306625
If two compounds have the same empirical formula but different molecular formula, they must have
1 different percentage composition.
2 different molecular weights.
3 same viscosity.
4 same vapour density.
Explanation:
Since, the molecular formula is \(n\) times the empirical formula, therefore, different compounds having the same empirical formula must have different molecular weights.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306626
An organic compound has an empirical formula \(\mathrm{CH}_{2} \mathrm{O}\), its vapour density is 45 . The molecular formula of the compound is
The acid with empirical formula \({\rm{C}}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}\) is formic acid, HCOOH.
KCET - 2013
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306628
An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is : [Atomic wt. of C is 12, H is 1]
1 \({\rm{C}}{{\rm{H}}_{\rm{2}}}\)
2 \({\rm{C}}{{\rm{H}}_{\rm{3}}}\)
3 \({\rm{C}}{{\rm{H}}_{\rm{4}}}\)
4 \({\rm{CH}}\)
Explanation:
Percentage of H = 100 – 78 = 22% Element Mass Mole Simple ratio C 78g \(\frac{{{\rm{78}}}}{{{\rm{12}}}}{\rm{ = 6}}{\rm{.5}}\) \(\frac{{{\rm{6}}{\rm{.5}}}}{{{\rm{6}}{\rm{.5}}}}{\rm{ = 1}}\) H 22g \(\frac{{{\rm{22}}}}{{\rm{1}}}{\rm{ = 22}}\) \(\frac{{{\rm{22}}}}{{{\rm{6}}{\rm{.5}}}}{\rm{ = 3}}{\rm{.38}}\) Empirical formula of organic compound \({\rm{ = C}}{{\rm{H}}_{\rm{3}}}\)
NEET - 2021
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306629
Find empirical formula of the compound, if \({\rm{M = 68}}\% \) (atomic mass M = 34) and remaining \({\rm{M = 32}}\% \) oxygen.
1 \(\rm{{\text{MO}}}\)
2 \(\mathrm{\mathrm{M}_{2} \mathrm{O}}\)
3 \(\mathrm{\mathrm{MO}_{2}}\)
4 \(\mathrm{\mathrm{M}_{2} \mathrm{O}_{3}}\)
Explanation:
Hence empirical formula of given compound is \(\mathrm{\mathrm{M}_{2} \mathrm{O}_{2}}\) or MO.
306625
If two compounds have the same empirical formula but different molecular formula, they must have
1 different percentage composition.
2 different molecular weights.
3 same viscosity.
4 same vapour density.
Explanation:
Since, the molecular formula is \(n\) times the empirical formula, therefore, different compounds having the same empirical formula must have different molecular weights.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306626
An organic compound has an empirical formula \(\mathrm{CH}_{2} \mathrm{O}\), its vapour density is 45 . The molecular formula of the compound is
The acid with empirical formula \({\rm{C}}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}\) is formic acid, HCOOH.
KCET - 2013
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306628
An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is : [Atomic wt. of C is 12, H is 1]
1 \({\rm{C}}{{\rm{H}}_{\rm{2}}}\)
2 \({\rm{C}}{{\rm{H}}_{\rm{3}}}\)
3 \({\rm{C}}{{\rm{H}}_{\rm{4}}}\)
4 \({\rm{CH}}\)
Explanation:
Percentage of H = 100 – 78 = 22% Element Mass Mole Simple ratio C 78g \(\frac{{{\rm{78}}}}{{{\rm{12}}}}{\rm{ = 6}}{\rm{.5}}\) \(\frac{{{\rm{6}}{\rm{.5}}}}{{{\rm{6}}{\rm{.5}}}}{\rm{ = 1}}\) H 22g \(\frac{{{\rm{22}}}}{{\rm{1}}}{\rm{ = 22}}\) \(\frac{{{\rm{22}}}}{{{\rm{6}}{\rm{.5}}}}{\rm{ = 3}}{\rm{.38}}\) Empirical formula of organic compound \({\rm{ = C}}{{\rm{H}}_{\rm{3}}}\)
NEET - 2021
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306629
Find empirical formula of the compound, if \({\rm{M = 68}}\% \) (atomic mass M = 34) and remaining \({\rm{M = 32}}\% \) oxygen.
