
decimal is not significant)

• Now, we can find the percentage composition

18

• Empirical Formula (Simplest Formula)
• The formula of a substance written with the
  smallest integer subscripts
• For example
• The empirical formula for N2O4 is NO2.
• The empirical formula for H2O2 is HO

19

• Determining the Empirical Formula
• Beginning with percent composition
• Assume exactly 100 g so percentages convert
  directly to grams.
• Convert grams to moles for each element.
• Manipulate the resulting mole ratios to obtain
  whole numbers.

20

• Manipulating the ratios
• Divide each mole amount by the smallest mole
  amount.
• If the result is not a whole number
• Multiply each mole amount by a factor.
• For example
• If the decimal portion is 0.5, multiply by 2.
• If the decimal portion is 0.33 or 0.67, multiply
  by 3.
• If the decimal portion is 0.25 or 0.75, multiply
  by 4.

21

• Benzene is composed of 92.3 carbon and 7.7
  hydrogen. What is the empirical formula of
  benzene?

Empirical formula CH
22

• Molecular Formula
• A formula for a molecule in which the subscripts
  are whole-number multiples of the subscripts in
  the empirical formula

23

• To determine the molecular formula
• Compute the empirical formula weight.
• Find the ration of the molecular weight to the
  empirical formula weight.
• Multiply each subscript of the empirical formula
  by n.

24

• Benzene has the empirical formula CH. Its
  molecular weight is 78.1 amu. What is its
  molecular formula?

Molecular formula C6H6
25
Determining Empirical Formulas from
Masses of Elements - I
Problem The elemental analysis of a sample
compound gave the following results 5.677g Na,
6.420 g Cr, and 7.902 g O. What is the empirical
formula and name of the compound? Plan First we
have to convert mass of the elements to moles of
the elements using the molar masses. Then we
construct a preliminary formula and name of the
compound. Solution Finding the moles of the
elements Moles of Na 5.678 g Na x
Moles of Cr
6.420 g Cr x
Moles of O 7.902 g O x
1 mol Na 22.99 g Na
1 mol Cr 52.00 g Cr
1 mol O 16.00 g O
26
Determining Empirical Formulas from
Masses of Elements - I
Problem The elemental analysis of a sample
compound gave the following results 5.677g Na,
6.420 g Cr, and 7.902 g O. What is the empirical
formula and name of the compound? Plan First we
have to convert mass of the elements to moles of
the elements using the molar masses. Then we
construct a preliminary formula and name of the
compound. Solution Finding the moles of the
elements Moles of Na 5.678 g Na x
0.2469 mol Na
Moles of Cr 6.420 g Cr x
0.12347 mol Cr Moles of O 7.902 g
O x 0.4939 mol O
1 mol Na 22.99 g Na
1 mol Cr 52.00 g Cr
1 mol O 16.00 g O
27
Determining Empirical Formulas from
Masses of Elements - II
Constructing the preliminary formula
Na0.2469 Cr0.1235 O0.4939
Converting to integer subscripts (dividing all by
smallest subscript)
Na1.99 Cr1.00 O4.02
Rounding off to whole numbers
Na2CrO4 Sodium Chromate
28
Determining the Molecular Formula from
Elemental Composition and Molar Mass - I
Problem The sugar burned for energy in cells of
the body is Glucose (M 180.16 g/mol), elemental
analysis shows that it contains 40.00 mass C,
6.719 mass H, and 53.27 mass O. (a)
Determine the empirical formula of glucose.
(b) Determine the molecular formula. Plan We are
only given mass , and no weight of the compound
so we will assume 100g of the compound,
and becomes grams, and we can do as
done previously with masses of the
elements. Solution Mass Carbon
40.00 x 100g/100 40.00 g C Mass
Hydrogen 6.719 x 100g/100 6.719g H
Mass Oxygen 53.27 x 100g/100 53.27 g
O
99.989 g Cpd
29
Determining the Molecular Formula from
Elemental Composition and Molar Mass - II
Converting from Grams of Elements to moles
Moles of C Mass of C x
3.3306 moles C Moles of H Mass of H x
6.6657 moles H Moles
of O Mass of O x 3.3294
moles O Constructing the preliminary formula
C 3.33 H 6.67 O 3.33 Converting to integer
subscripts, divide all subscripts by the
smallest C 3.33/3.33 H 6.667 / 3.33 O3.33 /
3.33 CH2O
1 mole C 12.01 g C
1 mol H 1.008 g H
1 mol O 16.00 g O
30
Determining the Molecular Formula
from Elemental Composition and Molar Mass - III
(b) Determining the Molecular Formula The
formula weight of the empirical formula is
1 x C 2 x H 1 x O 1 x 12.01 2 x 1.008 1
x 16.00 30.03
M of Glucose empirical formula mass
Whole-number multiple

