# What is the Mass of 170 Mol of Carbon-12

What is the Mass of 170 Mol of Carbon-12

### Learning Objectives

Past the end of this department, yous will exist able to:

• Compute the pct composition of a chemical compound
• Determine the empirical formula of a chemical compound
• Decide the molecular formula of a compound

The previous department discussed the human relationship between the bulk mass of a substance and the number of atoms or molecules information technology contains (moles). Given the chemical formula of the substance, i may determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, these same principles will be applied to derive the chemic formulas of unknown substances from experimental mass measurements.

### Percent Composition

The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the offset pace in the process of determining the formula experimentally. The results of these measurements let the calculation of the compound’south
percent limerick, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The pct composition of this chemical compound could be represented equally follows:



%

H
=

mass H

mass compound

×

100
%

%

H
=

mass H

mass compound

×

100
%



%

C
=

mass C

mass compound

×

100
%

%

C
=

mass C

mass compound

×

100
%

If assay of a 10.0-g sample of this gas showed it to contain 2.5 g H and vii.five g C, the percentage composition would exist calculated to be 25% H and 75% C:



%

H
=

2.5

one thousand H

10.0

g chemical compound

×

100
%
=
25
%

%

H
=

ii.5

g H

x.0

g compound

×

100
%
=
25
%



%

C
=

seven.5

g C

ten.0

thou compound

×

100
%
=
75
%

%

C
=

7.5

grand C

x.0

thou compound

×

100
%
=
75
%

### Case iii.nine

#### Calculation of Per centum Composition

Analysis of a 12.04-thou sample of a liquid chemical compound composed of carbon, hydrogen, and nitrogen showed it to incorporate 7.34 k C, ane.85 thou H, and two.85 g N. What is the percent composition of this compound?

#### Solution

To calculate percent limerick, divide the experimentally derived mass of each element past the overall mass of the compound, and and so convert to a percentage:



%

C
=

seven.34

g C

12.04

g compound

×

100
%
=
61.0
%

%

H
=

one.85

1000 H

12.04

×

100
%
=
15.iv
%

%

Northward
=

2.85

thou N

12.04

thousand chemical compound

×

100
%
=
23.vii
%

%

C
=

vii.34

yard C

12.04

×

100
%
=
61.0
%

%

H
=

1.85

g H

12.04

g compound

×

100
%
=
15.4
%

%

N
=

2.85

grand N

12.04

g compound

×

100
%
=
23.vii
%

The analysis results betoken that the compound is 61.0% C, 15.4% H, and 23.vii% N past mass.

#### Check Your Learning

A 24.81-g sample of a gaseous chemical compound containing only carbon, oxygen, and chlorine is adamant to contain three.01 g C, 4.00 g O, and 17.81 g Cl. What is this compound’s percent composition?

12.1% C, sixteen.one% O, 71.79% Cl

#### Determining Percentage Composition from Molecular or Empirical Formulas

Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. Every bit one example, consider the mutual nitrogen-containing fertilizers ammonia (NH3), ammonium nitrate (NHfourNO3), and urea (CH4NiiO). The element nitrogen is the active ingredient for agricultural purposes, so the mass per centum of nitrogen in the chemical compound is a practical and economic business concern for consumers choosing among these fertilizers. For these sorts of applications, the per centum composition of a compound is hands derived from its formula mass and the diminutive masses of its constituent elements. A molecule of NH3
contains i Northward atom weighing xiv.01 amu and three H atoms weighing a total of (3



×

×

1.008 amu) = 3.024 amu. The formula mass of ammonia is therefore (xiv.01 amu + 3.024 amu) = 17.03 amu, and its percent limerick is:



%

N
=

fourteen.01

amu N

17.03

amu

NH

3

×

100
%
=
82.27
%

%

H
=

3.024

amu H

17.03

amu

NH

3

×

100
%
=
17.76
%

%

N
=

14.01

amu Due north

17.03

amu

NH

three

×

100
%
=
82.27
%

%

H
=

3.024

amu H

17.03

amu

NH

3

×

100
%
=
17.76
%

This aforementioned approach may be taken because a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most user-friendly and would merely involve the use of molar masses instead of atomic and formula masses, as demonstrated Example 3.x. As long as the molecular or empirical formula of the chemical compound in question is known, the per centum composition may exist derived from the atomic or tooth masses of the chemical compound’s elements.

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### Example 3.10

#### Determining Percent Limerick from a Molecular Formula

Aspirin is a chemical compound with the molecular formula C9H8O4. What is its percentage composition?

