Home
Class 10
CHEMISTRY
PASSAGE-II : Empirical formula is the si...

PASSAGE-II : Empirical formula is the simplest formula of the compound which gives the atomic ratio of various elements present in one molecule of the compound. However, the molecular formula of the compound gives the number of atoms of various elements present in one molecule of the compound. Molacular formula = (Empirical formula) `xxn`
implies `n=("Molecular mass")/("Empirical formula mass")`
A compound may have same empirical and molecular formula. Both these formulae are calculated using percentage composition of constituent elements.
Which of the following compounds have the same empirical formula as that of formaldehyde?

A

Formic acid

B

Glucose

C

Sucrose

D

Ethanol

Text Solution

AI Generated Solution

The correct Answer is:
To determine which of the following compounds has the same empirical formula as that of formaldehyde (HCHO), we first need to identify the empirical formula of formaldehyde. ### Step 1: Identify the empirical formula of formaldehyde - The molecular formula of formaldehyde is HCHO. - To find the empirical formula, we look for the simplest whole number ratio of the elements in the compound. - In HCHO, we have: - Hydrogen (H): 2 atoms - Carbon (C): 1 atom - Oxygen (O): 1 atom - The simplest ratio of H:C:O is 2:1:1. - Therefore, the empirical formula of formaldehyde is CH₂O. ### Step 2: Check the empirical formulas of the given compounds Now, we will check the empirical formulas of the provided compounds to see which one matches CH₂O. 1. **Formic Acid (HCOOH)**: - Molecular formula: HCOOH - Simplifying HCOOH gives us: - H: 2 - C: 1 - O: 2 - The empirical formula is CH₂O. 2. **Glucose (C₆H₁₂O₆)**: - Molecular formula: C₆H₁₂O₆ - Simplifying C₆H₁₂O₆ gives us: - H: 12 / 6 = 2 - C: 6 / 6 = 1 - O: 6 / 6 = 1 - The empirical formula is CH₂O. 3. **Sucrose (C₁₂H₂₂O₁₁)**: - Molecular formula: C₁₂H₂₂O₁₁ - Simplifying C₁₂H₂₂O₁₁ gives us: - H: 22 / 11 = 2 - C: 12 / 12 = 1 - O: 11 / 11 = 1 - The empirical formula is C₁H₁. 4. **Ethanol (C₂H₅OH)**: - Molecular formula: C₂H₅OH - Simplifying C₂H₅OH gives us: - H: 6 / 2 = 3 - C: 2 / 2 = 1 - O: 1 / 1 = 1 - The empirical formula is C₂H₅O. ### Step 3: Conclusion The compounds that have the same empirical formula as formaldehyde (CH₂O) are: - Formic Acid (HCOOH) → CH₂O - Glucose (C₆H₁₂O₆) → CH₂O ### Final Answer The compounds that have the same empirical formula as that of formaldehyde are **Formic Acid and Glucose**. ---
To determine which of the following compounds has the same empirical formula as that of formaldehyde (HCHO), we first need to identify the empirical formula of formaldehyde. ### Step 1: Identify the empirical formula of formaldehyde - The molecular formula of formaldehyde is HCHO. - To find the empirical formula, we look for the simplest whole number ratio of the elements in the compound. - In HCHO, we have: - Hydrogen (H): 2 atoms - Carbon (C): 1 atom ...
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • MOLE CONCEPT, STOICHIOMETRY AND BEHAVIOUR OF GASES

    MTG IIT JEE FOUNDATION|Exercise EXERCISE (INTEGER/NUMERICAL VALUE TYPE) |5 Videos

  • MOLE CONCEPT, STOICHIOMETRY AND BEHAVIOUR OF GASES

    MTG IIT JEE FOUNDATION|Exercise EXERCISE (ASSERTION & REASON TYPE) |10 Videos

  • METALS AND NON METALS

    MTG IIT JEE FOUNDATION|Exercise Olympiad/HOTS Corner|20 Videos

  • PERIODIC CLASSIFICATION OF ELEMENTS

    MTG IIT JEE FOUNDATION|Exercise Olympiad/HOTS Corner|15 Videos

Similar Questions

Explore conceptually related problems

The empirical and molecular formula of sucrose is same

Empirical formula is the simplest formula of the compound which gives the atomic ratio of various elements present in one molecule of the compound. However, the molecular formula of the compound gives the number of atoms of various elements present in one molecule of the compound. Molecular formula=(Empirical formula) xxn n=("Molecular mass")/("Empirical formula mass") A compound may have same empirical and molecular formulae Both these formulae are calculated by using percentage composition of constituent elements. The pair of species having different percentage composition of carbon are:

Knowledge Check

  • Empirical formula is the simplest formula of the compound which gives the atomic ratio of various elements present in one molecule of the compound. However, the molecular formula of the compound gives the number of atoms of various elements present in one molecule of the compound. Molecular formula=(Empirical formula) xxn n=("Molecular mass")/("Empirical formula mass") A compound may have same empirical and molecular formulae Both these formulae are calculated by using percentage composition of constituent elements. Which of the following compounds have same empirical formula?