6.00 6
180.16 30.03
Therefore the Molecular Formula is
C 1 x 6 H 2 x 6 O 1 x 6 C6H12O6
31
Adrenaline Is a Very Important Compound in the
Body - I

• Analysis gives
• C 56.8
• H 6.50
• O 28.4
• N 8.28
• Calculate the Empirical Formula

32
Adrenaline - II

• Assume 100g!
• C 56.8 g C/(12.01 g C/ mol C) 4.73 mol C
• H 6.50 g H/( 1.008 g H / mol H) 6.45 mol H
• O 28.4 g O/(16.00 g O/ mol O) 1.78 mol O
• N 8.28 g N/(14.01 g N/ mol N) 0.591 mol N
• Divide by 0.591
• C 8.00 mol C 8.0 mol C or
• H 10.9 mol H 11.0 mol H
• O 3.01 mol O 3.0 mol O C8H11O3N
• N 1.00 mol N 1.0 mol N

33
Combustion Train for the Determination of the
Chemical Composition of Organic Compounds.
m 2
m 2
CnHm (n ) O2 n CO(g) H2O(g)
Fig. 3.4
34
Ascorbic Acid ( Vitamin C ) - I Contains C , H ,
and O

• Upon combustion in excess oxygen, a 6.49 mg
  sample yielded 9.74 mg CO2 and 2.64 mg H2O
• Calculate its Empirical formula!
• C 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2)
• 2.65 x 10-3 g C
• H 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O)
• 2.92 x 10-4 g H
• Mass Oxygen 6.49 mg - 2.65 mg - 0.30 mg
• 3.54 mg O

35
Vitamin C Combustion - II

• C 2.65 x 10-3 g C / ( 12.01 g C / mol C )
• 2.21 x 10-4 mol C
• H 0.295 x 10-3 g H / ( 1.008 g H / mol H )
• 2.92 x 10-4 mol H
• O 3.54 x 10-3 g O / ( 16.00 g O / mol O )
• 2.21 x 10-4 mol O
• Divide each by 2.21 x 10-4
• C 1.00 Multiply each by 3 3.00 3.0
• H 1.32
  3.96 4.0
• O 1.00
  3.00 3.0

C3H4O3
36
Determining a Chemical Formula from
Combustion Analysis - I
Problem Erthrose (M 120 g/mol) is an
important chemical compound as
a starting material in chemical synthesis, and
contains Carbon Hydrogen, and
Oxygen. Combustion analysis of
a 700.0 mg sample yielded 1.027 g CO2 and
0.4194 g H2O. Plan We find the masses
of Hydrogen and Carbon using the mass
fractions of H in H2O, and C in CO2. The mass of
Carbon and Hydrogen are subtracted from
the sample mass to get the mass of
Oxygen. We then calculate moles, and construct
the empirical formula, and from the
given molar mass we can calculate the
molecular formula.
37
Determining a Chemical Formula from Combustion
Analysis - II
Calculating the mass fractions of the elements
Mass fraction of C in CO2

0.2729 g C
/ 1 g CO2 Mass fraction of H in H2O


0.1119 g H / 1 g H2O Calculating masses of C
and H Mass of Element mass of compound x
mass fraction of element
mol C x M of C mass of 1 mol CO2
1 mol C x 12.01 g C/ 1 mol C 44.01 g
CO2
mol H x M of H mass of 1 mol H2O
2 mol H x 1.008 g H / 1 mol H 18.02
g H2O
38
Determining a Chemical Formula from
Combustion Analysis - III
0.2729 g C 1 g CO2
Mass (g) of C 1.027 g CO2 x
0.2803 g C Mass (g) of H 0.4194 g H2O x
0.04693 g H Calculating
the mass of O Mass (g) of O Sample mass -(
mass of C mass of H )
0.700 g - 0.2803 g C - 0.04693 g H 0.37277 g
O Calculating moles of each element C
0.2803 g C / 12.01 g C/ mol C 0.02334 mol C
H 0.04693 g H / 1.008 g H / mol H 0.04656 mol
H O 0.37277 g O / 16.00 g O / mol O
0.02330 mol O C0.02334H0.04656O0.02330 CH2O
formula weight 30 g / formula 120 g /mol / 30 g
/ formula 4 formula units / cpd C4H8O4
0.1119 g H 1 g H2O
39
Some Compounds with Empirical Formula CH2O
(Composition by Mass 40.0 C, 6.71 H, 53.3O)
Molecular M Formula
(g/mol) Name Use or Function
CH2O 30.03 Formaldehyde
Disinfectant Biological