#### Solution

To calculate the percent composition, the masses of C, H, and O in a known mass of C9H8O4
are needed. It is convenient to consider 1 mol of C9H8O4
and use its molar mass (180.159 grand/mole, adamant from the chemical formula) to calculate the percentages of each of its elements:



%

C
=

9

mol C

×

molar mass C

tooth mass

C
9

H

8

O
4

×

100
=

nine

×

12.01

g/mol

180.159

g/mol

×

100
=

108.09

g/mol

180.159

g/mol

×

100

%

C
=
60.00
%

C

%

C
=

ix

mol C

×

molar mass C

molar mass

C
nine

H

eight

O
4

×

100
=

ix

×

12.01

g/mol

180.159

1000/mol

×

100
=

108.09

g/mol

180.159

one thousand/mol

×

100

%

C
=
lx.00
%

C



%

H
=

8

mol H

×

molar mass H

molar mass

C
nine

H

8

O
4

×

100
=

8

×

1.008

g/mol

180.159

g/mol

×

100
=

8.064

g/mol

180.159

thou/mol

×

100

%

H
=
iv.476
%

H

%

H
=

viii

mol H

×

molar mass H

molar mass

C
nine

H

eight

O
4

×

100
=

8

×

i.008

1000/mol

180.159

×

100
=

8.064

one thousand/mol

180.159

k/mol

×

100

%

H
=
4.476
%

H



%

O
=

4

mol O

×

molar mass O

molar mass

C
9

H

8

O
4

×

100
=

4

×

xvi.00

g/mol

180.159

thou/mol

×

100
=

64.00

g/mol

180.159

g/mol

×

100

%

O
=
35.52
%

%

O
=

iv

mol O

×

molar mass O

molar mass

C
ix

H

8

O
four

×

100
=

4

×

16.00

m/mol

180.159

g/mol

×

100
=

64.00

thousand/mol

180.159

1000/mol

×

100

%

O
=
35.52
%

Notation that these percentages sum to equal 100.00% when accordingly rounded.

#### Check Your Learning

To three significant digits, what is the mass percentage of iron in the compound Fe2O3?

### Decision of Empirical Formulas

Equally previously mentioned, the about common approach to determining a compound’s chemical formula is to first measure the masses of its constituent elements. However, keep in mind that chemical formulas represent the relative
numbers, not masses, of atoms in the substance. Therefore, whatever experimentally derived data involving mass must be used to derive the corresponding numbers of atoms in the compound. This is accomplished using molar masses to convert the mass of each element to a number of moles. These tooth amounts are used to compute whole-number ratios that tin can exist used to derive the empirical formula of the substance. Consider a sample of compound determined to contain one.71 thousand C and 0.287 g H. The corresponding numbers of atoms (in moles) are:



1.71

g C

×

1

mol C

12.01

thousand C

=
0.142

mol C

0.287

g H

×

1

mol H

ane.008

one thousand H

=
0.284

mol H

1.71

grand C

×

1

mol C

12.01

g C

=
0.142

mol C

0.287

g H

×

1

mol H

1.008

g H

=
0.284

mol H

Thus, this compound may exist represented by the formula C0.142H0.284. Per convention, formulas incorporate whole-number subscripts, which can exist achieved by dividing each subscript by the smaller subscript:



C

0.142

0.142

H

0.284

0.142

or

CH

2

C

0.142

0.142

H

0.284

0.142

or

CH

2

(Recall that subscripts of “ane” are not written but rather causeless if no other number is present.)

The empirical formula for this chemical compound is thus CH2. This may or may not be the chemical compound’s
molecular formula
as well; however, additional information is needed to make that determination (every bit discussed later in this section).

Consider equally another example a sample of compound determined to contain v.31 g Cl and 8.forty g O. Post-obit the same approach yields a tentative empirical formula of:



Cl

0.150

O

0.525

=

Cl

0.150

0.150

O

0.525

0.150

=

ClO

3.5

Cl

0.150

O

0.525

=

Cl

0.150

0.150

O

0.525

0.150

=

ClO

3.five

In this example, dividing past the smallest subscript still leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, multiply each of the subscripts by two, retaining the same atom ratio and yielding Cl2Ovii
as the terminal empirical formula.

In summary, empirical formulas are derived from experimentally measured element masses by:

1. Deriving the number of moles of each element from its mass
2. Dividing each element’due south molar amount by the smallest tooth amount to yield subscripts for a tentative empirical formula
3. Multiplying all coefficients past an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained

Figure 3.eleven outlines this process in flow chart fashion for a substance containing elements A and Ten.

### Example 3.11

#### Determining a Compound’southward Empirical Formula from the Masses of Its Elements

A sample of the blackness mineral hematite (Figure three.12), an oxide of iron plant in many atomic number 26 ores, contains 34.97 g of iron and fifteen.03 yard of oxygen. What is the empirical formula of hematite?

