Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 3, Problem 3.79QP

(a)

Interpretation Introduction

Interpretation: The compound with greatest percentage of carbon by mass is to be determined. If any two of the given compounds have same empirical formula is to be stated.

Concept introduction: Empirical formula signifies the simplest or most reduced ratio of elements in a compound. Molecular formula is obtained by multiplying empirical formula with a multiple.

To determine: The percentage of carbon by mass in the given compound and its empirical formula.

(a)

Expert Solution
Check Mark

Answer to Problem 3.79QP

Solution

The percentage of carbon by mass in the given compound is 93.71%_ and its empirical formula is C5H4 .

Explanation of Solution

Explanation

The given compound is naphthalene ( C10H8 ).

The molar mass of naphthalene ( C10H8 ) is calculated by the formula,

Molar mass=(Number of C atoms×Molar mass of C)+(Number of H atoms×Molar mass of H)

There is 10 carbon atoms and 8 hydrogen atoms present in C10H8 .

Molar mass of C is 12.01 g/mol .

Molar mass of H is 1.008g/mol .

Substitute the values of number of carbon and hydrogen atoms and their respective masses in the above equation.

Molar mass=(10×12.01 g/mol)+(8×1.008 g/mol)=128.16g/mol

The mass percentage of carbon ( %C ) is calculated by the formula,

%C=Mass of 10CMolarMass of C10H8×100

Mass of 10 carbon atoms =10×12.01 g/mol=120.10 g/mol .

Substitute the value of mass of 10 carbon atom and molar mass of C10H8 in the above equation.

%C=120.10 g/mol128.16g/mol×100=93.71%

Therefore, the mass percentage of carbon ( %C ) in C10H8 is 93.71%_ .

The molecular formula of naphthalene is C10H8 . The ratio for C:H is 10:8 . The highest common multiple in the subscripts is 2 . Divide the ratio of C:H by 2 . It will become,

102:82=5:4

The ratio of C:H will be 5:4 .

Thus the empirical formula of naphthalene is C5H4 .

(b)

Interpretation Introduction

To determine: The percentage of carbon by mass in the given compound and its empirical formula.

(b)

Expert Solution
Check Mark

Answer to Problem 3.79QP

Solution

The percentage of carbon by mass in the given compound is 94.7%_ and its empirical formula is C3H2 .

Explanation of Solution

Explanation

The given compound is chrysene ( C18H12 ).

The molar mass of chrysene ( C18H12 ) is calculated by the formula,

Molar mass=(Number of C atoms×Molar mass of C)+(Number of H atoms×Molar mass of H)

There is 18 carbon atoms and 12 hydrogen atoms present in C18H12 .

Molar mass of C is 12.01 g/mol .

Molar mass of H is 1.008g/mol .

Substitute the values of number of carbon and hydrogen atoms and their respective masses in the above equation.

Molar mass=(18×12.01 g/mol)+(12×1.008 g/mol)=228.27g/mol

The mass percentage of carbon ( %C ) is calculated by the formula,

%C=Mass of 18CMolarMass of C18H12×100

Mass of 18 carbon atoms =18×12.01 g/mol=216.18 g/mol .

Substitute the value of mass of 18 carbon atom and molar mass of C18H12 in the above equation.

%C=216.18 g/mol228.27g/mol×100=94.7%

Therefore, the mass percentage of carbon ( %C ) in C18H12 is 94.7%_ .

The molecular formula of chrysene is C18H12 . The ratio for C:H is 18:12 . The highest common multiple in the subscripts is 6 . Divide the ratio of C:H by 6 . It will become,

186:126=3:2

The ratio of C:H will be 3:2 .

Thus the empirical formula of chrysene is C3H2 .

(c)

Interpretation Introduction

To determine: The percentage of carbon by mass in the given compound and its empirical formula.

(c)

Expert Solution
Check Mark

Answer to Problem 3.79QP

Solution

The percentage of carbon by mass in the given compound is 94.93%_ and its empirical formula is C11H7 .

Explanation of Solution

Explanation

The given compound is pentacene ( C22H14 ).

The molar mass of pentacene ( C22H14 ) is calculated by the formula,

Molar mass=(Number of C atoms×Molar mass of C)+(Number of H atoms×Molar mass of H)

There is 22 carbon atoms and 14 hydrogen atoms present in C22H14 .

Molar mass of C is 12.01 g/mol .

Molar mass of H is 1.008g/mol .

Substitute the values of number of carbon and hydrogen atoms and their respective masses in the above equation.

