QBManager

Some Basic Concepts of Chemistry

228913 Compound A contains 8.7\% Hydrogen, $74 %$ Carbone and $17.3 %$ nitrogen. The molecular formula of the compound is,
Given : Atomic masses of $C, H$ and $N$ are 12, 1 and 14 amu respectively.

1

C_{4} H_{6} N_{2}

2

C_{2} H_{3} N

3

C_{5} H_{7} N

4

C_{10} H_{14} N_{2}

Explanation:

| $\begin{matrix} Element of \\ compound \\ A \end{matrix}$ | $\begin{matrix} Mass \\ percentage \end{matrix}$ | $\begin{matrix} Atomic \\ mass \end{matrix}$ | Moles | Ratio |
| :---: | :---: | :---: | :---: | :---: |
| $H$ | 8.7 | 1 | $\frac{8.7}{1} = 8.7$ | $\frac{8.7}{1.236} = 7.03 \approx 7$ |
| $C$ | 74 | 12 | $\frac{74}{12} = 6.176$ | $\frac{6.167}{1.236} = 4.989 \approx 5$ |
| :--- | :--- | :--- | :--- | :--- |
| $N$ | 17.3 | 14 | $\frac{17.3}{14} = 1.236$ | $\frac{1.236}{1.236} = 1$ |
So, Empirical formula is $C_{5} H_{7} N$
Empirical mass $= 5 \times 12 + 7 \times 1 + 14 = 81 g$
Molar mass of compound A is $162 g / mol$
Multiplying factor $(n) = \frac{162}{81} = 2$
Molecular formula $= n \times$ Empirical formula $= 2 \times C_{5} H_{7} N$
Hence, molecular formula $= C_{10} H_{14} N_{2}$

Shift-II

Some Basic Concepts of Chemistry

228914 The elemental analysis of an organic compound gave $C : 38.71 %, H : 9.67 %$ What is the empirical formula of the compound?

1

{CH}_{2} O

2

{CH}_{3} O

3

{CH}_{4} O

4

CHO

5

{CH}_{5} O

Explanation:

Given,
$%$ of carbon $= 38.71$
$%$ of hydrogen $= 9.67$
Then,
$\begin{aligned} % of oxygen & = 100 - 38.71 - 9.67 \\ = 51.62 \end{aligned}$
| Element | \% abundance | $\begin{matrix} At. \\ wt. \end{matrix}$ | Molar ratio | simple ratio |
| :---: | :---: | :---: | :---: | :---: |
| $C$ | 38.71 | 12 | $\frac{38.71}{12} = 3.23$ | $\frac{3.23}{3.23} = 1$ |
| $H$ | 9.67 | 1 | $\frac{9.67}{1} = 9.67$ | $\frac{9.67}{3.23} = 3$ |
| $O$ | $\begin{matrix} 100 - 38.71 + \\ 9.67 = 51.62 \end{matrix}$ | 16 | $\frac{51.62}{16} = 3.23$ | $\frac{3.23}{3.23} = 1$ |
So, the empirical formula of the compound $= {CH}_{3} O$

Kerala CEE -03.07.2022

Some Basic Concepts of Chemistry

228915 Consider the following reaction:
${xAs}_{2} S_{3} + {yO}_{2} \to {zAs}_{2} O_{3} + {wSO}_{2}$
What is $y$ (the coefficient for $O_{2}$ ) when this equation is balanced using whole molecule coefficients?

1 5

2 7

3 9

4 11

Explanation:

Balance eq. of the following reaction is-
$\begin{aligned} x As {As}_{2} S_{3} + y O_{2} \to {zAs}_{2} O_{3} + w {SO}_{2} \\ 2 {As}_{2} S_{3} + 9 O_{2} \to 2 {As}_{2} O_{3} + 6 {SO}_{2} \end{aligned}$
Here,
$\begin{aligned} x = 2 \\ w = 6 \\ y = 9 \\ z = 2 \end{aligned}$
So, the value of $y$ after the balanced eq. is 9 .