Figure

iii.12

Hematite is an iron oxide that is used in jewelry. (credit: Mauro Cateb)

#### Solution

This problem provides the mass in grams of each element. Begin past finding the moles of each:



34.97

g Iron

(

mol Fe

55.85

g

)

=

0.6261

mol Fe

xv.03

one thousand O

(

mol O

16.00

g

)

=

0.9394

mol O

34.97

g Atomic number 26

(

mol Atomic number 26

55.85

g

)

=

0.6261

mol Fe

fifteen.03

g O

(

mol O

16.00

g

)

=

0.9394

mol O

Next, derive the atomic number 26-to-oxygen molar ratio by dividing past the bottom number of moles:



0.6261

0.6261

=
1.000

mol Iron

0.9394

0.6261

=
1.500

mol O

0.6261

0.6261

=
1.000

mol Fe

0.9394

0.6261

=
1.500

mol O

The ratio is ane.000 mol of iron to 1.500 mol of oxygen (Fe1O1.5). Finally, multiply the ratio by two to get the smallest possible whole number subscripts while yet maintaining the correct iron-to-oxygen ratio:



2

(

Fe

one

O

1.v

)

=

Fe

2

O
3

2

(

Atomic number 26

1

O

i.5

)

=

Fe

2

O
3

The empirical formula is FetwoO3.

#### Check Your Learning

What is the empirical formula of a compound if a sample contains 0.130 k of nitrogen and 0.370 g of oxygen?

#### Deriving Empirical Formulas from Percent Limerick

Finally, with regard to deriving empirical formulas, consider instances in which a compound’s pct composition is bachelor rather than the absolute masses of the compound’due south constituent elements. In such cases, the per centum composition can be used to summate the masses of elements nowadays in any convenient mass of compound; these masses tin and so be used to derive the empirical formula in the usual fashion.

### Example 3.12

#### Determining an Empirical Formula from Pct Composition

The bacterial fermentation of grain to produce ethanol forms a gas with a percentage composition of 27.29% C and 72.71% O (Effigy 3.xiii). What is the empirical formula for this gas?

Figure

3.13

An oxide of carbon is removed from these fermentation tanks through the large copper pipes at the top. (credit: “Dual Freq”/Wikimedia Commons)

#### Solution

Since the calibration for percentages is 100, it is most convenient to summate the mass of elements present in a sample weighing 100 g. The adding is “nearly user-friendly” because, per the definition for percent composition, the mass of a given element in grams is numerically equivalent to the element’s mass per centum. This numerical equivalence results from the definition of the “percentage” unit, whose name is derived from the Latin phrase
per centum
meaning “by the hundred.” Considering this definition, the mass percentages provided may be more conveniently expressed equally fractions:



27.29
%

C

=

27.29

g C

100

g compound

72.71
%

O

=

72.71

g O

100

g compound

27.29
%

C

=

27.29

g C

100

thou chemical compound

72.71
%

O

=

72.71

g O

100

g compound

The molar amounts of carbon and oxygen in a 100-g sample are calculated by dividing each element’s mass by its molar mass:



27.29

g C

(

mol C

12.01

m

)

=

2.272

mol C

72.71

g O

(

mol O

16.00

g

)

=

4.544

mol O

27.29

m C

(

mol C

12.01

g

)

=

2.272

mol C

72.71

m O

(

mol O

16.00

g

)

=

4.544

mol O

Coefficients for the tentative empirical formula are derived past dividing each molar corporeality past the lesser of the two:



2.272

mol C

2.272

=
1

4.544

mol O

2.272

=
2

2.272

mol C

2.272

=
1

four.544

mol O

2.272

=
ii

Since the resulting ratio is one carbon to two oxygen atoms, the empirical formula is CO2.

#### Check Your Learning

What is the empirical formula of a chemical compound containing xl.0% C, half-dozen.71% H, and 53.28% O?

### Derivation of Molecular Formulas

Recall that empirical formulas are symbols representing the
relative
numbers of a compound’s elements. Determining the
absolute
numbers of atoms that etch a single molecule of a covalent compound requires knowledge of both its empirical formula and its molecular mass or molar mass. These quantities may be determined experimentally by diverse measurement techniques. Molecular mass, for instance, is often derived from the mass spectrum of the chemical compound (see give-and-take of this technique in the previous chapter on atoms and molecules). Tooth mass tin be measured by a number of experimental methods, many of which volition be introduced in afterward chapters of this text.

