Chapter 6, Problem 6.28AP

(a)

Interpretation Introduction

Interpretation:

The number carbon presents in $CaC{O}_3$ and the grams of carbon present in one moles of it should be determined.

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $\text{12}\text{g}$ of ${}^{\text{12}}\text{C}$.

The mole is actually quantity of particles that is Avogadro’s number of particles present in elements or compounds.

The sum of mass of all atoms present in formula of chemical substance is referred as molecular or formula weight of that substance.

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$.

Molar mass: It is obtained by dividing the mass of substance with the amount of substance and the S.I. unit of molar mass is $\text{kg/mol}$. It is numerically equal to the molecular weight since molecular weight is sum of all individual atom weight present in molecule which is represented in $amu$.

$\text{Molar}\text{mass}\left(\text{g/mol}\right)=\frac{\text{Given}\text{mass}\text{of}\text{substance}\left(\text{g}\right)}{\text{Moles}\text{of}\text{substance}\left(mol\right)}$

(b)

Interpretation Introduction

Interpretation:

The number carbon presents in $\text{CO}{\left({\text{NH}}_{\text{2}}\right)}_{\text{2}}$ and the grams of carbon present in one moles of it should be determined.

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $\text{12}\text{g}$ of ${}^{\text{12}}\text{C}$.

The mole is actually quantity of particles that is Avogadro’s number of particles present in elements or compounds.

The sum of mass of all atoms present in formula of chemical substance is referred as molecular or formula weight of that substance.

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$.

Molar mass: It is obtained by dividing the mass of substance with the amount of substance and the S.I. unit of molar mass is $\text{kg/mol}$. It is numerically equal to the molecular weight since molecular weight is sum of all individual atom weight present in molecule which is represented in $amu$.

$\text{Molar}\text{mass}\left(\text{g/mol}\right)=\frac{\text{Given}\text{mass}\text{of}\text{substance}\left(\text{g}\right)}{\text{Moles}\text{of}\text{substance}\left(mol\right)}$

(c)

Interpretation Introduction

Interpretation:

The number carbon presents in ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$ and the grams of carbon present in one moles of it should be determined.

Concept introduction:

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $\text{12}\text{g}$ of ${}^{\text{12}}\text{C}$.

The mole is actually quantity of particles that is Avogadro’s number of particles present in elements or compounds.

The sum of mass of all atoms present in formula of chemical substance is referred as molecular or formula weight of that substance.

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$.

Molar mass: It is obtained by dividing the mass of substance with the amount of substance and the S.I. unit of molar mass is $\text{kg/mol}$. It is numerically equal to the molecular weight since molecular weight is sum of all individual atom weight present in molecule which is represented in $amu$.

$\text{Molar}\text{mass}\left(\text{g/mol}\right)=\frac{\text{Given}\text{mass}\text{of}\text{substance}\left(\text{g}\right)}{\text{Moles}\text{of}\text{substance}\left(mol\right)}$