1 \(\rm{{\text{MO}}}\)
2 \(\mathrm{\mathrm{M}_{2} \mathrm{O}}\)
3 \(\mathrm{\mathrm{MO}_{2}}\)
4 \(\mathrm{\mathrm{M}_{2} \mathrm{O}_{3}}\)
Explanation:
Hence empirical formula of given compound is \(\mathrm{\mathrm{M}_{2} \mathrm{O}_{2}}\) or MO.
NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306625
If two compounds have the same empirical formula but different molecular formula, they must have
1 different percentage composition.
2 different molecular weights.
3 same viscosity.
4 same vapour density.
Explanation:
Since, the molecular formula is \(n\) times the empirical formula, therefore, different compounds having the same empirical formula must have different molecular weights.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306626
An organic compound has an empirical formula \(\mathrm{CH}_{2} \mathrm{O}\), its vapour density is 45 . The molecular formula of the compound is
The acid with empirical formula \({\rm{C}}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}\) is formic acid, HCOOH.
KCET - 2013
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306628
An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is : [Atomic wt. of C is 12, H is 1]
1 \({\rm{C}}{{\rm{H}}_{\rm{2}}}\)
2 \({\rm{C}}{{\rm{H}}_{\rm{3}}}\)
3 \({\rm{C}}{{\rm{H}}_{\rm{4}}}\)
4 \({\rm{CH}}\)
Explanation:
Percentage of H = 100 – 78 = 22% Element Mass Mole Simple ratio C 78g \(\frac{{{\rm{78}}}}{{{\rm{12}}}}{\rm{ = 6}}{\rm{.5}}\) \(\frac{{{\rm{6}}{\rm{.5}}}}{{{\rm{6}}{\rm{.5}}}}{\rm{ = 1}}\) H 22g \(\frac{{{\rm{22}}}}{{\rm{1}}}{\rm{ = 22}}\) \(\frac{{{\rm{22}}}}{{{\rm{6}}{\rm{.5}}}}{\rm{ = 3}}{\rm{.38}}\) Empirical formula of organic compound \({\rm{ = C}}{{\rm{H}}_{\rm{3}}}\)
NEET - 2021
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306629
Find empirical formula of the compound, if \({\rm{M = 68}}\% \) (atomic mass M = 34) and remaining \({\rm{M = 32}}\% \) oxygen.
1 \(\rm{{\text{MO}}}\)
2 \(\mathrm{\mathrm{M}_{2} \mathrm{O}}\)
3 \(\mathrm{\mathrm{MO}_{2}}\)
4 \(\mathrm{\mathrm{M}_{2} \mathrm{O}_{3}}\)
Explanation:
Hence empirical formula of given compound is \(\mathrm{\mathrm{M}_{2} \mathrm{O}_{2}}\) or MO.
306625
If two compounds have the same empirical formula but different molecular formula, they must have
1 different percentage composition.
2 different molecular weights.
3 same viscosity.
4 same vapour density.
Explanation:
Since, the molecular formula is \(n\) times the empirical formula, therefore, different compounds having the same empirical formula must have different molecular weights.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306626
An organic compound has an empirical formula \(\mathrm{CH}_{2} \mathrm{O}\), its vapour density is 45 . The molecular formula of the compound is
The acid with empirical formula \({\rm{C}}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}\) is formic acid, HCOOH.
