228913
Compound A contains 8.7\% Hydrogen, $74 \%$ Carbone and $17.3 \%$ nitrogen. The molecular formula of the compound is, Given : Atomic masses of $\mathrm{C}, \mathrm{H}$ and $\mathrm{N}$ are 12, 1 and 14 amu respectively.
1 $\mathrm{C}_4 \mathrm{H}_6 \mathrm{~N}_2$
2 $\mathrm{C}_2 \mathrm{H}_3 \mathrm{~N}$
3 $\mathrm{C}_5 \mathrm{H}_7 \mathrm{~N}$
4 $\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{~N}_2$
Explanation:
| $\begin{array}{c}\text { Element of } \\ \text { compound } \\ \text { A }\end{array}$ | $\begin{array}{c}\text { Mass } \\ \text { percentage }\end{array}$ | $\begin{array}{c}\text { Atomic } \\ \text { mass }\end{array}$ | Moles | Ratio | | :---: | :---: | :---: | :---: | :---: | | $\mathrm{H}$ | 8.7 | 1 | $\frac{8.7}{1}=8.7$ | $\frac{8.7}{1.236}=7.03 \approx 7$ | | $\mathrm{C}$ | 74 | 12 | $\frac{74}{12}=6.176$ | $\frac{6.167}{1.236}=4.989 \approx 5$ | | :--- | :--- | :--- | :--- | :--- | | $\mathrm{N}$ | 17.3 | 14 | $\frac{17.3}{14}=1.236$ | $\frac{1.236}{1.236}=1$ | So, Empirical formula is $\mathrm{C}_5 \mathrm{H}_7 \mathrm{~N}$ Empirical mass $=5 \times 12+7 \times 1+14=81 \mathrm{~g}$ Molar mass of compound A is $162 \mathrm{~g} / \mathrm{mol}$ Multiplying factor $(\mathrm{n})=\frac{162}{81}=2$ Molecular formula $=\mathrm{n} \times$ Empirical formula $=2 \times \mathrm{C}_5 \mathrm{H}_7 \mathrm{~N}$ Hence, molecular formula $=\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{~N}_2$
Shift-II
Some Basic Concepts of Chemistry
228914
The elemental analysis of an organic compound gave $\mathrm{C}: 38.71 \%, \mathrm{H}: 9.67 \%$ What is the empirical formula of the compound?
1 $\mathrm{CH}_2 \mathrm{O}$
2 $\mathrm{CH}_3 \mathrm{O}$
3 $\mathrm{CH}_4 \mathrm{O}$
4 $\mathrm{CHO}$
5 $\mathrm{CH}_5 \mathrm{O}$
Explanation:
Given, $\%$ of carbon $=38.71$ $\%$ of hydrogen $=9.67$ Then, $\begin{aligned} \% \text { of oxygen } & =100-38.71-9.67 \\ & =51.62 \end{aligned}$ | Element | \% abundance | $\begin{array}{c}\text { At. } \\ \text { wt. }\end{array}$ | Molar ratio | simple ratio | | :---: | :---: | :---: | :---: | :---: | | $\mathrm{C}$ | 38.71 | 12 | $\frac{38.71}{12}=3.23$ | $\frac{3.23}{3.23}=1$ | | $\mathrm{H}$ | 9.67 | 1 | $\frac{9.67}{1}=9.67$ | $\frac{9.67}{3.23}=3$ | | $\mathrm{O}$ | $\begin{array}{c}100-38.71+ \\ 9.67=51.62\end{array}$ | 16 | $\frac{51.62}{16}=3.23$ | $\frac{3.23}{3.23}=1$ | So, the empirical formula of the compound $=\mathrm{CH}_3 \mathrm{O}$
Kerala CEE -03.07.2022
Some Basic Concepts of Chemistry
228915
Consider the following reaction: $\mathrm{xAs}_2 \mathrm{~S}_3+\mathrm{yO}_2 \rightarrow \mathrm{zAs}_2 \mathrm{O}_3+\mathrm{wSO}_2$ 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- $\begin{aligned} & \text { x As } \mathrm{As}_2 \mathrm{~S}_3+\mathrm{y} \mathrm{O}_2 \rightarrow \mathrm{zAs}_2 \mathrm{O}_3+\mathrm{w} \mathrm{SO}_2 \\ & 2 \mathrm{As}_2 \mathrm{~S}_3+9 \mathrm{O}_2 \rightarrow 2 \mathrm{As}_2 \mathrm{O}_3+6 \mathrm{SO}_2 \end{aligned}$ Here, $\begin{aligned} & x=2 \\ & w=6 \\ & y=9 \\ & z=2 \end{aligned}$ So, the value of $y$ after the balanced eq. is 9 .