    A
    Formaldehyde
    B
    Glucose
    C
    Sucrose
    D
    Acetic acid

  • Empirical formula is the simplest formula of the compound which gives the atomic ratio of various elements present in one molecule of the compound. However, the molecular formula of the compound gives the number of atoms of various elements present in one molecule of the compound. Molecular formula=(Empirical formula) xxn n=("Molecular mass")/("Empirical formula mass") A compound may have same empirical and molecular formulae Both these formulae are calculated by using percentage composition of constituent elements. Which of the following represents the formula of a substance which contains 50% oxygen?

    A
    `N_(2)O`
    B
    `CO_(2)`
    C
    `NO_(2)`
    D
    `CH_(3)OH`

  • PASSAGE-II : Empirical formula is the simplest formula of the compound which gives the atomic ratio of various elements present in one molecule of the compound. However, the molecular formula of the compound gives the number of atoms of various elements present in one molecule of the compound. Molacular formula = (Empirical formula) xxn implies n=("Molecular mass")/("Empirical formula mass") A compound may have same empirical and molecular formula. Both these formulae are calculated using percentage composition of constituent elements. 10 g of hydrofluoric acid gas occupies 5.6 litre of volume at STP. If the empirical formula of the gas is HF, then its molecular formula in the gaseous state will be

    A
    HF
    B
    `H_(2)F_(2)`
    C
    `H_(3)F_(3)`
    D
    `H_(4)F_(4)`

    • Similar Questions

    Explore conceptually related problems

    PASSAGE-II : Empirical formula is the simplest formula of the compound which gives the atomic ratio of various elements present in one molecule of the compound. However, the molecular formula of the compound gives the number of atoms of various elements present in one molecule of the compound. Molacular formula = (Empirical formula) xxn implies n=("Molecular mass")/("Empirical formula mass") A compound may have same empirical and molecular formula. Both these formulae are calculated using percentage composition of constituent elements. An oxide of iodine (Atomic mass of I = 127 u) contains 25.4 g of iodine and 8 g of oxygen. Its formula could be :

    Empirical formula is the simplest formula of the compound which gives the atomic ratio of various elements present in one molecule of the compound. However, the molecular formula of the compound gives the number of atoms of various elements present in one molecule of the compound. Molecular formula=(Empirical formula) xxn n=("Molecular mass")/("Empirical formula mass") A compound may have same empirical and molecular formulae Both these formulae are calculated by using percentage composition of constituent elements. An oxide of iodine (I=127) contains 25.4 g of iodine and 8 g of oxygen. its formula could be:

    Empirical formula is the simplest formula of the compound which gives the atomic ratio of various elements present in one molecule of the compound. However, the molecular formula of the compound gives the number of atoms of various elements present in one molecule of the compound. Molecular formula=(Empirical formula) xxn n=("Molecular mass")/("Empirical formula mass") A compound may have same empirical and molecular formulae Both these formulae are calculated by using percentage composition of constituent elements. 10 g of hydrofluoric acid gas occupies 5.6 litres of volume at STP. if the empiriccal formula of the gas is HF, then its molecular formula in the gaseous state will be:

    Empirical formula is the simplest formula of the compound which gives the atomic ratio of various elements present in one molecule of the compound. However, the molecular formula of the compound gives the number of atoms of various elements present in one molecule of the compound. Molecular formula=(Empirical formula) xxn n=("Molecular mass")/("Empirical formula mass") A compound may have same empirical and molecular formulae Both these formulae are calculated by using percentage composition of constituent elements. A compound of Na, C and O contains 0.0887 mol Na, 0.132 mol O and 2.65xx10^(22) atoms of carbon . the empirical formula of the compound is:

    Empirical formula is the simplest formula of the compound which gives the atomic ratio of various elements present in one molecule of the compound. However, the molecular formula of the compound gives the number of atoms of various elements present in one molecule of the compound. Molecular formula=(Empirical formula) xxn n=("Molecular mass")/("Empirical formula mass") A compound may have same empirical and molecular formulae Both these formulae are calculated by using percentage composition of constituent elements. Two metallic oxides contain 27.6% and 30% oxygen respectively. if the formulae of first oxide is M_(3)O_(4) , that of second will be:

    Home
    Profile