Molar mass=(22×12.01 g/mol)+(14×1.008 g/mol)=278.33g/mol

The mass percentage of carbon ( %C ) is calculated by the formula,

%C=Mass of 22CMolarMass of C22H14×100

Mass of 22 carbon atoms =22×12.01 g/mol=264.22 g/mol .

Substitute the value of mass of 22 carbon atom and molar mass of C22H14 in the above equation.

%C=264.22 g/mol278.33g/mol×100=94.93%

Therefore, the mass percentage of carbon ( %C ) in C22H14 is 94.93%_ .

The molecular formula of pentacene is C22H14 . The ratio for C:H is 22:14 . The highest common multiple in the subscripts is 2 . Divide the ratio of C:H by 2 . It will become,

222:142=11:7

The ratio of C:H will be 11:7 .

Thus the empirical formula of pentacene is C11H7 .

(d)

Interpretation Introduction

To determine: The percentage of carbon by mass in the given compound and its empirical formula.

(d)

Expert Solution
Check Mark

Answer to Problem 3.79QP

Solution

The percentage of carbon by mass in the given compound is 95.01%_ and its empirical formula is C8H5 .

Explanation of Solution

Explanation

The given compound is pyrene ( C16H10 ).

The molar mass of pyrene ( C16H10 ) is calculated by the formula,

Molar mass=(Number of C atoms×Molar mass of C)+(Number of H atoms×Molar mass of H)

There is 16 carbon atoms and 10 hydrogen atoms present in C16H10 .

Molar mass of C is 12.01 g/mol .

Molar mass of H is 1.008g/mol .

Substitute the values of number of carbon and hydrogen atoms and their respective masses in the above equation.

Molar mass=(16×12.01 g/mol)+(10×1.008 g/mol)=202.24g/mol

The mass percentage of carbon ( %C ) is calculated by the formula,

%C=Mass of 16CMolarMass of C16H10×100

Mass of 16 carbon atoms =16×12.01 g/mol=192.16 g/mol .

Substitute the value of mass of 16 carbon atoms and molar mass of C16H10 in the above equation.

%C=192.16 g/mol202.24g/mol×100=95.01%

Therefore, the mass percentage of carbon ( %C ) in C16H10 is 95.01%_ .

The molecular formula of pyrene is C16H10 . The ratio for C:H is 16:10 . The highest common multiple in the subscripts is 2 . Divide the ratio of C:H by 2 . It will become,

162:102=8:5

The ratio of C:H will be 8:5 .

Thus the empirical formula of pyrene is C8H5 .

Therefore, the compound with greatest percentage of carbon by mass is pyrene ( C16H10 ). None of the compounds have same empirical formula.

Conclusion

The compound with greatest percentage of carbon by mass is pyrene ( C16H10 ). None of the compounds have same empirical formula