228913
Compound A contains 8.7\% Hydrogen, $74 \%$ Carbone and $17.3 \%$ nitrogen. The molecular formula of the compound is, Given : Atomic masses of $\mathrm{C}, \mathrm{H}$ and $\mathrm{N}$ are 12, 1 and 14 amu respectively.
1 $\mathrm{C}_4 \mathrm{H}_6 \mathrm{~N}_2$
2 $\mathrm{C}_2 \mathrm{H}_3 \mathrm{~N}$
3 $\mathrm{C}_5 \mathrm{H}_7 \mathrm{~N}$
4 $\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{~N}_2$
Explanation:
| $\begin{array}{c}\text { Element of } \\ \text { compound } \\ \text { A }\end{array}$ | $\begin{array}{c}\text { Mass } \\ \text { percentage }\end{array}$ | $\begin{array}{c}\text { Atomic } \\ \text { mass }\end{array}$ | Moles | Ratio | | :---: | :---: | :---: | :---: | :---: | | $\mathrm{H}$ | 8.7 | 1 | $\frac{8.7}{1}=8.7$ | $\frac{8.7}{1.236}=7.03 \approx 7$ | | $\mathrm{C}$ | 74 | 12 | $\frac{74}{12}=6.176$ | $\frac{6.167}{1.236}=4.989 \approx 5$ | | :--- | :--- | :--- | :--- | :--- | | $\mathrm{N}$ | 17.3 | 14 | $\frac{17.3}{14}=1.236$ | $\frac{1.236}{1.236}=1$ | So, Empirical formula is $\mathrm{C}_5 \mathrm{H}_7 \mathrm{~N}$ Empirical mass $=5 \times 12+7 \times 1+14=81 \mathrm{~g}$ Molar mass of compound A is $162 \mathrm{~g} / \mathrm{mol}$ Multiplying factor $(\mathrm{n})=\frac{162}{81}=2$ Molecular formula $=\mathrm{n} \times$ Empirical formula $=2 \times \mathrm{C}_5 \mathrm{H}_7 \mathrm{~N}$ Hence, molecular formula $=\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{~N}_2$
Shift-II
Some Basic Concepts of Chemistry
228914
The elemental analysis of an organic compound gave $\mathrm{C}: 38.71 \%, \mathrm{H}: 9.67 \%$ What is the empirical formula of the compound?
1 $\mathrm{CH}_2 \mathrm{O}$
2 $\mathrm{CH}_3 \mathrm{O}$
3 $\mathrm{CH}_4 \mathrm{O}$
4 $\mathrm{CHO}$
5 $\mathrm{CH}_5 \mathrm{O}$
Explanation:
Given, $\%$ of carbon $=38.71$ $\%$ of hydrogen $=9.67$ Then, $\begin{aligned} \% \text { of oxygen } & =100-38.71-9.67 \\ & =51.62 \end{aligned}$ | Element | \% abundance | $\begin{array}{c}\text { At. } \\ \text { wt. }\end{array}$ | Molar ratio | simple ratio | | :---: | :---: | :---: | :---: | :---: | | $\mathrm{C}$ | 38.71 | 12 | $\frac{38.71}{12}=3.23$ | $\frac{3.23}{3.23}=1$ | | $\mathrm{H}$ | 9.67 | 1 | $\frac{9.67}{1}=9.67$ | $\frac{9.67}{3.23}=3$ | | $\mathrm{O}$ | $\begin{array}{c}100-38.71+ \\ 9.67=51.62\end{array}$ | 16 | $\frac{51.62}{16}=3.23$ | $\frac{3.23}{3.23}=1$ | So, the empirical formula of the compound $=\mathrm{CH}_3 \mathrm{O}$
Kerala CEE -03.07.2022
Some Basic Concepts of Chemistry
228915
Consider the following reaction: $\mathrm{xAs}_2 \mathrm{~S}_3+\mathrm{yO}_2 \rightarrow \mathrm{zAs}_2 \mathrm{O}_3+\mathrm{wSO}_2$ 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- $\begin{aligned} & \text { x As } \mathrm{As}_2 \mathrm{~S}_3+\mathrm{y} \mathrm{O}_2 \rightarrow \mathrm{zAs}_2 \mathrm{O}_3+\mathrm{w} \mathrm{SO}_2 \\ & 2 \mathrm{As}_2 \mathrm{~S}_3+9 \mathrm{O}_2 \rightarrow 2 \mathrm{As}_2 \mathrm{O}_3+6 \mathrm{SO}_2 \end{aligned}$ Here, $\begin{aligned} & x=2 \\ & w=6 \\ & y=9 \\ & z=2 \end{aligned}$ So, the value of $y$ after the balanced eq. is 9 .