NDA (II)-2014

Some Basic Concepts of Chemistry

228913 Compound A contains 8.7\% Hydrogen, $74 %$ Carbone and $17.3 %$ nitrogen. The molecular formula of the compound is,
Given : Atomic masses of $C, H$ and $N$ are 12, 1 and 14 amu respectively.

1

C_{4} H_{6} N_{2}

2

C_{2} H_{3} N

3

C_{5} H_{7} N

4

C_{10} H_{14} N_{2}

Explanation:

| $\begin{matrix} Element of \\ compound \\ A \end{matrix}$ | $\begin{matrix} Mass \\ percentage \end{matrix}$ | $\begin{matrix} Atomic \\ mass \end{matrix}$ | Moles | Ratio |
| :---: | :---: | :---: | :---: | :---: |
| $H$ | 8.7 | 1 | $\frac{8.7}{1} = 8.7$ | $\frac{8.7}{1.236} = 7.03 \approx 7$ |
| $C$ | 74 | 12 | $\frac{74}{12} = 6.176$ | $\frac{6.167}{1.236} = 4.989 \approx 5$ |
| :--- | :--- | :--- | :--- | :--- |
| $N$ | 17.3 | 14 | $\frac{17.3}{14} = 1.236$ | $\frac{1.236}{1.236} = 1$ |
So, Empirical formula is $C_{5} H_{7} N$
Empirical mass $= 5 \times 12 + 7 \times 1 + 14 = 81 g$
Molar mass of compound A is $162 g / mol$
Multiplying factor $(n) = \frac{162}{81} = 2$
Molecular formula $= n \times$ Empirical formula $= 2 \times C_{5} H_{7} N$
Hence, molecular formula $= C_{10} H_{14} N_{2}$

Shift-II

Some Basic Concepts of Chemistry

228914 The elemental analysis of an organic compound gave $C : 38.71 %, H : 9.67 %$ What is the empirical formula of the compound?

1

{CH}_{2} O

2

{CH}_{3} O

3

{CH}_{4} O

4

CHO

5

{CH}_{5} O

Explanation:

Given,
$%$ of carbon $= 38.71$
$%$ of hydrogen $= 9.67$
Then,
$\begin{aligned} % of oxygen & = 100 - 38.71 - 9.67 \\ = 51.62 \end{aligned}$
| Element | \% abundance | $\begin{matrix} At. \\ wt. \end{matrix}$ | Molar ratio | simple ratio |
| :---: | :---: | :---: | :---: | :---: |
| $C$ | 38.71 | 12 | $\frac{38.71}{12} = 3.23$ | $\frac{3.23}{3.23} = 1$ |
| $H$ | 9.67 | 1 | $\frac{9.67}{1} = 9.67$ | $\frac{9.67}{3.23} = 3$ |
| $O$ | $\begin{matrix} 100 - 38.71 + \\ 9.67 = 51.62 \end{matrix}$ | 16 | $\frac{51.62}{16} = 3.23$ | $\frac{3.23}{3.23} = 1$ |
So, the empirical formula of the compound $= {CH}_{3} O$

Kerala CEE -03.07.2022

Some Basic Concepts of Chemistry

228915 Consider the following reaction:
${xAs}_{2} S_{3} + {yO}_{2} \to {zAs}_{2} O_{3} + {wSO}_{2}$
What is $y$ (the coefficient for $O_{2}$ ) when this equation is balanced using whole molecule coefficients?

1 5

2 7

3 9

4 11

Explanation:

Balance eq. of the following reaction is-
$\begin{aligned} x As {As}_{2} S_{3} + y O_{2} \to {zAs}_{2} O_{3} + w {SO}_{2} \\ 2 {As}_{2} S_{3} + 9 O_{2} \to 2 {As}_{2} O_{3} + 6 {SO}_{2} \end{aligned}$
Here,
$\begin{aligned} x = 2 \\ w = 6 \\ y = 9 \\ z = 2 \end{aligned}$
So, the value of $y$ after the balanced eq. is 9 .