preservative C2H4O2 60.05
Acetic acid Acetate polymers vinegar

( 5 solution) C3H6O3
90.08 Lactic acid Causes milk
to sour forms
in muscle
during exercise C4H8O4 120.10
Erythrose Forms during sugar

metabolism C5H10O5
150.13 Ribose Component of
many nucleic
acids and
vitamin B2 C6H12O6 180.16
Glucose Major nutrient for energy

in cells
40
Two Compounds with Molecular Formula C2H6O
Property Ethanol
Dimethyl Ether
M (g/mol) 46.07
46.07 Color
Colorless
Colorless Melting point - 117oC
- 138.5oC Boiling point
78.5oC -
25oC Density (at 20oC) 0.789 g/mL
0.00195 g/mL Use
Intoxicant in In
refrigeration
alcoholic beverages
H H
H H H C C O
H H C O C H
H
H H
H
Table 3.4
41
Molecular Formula
Molecules
Atoms
Avogadros Number
6.022 x 1023
Moles
Moles
42
Chemical Equations
Qualitative Information
Reactants
Products
States of Matter (s) solid (l)
liquid (g) gaseous (aq) aqueous
2 H2 (g) O2 (g) 2 H2O (g)
43
Chemical Equation Calculation - I
Atoms (Molecules)
Avogadros Number
6.02 x 1023
Molecules
Reactants
Products
44
Chemical Equation Calculation - II
Mass
Atoms (Molecules)
Molecular Weight
Avogadros Number
g/mol
6.02 x 1023
Molecules
Reactants
Products
Moles
45

• Stoichiometry
• The calculation of the quantities of reactants
  and products involved in a chemical reaction
• Interpreting a Chemical Equation
• The coefficients of the balanced chemical
  equation may be interpreted in terms of either
  (1) numbers of molecules (or ions or formula
  units) or (2) numbers of moles, depending on your
  needs.

46
Information Contained in a Balanced Equation
Viewed in Reactants
Products terms of 2 C2H6 (g)
7 O2 (g) 4 CO2 (g) 6 H2O(g) Energy
Molecules 2 molecules of C2H6 7 molecules of
O2
4 molecules of CO2 6 molecules of
H2O Amount (mol) 2 mol C2H6 7 mol O2
4 mol CO2 6 mol H2O Mass (amu) 60.14 amu
C2H6 224.00 amu O2
176.04 amu CO2
108.10 amu H2O Mass (g) 60.14 g C2H6
224.00 g O2 176.04 g CO2 108.10 g H2O Total
Mass (g) 284.14g
284.14g
47

• To find the amount of B (one reactant or product)
  given the amount of A (another reactant or
  product)

• 1. Convert grams of A to moles of A
• ? Using the molar mass of A
• 2. Convert moles of A to moles of B
• ? Using the coefficients of the balanced chemical
  equation
• 3. Convert moles of B to grams of B
• ? Using the molar mass of B

48

• Propane, C3H8, is normally a gas, but it is sold
  as a fuel compressed as a liquid in steel
  cylinders. The gas burns according to the
  following equation
• C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(g)
• How many grams of CO2 are produced when 20.0 g of
  propane is burned?

49
Molar masses C3H8 3(12.01) 8(1.008) 44.094
g CO2 1(12.01) 2(16.00) 44.01 g
59.9 g CO2 (3 significant figures)
50

• Propane, C3H8, is normally a gas, but it is sold
  as a fuel compressed as a liquid in steel
  cylinders. The gas burns according to the
  following equation
• C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(g)
• How many grams of O2 are required to burn 20.0 g
  of propane?

51

• Molar masses
• O2 2(16.00) 32.00 g
• C3H8 3(12.01) 8(1.008) 44.094 g

72.6 g O2 (3 significant figures)
52

• Limiting Reactant
• The reactant that is entirely consumed when a
  reaction goes to completion
• Once one reactant has been completely consumed,
  the reaction stops.
• Any problem giving the starting amount for more
  than one reactant is a limiting reactant problem.

53

• All amounts produced and reacted are determined
  by the limiting reactant.
• How can we determine the limiting reactant?
• Use each given amount to calculate the amount of
  product produced.
• The limiting reactant will produce the lesser or
  least amount of product.

54

• Magnesium metal is used to prepare zirconium
  metal, which is used to make the container for
  nuclear fuel (the nuclear fuel rods)
• ZrCl4(g) 2Mg(s) ? 2MgCl2(s) Zr(s)
• How many moles of zirconium metal can be produced
  from a reaction mixture containing 0.20 mol ZrCl4
  and 0.50 mol Mg?

