Arrange the following substances in order of increasing mass percent of carbon. a. caffeine. C 8 H 10 N 4 O 2 b. sucrose, C 12 H 22 O 11 c. ethanol, C 2 H 5 OH

Interpretation: The compounds C8H10N4O2,C12H22O11 and C2H5OH are to be arranged in increasing order of mass percent of carbon.

Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons.

Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound.

The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.

To determine: The arrangement of compounds C8H10N4O2,C12H22O11 and C2H5OH in order of increasing mass percent of carbon.

To determine: The percentage composition (by mass) of carbon in the compound C8H10N4O2 .

The atomic weight of oxygen (O) is 15.999 g/mol . In a compound C8H10N4O2 , two oxygen atoms are present. Hence, mass of oxygen in C8H10N4O2 is,

2×15.999 g/mol=31.998 g/mol

The atomic weight of hydrogen (H) is 1.008 g/mol . In a compound C8H10N4O2 , ten hydrogen atoms are present. Hence, mass of hydrogen in C8H10N4O2 is,

10×1.008 g/mol=10.08 g/mol

The atomic weight of nitrogen (N) atom is 14.006 g/mol . In a compound C8H10N4O2 , four nitrogen atoms are present. Hence, mass of nitrogen in C8H10N4O2 is,

4×14.006 g/mol=56.024 g/mol

The molar mass is the sum of mass of individual atoms present in it. Hence, molar mass of C8H10N4O2 is calculated as,

(96.08+31.998+10.08+56.024) g/mol=194.182 g/mol

Formula

The mass percent of element is calculated as,

Mass% of element=Mass of elementMolar mass of compound

Substitute the values of mass of carbon and molar mass of C8H10N4O2 in the above equation.

Mass% of C atom=Mass of C atomMolar mass of C8H10N4O2=(96.08 g/mol194.182 g/mol)×100=49.47%_

The percentage composition (by mass) of carbon in C8H10N4O2 is 49.47%_ .

The atomic weight of carbon (C) is 12.01 g/mol . In the compound C8H10N4O2 , eight carbon atoms are present. Hence, mass of carbon in C8H10N4O2 is,

8×12.01 g/mol=96.08 g/mol

To determine: The percentage composition (by mass) of carbon in compound C12H22O11 .

The atomic weight of carbon (C) is 12.01 g/mol . In a compound C12H22O11 , twelve carbon atoms are present. Hence, mass of carbon in C12H22O11 is,

12×12.01 g/mol=144.12 g/mol

The atomic weight of oxygen (O) is 15.999 g/mol . In a compound C12H22O11 , eleven oxygen atoms are present. Hence, mass of oxygen in C12H22O11 is,

11×15.999 g/mol=175.989 g/mol

The atomic weight of hydrogen (H) is 1.008 g/mol . In a compound C12H22O11 , twenty-two hydrogen atoms are present. Hence, mass of hydrogen in C12H22O11 is,

22×1