306639 An organic compound contains carbon hydrogen and oxygen. Its elemental analysis gave C, 38.71%, and H, 9.67%. The empirical formula of the compound would be

306640 A hydrocarbon contains 82.8% carbon. The empirical formula is

306641 0.30 g of an organic compound containing C, H and O on combustion yields 0.44 g \({\rm{C}}{{\rm{O}}_{\rm{2}}}\) and 0.18 g \({{\rm{H}}_{\rm{2}}}{\rm{O}}\). If one mol of compound weight 60, then the molecular formula of the compound is

306642 Statement A :
Molecular formula shows the exact number of different types of atoms present in a compound.
Statement B :
Molecular formula can be obtained directly from empirical formula which represents the simplest whole number ratio of various atoms present in a compound.

306639 An organic compound contains carbon hydrogen and oxygen. Its elemental analysis gave C, 38.71%, and H, 9.67%. The empirical formula of the compound would be

306640 A hydrocarbon contains 82.8% carbon. The empirical formula is

306641 0.30 g of an organic compound containing C, H and O on combustion yields 0.44 g \({\rm{C}}{{\rm{O}}_{\rm{2}}}\) and 0.18 g \({{\rm{H}}_{\rm{2}}}{\rm{O}}\). If one mol of compound weight 60, then the molecular formula of the compound is

306642 Statement A :
Molecular formula shows the exact number of different types of atoms present in a compound.
Statement B :
Molecular formula can be obtained directly from empirical formula which represents the simplest whole number ratio of various atoms present in a compound.

306639 An organic compound contains carbon hydrogen and oxygen. Its elemental analysis gave C, 38.71%, and H, 9.67%. The empirical formula of the compound would be

306640 A hydrocarbon contains 82.8% carbon. The empirical formula is

306641 0.30 g of an organic compound containing C, H and O on combustion yields 0.44 g \({\rm{C}}{{\rm{O}}_{\rm{2}}}\) and 0.18 g \({{\rm{H}}_{\rm{2}}}{\rm{O}}\). If one mol of compound weight 60, then the molecular formula of the compound is

306642 Statement A :
Molecular formula shows the exact number of different types of atoms present in a compound.
Statement B :
Molecular formula can be obtained directly from empirical formula which represents the simplest whole number ratio of various atoms present in a compound.

306639 An organic compound contains carbon hydrogen and oxygen. Its elemental analysis gave C, 38.71%, and H, 9.67%. The empirical formula of the compound would be

306640 A hydrocarbon contains 82.8% carbon. The empirical formula is

306641 0.30 g of an organic compound containing C, H and O on combustion yields 0.44 g \({\rm{C}}{{\rm{O}}_{\rm{2}}}\) and 0.18 g \({{\rm{H}}_{\rm{2}}}{\rm{O}}\). If one mol of compound weight 60, then the molecular formula of the compound is

306642 Statement A :
Molecular formula shows the exact number of different types of atoms present in a compound.
Statement B :
Molecular formula can be obtained directly from empirical formula which represents the simplest whole number ratio of various atoms present in a compound.