55
ZrCl4 is the limiting reactant. 0.20 mol Zr will
be produced.
56

• Urea, CH4N2O, is used as a nitrogen fertilizer.
  It is manufactured from ammonia and carbon
  dioxide at high pressure and high temperature
• 2NH3 CO2(g) ? CH4N2O H2O
• In a laboratory experiment, 10.0 g NH3 and 10.0 g
  CO2 were added to a reaction vessel. What is the
  maximum quantity (in grams) of urea that can be
  obtained? How many grams of the excess reactant
  are left at the end of the reactions?

57
Molar masses NH3 1(14.01) 3(1.008) 17.02
g CO2 1(12.01) 2(16.00) 44.01
g CH4N2O 1(12.01) 4(1.008) 2(14.01)
1(16.00) 60.06 g
CO2 is the limiting reactant. 13.6 g CH4N2O will
be produced.
58
To find the excess NH3, we find how much NH3
reacted
Now subtract the amount reacted from the starting
amount
10.0 at start -7.73 reacted 2.27 g remains
2.3 g NH3 is left unreacted. (1 decimal place)
59

• Theoretical Yield
• The maximum amount of product that can be
  obtained by a reaction from given amounts of
  reactants. This is a calculated amount.

60

• Actual Yield
• The amount of product that is actually obtained.
  This is a measured amount.
• Percentage Yield

61

• 2NH3 CO2(g) ? CH4N2O H2O
• When 10.0 g NH3 and 10.0 g CO2 are added to a
  reaction vessel, the limiting reactant is CO2.
  The theoretical yield is 13.6 of urea. When this
  reaction was carried out, 9.3 g of urea was
  obtained. What is the percent yield?

Theoretical yield 13.6 g Actual yield 9.3 g
68 yield (2 significant figures)
62
Other Resources

• Visit the student website at college.hmco.com/pic/
  ebbing9e

63

• Calcite is a mineral composed of calcium
  carbonate, CaCO3. A sample of calcite composed of
  pure calcium carbonate weighs 23.6 g. How many
  moles of calcium carbonate is this?

First, find the molar mass of HNO3 1
Ca 1(40.08) 40.08 1 C 1(12.01) 12.01 3
O 3(16.00) 48.00
2 decimal places 100.09 g/mol
100.09
64
Next, find the number of moles in 23.6 g
65

• The daily requirement of chromium in the human
  diet is 1.0 10-6 g. How many atoms of chromium
  does this represent?

66
First, find the molar mass of Cr 1 Cr 1(51.996)
51.996
Now, convert 1.0 x 10-6 grams to moles
1.157781368 x 1016 atoms
1.2 x 1016 atoms (2 significant figures)
67

• Lead(II) chromate, PbCrO4, is used as a paint
  pigment (chrome yellow). What is the percentage
  composition of lead(II) chromate?

First, find the molar mass of PbCrO4 1
Pb 1(207.2) 207.2 1 Cr 1(51.996) 51.996 4
O 4(16.00) 64.00
(1 decimal place) 323.2 g/mol
323.196
68
Now, convert each to percent composition
Check 64.11 16.09 19.80 100.00
69

• The chemical name of table sugar is sucrose,
  C12H22O11. How many grams of carbon are in 68.1 g
  of sucrose.

First, find the molar mass of C12H22O11 12
C 12(12.01) 144.12 11 O 11(16.00) 176.00 22
H 22(1.008) 22.176
(2 decimal places) 342.30 g/mol
342.296
70
Now, find the mass of carbon in 61.8 g sucrose
71
Determining Empirical Formulas from
Masses of Elements - I
Problem The elemental analysis of a sample
compound gave the following results 5.677g Na,
6.420 g Cr, and 7.902 g O. What is the empirical
formula and name of the compound? Plan First we
have to convert mass of the elements to moles of
the elements using the molar masses. Then we
construct a preliminary formula and name of the
compound. Solution Finding the moles of the
elements Moles of Na 5.678 g Na x
Moles of Cr
6.420 g Cr x
Moles of O 7.902 g O x
1 mol Na 22.99 g Na
72

• Sodium pyrophosphate is used in detergent
  preparations. It is composed of 34.5 Na, 23.3
  P, and 42.1 O. What is its empirical formula?

Empirical formula Na4P2O7
73

• Hexamethylene is one of the materials used to
  produce a type of nylon. It is composed of 62.1
  C, 13.8 H, and 24.1 N. Its molecular weight is
  116 amu. What is its molecular formula?

Empirical formula C3H8N
74

• The empirical formula is C3H8N.
• Find the empirical formula weight
• 3(12.01) 8(1.008) 1(14.01) 58.104 amu
• Molecular formula C6H16N2