228913
Compound A contains 8.7\% Hydrogen, $74 \%$ Carbone and $17.3 \%$ nitrogen. The molecular formula of the compound is, Given : Atomic masses of $\mathrm{C}, \mathrm{H}$ and $\mathrm{N}$ are 12, 1 and 14 amu respectively.
1 $\mathrm{C}_4 \mathrm{H}_6 \mathrm{~N}_2$
2 $\mathrm{C}_2 \mathrm{H}_3 \mathrm{~N}$
3 $\mathrm{C}_5 \mathrm{H}_7 \mathrm{~N}$
4 $\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{~N}_2$
Explanation:
| $\begin{array}{c}\text { Element of } \\ \text { compound } \\ \text { A }\end{array}$ | $\begin{array}{c}\text { Mass } \\ \text { percentage }\end{array}$ | $\begin{array}{c}\text { Atomic } \\ \text { mass }\end{array}$ | Moles | Ratio | | :---: | :---: | :---: | :---: | :---: | | $\mathrm{H}$ | 8.7 | 1 | $\frac{8.7}{1}=8.7$ | $\frac{8.7}{1.236}=7.03 \approx 7$ | | $\mathrm{C}$ | 74 | 12 | $\frac{74}{12}=6.176$ | $\frac{6.167}{1.236}=4.989 \approx 5$ | | :--- | :--- | :--- | :--- | :--- | | $\mathrm{N}$ | 17.3 | 14 | $\frac{17.3}{14}=1.236$ | $\frac{1.236}{1.236}=1$ | So, Empirical formula is $\mathrm{C}_5 \mathrm{H}_7 \mathrm{~N}$ Empirical mass $=5 \times 12+7 \times 1+14=81 \mathrm{~g}$ Molar mass of compound A is $162 \mathrm{~g} / \mathrm{mol}$ Multiplying factor $(\mathrm{n})=\frac{162}{81}=2$ Molecular formula $=\mathrm{n} \times$ Empirical formula $=2 \times \mathrm{C}_5 \mathrm{H}_7 \mathrm{~N}$ Hence, molecular formula $=\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{~N}_2$
Shift-II
Some Basic Concepts of Chemistry
228914
The elemental analysis of an organic compound gave $\mathrm{C}: 38.71 \%, \mathrm{H}: 9.67 \%$ What is the empirical formula of the compound?
1 $\mathrm{CH}_2 \mathrm{O}$
2 $\mathrm{CH}_3 \mathrm{O}$
3 $\mathrm{CH}_4 \mathrm{O}$
4 $\mathrm{CHO}$
5 $\mathrm{CH}_5 \mathrm{O}$
Explanation:
Given, $\%$ of carbon $=38.71$ $\%$ of hydrogen $=9.67$ Then, $\begin{aligned} \% \text { of oxygen } & =100-38.71-9.67 \\ & =51.62 \end{aligned}$ | Element | \% abundance | $\begin{array}{c}\text { At. } \\ \text { wt. }\end{array}$ | Molar ratio | simple ratio | | :---: | :---: | :---: | :---: | :---: | | $\mathrm{C}$ | 38.71 | 12 | $\frac{38.71}{12}=3.23$ | $\frac{3.23}{3.23}=1$ | | $\mathrm{H}$ | 9.67 | 1 | $\frac{9.67}{1}=9.67$ | $\frac{9.67}{3.23}=3$ | | $\mathrm{O}$ | $\begin{array}{c}100-38.71+ \\ 9.67=51.62\end{array}$ | 16 | $\frac{51.62}{16}=3.23$ | $\frac{3.23}{3.23}=1$ | So, the empirical formula of the compound $=\mathrm{CH}_3 \mathrm{O}$
Kerala CEE -03.07.2022
Some Basic Concepts of Chemistry
228915
Consider the following reaction: $\mathrm{xAs}_2 \mathrm{~S}_3+\mathrm{yO}_2 \rightarrow \mathrm{zAs}_2 \mathrm{O}_3+\mathrm{wSO}_2$ 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- $\begin{aligned} & \text { x As } \mathrm{As}_2 \mathrm{~S}_3+\mathrm{y} \mathrm{O}_2 \rightarrow \mathrm{zAs}_2 \mathrm{O}_3+\mathrm{w} \mathrm{SO}_2 \\ & 2 \mathrm{As}_2 \mathrm{~S}_3+9 \mathrm{O}_2 \rightarrow 2 \mathrm{As}_2 \mathrm{O}_3+6 \mathrm{SO}_2 \end{aligned}$ Here, $\begin{aligned} & x=2 \\ & w=6 \\ & y=9 \\ & z=2 \end{aligned}$ So, the value of $y$ after the balanced eq. is 9 .