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Chapter 3 Solutions

Chemistry

Ch. 3.5 - Prob. 12PECh. 3.5 - Prob. 13PECh. 3.6 - Prob. 14PECh. 3.6 - Prob. 16PECh. 3.7 - Prob. 17PECh. 3.8 - Prob. 18PECh. 3.8 - Prob. 19PECh. 3.9 - Prob. 20PECh. 3.9 - Prob. 21PECh. 3.9 - Prob. 22PECh. 3.9 - Prob. 23PECh. 3 - Prob. 3.1VPCh. 3 - Prob. 3.2VPCh. 3 - Prob. 3.3VPCh. 3 - Prob. 3.4VPCh. 3 - Prob. 3.5VPCh. 3 - Prob. 3.6VPCh. 3 - Prob. 3.7VPCh. 3 - Prob. 3.8VPCh. 3 - Prob. 3.9QPCh. 3 - Prob. 3.10QPCh. 3 - Prob. 3.11QPCh. 3 - Prob. 3.12QPCh. 3 - Prob. 3.13QPCh. 3 - Prob. 3.14QPCh. 3 - Prob. 3.15QPCh. 3 - Prob. 3.16QPCh. 3 - Prob. 3.17QPCh. 3 - Prob. 3.18QPCh. 3 - Prob. 3.19QPCh. 3 - Prob. 3.20QPCh. 3 - Prob. 3.21QPCh. 3 - Prob. 3.22QPCh. 3 - Prob. 3.23QPCh. 3 - Prob. 3.24QPCh. 3 - Prob. 3.25QPCh. 3 - Prob. 3.26QPCh. 3 - Prob. 3.27QPCh. 3 - Prob. 3.28QPCh. 3 - Prob. 3.29QPCh. 3 - Prob. 3.30QPCh. 3 - Prob. 3.31QPCh. 3 - Prob. 3.32QPCh. 3 - Prob. 3.33QPCh. 3 - Prob. 3.34QPCh. 3 - Prob. 3.35QPCh. 3 - Prob. 3.36QPCh. 3 - Prob. 3.37QPCh. 3 - Prob. 3.38QPCh. 3 - Prob. 3.39QPCh. 3 - Prob. 3.40QPCh. 3 - Prob. 3.41QPCh. 3 - Prob. 3.42QPCh. 3 - Prob. 3.43QPCh. 3 - Prob. 3.44QPCh. 3 - Prob. 3.45QPCh. 3 - Prob. 3.46QPCh. 3 - Prob. 3.47QPCh. 3 - Prob. 3.48QPCh. 3 - Prob. 3.49QPCh. 3 - Prob. 3.50QPCh. 3 - Prob. 3.51QPCh. 3 - Prob. 3.52QPCh. 3 - Prob. 3.53QPCh. 3 - Prob. 3.54QPCh. 3 - Prob. 3.55QPCh. 3 - Prob. 3.56QPCh. 3 - Prob. 3.57QPCh. 3 - Prob. 3.58QPCh. 3 - Prob. 3.59QPCh. 3 - Prob. 3.60QPCh. 3 - Prob. 3.61QPCh. 3 - Prob. 3.62QPCh. 3 - Prob. 3.63QPCh. 3 - Prob. 3.64QPCh. 3 - Prob. 3.65QPCh. 3 - Prob. 3.66QPCh. 3 - Prob. 3.67QPCh. 3 - Prob. 3.68QPCh. 3 - Prob. 3.69QPCh. 3 - Prob. 3.70QPCh. 3 - Prob. 3.71QPCh. 3 - Prob. 3.72QPCh. 3 - Prob. 3.73QPCh. 3 - Prob. 3.74QPCh. 3 - Prob. 3.75QPCh. 3 - Prob. 3.76QPCh. 3 - Prob. 3.77QPCh. 3 - Prob. 3.78QPCh. 3 - Prob. 3.79QPCh. 3 - Prob. 3.80QPCh. 3 - Prob. 3.81QPCh. 3 - Prob. 3.82QPCh. 3 - Prob. 3.83QPCh. 3 - Prob. 3.84QPCh. 3 - Prob. 3.85QPCh. 3 - Prob. 3.86QPCh. 3 - Prob. 3.87QPCh. 3 - Prob. 3.88QPCh. 3 - Prob. 3.89QPCh. 3 - Prob. 3.90QPCh. 3 - Prob. 3.91QPCh. 3 - Prob. 3.92QPCh. 3 - Prob. 3.93QPCh. 3 - Prob. 3.94QPCh. 3 - Prob. 3.95QPCh. 3 - Prob. 3.96QPCh. 3 - Prob. 3.97QPCh. 3 - Prob. 3.98QPCh. 3 - Prob. 3.99QPCh. 3 - Prob. 3.100QPCh. 3 - Prob. 3.101QPCh. 3 - Prob. 3.102QPCh. 3 - Prob. 3.103QPCh. 3 - Prob. 3.104QPCh. 3 - Prob. 3.105QPCh. 3 - Prob. 3.106QPCh. 3 - Prob. 3.107QPCh. 3 - Prob. 3.108QPCh. 3 - Prob. 3.109QPCh. 3 - Prob. 3.110QPCh. 3 - Prob. 3.111QPCh. 3 - Prob. 3.112QPCh. 3 - Prob. 3.113QPCh. 3 - Prob. 3.114QPCh. 3 - Prob. 3.115QPCh. 3 - Prob. 3.116QPCh. 3 - Prob. 3.117APCh. 3 - Prob. 3.118APCh. 3 - Prob. 3.119APCh. 3 - Prob. 3.120APCh. 3 - Prob. 3.121APCh. 3 - Prob. 3.122APCh. 3 - Prob. 3.123APCh. 3 - Prob. 3.124APCh. 3 - Prob. 3.125APCh. 3 - Prob. 3.126APCh. 3 - Prob. 3.127APCh. 3 - Prob. 3.128APCh. 3 - Prob. 3.129APCh. 3 - Prob. 3.130APCh. 3 - Prob. 3.131APCh. 3 - Prob. 3.132APCh. 3 - Prob. 3.133APCh. 3 - Prob. 3.134APCh. 3 - Prob. 3.135APCh. 3 - Prob. 3.136APCh. 3 - Prob. 3.137APCh. 3 - Prob. 3.138APCh. 3 - Prob. 3.139APCh. 3 - Prob. 3.140APCh. 3 - Prob. 3.141APCh. 3 - Prob. 3.142APCh. 3 - Prob. 3.143APCh. 3 - Prob. 3.144AP
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