KCET - 2013
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306628
An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is : [Atomic wt. of C is 12, H is 1]
1 \({\rm{C}}{{\rm{H}}_{\rm{2}}}\)
2 \({\rm{C}}{{\rm{H}}_{\rm{3}}}\)
3 \({\rm{C}}{{\rm{H}}_{\rm{4}}}\)
4 \({\rm{CH}}\)
Explanation:
Percentage of H = 100 – 78 = 22% Element Mass Mole Simple ratio C 78g \(\frac{{{\rm{78}}}}{{{\rm{12}}}}{\rm{ = 6}}{\rm{.5}}\) \(\frac{{{\rm{6}}{\rm{.5}}}}{{{\rm{6}}{\rm{.5}}}}{\rm{ = 1}}\) H 22g \(\frac{{{\rm{22}}}}{{\rm{1}}}{\rm{ = 22}}\) \(\frac{{{\rm{22}}}}{{{\rm{6}}{\rm{.5}}}}{\rm{ = 3}}{\rm{.38}}\) Empirical formula of organic compound \({\rm{ = C}}{{\rm{H}}_{\rm{3}}}\)
NEET - 2021
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306629
Find empirical formula of the compound, if \({\rm{M = 68}}\% \) (atomic mass M = 34) and remaining \({\rm{M = 32}}\% \) oxygen.
1 \(\rm{{\text{MO}}}\)
2 \(\mathrm{\mathrm{M}_{2} \mathrm{O}}\)
3 \(\mathrm{\mathrm{MO}_{2}}\)
4 \(\mathrm{\mathrm{M}_{2} \mathrm{O}_{3}}\)
Explanation:
Hence empirical formula of given compound is \(\mathrm{\mathrm{M}_{2} \mathrm{O}_{2}}\) or MO.
306625
If two compounds have the same empirical formula but different molecular formula, they must have
1 different percentage composition.
2 different molecular weights.
3 same viscosity.
4 same vapour density.
Explanation:
Since, the molecular formula is \(n\) times the empirical formula, therefore, different compounds having the same empirical formula must have different molecular weights.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306626
An organic compound has an empirical formula \(\mathrm{CH}_{2} \mathrm{O}\), its vapour density is 45 . The molecular formula of the compound is
The acid with empirical formula \({\rm{C}}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}\) is formic acid, HCOOH.
KCET - 2013
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306628
An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is : [Atomic wt. of C is 12, H is 1]
1 \({\rm{C}}{{\rm{H}}_{\rm{2}}}\)
2 \({\rm{C}}{{\rm{H}}_{\rm{3}}}\)
3 \({\rm{C}}{{\rm{H}}_{\rm{4}}}\)
4 \({\rm{CH}}\)
Explanation:
Percentage of H = 100 – 78 = 22% Element Mass Mole Simple ratio C 78g \(\frac{{{\rm{78}}}}{{{\rm{12}}}}{\rm{ = 6}}{\rm{.5}}\) \(\frac{{{\rm{6}}{\rm{.5}}}}{{{\rm{6}}{\rm{.5}}}}{\rm{ = 1}}\) H 22g \(\frac{{{\rm{22}}}}{{\rm{1}}}{\rm{ = 22}}\) \(\frac{{{\rm{22}}}}{{{\rm{6}}{\rm{.5}}}}{\rm{ = 3}}{\rm{.38}}\) Empirical formula of organic compound \({\rm{ = C}}{{\rm{H}}_{\rm{3}}}\)
NEET - 2021
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306629
Find empirical formula of the compound, if \({\rm{M = 68}}\% \) (atomic mass M = 34) and remaining \({\rm{M = 32}}\% \) oxygen.
1 \(\rm{{\text{MO}}}\)
2 \(\mathrm{\mathrm{M}_{2} \mathrm{O}}\)
3 \(\mathrm{\mathrm{MO}_{2}}\)
4 \(\mathrm{\mathrm{M}_{2} \mathrm{O}_{3}}\)
Explanation:
Hence empirical formula of given compound is \(\mathrm{\mathrm{M}_{2} \mathrm{O}_{2}}\) or MO.