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Molecular formulas are derived by comparison the compound’s molecular or tooth mass to its
empirical formula mass. Every bit the proper noun suggests, an empirical formula mass is the sum of the average diminutive masses of all the atoms represented in an empirical formula. If the molecular (or molar) mass of the substance is known, it may be divided by the empirical formula mass to yield the number of empirical formula units per molecule (north):



molecular or molar mass

(

amu or

g

mol

)

empirical formula mass

(

amu or

g

mol

)

=
northward

formula units/molecule

molecular or molar mass

(

amu or

one thousand

mol

)

empirical formula mass

(

amu or

g

mol

)

=
n

formula units/molecule

The molecular formula is and then obtained by multiplying each subscript in the empirical formula by
n, as shown by the generic empirical formula A10By:



(

A

x

B
y

)

n

=

A

nx

B

ny

(

A

x

B
y

)

n

=

A

nx

B

ny

For example, consider a covalent compound whose empirical formula is adamant to be CHtwoO. The empirical formula mass for this chemical compound is approximately xxx amu (the sum of 12 amu for ane C atom, 2 amu for two H atoms, and xvi amu for one O atom). If the chemical compound’s molecular mass is determined to be 180 amu, this indicates that molecules of this compound contain half dozen times the number of atoms represented in the empirical formula:



180

amu/molecule

30

amu

formula unit of measurement

=
vi

formula units/molecule

180

amu/molecule

30

amu

formula unit

=
half-dozen

formula units/molecule

Molecules of this chemical compound are then represented by molecular formulas whose subscripts are half dozen times greater than those in the empirical formula:



(CH

2

O)

six

=

C
half dozen

H

12

O
6

(CH

2

O)

vi

=

C
6

H

12

O
6

Annotation that this same approach may be used when the molar mass (k/mol) instead of the molecular mass (amu) is used. In this case,
one mole
of empirical formula units and molecules is considered, as opposed to single units and molecules.

### Case iii.13

#### Determination of the Molecular Formula for Nicotine

Nicotine, an alkaloid in the nightshade family of plants that is mainly responsible for the addictive nature of cigarettes, contains 74.02% C, 8.710% H, and 17.27% N. If forty.57 grand of nicotine contains 0.2500 mol nicotine, what is the molecular formula?

#### Solution

Determining the molecular formula from the provided information will require comparing of the compound’s empirical formula mass to its tooth mass. As the first stride, use the percent composition to derive the compound’s empirical formula. Bold a convenient, a 100-m sample of nicotine yields the following molar amounts of its elements:



(

74.02

g C

)

(

1

mol C

12.01

g C

)

=

6.163

mol C

(

8.710

grand H

)

(

one

mol H

1.01

g H

)

=

8.624

mol H

(

17.27

k North

)

(

1

mol Due north

14.01

thou N

)

=

1.233

mol N

(

74.02

)

(

ane

mol C

12.01

thousand C

)

=

vi.163

mol C

(

8.710

m H

)

(

one

mol H

1.01

g H

)

=

eight.624

mol H

(

17.27

m North

)

(

1

mol N

14.01

)

=

one.233

mol N

Adjacent, calculate the molar ratios of these elements relative to the least abundant chemical element, North.



vi.163

mol

C
/

1.233

mol

Due north
=
5

6.163

mol

C
/

1.233

mol

Northward
=
five



8.264

mol

H
/

ane.233

mol

N
=
7

8.264

mol

H
/

one.233

mol

N
=
7



i.233

mol

N
/

i.233

mol

N
=
one

1.233

mol

Due north
/

1.233

mol

N
=
1



one.233

ane.233

=

1.000

mol N

half-dozen.163

one.233

=

4.998

mol C

eight.624

i.233

=

half dozen.994

mol H

i.233

1.233

=

1.000

mol N

6.163

ane.233

=

four.998

mol C

8.624

ane.233

=

6.994

mol H

The C-to-North and H-to-N molar ratios are adequately close to whole numbers, and so the empirical formula is C5H7N. The empirical formula mass for this chemical compound is therefore 81.thirteen amu/formula unit of measurement, or 81.13 g/mol formula unit of measurement.

Summate the molar mass for nicotine from the given mass and molar amount of compound:



40.57

g nicotine

0.2500

mol nicotine

=

162.iii

g

mol

forty.57

thou nicotine

0.2500

mol nicotine

=

162.3

g

mol

Comparing the molar mass and empirical formula mass indicates that each nicotine molecule contains ii formula units:



162.3

grand/mol

81.xiii

g

formula unit

=

2

formula units/molecule

162.3

grand/mol

81.xiii

g

formula unit of measurement

=

two

formula units/molecule

Finally, derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by ii:



(C

5

H
vii

N)
2

=

C

10

H

14

N
2

(C

5

H
vii

Northward)
2

=

C

10

H

xiv

N
2

#### Check Your Learning

What is the molecular formula of a chemical compound with a per centum composition of 49.47% C, five.201% H, 28.84% N, and 16.48% O, and a molecular mass of 194.2 amu?

## What is the Mass of 170 Mol of Carbon-12

Source: https://openstax.org/books/chemistry-2e/pages/3-2-determining-empirical-and-molecular-formulas