NDA (II)-2014

Some Basic Concepts of Chemistry

228913 Compound A contains 8.7\% Hydrogen, $74 %$ Carbone and $17.3 %$ nitrogen. The molecular formula of the compound is,
Given : Atomic masses of $C, H$ and $N$ are 12, 1 and 14 amu respectively.

1

C_{4} H_{6} N_{2}

2

C_{2} H_{3} N

3

C_{5} H_{7} N

4

C_{10} H_{14} N_{2}

Explanation:

| $\begin{matrix} Element of \\ compound \\ A \end{matrix}$ | $\begin{matrix} Mass \\ percentage \end{matrix}$ | $\begin{matrix} Atomic \\ mass \end{matrix}$ | Moles | Ratio |
| :---: | :---: | :---: | :---: | :---: |
| $H$ | 8.7 | 1 | $\frac{8.7}{1} = 8.7$ | $\frac{8.7}{1.236} = 7.03 \approx 7$ |
| $C$ | 74 | 12 | $\frac{74}{12} = 6.176$ | $\frac{6.167}{1.236} = 4.989 \approx 5$ |
| :--- | :--- | :--- | :--- | :--- |
| $N$ | 17.3 | 14 | $\frac{17.3}{14} = 1.236$ | $\frac{1.236}{1.236} = 1$ |
So, Empirical formula is $C_{5} H_{7} N$
Empirical mass $= 5 \times 12 + 7 \times 1 + 14 = 81 g$
Molar mass of compound A is $162 g / mol$
Multiplying factor $(n) = \frac{162}{81} = 2$
Molecular formula $= n \times$ Empirical formula $= 2 \times C_{5} H_{7} N$
Hence, molecular formula $= C_{10} H_{14} N_{2}$

Shift-II

Some Basic Concepts of Chemistry

228914 The elemental analysis of an organic compound gave $C : 38.71 %, H : 9.67 %$ What is the empirical formula of the compound?

1

{CH}_{2} O

2

{CH}_{3} O

3

{CH}_{4} O

4

CHO

5

{CH}_{5} O

Explanation:

Given,
$%$ of carbon $= 38.71$
$%$ of hydrogen $= 9.67$
Then,
$\begin{aligned} % of oxygen & = 100 - 38.71 - 9.67 \\ = 51.62 \end{aligned}$
| Element | \% abundance | $\begin{matrix} At. \\ wt. \end{matrix}$ | Molar ratio | simple ratio |
| :---: | :---: | :---: | :---: | :---: |
| $C$ | 38.71 | 12 | $\frac{38.71}{12} = 3.23$ | $\frac{3.23}{3.23} = 1$ |
| $H$ | 9.67 | 1 | $\frac{9.67}{1} = 9.67$ | $\frac{9.67}{3.23} = 3$ |
| $O$ | $\begin{matrix} 100 - 38.71 + \\ 9.67 = 51.62 \end{matrix}$ | 16 | $\frac{51.62}{16} = 3.23$ | $\frac{3.23}{3.23} = 1$ |
So, the empirical formula of the compound $= {CH}_{3} O$

Kerala CEE -03.07.2022

Some Basic Concepts of Chemistry

228915 Consider the following reaction:
${xAs}_{2} S_{3} + {yO}_{2} \to {zAs}_{2} O_{3} + {wSO}_{2}$
What is $y$ (the coefficient for $O_{2}$ ) when this equation is balanced using whole molecule coefficients?

1 5

2 7

3 9

4 11

Explanation:

Balance eq. of the following reaction is-
$\begin{aligned} x As {As}_{2} S_{3} + y O_{2} \to {zAs}_{2} O_{3} + w {SO}_{2} \\ 2 {As}_{2} S_{3} + 9 O_{2} \to 2 {As}_{2} O_{3} + 6 {SO}_{2} \end{aligned}$
Here,
$\begin{aligned} x = 2 \\ w = 6 \\ y = 9 \\ z = 2 \end{aligned}$
So, the value of $y$ after the balanced eq. is 9 .

NDA (II)-2014

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